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Dasar-dasr pengukuran tanah untuk wilayah tambang
Problems
X-
Solutions
Shepherd
Surveying
Problems and Solutions
52-
F.
ARNOLD »ii^,^iiA
*
A. Shepherd
£
\
Thfs new book gives a presentation concentrating on mathematical problems, an aspect of the subject which usually
causes most
difficulty.
Summaries of basic theory are followed by worked examples and selected exercises. The book covers three main branches of surveying: measurement, surveying techniques and industrial applications. It is a book concerned mainly with engineering surveying as applied, for example, in the construction and mining industries.
Printed in Great Britain by Bookprint Ltd., Crawley, Sussex
|
!
PREFACE This book is an attempt to deal with the basic mathematical aspects of 'Engineering Surveying', i.e. surveying applied to construction and
mining engineering projects, and to give guidance on practical methods of solving the typical problems posed in practice and, in theory, by the various examining bodies.
The general approach adopted is to give a theoretical analysis of each topic, followed by worked examples and, finally, selected exercises for private study. Little claim is made to new ideas, as the ground covered is elementary and generally well accepted. It is hoped that the mathematics of surveying, which so often causes trouble to beginners, is presented in as clear and readily understood a manner as possible. The main part of the work of the engineering surveyor, civil
and mining engineer, and all workers in the construction industry is confined to plane surveying, and this book is similarly restricted. It is hoped that the order of the chapters provides a natural sequence, viz.:
(a)
Fundamental measurement Linear measurement in the horizontal plane, (ii) Angular measurement and its relationship to linear values, (i)
i.e. (iii)
(b)
trigonometry,
Co-ordinates as a graphical and mathematical tool.
Basic trigonometry is included, to provide a fundamental mathematical tool for the surveyor. It is generally found that there is a deficiency in the student's ability to apply numerical values to trigonometrical problems, particularly in the solution of triangles, and it is
hoped that the chapter in question shows that more is required than the sine and cosine formulae. Many aspects of surveying, e.g. errors in surveying, curve ranging, etc. require the use of small angles, and the application of radians is suggested. Few numerical problems are posed relating to instrumentation, but
it
is felt that a
knowledge of basic
physical properties affords a more complete understanding of the construction and use of instruments. ject, the effects of errors are
To
facilitate a real grasp of the sub-
analysed
in all sections.
This may
appear too advanced for students who are not familiar with the elementary calculus, but it is hoped that the conclusions derived will be beneficial to
all.
With the introduction of the Metric System in the British Isles and
elsewhere, its effect on all aspects of surveying is pin-pointed and conversion factors are given. Some examples are duplicated in the proposed units based on the International System (S.I.) and in order to give a 'feel' for the new system, during the difficult transition period, equivalent S.I. values are given in brackets for a few selected examples. The book is suitable for all students in Universities and Technical Colleges, as well as for supplementary postal tuition, in such courses as Higher National Certificates, Diplomas and Degrees in Surveying, Construction, Architecture, Planning, Estate Management, Civil and
Mining Engineering, as well as for professional qualification for the Royal Institution of Chartered Surveyors, the Institution of Civil Engineers, the Incorporated Association of Architects and Surveyors, the Institute of Quantity Surveyors, and the Institute of Building.
ACKNOWLEDGMENTS of
I am greatly indebted to the Mining Qualifications Board (Ministry Power) and the Controller of H.M. Stationery Office, who have given
permission
for the reproduction of
examination questions. My thanks
are also due to the Royal Institution of Chartered Surveyors, the Institution of Civil Engineers, to the Senates of the Universities of
London and Nottingham,
to the
East Midlands Educational Union and whom have
the Nottingham Regional College of Technology, all of
allowed their examination questions to be used. My special thanks are due to many of my colleagues at Nottingham, but especially to Messrs. J. H. Ball, A.R.I.C.S., A.I.A.S., A.M.I.Min.E., A. Eaton, B.Sc., C.Eng., A.M.I.C.E., A.M.B.I.M., G. M. Lewis, B.Sc, Ph.D., M. B. Pate, M.Sc, A. A. Payne, B.Sc, C. Rayner, B.Sc, A.R.I.C.S., R. Robb, A.R.I.C.S., A.M.I.Min.E.,
D.B. Shaw, B.Sc, and
whom have
J. P. Withers,
B.Sc, C.Eng., A.M.I.C.E.,
all of
offered advice and help in checking the text
The ultimate responsibility I
am
for the accuracy is, of course, my own. very conscious that, even with the most careful checking, it is not
5.82 The use of sight rails and boning (or travelling) rods 5.83
The setting of slope stakes Exercises 5(g) (Construction levelling) Exercises 5 (h) (General)
TRAVERSE SURVEYS 6. 1
Types
6.11
of traverse
Open
6.2 Methods of traversing
Compass traversing
6.22 Continuous azimuth method 6.23 Direction method 6.
6.3
24
Separate angular measurement Exercises 6(a)
Office tests for locating mistakes in traversing
6.31
6.32 6.33
A A
mistake in the linear value of one line mistake in the angular value at one station
When the traverse is closed on to fixed points and a mistake in the bearing is known to exist
6.4 Omitted measurements in closed traverses
6.41
6.42 6-43
6.44 6.45 6.46
6.5
Where the bearing of one line is missing Where the length of one line is missing Where the length and bearing of a line are missing Where the bearings of two lines are missing Where two lengths are missing Where the length of one line and the bearing
289 298 298
299 300 301
302 304 304
306 306 307 307 308 308 309
309 309 314
of another line are missing
315
Exercises 6(b) (Omitted values)
316
The adjustment
6.51
288
298 298
6.12 Closed
6.21
284 286
of closed traverses
317
Where the start and finish of a traverse are fixed
6-52 Traverses which return to their starting point 6.53 Adjusting the lengths without altering the bearings 6.54 Adjustment to the length and bearing 6.55 Comparison of methods of adjustment
Determination of the tacheometric constants m and K
By physical measurement 7.22 By field measurement 7.21
7.3
359
fixed stadia
of the instrument
Inclined sights
362
7.31
Staff normal to the line of sight
7.32
Staff vertical
7-4
The
362 363
effect of errors in stadia tacheometry
7.41
Staff tilted from the normal
7.42
Error in the angle of elevation
367
367 368
Staff tilted from the vertical
7.44 Accuracy of the vertical angle to the overall
to conform
371
accuracy
7.45 The effect of the stadia intercept assumption Exercises 7(a) 7.5
Subtense systems
372 380 383
7.51
Tangential method
383
7.52
Horizontal subtense bar system
388
7.6
Methods used
392
in the field
measurement 7.62 Auxiliary base measurement 7.63 Central auxiliary base 7.64 Auxiliary base perpendicularly bisected by 7.61
Serial
the traverse line
7.65 7-66
8
367
with the
staff normal
7.43
360
360 361
Two
392 393 395
397 398
auxiliary bases
The auxiliary base used
in
between two
traverse lines
400
Exercises 7(b)
403
AND FAULT PROBLEMS
411
8.1
Definitions
8.2
Dip problems
411 413
DIP
8.21
Given the rate and direction of
full dip, to find
the apparent dip in any other direction
8.22
Given the direction of
full dip
413
and the rate and
direction of an apparent dip, to find the rate of full dip
8.23
Given the rate and direction of the bearing of an apparent dip
413 full dip, to find
415
XV 8.24 Given two apparent dips, to find the rate and direction of full dip
8.25
8.26
8.3
416
Given the rate of full dip and the rate and direction of an apparent dip, to find the direction of full dip
421
Given the levels and relative positions of three points in a plane (bed or seam), to find the direction and rate of full dip
422
Problems
in
which the inclinations are expressed
as angles and a graphical solution
427
required
is
8.31 Given the inclination and direction of full dip, to find the rate of apparent dip in a given direction 8.32
8.33
Given the inclination and direction of
to find the direction of a given apparent dip
428
Given the inclination and direction of two apparent dips, to find the inclination and direction of full dip Exercises 8(a)
429 429
8.4
The
8.5
Fault problems
8.51
8.52
rate of approach
method
for convergent lines
To
find the relationship
To
437 between the true and 443
when the throw of the fault opposes the dip of the seam 8.54 Given the angle 8 between the full dip of the seam find the true bearing of a fault
and the true bearing of the
fault, to find the
To
when the downthrow of the fault is in the same general direction as the dip of the seam 8.56 Given the angle 8 between the full dip of the seam and the true bearing of the fault, to find the
446
find the true bearing of a fault
bearing of the line of contact 8.6
444
bearing
of the line of contact
8.55
432
437
Definitions
apparent bearings of a fault 8.53
427
full dip,
To find the bearing and inclination of the line of intersection (AB) of two inclined planes Exercises 8 (b) (Faults) Exercises 8 (c) (General)
449
449
450 452 454
AREAS
457
9.1 Areas of regular figures
457
Areas bounded by straight lines 9. 12 Areas involving circular curves 9. 13 Areas involving non-circular curves 9.11
457 459 460
XVI
9.14 Surface areas 9.2
461 471
Areas of irregular figures
9.21
Equalisation of the boundary to give straight lines
9.22 The mean ordinate rule
471 472
9.23 The mid-ordinate rule
473
9.24 The trapezoidal rule
473 474 477
9.25 Simpson's rule 9.26 The planimeter 9.3
Plan areas
481
9.31 Units of area
481
9.32 Conversion of planimetric area in square inches into acres
482 482 488
9.33 Calculation of area from co-ordinates 9.34 Machine calculations with checks 9.4
Subdivisions of areas
490
9.41
The subdivision
of
an area into specified parts from a point on the boundary
9.42
The subdivision
of an area by a line of
490
known
bearing 9.43
The sub-division of an area by a a known point inside the figure
10.31 Calculation of volumes from cross-sectional areas
513
Exercises 10 (b) (Cross- sectional areas) 10.32 Alternative formulae for the calculation of volumes from the derived cross-sectional areas
523
10.33 Curvature correction
10.34 Derivation of the eccentricity e of the centroid
G
525 535 537
10.4 Calculation of volumes from contour maps 10.5 Calculation of volumes from spot-heights
543 543
10.6 Mass-haul diagrams
544
10.61 Definitions
544
10.62 Construction of the mass-haul diagram 10.63 Characteristics of the mass-haul diagram
545 546
1
XV11
10.64
11
Free-haul and overhaul Exercises 10 (c) (Earthwork volumes)
CIRCULAR CURVES 11.1
546 552
559 559 559 560 560
Definition
11.2 Through chainage
Length of curve L Geometry of the curve 11.5 Special problems 11.3
11.4
561
11.51
To pass a curve tangential
11.52
To pass a curve through
11.53
To pass
a curve through a given point P Exercises 11(b) (Curves passing through
567
a given point)
571
to three given
straights three points
Exercises 11(a)
561 563 566
11.54 Given a curve joining two tangents, to find the change required in the radius for an
assumed change
in the tangent length
11.6 Location of tangents and curve
11.7 Setting out of curves
11.71
By
linear equipment only
11.72 By linear and angular equipment 11.73
11.8
By angular equipment only Exercises 11(c)
Compound curves
11.9 Reverse curves
Exercises 11(e) (Reverse curves)
VERTICAL AND TRANSITION CURVES 12.
575 576 576
580 580 588 591
Exercises 11(d) (Compound curves;
12
572
Vertical curves
12.2 Properties of the simple parabola 12.3 Properties of the vertical curve
12.4 Sight distances
summits
12.41
Sight distances for
12.42
Sight distances for valley curves
599 600 605
607 607 608
609 611 611
613
12.43 Sight distance related to the length of the
beam
of a vehicle's
12.5 Setting-out data
Exercises 12(a)
headlamp
615 616 624
XV111
Transition curves
12.6
12.61
Superelevation
12.62
Cant
12.63 Minimum curvature for standard velocity 12.64 Length of transition 12.65 Radial acceleration 12.7 12.8
The ideal transition curve The clothoid
12.81
To
find Cartesian co-ordinates
The tangential angle 12.83 Amount of shift 12.82
12.9
The Bernouilli lemniscate
12.91
Setting out using the lemniscate
12. 10 The cubic parabola 12.11 The insertion of transition curves
12.12 Setting-out processes 12. 13 Transition
curves applied to compound curves Exercises 12(b)
627
627 628 628 629 629
630 632
632 633 633
634
635 636 637 640 644 649
Abbreviations used for Examination Papers
E.M.E.U.
East Midlands Educational Union
I.C.E.
Institution Of Civil Engineers
L.U.
London University B.Sc. (Civil Engineering) London University B.Sc. (Estate Management)
L.U./E M.Q.B./S M.Q.B./M M.Q.B./UM
R.I.C.S./G
Mining Qualifications Board (Mining Surveyors) Mining Qualifications Board (Colliery Managers) Mining Qualifications Board (Colliery Undermanagers) Nottingham Regional College of Technology Nottingham University Royal Institution of Chartered Surveyors (General)
R.I.C.S./M R.I.C.S./ML R.I.C.S./Q
Royal Institution of Chartered Surveyors (Mining) Royal Institution of Chartered Surveyors (Mining/Land) Royal Institution of Chartered Surveyors (Quantity)
N.R.C.T. N.U.
LINEAR MEASUREMENT The Basic Principles
1.1
of Surveying
Fundamental rule 'Always work from the whole to the part*. This implies 'precise control surveying' as the first consideration, followed by 'subsidiary detail surveying'.
A
C
point
in a
plane may be fixed relative to a given line
AB
in
one of the following ways: 1.
Triangulation Angular measurement from a fixed base line. AB is known. The angles a and /3 are measured.
The
length
ȣ a.
Xe .V li
B 2.
Fig. 1.1(a)
Trilateration Linear
measurement only. The lengths AC and BC The position of C is always fixed provid-
are measured or plotted.
ed
AC
+
BC
> AB.
Uses: (a) Replacing triangulation with the use of microwave measuring equipment. (b)
Chain surveying. A
Bt Fig. 1.1(b) 1
SURVEYING PROBLEMS AND SOLUTIONS 3.
Polar co-ordinates Linear and angular measurement. Uses: (a) Traversing. (b) Setting out. (c) Plotting
by protractor.
,-° c (s,6)
BhT Fig. 1.1(c) 4.
Rectangular co-ordinates Linear measurement only
at right-angles.
Uses: (a) Offsets. (b) Setting out. (c) Plotting.
A
A
90"
OC
Bit
Fig. 1.1(d)
1.2
General Theory of Measurement
The following points should be (1)
There
(2)
As
noted:
no such thing as an exact measurement. All measurements contain some error, the magnitude of the error being dependent on the instruments used and the ability of the observer. is
the true value is never known, the true error is never deter-
LINEAR MEASUREMENT
3
mined. (3) The degree of accuracy, or its precision, can only be quoted as a relative accuracy, i.e. the estimated error is quoted as a fraction
of the
measured quantity. Thus 100
error of 1 inch represents a relative
lcm
in
100
measured with an estimated accuracy of 1/1200. An error of ft
m = 1/10000.
(4) Where readings are taken on a graduated scale to the nearest subdivision, the maximum error in estimation will be ± l/2 division. (5) Repeated measurement increases the accuracy by y/n, where n is the number of repetitions. N.B. This cannot be applied indefinitely-
(6) Agreement between repeated measurements does not imply accuracy but only consistency.
1.3
Significant Figures in Measurement and Computation
If a measurement is recorded as 205 ft to the nearest foot, its most probable value is 205 ±0*5 ft, whilst if measured to the nearest 0*1 ft its most probable value is 205-0 ± 0-05 ft. Thus the smallest recorded
digit is subject to a
maximum
error of half its value.
In computation, figures are
rounded off to the required degree of precision, generally by increasing the last significant figure by 1 if the following figure is 5 or more. (An alternative is the rounding off with 5 to the nearest
Thus
even number.)
205-613 becomes 205-61 to 2 places,
whilst
205-615 becomes 205-62 to 2 places,
or
205-625 "may also be 205*62, giving a less biased value.
It is generally better to work to 1 place of decimals more than is required in the final answer, and to carry out the rounding-off process at the end.
In multiplication the
number of significant figures depends on the
accuracy of the individual components, e.g.,
P
if
P + 8P
then
=
x.y,
= (x + 8x)(y + 8y) = xy + x8y + y8x + 8x8y
Neglecting the last term and substracting equation,
~ P
8p = gives
P
from both sides of the
x8y + ySx
S£ = ^§X + y8x = 8y + 8x
P
xy
sp - p
(f
xy
+
t)
(11)
SURVEYING PROBLEMS AND SOLUTIONS
4
Thus the
relative accuracy of the product is the
sum
of all the
relative accuracies involved in the product.
Example ± 0*005
A rectangle measures 3-82 in. and 7-64 in. with errors of Express the area to the correct number of significant
1.1
in.
figures.
P
= 3-82 x 7-64 = 29*184 8 in2
relative accuracies
° 3-82
~ ~
0-005
..
_i_ 750 1
7-64
SP = = .-.
1500
290-
+ -L-)
\750
1500/
™
=
500
± 0-06
the area should be given as 29-2in 2
.
As a general rule the number of significant figures in the product should be at least the same as, or preferably have one more significant figure than, the least significant factor.
The area would thus be quoted as 29-18 in 2 In division the same rule applies.
Q =
y
x + 8x = * + £f - rf^ + 2 y + 8y y y y
Q + 8Q = Subtracting
Q
from both sides and dividing by
SQ = Q (?I Powers
R
gives
*)
(1.2)
= xn
R + 8R =
(x + 8x)
n
= x n + n8x +
—
Q
...
8R = n8x —
—
i.e.
...
.
nx
,
..
relative accuracy of
single value.
8R = n8x Roots
This
(1 3) .
is the opposite relationship
R = ^x From the above
R
n
.'.
Rn =
+ n8R = x + 8x
x
LINEAR MEASUREMENT
5
nSR = Sx
8R _ 8x_ R n ~ nx
8R = -8x Example 1.2
R
If
= (5-01 ± 0-005)
5-01 2
R = V 25
If
v'25-10 =
8R R
= 0-01
2 x 0-005
should be given as 25*10
Example 1.3
.*.
2
= 25-1001
8R =
R
.'.
(1.4)
*
10 ± °* 01
5-009 9
^
=
= 0-005
should be given as 5-01
Example 1.4 A rectangular building has sides approximately 480 metres and 300 metres. If the area is to be determined to the nearest 2 10 m what will be the maximum error permitted in each line, assuming equal precision ratios for each length? To what degree of accuracy should the lines be measured?
A =
480 x 300 -
144 000
8A =
10
8A = _1
A
8x = 8y
but
x 8x_
x i.e.
§x
=
1
Sy 8y _ 28x y ~ x
x 1
=
2 x 14400
m2
y
8x
.
y
=
+
x
14400
m2
28 800
the precision ratio of each line is
*„ 28 800
480
m
of
and in 300
m
of -i9p-
This represents a maximum
in
-
Zq 800
= 0*016 7
m
= 0-0104
m
2o 800 the number of significant figures in the area is 5, i.e. to the m 2 , then each line also must be measured to at least 5 significant figures, i.e. 480-00 m and 300-00m. If
nearest 10
SURVEYING PROBLEMS AND SOLUTIONS Chain Surveying
1.4
The chain There are two types (a) Gunter's 1
Its
chain
chain* = 100 links = 66 ft = 0-66 1 link
ft
= 7-92 in.
advantage lies in its relationship to the acre 10 sq chains = 100 000 sq links = 1 acre. (b)
Engineer's chain
100 links =
100
ft
(Metric chain
100 links =
20
m
= 0-2 m) 1 link Basic figures There are many combinations of chain lines linear dimensions forming trilateration, Fig. 1 .2.
all
dependent on the
Tie line
C A
Tie lines
Fig.
1.41
1 .2
Basic figures
in chain surveying
Corrections to the ground measurements
Standardisation
Where the length of the chain or tape does not agree with *
See conversion factors, pp. v —
vii.
its
nom-
LINEAR MEASUREMENT inal value, a correction
must be made
7
to the recorded value of a
meas-
ured quantity.
The following
rules apply
(1) If the tape is too long, the
measurement
will be too short
— the
correction will be positive. (2) If the tape is too short, the
measurement will be too long — the
correction will be negative. If
the length of tape of nominal length
/
is
/
±
the error per unit length = If
the
measured length
is
81,
dm and the true length d t = dm ± dm
=
— fil
±
is
dt then ,
—
^(l±f)
d-5)
Alternatively,
1
dm
+ 81
(1.6)
nominal length of tape
I
d > = *» 1 ± j
Example 1.5
actual length of tape
(1 5 >
t)
-
A
chain of nominal length 100 links, when compared with If this chain is used to measure a line and the recorded measurement is 653 links, what is the true
a standard, measures 101 links.
AB
length
AB? Error per link =
.'.
true length
-i— = 0*01 100
= 653(1 + 0-01) = 653 + 6-53 = 659-53 links
.
Alternatively, true length
= 653 x
^
= 659-53 links
.
Effect of standardisation on areas Based on the principle of similar figures, true a,ea (.,) .
apparent area (,„) x
(,££££ 5%.)'
SURVEYING PROBLEMS AND SOLUTIONS
A T = A M (l
or
±y)
(1.8)
Effect of standardisation on volumes
Based on the
principle of similar volumes, ,
true
volume VT = apparent volume x
/
true length of tape
( apparent length of tap
V
J
ue.
Vr = V„(l ±^)° Where the
N.B.
of the area, the
(110)
error in standardisation is small
%
compared to the size
error in area is approximately 2 x
%
error in length.
Example 1.6 A chain is found to be 0*8 link too long and on using an area of 100 acres is computed.
it
2
™. The
-TqTT)\ = inn 1UU A00-8
.
true area
I
= 100 x 1-008 2 = 101-61 acres alternatively,
= 0*8%
linear error
= 2 x 0*8 = 1*6%
area error
••
= 100 + 1*6 acres = 101*6 acres
acreage
This
derived from the binomial expansion of (1 + x) z
is
= i.e .if /.
x
(1
+
is
x)
1
+ 2x + x 2
small x z may be neglected 2
a
1
+ 2x
Correction for slope (Fig. 1.3)
This may be based on (1) the angle of inclination, between the ends of the line.
in level
Fig. 1.3 (page 9)
Length
AC
measured
Horizontal length
AB
(/)
required (h)
Difference in level between
Angle of inclination
A and C
(a)
Correction to measured length (c)
(d)
(2) the difference
LINEAR MEASUREMENT h
Fig. 1.3 (1)
a
Given the angle of inclination
AB i.e.
N.B.
The
=
AC
cos a
a
h =
/
cos
c =
I
- h
=
I
-
=
/(1-cosa) =
I
(1.11)
cos
a
latter equation is a better
Example 1.7
AC = AB =
If
byEq.(l.ll)
126-3 m,
by Eq. (1.12)
c =
AB Example 1.8
versine a
(1-12)
computation process.
a = 2°34\
126-3 cos 2°34'
= 126-3 x 0-999 or
/
126-3
(1
=
126-174
m
- 0-999)
=
126-3 x 0-001
-
126-3 - 0-126 =
= 0-126
m
126-174
m
In chaining, account should be taken of any significant
effect of the slope of the ground on the accuracy of the horizontal
minimum angle
length. Calculate the
of inclination that gives rise to
relative accuracies of 1/1000 and 1/3000.
From Eq.
(1.12),
c = If
c
T
=
I
- h =
_J_ 1000
1(1 1
- cos a)
- cos a
SURVEYING PROBLEMS AND SOLUTIONS
10
cos a =
a = Also,
if
- 0-001 = 0-999
1
2°34'
1 in
22)
=
1
- cos a
cos a =
1
- 0-00033
==
-j
(i.e.
3000
= 0-99967
a = If the difference in level
(I
I
=
(/
=
I
2
39)
- d 2 y = j(/-d)x (/+ d)}*
(1.13)
- cf + d 2 - 2lc + c 2 + d 2
-d 2
c 2 - 2lc =
.-.
1 in
= h2 + d 2
2
or
2
(i.e.
known
d, is
,
h =
1°29'
2 c(c-2l) = -d
-d 2
c =
c-2l z
—d
~
c
as c is small compared
Rigorously, using the binomial expansion, c -
2
-
I
- d2 y
(I
-'-<-£)'
=
The use
d
2
Tl
d'
~ gji
(1.15)
of the first term only gives the following relative accura-
cies (the units may be
Gradient
ft
or metres).
Error per 100
lin4
0-051
ft
1 in
8
0-003
1 ft
1 in
10 20
0-001 3 0-000 1
1 in
•••
Thus the approximation
is
Chain surveying under
ft
Relative accuracy
m) (or m) (or m) (or m)
(or
ft ft
acceptable all
m
or
1/2000 1/30 000
1/80000 1/1 000 000
for:
general conditions.
Traversing, gradients up to 1 in 10. Precise measurement (e.g. base lines), gradients up to 1 in 20.
LINEAR MEASUREMENT For setting out purposes Here the horizontal length
11
given and the slope length
(h) is
(/) is
required.
= h sec a
/
a -
h
= h(sec a -
1)
c = h sec
Writing sec
a
as a series 1 + ^- +
Example 1.9 by Eq. (1.16)
If
where a
is in radians,
see
x
-s-
~
^(0-017 45a)
(«•
in radians)
4 x 10~ x
1-53
-
2 x 1*53 x 10"
ft
h = 100 ft (orm),
100(1-003
(1.17)
2
2i
a=
a2
(a in degrees) (1.18)
a2
per 100
ft (or
5°,"
820-1)
= 0-382 Oft (orm) per 100 ft (orm) 4 2 1-53 x 100 x 10" x 5
or by Eq. (1.18) c =
= 1-53 x 25 x 10~ 2 = 0-382 5 ft
(or
m) per 100 ft
(or
m)
Correction per 100ft (orm)
If
1°
0-015
ft
(orm)
6°
0-551
ft
(orm)
2°
0-061
ft
(orm)
7°
0-751
ft
(orm)
3°
0-137 ft (orm)
8°
0-983
ft
(orm)
4°
0-244
ft
(orm)
9°
1-247 ft (orm)
5°
0-382
ft
(orm)
10°
1-543
the difference in level, d, is given, 2 I
(h
h
2
p. 72.
2
—
c =
•••,
„
i
ho.
^+
(1.16)
+ cf
= h 2 + d2 = h2 + d2
+ 2hc + c 2 = h 2 + d 2 c(2h + c)
= d2
c- -*2h + c
ft
(orm)
m)
(1.19)
SURVEYING PROBLEMS AND SOLUTIONS
12
(1.20)
2h or rigorously
+
8h
2h N.B.
If
known as
the gradient of the ground is
tal the angle of inclination
e.g.
1 in
a~
(1.21)
3
Sn
~
(1 rad
57
10 gives
1 vertical to
n horizon-
57" 3°)
5-7 c
10
To find the horizontal length h given measured length I
the gradient 1 in n
and the
z
n\fn + 1
V« 2 +
/
lny/n z + 1
h
-.
«2
As an
n2 +
i
1
(1.22)
+l
alternative to the above,
h
Fig. 1.4
=
I
but
if
- c
'"IT
the gradient is given as 1 in n, then
d
c*
h
^
— n I
2nH (1.23)
2n 2 J
\
This
is
only applicable where n > 20.
Example 1.10
a length of 300
If
ft
(orm)
is
measured on a slope of
in 3, the horizontal length is given as:
by Eq. (1.22)
h =
300x3^10
= 9Q x 3 1623 .
10
- 284-61
ft
(orm)
1
.
LINEAR MEASUREMENT To find
the inclined length
I
13
given the horizontal length h and the
gradient (1 in n)
±
=
2 V" +
h ,
1
n __
hy/n z +
1
(1-24)
Example 1.11 by Eq. (1.24)
If
h = 300 /
=
ft
3Q0
(orra) and the gradient is 1 in 6,
^
= 50 x 6-083
6
= 304 -15
The maximum length
1.42
A B,
P
ft
(or
of offsets from chain lines
P is measured from a chain line ABC measured instead of BP, due to an error a
point
is
m)
in
such a way that
in estimating the
perpendicular, Fig. 1.5.
Fig. 1.5
On plotting, P, is fixed from B, Thus the displacement on the plan due PP,
_ = (N.B.
1
to the error in direction
= B,P a(radians) /a,
206265
radian = 206265 seconds of arc)
a
x
SURVEYING PROBLEMS AND SOLUTIONS
14 If
i.e.
the
PP
maximum length
represents the minimum plotable point,
%
0*01 in which represents
xft, ^— 12
where x
is
the representative
fraction 1/x, then
la
0-000 83 x =
206 265 171-82 x I
a Assuming the maximum
error
4°,
i.e.
171-82* 14400
=
If
a=
14400
,
o-012 x
(1.25)
the scale is 1/2500, then x = 2500, and /
= 2500 x 0-012 = 30ft (^10 m)
If the point P lies on a fence approximately parallel to Fig. 1.6, then the plotted point will be in error by an amount
=
1(1- cos a).
(Fig. 1.5).
Boundary
line
C
fl
Fig.
,
If
a=
4°,
1 .6
0-01
= 12
Example 1.12
ABC, P^P2
(1
- cos a)
(1>26)
by Eq. 1.26 0-01 x
/
12 x
(1-0-9976)
0-35 x
Thus,
if
x=
(1.27)
2500, /
= 875 ft (267 m)
error due to this source is almost negligible and the offset is only limited by practical considerations, e.g. the length of the tape. that is It is thus apparent that in fixing the position of a point
The
critical, e.g. the corner of a building, the length of a perpendicular offset is limited to 0*012 x ft, and beyond this length tie lines are required,
LINEAR MEASUREMENT
15
the direction of the measurement being ignored, Fig. 1.7.
Fig. 1.7
1.43
Setting out a right angle by chain
From a point on
BA = BC (ii) From A and C measure triangles ADB and DCB are
(a)
(i)
(Proof:
as
the chain line (Fig. 1.8)
ABC
Measure
off
/
Z
\ \
ABD
=
DBC
= 90c
8
B Fig. 1.9
Using the principle of Pythagoras, zz =
By choosing
x z + y2
(Fig. 1.9)
suitable values the right angle
The basic relationship
x:y:z If
CD
=
7k / \
/
n =
AD
is a straight line)
Fig. 1.8
(b)
off
congruent, thus
::
set out.
is
2n + 2n +
I,
may be
l
:
2n(rc
+ l)
:
2n(n + l)+l.
l=3
2n(n + 1) = 4 2n(n + 1) +
Check: (2n +
1
\2n(n + 1) + l}
if +
\2n(n
+
1)|
=
5.
2
= (2n 2 + 2n + l)
2
= 4n 2 + 4n +
1
2
+ 4n 4 + 8n 3 + 4n 2
= 4n z + 8n 3 + 8n 2 + 4n +
1
(1.28)
SURVEYING PROBLEMS AND SOLUTIONS
16
= (2n 2 + 2n + If. 2
Check:
5
- 32 + 42
25 - 9 + 16.
or
Similarly, if n = 3/4,
6 In +
1
10
,
= 4
2n(rc
+
1)
2n(n + 1) + 1
6/3
=
=
+1 \ ,
4U
=
=
6
X
4
j
40 16
4
16
7
=
42
4
16
16
58 16
16
Thus the ratios become 40 42 58 and this is probably the best combination for 100 unit measuring equipment; e.g. on the line ABC, Fig. 1.10, set out BC = 40 units. :
:
Then holding the ends of the chain at B and C the position of D is fixed by pulling taut at the 42/58
on the chain.
Alternative values for n give the following:
1.44
n = 2
5, 12,
13
n = 3
7, 24,
25
n = 4
9, 40, 41.
To
(Probably the best ratio for 30
find the point on the chain line
ular from a point outside the line
(1)
When
the point is accessible
From the
(Fig. 1.11).
the chain of length >
chain line position ab.
B
at is
point
DB
D swing
to cut the
a and b. The required
then the mid-point of
m
tapes)
which produces a perpendic-
LINEAR MEASUREMENT
(2)
When
the point is not accessible
(Fig. 1.12).
Da
and
From D
set out lines
Db
and, from these lines, perpendicular ad and be. The inter-
section of these lines at the line
DX
X
gives the
which when produced
gives B, the required point.
To set out a line through a given point parallel to the given chain line (Fig. 1.13).
Given the chain and the given point C. From the given point C bisect the line
AB
line CB at X. Measure AX and produce the line to D such that AX
= XD.
CD
1.45
Obstacles in chain surveying
will then
be parallel to AB.
17
SURVEYING PROBLEMS AND SOLUTIONS
18 (1)
Obstacles to ranging
(a) Visibility from intermediates (Fig. 1.14). Required to line
C
and
D
on the line AB. Place ranging pole at d, and line in c, on line Ad From B observe c, and move d 2 on to line Be, Repetition will produce c2 c 3 and d 2 d 3 etc until C and D lie on the line AB. .
}
.
,
,
(b) Non-visibility from intermediates (Fig. 1.15).
Required to measure a long line AB in which A and B are not intervisible and intermediates on these lines are not possible.
Set out a 'random line' AC approximately on the line AB.
From B
B Fig. 1.15
AC as above. Measure and BC. Calculate AB.
BC AC (2)
o*= A
find the perpendicular
to line
Obstacles
No
(a)
to chaining
obstacle to ranging
Obstacle can be chained around. There are many possible variations depending on whether a right angle is set out or not. (i)
A6
Fig. 1.16
By
(m)
setting out right angles
I
Set out equal perpendiculars
II
Set out Bb. Measure
Bb and
Bb and Cc; then be - BC. bC. Compute BC.
HI Set out line Bb. At b set out the right angle to give C on the chain line. Measure Bb and bC. Compute BC. IV and V Set out parallel lines be as described above to give similar figures, triangles BCX and bcX.
Then
BC =
be x
bX
BX
(1.29)
LINEAR MEASUREMENT
19
Fig. 1.17
VI
Set out line be so that
BC* =
CO
2
bB = Be. Compute BC
thus,
Be + jCcf bB _
(1.30)
be but .
-
bB = Be,
RC ^
z
=
2 BbjbC - Cc 2)
2fib
=
UbC z +
_. DK -BbxBb
Cc 2) - Bb 2
(1.31)
Proof. In Fig. 1.18 using the cosine rule ,2
?
2
_ xv
2
+ d
2
(assuming 6 > 90°), see
+ 2xd cos0
and
= y 2 + d 2 - 2yd cosO
2d cos 6 = p _ y2
JT
+d 2_
?
2
2 2 2 2 2 2 P y - * y - d y = xy + d x - q x 2 d (x + y) =
2
2 x + p y - xy(x +
y)
p.
81
SURVEYING PROBLEMS AND SOLUTIONS
20 z
d
2
=
2
q x + p y
- xy
x + y
+ p y - xy x + y
Q x
--/ If
x =
(1.32)
y,
Obstacle cannot be chained A river or stream represents this type of obstacle. Again there are many variations depending on
(ii)
around. .
whether a right angle
is set out or
not.
By
setting out right angles (Fig. 1.19).
A random line DA^ is set out C and B are obtained.
and from perpendiculars
points
By
similar triangles
DC,C and C B B 2
DC CB DC
CC Bfi,
,
1
1
-
X
CC y
CB
x CC,
BB. - CC,
Fig. 1.20
at
C and B
.
LINEAR MEASUREMENT
21
Without setting out a right angle (Fig. 1.20). A point F is chosen. From points B and C on line AE, BF and CF are measured and produced to G and H. BF = FG and CF = FH. The intersection of DF and GH produce to intersect at J. Then
HJ = CD. (iii) Obstacles which obstruct ranging and chaining. The obstruction, e.g. a building, prevents the line from being ranged and thus produced
beyond the obstacle.
By
setting out right angles (Fig. 1 .21) On line ABC right angles are set out at
and
C,
,
B, C,
B
and
C
to produce
B '
where SB, = CC, is
now produced
D and line ABC
to give D,
and
E
t
where right angles are = CC, D and E are
D^D=E^E=BB
set out to give
E, where
thus on the
produced and 0,0, = DC.
.
y
A
A
Fi B-l.21
Fig. 1.22
Without setting out right angles (Fig. 1.22) On line ABC, CB is measured and G set out to form an equilateral triangle,
An
i.e.
CB= CG =
BG is produced to J. HKJ sets out the line JE
BG.
equilateral triangle
such that
JE =
BJ.
A duced
further equilateral triangle
ELD
will restore the line
.
The missing
length
BE
= BJ = EJ
.
ABC
pro-
SURVEYING PROBLEMS AND SOLUTIONS
22
Exercises
The
1.
1 (a)
following measurements were
Reduce the slope distances (a) 200-1 yd
at 1 in
2^
(b) 485*5 links
at 1 in
5-75
at 1 in
10-25
1/24 th of a mile
(c)
made on
inclined ground.
to the horizontal giving the answer in feet.
(a) 557-4 ft
(Ans.
(b) 315-7 ft
(c) 218-9 ft)
Calculate the acreage of an area of 4 in 2 on each of the plans drawn to scale, 2 chains to 1 in., 1/63 360, 1/2500 and 6 in. to 1 mile 2.
respectively. 1-6, 2560, 3-986, 71-1 acres)
(Ans. 3.
A
field
was
area thus found
measured with a chain 0-3 of a link too long.
was 30
The
acres. What is the true area? (I.C.E. Ans. 30-18 acres)
4.
State in acres and decimals thereof the area of an enclosure meain. square on each of three plans drawn to scale of 1/1584,
A survey line was measured on sloping ground and recorded as 5. 386-6 ft (117-84 m). The difference of elevation between the ends was 19-3
ft
(5-88 m).
The tape used was
later
found to be 100-6 ft (30-66 m) when com-
pared with a standard of 100 ft (30-48 m). Calculate the corrected horizontal length of the line. 388-4 ft (118-38 m)) (Ans. 6.
A
plot of land in the form of a rectangle in
twice the width has an area of 180000 ft
which the length
is
2 .
Calculate the length of the sides as drawn on plans of the following scales. (a) 2 chains to 1 inch,
(Ans. 7.
(a) 4-55
(b)
1/25000.
(c) 6 inch to 1 mile.
x 2-27 in. (b) 0-29 x 0-14
(c) 0-68
in.
x 0-34
in.)
Express the following gradients in degrees to the horizontal: 1 in 0-5, being vertical to horizontal in each case. (b) Express the following scales as fractions: 6 in. to 1 mile, 1 in. (a)
1 in 3,
1 in 200,
to 1 mile, 1 in. to 1 chain, 1/8 in. to 1 (c)
ft.
Express the following scales as inches to
1 mile:
1/2500,
1/500, 1/1080.
(M.Q.B./UM Ans.
(a) 18°26', 0°17', (b)
63°26'
1/10560, 1/63360, 1/792, 1/96
(c) 25-34, 126-72, 58-67)
LINEAR MEASUREMENT
23
8. Find, without using tables, the horizontal length in feet of a line recorded as 247*4 links when measured
On ground sloping 1 in 4, (b) on ground sloping at 18°26' (tanl8°26' (a)
(Ans.
Show
9.
(a)
= 0-333). 158-40 (b) 154-89
that for small angles of slope the difference
ft)
between the
horizontal and sloping lengths is h z /2l (where h is the difference of vertical height of the two ends of a line of sloping length /). If errors in chaining are not to exceed 1 part in 1000, what is the greatest slope that can be ignored?
(L.U./E Ans. 1.5
1 in 22*4)
Corrections to be Applied to Measured Lengths
For every linear measurement the following corrections must be considered, the need for their application depending on the accuracy required. 1. In all
cases
(a) Standardisation. 2.
(c) Sag. 3.
(b) Slope.
For relative accuracies of 1/5000 plus (a) Temperature. (b) Tension, (where applicable)
For special cases, 1/50000 plus (a) Reduction to mean sea level.
(b)
Reduction to
grid.
Consideration has already been given, p. 6/9, to both standardisation and reduction to the horizontal as they apply to chain surveying but more care must be exercised in precise measurement reduction. 1.51
Standardisation
The measuring band in the form of a tape or wire must be compared with a standard under specified conditions of temperature (t s ) and tension (Ts ) If there is any variation from the nominal length then a standardisation correction is needed as already shown. A combination of temperature and standardisation can be seen under correction for tem.
perature.
1.52
Correction for slope
Where the inclination of the measured length is obtained by measurement of the vertical angle the following modification should be noted.
Let the height of the instrument be the height of the target
ft,
h2
SURVEYING PROBLEMS AND SOLUTIONS
24
the measured vertical angle the slope of the measured line the length of the measured line
6
a /
Fig. 1.23
a = d + 8$. A B Z B^ by the
In Fig. 1.23,
In triangle
sine rule
(see
A
(h,
-h z )
(fc,
- h 2 ) cos 6
sin 86
p. 80),
sin (90 + 0) I
(1.34)
/
206 265 (h
86"
N.B.
}
_
-h z ) cos 6
(1.35)
The sign
of the correction conforms precisely to the equation.
(1) If
= h z Z < h z and 6
is
h,> h z and 6 a if < h z and 6
is
/i
(2) If h\ (3) If
(4) If
+ve,
a = 6 86 = -ve (Fig. 1.24a) is 86
is
-ve,
86
is
is
+ve,
86
is
-ve,
86
is
,
i
/i,
-ve +ve +ve
(Fig. 1.24d) (Fig. 1.24b)
(Fig. 1.24c)
Example 1.13 If
^ /
= 4-5ft(l-37m),
= 5-5ft (1.68m),
h z
= 350
then 86 =
ft
(106-68 m)
206 265 (4-5 - 5-5) cos4°30'
350 - -588" = -0°09'48"
a = +4 o 30'00" - 0°09'48" = +4° 20 '12"
6 = +4°30'
LINEAR MEASUREMENT
25 /> 4-/>,
(a)
(b)
(c)
(d)
Fig. 1.24
Correction to measured length (by Eq. 1.12), c = -1(1
-cos0°)
= -350(1
-cos 4° 20'
12")
= -350(1-0-99714) = -350 x 0-00286 = •••
If
the effect
Horizontal length =
was
- 1-001
ft
(0-3051 m)
348-999 ft (106-3749 m)
ignored;
Horizontal length = 350 cos 4° 30'
= 348-922 •••
Error =
0-077
ft
(106-3514 m)
ft
(0-0235 m)
:
SURVEYING PROBLEMS AND SOLUTIONS
26
Correction for temperature
1.53
The measuring band If
standardised at a given temperature
is
in the field the temperature of the
band
recorded as
is
m)
(*
(t8 ).
then the
band will expand or contract and a correction to the measured length is
given as
c =
where
/
-
la(tm
(1.36)
ta )
= the measured length
a=
the coefficient of linear expansion of the band metal.
The coefficient
of linear expansion (a) of a solid is defined as
'the increase in length per unit length of the solid
when
its
temper-
ature changes by one degree'.
a
For steel the average value of
is
given as
6-2 x 10" 6 per
Since a change of 1
°F =
°F
a change of 5/9 °C, using the value above
gives
a =
6 6-2 x 10" per 5/9
10" 6
= 11-2 x
The range
a
of linear coefficients
per
°C
is thus
per
per
To find the new standard temperature
1°C
10-6 to 12-2 (x 10 5-4 to
to 4
3
Invar
given as
1°F
5-9 to 6-8
Steel
°C
t' 8
7-2
-6
(xl0~
)
7 )
which will produce the
nominal length of the band. Standard length at
To
ta
=
reduce the length by 81
where
t
t
I
± 81
5/:
=
(/
= number of degrees of temperature change required 81
= (I
';
As
± 5/).a.r
-
81 is small compared with
K
»•
/,
=
±8t)a *
for practical
ts
(137>
orkz ± 81 £f la
purposes (1.38)
LINEAR MEASUREMENT
27
Example 1.14 A traverse line is 500 ft (152*4 m) long. If the tape used in the field is 100 ft (30-48 ra) when standardised at 63 °F (17-2 °C), what correction must be applied if the temperature at the time of measurement is 73 °F (22-8 °C)? 6 6-2 x 10~ per deg F = 11-2 x 10~ 6 per deg C)
(Assume a =
From Eq.
(1.36) 6 = 500 x 6-2 x 10~ x (73 - 63)
m
c
= +0-031 Oft or
= 152-4 x 11-2 x 10" 6 x (22-8 - 17-2)
c(m)
= +0-009 6 m
Example 1.15
If a field tape when standardised at 63 °F measures 100-005 2 ft, at what temperature will it be exactly the nominal value?
10" 6 per
(Assume a = 6-5 x SI = +0-0052 •'.
from Eq. (1.37)
t'
ft
0-0052 6 100 x 6-5 x 10"
=63 =
deg F)
63°F-8°F
= 55
°F
In its metric form the above problem becomes: If a field tape when standardised at 17-2 °C measures 100-005 2 m, at what temperature will it be exactly the nominal value?
10" 6 per
(Assume a = 11-2 x 81 .'.
from Eq.(1.37)
t's
deg C)
= +0-005 2 m = 17#2 _
0-0052 100 x 11-2 x 10" 6
=
17-2
°C -
=
12-6
°C (=54-7°F)
4-6
°C
Correction for tension
1.54
The measuring band
is standardised at a given tension
(Ts ).
If in
the field the applied tension is (T ) then the tape will, due to its m elasticity, expand or contract in accordance with Hooke's Law.
A
own
correction factor is thus given as c =
L(Tm - Ts )
A E _
(1.39)
SURVEYING PROBLEMS AND SOLUTIONS
28
where
L =
the measured length (the value of c is in the as L),
A =
cross-sectional area of the tape,
E The
= Young's modulus of elasticity
A
units used for T,
or
or
4, (cm
T2 (N)
A,(m
(lbf/in
)
2
(metric)
)
2
E 2 (N/m
)
stress/strain.
i.e.
2
E, (kgf/cm
)
2
unit
must be compatible, e.g.
E
(in ) 2
T,(kgf)
E
and
2
A
T(lbf)
same
(new
)
S.I.
units)
Conversion factors lib = 0-453592 kg 1 in .'.
2
lib/in 2
Based on the proposed use
The force
1 lbf
1
m2
703-070 kg/m 2
=
of the International
units) the unit of force is the
accelerate a mass of 1 kg
4 = 6-451 6 x 10~
Newton
System
of Units (S.I.
(N), i.e. the force required to
metre per second per second
= mass x gravitational acceleration = 0-453592 x 9-80665 m/s 2 (assuming standard value) = 4-448 22N
lkgf = 9-806 65 N 2
whilst for stress 1 lbf/in =
For steel,
E ~ ~ ot
For invar,
N.B.
E ~
(1) If
kg = 2-204 62
(1
N/m
6894-76
lb)
2
28 to 30 x 10 6 lbf/in 2
(British units)
5 20 to 22 x 10 kgf/cm 2
(Metric units)
10
19-3 to 20-7 x 10
20 to 22 x 10
6
N/m 2
lbf/in
^
14 to 15-5 x 10
~
13-8 to 15*2 x 10
5
(S.I. units)
2
kgf/cm 2
Tm = Ts no correction
,0
N/m 2
is
necessary.
considered good practice to over tension to minimise deformation of the tape, the amount of tension being strictly recorded and the correction applied.
(2) It is generally
(3)
The cross-sectional area
of the tape
may be physically or it may be com-
measured using a mechanical micrometer,
puted from the total weight W of the tape of length a value p for the density of the material.
L
and
W
A - j-
(1.40)
LINEAR MEASUREMENT
29
Example 1.16 A tape is 100 ft at a standard tension of 251bf and measures in cross-section 0-125 in. x 0-05 in. If the applied tension is 20 lbf and E = 30 x 10 5 lbf/in2 calculate the correction to be applied. ,
Rv F„1C.Q By t-q. l.iy
10 ° X (20 - 25) .
. = c
i
(0-125 x 0-05) x (30 x 10 6 )
= -0*009 7 ° °° 27
ft
"
Converting the above units to the metric equivalents gives 30-48 m x (9-072 - 11-340) kgf (40-32 x 10~ 7 )m2 x (21-09 x 10 9 ) kgf/m2
c =
= -0-008 13
Based on the
m
(i.e.
-0-002 7 ft)
International System of Units,
2-268 kgf = 2-268 x 9-806 65 or
5 lbf =
x
5
N 4-448 22 N
= 22-241 = 22-241 N.
For stress, 10 (21-09 x 10 ) kgf/m 2
= 21-09 x 10 9 x 9-80665 = 20-684 x 10 ,o N/m 2
or
(30 x 10
6 )
Thus, in
lbf/in2
30 x 10
=
6
x 6894-76 = 20-684 x 10'°
N/m2
S.I. units,
30-48 x 22-241 c = (40-32 x 10
= - 0-008 13
-7 )
x (20-684 x 10 ,o )N/m2
m
Measurement in the vertical plane Where a metal tape is freely suspended
it will elongate due to the applied tension produced by its own weight. The tension is not uniform and the stress varies along its length. Given an unstretched tape AB and a stretched tape AB, Fig. 1.25,' , let P and Q be two close points on the tape which become ^Q, under tension.
If
Let
AP =
PQ
x,
AP
t
= dx, then
= x + P,
s,
where s
is the
amount of elongation of
Q, = dx + ds and the strain in
%Q,= —
AP
as ds
is the
increase in length and dx is the original length. If T is the tension at P, in a tape of cross-section A, and Young's modulus, then ,
T = Given that the load per unit length
at P,
in tension
is
m
EA*1
(!)
dx = w.dx being the difference
E
is
w
between
then in P,
P, Q, the load and Q,
SURVEYING PROBLEMS AND SOLUTIONS
30 if
the tension at
Q
is
T + dT
T-(T + i.e.
wdx dT = -wdx
dT) =
(2)
x+s
dx+ds
Fig. 1.2 5 Elongation in a suspended tape In practice the value w is a function of x and by integrating the two equations the tension and extension are derived.
Assuming the weight per pended weight W, then from
unit length of the tape is
w
with a sus-
(2)
dT = -wdx T = -wx + c T =
and (1)
EA
ds = dx
-t—
EAs = When T =
x=
W,
I
and
d
EA^. dx
-wx
+ c
iwx 2 + ex + d
and when x =
W = = 0.
wl + c
(3)
0, s
i.e.
(4)
=
c =
W +
wl
LINEAR MEASUREMENT
31
Therefore putting constants into equations (3) and (4) gives
T = -wx + W + wl T = W + w(l-x).
EAs =
and
- 1 wx
2
(i.4i)
+ Wx + wlx
= Wx + ±w(2lx- x2 )
ml Wx +
= If
x =
/,
if
W
w & lx -* 2 )]
d-42)
then
S= and
i
=
A[^ +
I
w/2
(1-45)
]
0,
S
~
(1.44)
2E4
Example 1.17 Calculate the elongation at (1) 1000 ft and (2) 3000 ft of a 3000 ft mine-shaft measuring tape hanging vertically due to its own weight. The modulus of elasticity is 30 x 10 6 lbf/in 2 the weight of the ;
tape is 0*05 lbf/ft and the cross-sectional area of the tape is 0*015 in 2
From Eq.(1.42) s =
As
W =
s= when
ljz[
w*+ i"(2lx-x z )]
0,
m
[2lx - x * ]
x
= 1000 ft
/
= 3000 ft 0-05
S =
2 x 30 x 10* x 0-015
~
when
x =
C2
X 300° X 100 ° " 100(f]
^xl0«
°- 05 2 x 30 x 10° x 0-015 /
= 3000 ft.
2
From Eq. (1.44)
s =
—^L 2EA
0-05x3000 2 2 x 30 x 10 6 x 0-015
„„„„,ft = 0-500
.
SURVEYING PROBLEMS AND SOLUTIONS
32
Example 1.18 If the same tape is standardised as 3000 ft at 451bf tension what is the true length of the shaft recorded at 2998*632 ft?
s- -S? -i™: 2EA EA
I„Eq.(1.44)' i.e.
where
W=
T =
total weight of tape
\W = 3000 x 0-05 = 150
Ibf
Applying the tension correction, Eq.(1.39), C
L(Jm - T s ) =
EA 3000(75-45)
=
6
30 x 10 x 0-015 .'.
2 30 x 10 x 30 = 6 30 x 10 x 0-015
.
200ft
true length
= 2998-632 + 0-2
= 2998-832
ft
Correction for sag
1.55
The measuring band may be standardised flat or (b) in If
in
two ways,
(a)
the band is used in a manner contrary to the standard conditions
some correction (1) //
is
necessary.
standardised on the flat and used in catenary the general
equation for correction
is applied, viz.
c = -
w2
3 /
r-^2
d-45)
24 T
W2
l
(1.46)
24 T
where
w W
= weight of tape or wire per unit length = wl = total weight of tape in use,
T = applied N.B.
on the
catenary.
the units
w
tension.
and T must be compatible
(lbf,
kgf or N)
Measuring Catenary curve head Fig. 1.26 Measurement in catenary (2) //
standardised in catenary
(a)
The
length of the chord
may be given
relative to the length
.
LINEAR MEASUREMENT
33
of the tape or (b) the length of the tape in catenary (i)
If
may be
given.
the tape is used on the flat a positive sag correction must be
applied If the tape is used in catenary at a tension T m which is different from the standard tension T the correction will be the difference s between the two relative corrections, i.e.
(ii)
,
C If
Tm >
Ts
Wz l
IT
r 1
(1-47)
--24[n;-Tl\
the correction will be positive.
If standardised in catenary using a length l a and then applied in the field at a different length l m , the correction to be applied is
(iii)
given as
3
L. //,
c = ?i
w2
h \24I
HW'
2
24T
Alternatively, the equivalent tape length on the flat may be computed for each length and the subsequent catenary correction applied for the new supported condition, i.e. if /, is the standard length in catenary, the equivalent length on the ground = /. + c„ where c = the s catenary correction. If
=
flat
W l
m
is
the applied field length, then its equivalent length on the
+ cs )
Applying the catenary correction to this length gives lm + C =
=
Thus
£
(l s
+ C j _ Cm
+^
L
the required correction
'm^s i
/'> 2 \
I l
— Cm
s
\24T
j-^yi
3 l
Z
241'
I
( Z*
w2
~
l
D
as Eq.
(1 .48)
above
SURVEYING PROBLEMS AND SOLUTIONS
34
an acceptable approximation based on the assumption that the measuring heads are at the same level. If the heads are at considerably different levels, Fig. 1.27, the correction should be
The sag correction
is
c =
c,
(l±^sin*)
cos 2
(1.49)
the sign depending on whether the tension is applied at the upper or
lower end of the tape.
Measuring head
Fig. 1.27
For general purposes c =
c,
cos 2
w 2 Pcos 2 g 24 T
(1.50)
2
The weight of the tape determined in the field The catenary sag of the tape can be used to determine the weight
of
of the tape, Fig. 1.28
Fig. 1.28 Weight of the tape determined in the field If
y
is the
measured sag
at the mid-point, then the
weight per unit
length is given as
w
=
STy 2
(151)
i
or the
amount
where
w
of
sag
= weight/unit length, T = applied tension,
y =
wl
z
ST
(1.52)
y = vertical sag at the mid-point, / = length of tape between supports.
LINEAR MEASUREMENT
35
Example 1.19 Calculate the horizontal length between two supports, approximately level, if the recorded length is 100.237 ft, the tape weighs 15 ozf and the applied tension is 20 lbf
From Eq.(1.46)
W* 1
c = -
24T 2 The value
of
/
is
assumed to be 100 5\
for
ease of computation.
2
jjx 100
Then 24 x 20 2
= -0-0092 True length
ft
= 100-2370 - 0-0092
= 100-227 8 ft
A
Example 1.20
100 ft tape standardised in catenary at 25 lbf is used 20 lbf. Calculate the sag correction if
in the field with a tension of
w
= 0-021
lbf/ft.
From Eq. (1.47) c = -[1^1
-
100
3
T2
T2
)
x 0-021 2 / 1 5 24 (20
= -0-01656
i.e.
_1_
"25 1
-0-016 6 ft.
Example 1.21 A tape 100 ft long is suspended in catenary with a tension of 30 lbf. At the mid-point the sag is measured as 0-55 ft. Calculate the weight per ft of the tape.
From Eq. (1.51),
8Ty
T
Based on 1-
S.I. units
8 x 30 x 0-55
-
10000
= 0-0132 IWft.
these problems become
19(a)
Calculate the horizontal length between two supports approximately level if the recorded length is 30-552 2 m; the tape weighs 0-425 kgf and the applied tension is 9-072 kgf.
Converting the weight and tension into units of force, V2 30-5522 VJA Z, \VftZ.D JV (0-425 x X. 9-806 3'OUD 03 65)
24
(9-072 x 9-806 65) 2
SURVEYING PROBLEMS AND SOLUTIONS
36
Thus there
is
no significance in changing the weight
W
and tension T
into units of force, though the unit of tension must be the newton.
30-552 2
/
0-425 \ 2
24
\
9-072/
= -0-002 8 m 1.20(a)
A
in the field
w=
30*48
m
(-0-009 2 ft).
tape standardised in catenary at 111-21
N
is
used
Calculate the sag correction
with a tension of 88-96 N.
0-0312kgf/m. Conversion of the mass/ unit length
w
if
into a total force gives
30-48 x 0-0312 x 9-80665 = 9-326 N.
Eq.
becomes
(1 .47)
c =
24
'-- -) 2 z T
T
-30-48 x 9-326
=
\88-96
24
-0-00504 m
= 1.21(a)
A
tape 30-48
of 133-446 N.
m
2
111-21
2
/
(-0-016 6 ft).
long is suspended in catenary with a tension
At the mid-point the sag
is
measured as 0-168 m. Cal-
culate the weight per metre of the tape.
Eq. (1.51) becomes (/ , w n(kgf/m)
=
8 x
—
0-816 Ty T x y = 2
9-80665
2
/
I
0-816 x 133-446 x 0-168
30-48
= 0-019 6 kgf/m
2
(0-013 2 lbf/ft)
A tape nominally 100 ft is standardised in catenary at lOlbf and is found to be 99-933 ft. If the weight per foot is 0-01 lbf, calculate the true length of a span recorded as 49*964 ft. Example 1.22
= 99-933
Standardised length
Sag correction
for
100 ft
—
Q-01
c1 i
ft
=
True length on the
2
T^T 7Ta 24 x 10
flat
3
x 100-
0-042 ft = rtftyIoft .
= 99-975
ft
LINEAR MEASUREMENT True length of sub-length on =
37
flat
49-964
x 99-9747 = 49-952
1Q0
Sag correction for 50
ft
(c
3
oc
/
)
= l/8c,
= -0-005
True length between supports
=
49- 947 ft
Alternatively, by Eq.(1.48) 2
°
50 x 0-01 100 x 24 x 10»
=
(5 °
° " 10 ')
= -0-018 ft ."•
true length
between supports = 49-964 - 0-018
= 49-946 ft.
Example 1.23 A copper transmission line, %in. diameter, is stretched between two points, 1000ft apart, at the same level, with a tension of
ftton,when the temperature
90 °F.
necessary to define its limiting positions when the temperature varies. Making use of the corrections for sag, temperature and elasticity normally applied to base line measurements by tape in catenary, find the tension at a temperature of 10 °F and the sag in the two cases. Young's modulus for copper 10 x 10 6 lbf/in 2 , its density 555 lb/ft 3 and its coefficient of linear expansion 9-3 x 10~ 6 per °F. is
and h = height of line above or below mean sea level, then
is
L L
-
L =
R
R±h R±h Fig. 1.29 Reduction to
If
L = lm T
c,
mean sea
level
then
c ~
l„R
/mT /f±T (1.53)
= T
R±h
LINEAR MEASUREMENT
As h
is
small compared with R,
c
39
= ± ~^—
(1.54)
R If
R ^ 3960
miles, 100ft
3960 x 5280 4-8/j
xl0~ 6 per 100
ft
(1.55)
Reduction of ground length to grid length
1.57
The
local scale factor depends on the properties of the projection. Here we will consider only the Modified Transverse Mercator projection as adopted by the Ordnance Survey in the British Isles.
Local scale factor (F)
(•=)
F = Fo\l + where
(1.56)
„
FQ =
the local scale factor at the central meridian,
E = p =
the Easting in metres from the true origin,
v =
the radius of curvature at right angles to the meridian.
the radius of curvature to the meridian,
Assuming
p~v
= R, then
F
=
F
1
+
I
(1.57)
~w)
For practical purposes,
F ~ F
N.B.
E=
(l
2 + 1-23 E x 10" 8 )
(158)
0-9996013(1 + 1-23E 2 x 10~ 8 )
(1.59)
Eastings - 400 km.
Central meridian
L.S.F.
LS E -=^) -
|
(
Local scale
error
T
0-9996 Sub-parallel
0-9996
180km
Sub-parallel
180km
Distance from
CM. Fig. 1.30
1-0
0.9996
in
km
SURVEYING PROBLEMS AND SOLUTIONS
40
The
local scale error as
shown on
the graph approximates to
2R'<
Example 1.24
Calculate (a) the local scale factors for each corner of
the grid square
TA
(i.e. grid
co-ordinates of S.W. corner 54), (b) the
local scale factor at the centre of the square, (c) the percentage error in
Calculate the local scale factors applicable to a place and to coal seams there at depths of 500 ft, 1000 ft, 1500 ft
and 2000 ft respectively.
Local radius of the earth = 6362-758 km L.S.F.
= 0-9996013(1 + E 2 x 1-23 x 10" 8 )
Correction of length to mean sea level
~
at
500ft
R -
L =
m
h
6362-758
/
6362-758 - (500 x 0-3048 x 10~ 3 ) 6362-758
=
/,
6362-758 - 0-152 .
6362-758
L 6362^606 at
at
1000
1500
ft T
ft
L =
,
2000 ft
L =
,
6362-758
m 6362-758 -
0-304
6362 758 '
I
m 6362-758 -
,
at
nnnnn 1'MOMH „
'
/
0-457
6362-758
6362-758 6362-758 - 0-610
6362-758 000096 '""6362048 " 1 ,
-
At mean sea level, Easting 415 km,
''
(Eq. 1.59)
.
SURVEYING PROBLEMS AND SOLUTIONS
42
2 8 L.S.F. = 0-9996013 [1 + (415 - 400) x 1-23 x 10" ]
= 0-9996040 at
500 ft below, L.S.F.
at
x 1-000024 = 0-999628
= 0-999604
1000 ft below, L.S.F. = 0-9996040 x 1-000048 = 0-999652
at
1500 ft below, L.S.F. = 0-9996040 x 1-000072 = 0-999676
at
2000 ft below, x 1-000096 = 0-999700.
L.S.F. = 0-999604
Example 1.26 An invar reference tape was compared with standard on the flat at the National Physical Laboratory at 68 °F and 201bf tension and found to be 100-024 Oft in length.
The first bay of a colliery triangulation base line was measured in catenary using the reference tape and then with the invar field tape at
F and with 20 lbf tension. The means of these measurements were 99*876 3 ft and 99*912 1 ft respectively. The second bay of the base line was measured in catenary using the field tape at 56 °F and 20 lbf tension and the resulting mean measurement was 100*213 5 ft. a temperature of 60°
Given: (a) the coefficient of
expansion
for invar
(b) the weight of the tape per foot run (c) the inclination of the (d) the
mean height
Assuming the radius
second bay second bay
of the
of the earth to
7 = 3-3 x 10"
= 0-00824
,
lbf,
= 3° 15' 00", = 820 ft A.O.D.
be 20 890 000 ft, calculate the
horizontal length of the second bay reduced to Ordnance Datum.
(M.Q.B./S)
To
find the standardised length of the field tape
Reference tape on the
flat at
68 °F = 100-024 Oft.
Temperature correction 7 = 100 x 3-3 x 10" x (60 - 68)
= -0-000 264 .*.
Reference tape
at
i.e.
-0-0003
60 °F
= 100-024
- 0-0003
= 100-023 7 ft.
LINEAR MEASUREMENT Thus the standardisation correction
-w
Sag correction
is
43
+ 0*023 7 ft per 100 ft.
2 3 /
24T
2
2 -(8-24 x 10~ 3 ) x 100 3
24 x 202
- 8-24
2
9600
The length
of 100
ft
is
= -0*0071
ft
acceptable in all cases due to the close approxi-
mation.
The
true length of the first bay thus
becomes
99-8763 - 0-0071 + 0-0237 = 99-892 9 ft.
The
field tape applied under the
same conditions when corrected
for
sag gives 99-9121 - 0-0071 = 99-905 Oft. The difference represents the standardisation correction
99-9050 - 99-8929 = +0-022
The corrections may now be applied
lft.
to the second bay
Standardisation
+
c = + 0-0221 x
100-21 99-91
0-0221
(the proportion is not necessary because of the
close proximity) Temperature, c =
L.a.(tm -
7 100 x 3-3 x 10"
ta)
x (56 - 60)
0-0001
Sag,
As before as Slope,
L (1 -
length
a 100 ft
0-0071
cos 6)
c = - 100-21(1- cos 3° 15')
= - 100-21 x 0-00161
Sea
level,
c =
0-1613
- Ih/R
- 100 x 820/20890000
0-003 9
0-022
0-1724 0-022
0-1503
1
.
SURVEYING PROBLEMS AND SOLUTIONS
44
Horizontal length reduced to sea level
= 100-2135 - 0-1503 = 100-063 2
The
Example 1.27
ft.
details given below refer to the
measurement of
the first '100ft' bay of a base line. Determine the correct length of the
bay reduced to mean sea level. With the tape hanging in catenary at a tension of 20 lbf and at a mean temperature of 55° F, the recorded length was 100-082 4 ft. The difference in height between the ends
was 1-52 ft and the
site
was
1600ft above m.s.l.
The tape had previously been standardised
in catenary at a tension
60 °F, and the distance between zeros was 100-042 ft. R = 20890000 ft. Weight of tape/ft = 0-013 lbf. SecE = 30 x 10 6 lbf/ in 2 Temperature tional area of tape = 0-005 6 in 2 coefficient of expansion of tape = 0-00000625 per 1°F. of 15 lbf and at a temperature of
.
,
(I.C.E.)
Correction
+
Corrections Standardisation
Tape
100-042
is
c = 0-042
.-.
ft
at 15 lbf tension
ft
per 100
and 60 °F.
0-0420
ft
Temperature
L a
c =
.
.
m-
(t
ts )
6 100 x 6-25 x 10" x (55-60) = -0-0031
=
0-003
Tension _
L(Tm
-T9
)
A.E 100 x (20 - 15) 6 0-0056 x 30 x 10
= +0-003
0-003
Slope c = 21
1-52
2
0-0116 200 Sag c = difference between the corrections for field
and standard tensions
0-0116
LINEAR MEASUREMENT
W2
C = -
I
1
TJ
24
W
wl = 0-013 x 100 = 1-3 lbf 1-32 2 x 100
=
c
45
24
15
2
- 20 2
15
2
x 20 2
= +0-013 7
0-013 7
Height c =
hl_
R 1600 x 100
20890000
= -0-007 7
0-0077
+ 0-0587 -0-0224
0-022 4
+ 0-0363 Measured length
+ 100-0824
Total correction
+
Corrected length 1.6 If
The Effect
0-0363 100-118 7 ft
of Errors in Linear
Measurement
the corrections previously discussed (pp. 23-40) are not applied Any errors within the for-
correctly, then obviously errors will occur. mulae produce the following effects.
1.61
Standardisation
Where a tape is found to deviate from standard, the error SI can be corrected in the normal way or by altering the standard temperature as previously suggested. 1.62
Malalignment and deformation of the tape (Figs. 1.32 and 1.33) (a) Malalignment. If the end of the tape is out of line by an amount d in a length /, the error will be
21 e.g.,
if
d=
3
in.
and /= 100ft,
(1.60)
v
SURVEYING PROBLEMS AND SOLUTIONS
46
=
-
1 in
i.e.
^L 200
JL
= 0-000 3 ft,
330000. e Fig. 1.32 Malalignment of the tape
(b)
Deformation in the horizontal plane.
If
the tape is not pulled
straight and the centre of the tape is out of line by d, then
2d d d = TTv"" -77 = T" '
ez
&*=.
4X1)
e.g., if
d= 3
in.
e2
1/_|
,„ _„ N < L61)
,.
Jtl
"Sp 1
&L =**o
Fig. 1.33 Deformation of the tape
and 1= 100ft, = 4 x
e,
= 0-00123,
(c) Deformation in the vertical plane.
i.e. 1 in
This
is the
80000.
same as
(b) but
more
Any obvious change in gradient can be allowed for by grading the tape or by measuring in smaller bays between these
difficult to detect.
points. In (a) and (b) alignment
N.B.
by eye
is
acceptable for all purposes
except very precise work.
Reading or marking the tape
1.63
Tapes graduated to 0-01 ft can be read by estimation to give a probable error of ± 0-001 ft. Thus if both ends o f the tape are read simultaneously the probable 2 be V(2 x 0001 ), i.e. 0-001 x y/2, i.e. ±0-001 4 ft. Professor Briggs suggests that the error in setting or marking of the end of the tape is 3 times that of estimating the reading, i.e. ±0*003 ft per observation.
error in length will
Errors due to wrongly recorded temperature
1.64
From the correction formula c =
and
I.
a.
m-
(t
t^),
dc = ladtn
(1.62)
j. = aStn
(1.63)
has been suggested from practical observation that errors in recording the actual temperature of the tape for ground and catenary meaIt
surement are ±
5°F
and ±
3°F
respectively.
.
LINEAR MEASUREMENT If
the error is not to exceed 1/10000, then from Eq. 1.63
8c
1
a 8t,
•=
10000 1UUUU
I
8t
i.e.
47
1
=
10000 a
a=
If
6 6-5 x 10" per deg F, then
8tm
Thus 5° produces an
— 10
=
~
15.4
6'5 X 10*
~ ^
error of
3° produces an error of 1.65
6
1/30000, 1/50000.
Errors due to variation from the recorded value of tension
These may arise from two sources: (a) Lack of standardisation of tensioning apparatus. (b) Variation in the applied tension during application (this is sig-
nificant in ground taping).
From the correction formula
c
8c
i.e.
LT
~ T*) _
AE
~~AE
T
AE
(1.64)
"
r
(1.65)
=
8Tm
IE
<
166 >
the error is not to exceed say 1 in 10 000, then
AE
L
oiTm
•
i-e-
If
H Tm
L8T, gc =
differentiation gives
If
c =
(1.39)
A = 0-003 in 2
,
E= -
i.e.,
a
—w—
30 x 10 lbf/in 2
8 Tm = an error of 1/10000 an error of 1/30 000
,
10000 -
then
0-003 x 30 x 10 6
is is
AE 1O(M)0
'~
= 91bf
-
produced by a variation of 91bf, produced by a variation of 3 Ibf
The tape cross-section is x/2 in. wide to give A = 0*003 in2 If the width of the tape be reduced to Kin. then, if the other dimensions .
SURVEYING PROBLEMS AND SOLUTIONS
48
2 remain constant, the cross-sectional area is reduced to %A = 0*000 8 in error of 1/10 000 In this case a variation of 3 lbf will produce an
and the accuracy will be reduced as the cross-sectional area diminishes. Errors from sag
1.66
Where the tape has been standardised on the flat and is then used measuring heads at different levels, the approximation formula is given as in catenary with the
-/ 3
w 2 cos 2
c =
where 6
is the
6 (1.50)
24I 2
angle of inclination of the chord between measuring becomes negligible when 6 is small.
2 heads. The value of cos
The sources (a)
of error are derived from:
an error in the weight of the tape per unit length, w,
(b) an error in the angular value, B, (c)
an error in the tension applied, T.
By successive
differentiation,
=
dcw
- 2/ 3 wcos 2 flg W 24 T
(1 67)
2
2cdW (1.68)
w 8c„,
28w
c
w
i.e.
(1.69)
This may be due to an error in the measurement of the weight of the tape or due to foreign matter on the tape, e.g. dirt.
8c e =
See
—
-l 3 w 2 2
sin20S0
(1.70)
= 2c tan 6 86
(1.71)
= 2tan0S0
(1.72)
c 2/
Sc T =
3
w2
24T 2c8T T
8c T _ " c
cos
-28T T
2
0ST
(1.73)
:
(1.74)
(1.75)
.
LINEAR MEASUREMENT The compounded effect of a variation 2l*w 2 cos 2 (9
in
Example 1.28 T = 10 ± 1 lbf, C =
Then~ i.e.
w =
10%
If
/
100 3 x 0-012
+
2c 8w
error in
2
= 2 x
18T
°* 04161
=
«
2c tan fl 86"
cos 2 2° -
x O'l = 0-008 32 ft
°-° 524 x 10 "2^6167 x ft.
obviously negligible.
Sc t = 10%
= 0>04161ft
~24 ^lf~
= 0-000 000 21
i.e.
$ = 2° ± 10"
ft,
0-08322 =
206 265
is
L76 )
weight produces an error of 1/12000.
806 =
This
<
~AE
0-01 ± 0-001 lbf per
x cos 2 2°
24xl0
~w~
w=
= 100 ft,
tension gives
8T
24T 2
49
~y-
= 0-083 22
xO-U
0-008 322
ft
error in tension produces an error of 1/12000.
Example 1.29 A base line is measured and subsequent calculations show that its total length is 4638-00 ft. It is later discovered that the tension was recorded incorrectly, the proper figure being
10 lbf less than that stated in the field book, extracts from which are given below. Assuming that the base line was measured in 46 bays of nominal length 100 ft and one bay of nominal length 38 ft, calculate the error incurred in
ft.
Extract from field notes Standardisation temperature
= 50 °F
Standardisation tension
= 20 lbf = 45 op _ 40 lbf = 30 x 10 6 lbf/in 2
Measured temperature Measured tension Young's modulus of tape Cross-sectional area of tape Weight of lin 3 of steel
= 0-125 = 0-28
in.
x 0-05
in.-
lbf.
(N.U.)
Weight of steel tape per
ft
= 0-125 x 0-05 x 12 x 0-28 = 0-021
lbf.
SURVEYING PROBLEMS AND SOLUTIONS
50
From Eq.
(1.39)
Then the
error
c =
L(Jm
'
J
s)
due to wrongly applied tension = c - c'
where
=
L(Tm -
=
M.
Tm = T'm
r.)
_ L(T^ - Ts )
AE
AE
AE
(T K m) m - T')
true applied tension,
= assumed applied tension.
Error
4638 ( 30 - 40 > 0-125 x 0-05 x 30 x 10
From Eq. .'.
=
(1.46) correction for sag c
. 6
. -0-24736ft.
—
z —-W -ji
Error due to wrongly applied tension
= c -
c,
w!i/i 24 \T
W = 100 x Error for 100
ft
Z
_ i_\ Tfj
0-021 = 2-1 lbf
bay 2-1
2
x 100 /Jl_ _ _1_\ 40 2 / 24 Uo2
441 / l600 - 900 \ = ~ 24 \l600 x 900/
_ 441 24
700
/
\1
\
440 000/
= -0-00893 Error for 46 bays =
-0-410 78 ft
Error for 38 ft bay
W = 38 n0t
x 0-021 = 0-798 lbf
\ 0-798 2 x 38 / 700 _ ~ _ 440 \l 24 000/
= - 0-000 49 ft .-.
Total error for sag
= -0-411 27 ft
Total error for tension
= - 0-247 36 ft
f
.
LINEAR MEASUREMENT Total error = - 0-658 63
Apparent reduced length
i.e.
0*658 6
is
51 ft
too large.
ft
Inaccurate reduction to the horizontal
1.67
The inclined length may be reduced by obtaining measuring heads or
(a) the difference in level of the
(b) the angle of inclination of the tape,
(a)
The approximation formula (Eq. 1.15)
Adopting the
C first
given as
is
d
2
...)
=2l
term only, from the differentiation
d8d
8c _
(1.77)
I
2c 8d (1.78)
d
28d d
8c c If
8d/d= 1%, when
/
= 100 and
d=
(1.79)
5 ± 0-05
ft,
2
* = 8c
As ± 0-01
2 x 5 x 0-01 o T7^ 2 x 100
=
± 0-002 5
By
i .e
.
1/200 000
trigonometrical observations (Eq. 1.12),
Then
i.e. /
40000.
the difference in level can be obtained without difficulty to
c =
If
i.e. 1 in
ft,
8c = ± 0-000 5 ft, (b)
t
= 100, B = 30° ±
8c = ±
1(1
- cos0)
8c = /sin0S0
8c
(1.80)
_ "
/sinflgfl"
206265
C1,81)
oc
sin0 86
(1.82)
20",
10 ° * 5 2° °; * -
±0-0048ft
i.e.
2Uo 265
The accuracy obviously improves as 6 is reduced. As the angle of inclination increases the accuracy ment of 6 must improve.
1/20000.
in the
measure-
SURVEYING PROBLEMS AND SOLUTIONS
52
Errors in reduction from height above or below
1.68
From
mean sea
level
the formula
Ih
8c =
by differentiating
(1.83) R.
=
The % or
cSh T£-
(1.84)
%
error in the correction is equal to the
error in the height
above
below M.S.L.
1.69
Errors due to the difference between ground and grid distances
Local scale factor
is
given by Eq.(1.59)
2 8 0-9996013(1 + 1-23 E x 10~ )
where E is the distance in km from the central meridian (i.e. the Eastings - 400 km). As this amounts to a maximum of 0*04% it is only effective in precise surveys Exercises 1(b) 10.
A
300ft tape has been standardised at 80 °F and
A
its true
length
measured at 75 °F and reat this temperature is corded as 3486*940 ft. Find its true length assuming the coefficient of -6 per deg F. linear expansion is 6*2 x 10 300*023
ft.
line is
(Ans.
A base
3487- 10
ft)
10560 ft long when measured in catenary using a tape 300ft long which is standard without tension at 60 °F. The tape in cross-section is 1/8 x 1/20 in. 11.
line is found to be
If one half of the line is measured at 70 °F and the other half at 80 °F with an applied tension of 501bf, and the bays are approximately
equal, find the total correction to be applied to the measured length.
Coefficient of linear expansion =
Weight of
3 of steel 1 in
Young's modulus
12.
A
100
ft
6*5 x 10
= 0-28
-6
per deg F.
lbf 6
= 29 x 10 lbf/in 2 (Ans. .
-3 -042 ft) when The cross-
steel tape without tension is of standard length
placed on the ground horizontally
at a
temperature of 60 °F.
sectional area is 0*0103 in 2 and its weight 3*49 lbf, with a coefficient 6 of linear expansion of 6*5 x 10~ per deg F.
The tape
is
used in the field in catenary with a middle support such
LINEAR MEASUREMENT that all the supports are at the
same
53
level.
Calculate the actual length between the measuring heads if the temperature is 75 °F and the tension is 20 lbf. (Assume Young's mod6 ulus 30 x 10 lbf/in 2 ).
99-985 lft)
(Ans.
A nominal distance of 100 ft was set out with a 100 ft steel tape 13. from a mark on the top of one peg to a mark on the top of another, the tape being in catenary under a pull of 20 lbf and at a mean temperature of 70 °F. The top of one peg was 0*56 ft below the top of the other. The tape had been standardised in catenary under a pull of 25 lbf at a
temperature of 62 °F.
Calculate the exact horizontal distance between the marks on the two pegs and reduce it to mean sea level. The top of the higher peg was 800ft above mean sea level. 6 (Radius of earth = 20:9 x 10 ft density of tape 0*28 lbf/in 3 ; section of tape = 0*125 x 0*05 in.; Young's modulus 30 x 10 6 lbf/in 2 ; ;
coefficient of expansion 6' 25 x 10" 6 per 1° F)
Ans.
(I.C.E.
A
14.
steel tape is found to be 299-956
sion of 12
lbf.
The tape has
99-980 4
ft)
long at 58 °F under a ten-
ft
the following specifications:
Width
0-4 in.
Thickness
0-018
in. 5
Young's modulus of elasticity
30 x 10 lbf/in 2 Coefficient of thermal expansion 6-25 x 10" 6 per deg F. Determine the tension to be applied to the tape to give a length of precisely 300 ft at a temperature of 68 °F. (N.U. Ans. 30 lbf) 15. (a) Calculate to three decimal places the sag correction for a 300 ft tape used in catenary in three equal spans if the tape weighs 1 lb/100 ft and it is used under a tension of 20 lbf. (b) It is desired to find the
when suspended
weight of a tape by measuring
in catenary with both
long and the sag amounts to 9-375
20
lbf,
what
in.
its sag ends level. If the tape is 100 ft at mid-span under a tension of
is its weight in ozf per 100 ft?
(N.U.
Ans.
0-031
ft,
20 ozf)
Describe the methods used for the measurement of the depth of vertical mine shafts and discuss the possible application of electronic distance measuring equipment. Calculate the elongation of a shaft measuring tape due to its own weight at (1) 1000 ft and (2) 3000 ft, given that the modulus of elastici16.
6
30 x 10 lbf/in 2 } weight of the tape 0-051bf/ft run, and the crosssectional area of the tape 0-015 in 2
ty is
.
(N.U.
Ans.
0-055 6
ft,
0-500
ft)
.
SURVEYING PROBLEMS AND SOLUTIONS
54
(a) Describe the measuring and straining tripods used in geodelic 17. base measurement. (b) The difference between the readings on a steel tape at the terminals of a bay between which it is freely suspended was 94*007 ft, the tension applied being 201bf, the temperature 39*5 °F, and the height difference between the terminals 5*87 ft. The bay was 630 ft above mean sea level. If the tape, standardised on the flat, measured correctly at 68 °F under lOlbf tension, and its weight was 0*0175 lbf per ft, its coefficient 8 of expansion 0*62 x 10~ per deg F and its coefficient of extension 5 0*67 x 10" per lb, calculate the length of the bay reduced to mean sea 6 level. (Radius of earth = 20-9 x 10 ft). Ans. 93-784 ft) (L.U.
The steel band of nominal length 100ft used in the catenary measurement of a colliery base line, has the following specification: (i) Length 100*025 ft at 10 lbf tension and 68 °F. 18.
(ii)
(iii)
2 Sectional area 0*004 in
Weight 22 ozf
(iv) Coefficient of linear
(v)
6 expansion 6*25 x 10~ per deg F.
Modulus of elasticity 30 x 10* lbf/in 2
was measured
.
10 bays and the undernoted obserthe first five which were of average respect of recorded in were vations height 625 ft above Ordnance Datum.
The base
line
in
Observed Length
Bay
Bay Level
Bay
Temperature
Difference
Tension Applied
1
100*005
52 °F
0*64
20 lbf
2
99*983
54°F 54 °F 58 °F 60°F
1*23
20 lbf
0*01
20 lbf
0*79
20 lbf
2*14
20 lbf
3
100*067
4
100*018
5
99*992
Correct the bays for standard, temperature, tension, sag, slope, and height above Ordnance Datum and compute the corrected length of this part of the base line. Take the mean radius of the earth to be
20890000 ft. (M.Q.B./S Ans. 500'044ft) 19.
The
steel band used in the catenary measurement of the base line
has the undernoted specification: a tension of 25 lbf at 60° F.
of a colliery triangulation survey (i)
(ii) (iii)
length 50*000 3 m weight 2*5 lbf
at
8 coefficient of linear expansion 6*25 x 10~ per deg F.
LINEAR MEASUREMENT The undernoted data apply base
to the
55
measurement of one bay of the
line:
(i)
(ii) (iii)
length 50-002 7
m
mean temperature 53 °F tension applied 25 lbf
(iv) difference in level
(v)
between ends of bay 0*834
m
mean height of bay above mean sea level 255*4 m.
Correct the measured length of the bay for standard, temperature, sag, slope, and height above of the earth is 6-37 x 10
6
mean sea
level.
Assume
the
mean radius
m.
(M.Q.B./S
Ans.
49-971
m)
A base line was measured with an invar tape 100 ft long which 20. had been standardised on the flat under a tensile load of 15 lbf and at a temperature of 60 °F. Prior to the measurement of the base line the tape was tested under these conditions and found to record 0*015 ft too much on the standard length of 100 ft. The base line was then divided into bays and the results obtained from the measurement of the bays with the tape suspended are shown below:
Length
Difference in level
Air temperature
(ft)
between supports (ft)
(°F)
Bay 1
99*768
2*15
2
99*912
1*62
49*6
3
100*018
3*90
49*8
4
100*260
4*28
50*2
5
65*715
0*90
50*3
49*6
6
= 22 x 10 lbf/in 2 Coefficient of linear expansion of invar = 5*2 x 10~ 7 per deg F. Field pull = 25 lbf.
Modulus of elasticity (E)
for invar
.
Cross-sectional area of tape = 0*004 in 2
Weight per
ft
.
run of tape = 0*010 2 lbf.
Average reduced level of base line site = 754*5 ft. 6 Radius of earth = 20*8 x 10 ft. Correct the above readings and determine to the nearest 0*001 the length of the base line at
mean sea
(I.C.E. 21.
ft
level.
The following readings were taken
in
Ans.
465*397
ft)
measuring a base line with
a steel tape suspended in catenary in five spans:
.
SURVEYING PROBLEMS AND SOLUTIONS
56
Mean reading
pan
of tape (ft)
Tension
Difference in level
between Index Marks
(ft)
Mean
(Ibf)
temperature (°F)
1
100.155
3-1
25
73
2
100*140
0*9
50
76
3
100«060
1-2
25
78
4
100» 108
3-1
25
80
5
100-182
2-0
25
80
The tape reading was 100*005
ft
when calibrated
in
catenary under
a tension of 25 lbf at a temperature of 65 °F between two points at the
same level precisely 100 ft
apart.
Other tape constants are:
width of tape = 0*250
in;
thickness of tape = 0*010
weight of steel = 0*283 lbf/in 3
;
E
for steel
6
coefficient of expansion of steel = 6*2 x 10"
Compute the length
base
of the
22.
A in
per
;
deg F.
line.
(I.C.E.
band
in;
= 30 x 10 lbf/in 2
Ans.
500*568
ft)
short base line is measured in four bays with a 100ft invar catenary under a pull of 20 lbf with the following field readings:
Bay
1
2
3
4
t°F
65*2
64*0
65*5
63*8
hit I
ft
5*08
99*6480
2*31
1*31
99*541 7
99*751 7
2*13
99*9377
where t is the field temperature, h is the difference in level between the ends of each bay and / is the mean reading of the invar band When standardised in catenary under a pull of 20 lbf at 68*5°F the standard length of the invar band was 99*999 ft and the mean altitude of the base is 221 ft above sea level. If the coefficient of expansion of invar is 0*000 0003 per deg F and the radius of the earth is 20*9 x 10* ft
what
is
the length of the base line reduced to sea level? (I.C.E. Ans.
398*6829 ft)
Bibliography
CLARK, D., Plane and Geodetic Surveying, Vols. I and II (Constable) HOLLAND, J.L., WARDELL, K. and WEBSTER, A.G., Surveying, Vols. I and
SMIRNOFF,
M.V.,
Measurement
II.
Coal Mining Series (Virtue) and Other Surveys
for Engineering
(Prentice-Hall)
BANNISTER, A. and RAYMOND, S., Surveying (Pitman) THOMAS, W.N., Surveying, 5th ed. (Edward Arnold) BRIGGS, N., The Effects of Errors in Surveying (Griffin) MINISTRY OF TECHNOLOGY, Changing to the Metric System (H.M.S.O.) ORDNANCE SURVEY publication, Constants, Formulae and Methods used in Transverse Mercator (H.M.S.O.)
2 SURVEYING TRIGONOMETRY 'Who conquers the triangle half conquers his subject' M.H. Haddock
Of
all
the branches of mathematics, trigonometry is the most im-
portant to the surveyor, forming the essential basis of all calculations
and computation processes. It is therefore essential that a thorough working knowledge is acquired and this chapter is an attempt to summarize the basic requirements.
2.1
Angular Measurement
There are two ways of dividing the (a) the
circle:
degree system,
(b) the continental 'grade'
system.
The latter divides the circle into 4 quadrants of 100 grades each and thereafter subdivides on a decimal system. It has little to commend it apart from its decimalisation which could be applied equally to the degree system. It has found little favour and will not be considered here.
2.11
The degree system
180
Clockwise rotation used by
Anticlockwise rotation used by mathematicians
surveyors
Fig. 2.1
Comparison of notations 57
58
SURVEYING PROBLEMS AND SOLUTIONS
The
circle is divided into 360 equal parts or degrees, each degree
into 60 minutes, and each minute into 60 seconds.
The following sym-
bols are used:—
degrees (°)
( ')
minutes
seconds
(")
so that 47 degrees 26 minutes 6 seconds is written as
47° 26' 06" N.B. The use of 06" is preferred in surveying so as to remove any doubts in recorded or computed values. In mathematics the angle is assumed to rotate anti-clockwise whilst in surveying the direction of rotation is assumed clockwise. This variance in no way alters the subsequent calculations but is merely a different notation. 2.12
cos240° = - cos (240 - 180) = -cos60° = -0-5 tan 225°
= +tan (225 - 180) =
tan 45° =
1-0
sin330° = - sin (360 - 330) = -sin30° = -0-5 cos 315° = +cos(360-315) = + cos 45° =
0-7071
tan 300° = - tan (360 - 300) = -tan 60° = -1-7320
17
Points of the compass (Fig. 2.9)
These
are not used in Surveying but are replaced by Quadrant (or
uadrantal) Bearings where the prefix is always or W, Fig. 2.10. g.
NNE ENE
=
ESE
=
67° 30' E. 30' E. 67° S
SSE SW
=
S 22° 30' E.
=
S 45° 00' W.
NW
=
N 45°
=
N N
22° 30' E.
00' W.
N
or S with the suffix
66
SURVEYING PROBLEMS AND SOLUTIONS
Fig. 2.9
Points of the compass
NNE = N22°
30' E
NW»N45°W ENE = N67° 30'E
ESE =S67° 30'E
SW = S45° W SSE =S22° 30'E
SURVEYING TRIGONOMETRY 2.18
Easy problems based on the solution of the right-angled
67 triangle
N.B. An angle of elevation is an angle measured in the vertical plane is above eye level, i.e. a positive vertical angle,
where the object Fig. 2.11.
An angle
of depression is an angle measured in the vertical plane
where the object
is
below eye level,
Fig.
2.
11
i.e.
a negative vertical angle.
Vertical angles
In any triangle there are six parts, 3 sides
and 3 angles
The usual notation is to let the side opposite the angle A be a etc, as shown in Fig. 2.12. The following facts are thus known about the given right-angled triangle ABC.
C = 90° Angle A + Angle Angle
c sin
2
= a
A =
2
+ b
fi
z
= 90° (by Pythagoras)
SL
c
cos A
=
k c
tan
A = £ b
Fig. 2.12
SURVEYING PROBLEMS AND SOLUTIONS
68
tan
a -
A =
c sin
i.e.
A
~-
x
c b
-r
cos A.
To find the remaining parts of the triangle it is necessary to know 3 parts (in the case of the right-angled triangle, one angle = 90° and therefore only 2 other facts are required).
Example
In a right-angled triangle
2.1.
10 metres long, whilst angle
A
is 70°.
ABC,
the hypotenuse
AB
of the triangle.
As
a c
then
AB
the hypotenuse is
C
the right angle is at
a
b c b
=
sin 70°
(c)
(Fig. 2.13).
=
c sin 70°
=
10 sin 70°
=
10 x 0-939 69
=
9-397 metres
=
cos 70°
=
c cos 70°
=
10 x 0-34202
=
3-420 metres
Fig.
2.
13
AngleS = 90° - 70° = 20°
Aa
Check
9
angle
i.e.
Example 2.2
tanS = l'*®l = 0-363 97 9-396
=
20° 00'
B =
necessary to climb a vertical wall 45 ft (13-7 m) ft (15-2 m) long, Fig. 2.14. Find from the foot of the wall the ladder must be placed,
It is
high with a ladder 50 (a)
How
far
(b) the inclination of the ladder
||
=
sin^
=
0-9
50
thus
thus
Ans.
angle
A = 64°
angle
B
=
=
cos A
b
=
50 cos 64° 09' 30" =
ft
(b)
64° 09' 30"
09' 30"
25° 50' 30"
£ (a) 21-79
is
Calculate the remaining parts
21-79
ft
(6-63 m)
from the horizontal.
SURVEYING TRIGONOMETRY
69
B
45'
Fig. 2.14
A ship sails 30 miles (48-28 km) on a bearing N 30° E. Example 2.3. then changes course and sails a further 50 miles (80'4 km) N 45° W. Find (a) the bearing back to its starting point, (b) the distance back to its starting point.
It
N.B. See chapter 3 on bearings To solve this problem two triangles, to form a resultant third ACF (Fig. 2.15). In triangle
ADB, AB
is
N30°E
30 miles (48-28 km). The distance travelled N = AD. but
4§ AB
=
cos 30°
then
AD
=
30 cos 30°
= 25-98 miles (41 -812 km)
The distance but .-.
M
AB DB
=
sin 30°
=
30 sin 30°
=
15-00 miles (24 -140 km)
Similarly in triangle
tance travelled but
BE
E = DB
travelled
=
BC
BCE
the dis-
N = BE cos 45°
= 50 cos 45° = 35-35 miles (56-890 km)
The distance The distance
travelled
travelled
W (CE) = N = 35-35
miles as the bearing = 45° (sin 45° = cos 45°). In resultant triangle
ACF,
ADB
and BCE, are joined
SURVEYING PROBLEMS AND SOLUTIONS
70
CF = CE - DB = AF = AD + BE = tantf
=
CF AF
6
=
18° 21 20"
2.4.
20-35 miles (32-750)
=
61-33 miles (98-702)
.-.
0-33181 bearing A C
—
AF
61-33 cos 18° 21' 20 '-,
cos 6
Example
=
25-98 + 35-35
= JP-^I = 61-33 '
AC
35-35 - 15-00
An angle
Tl
=
N
18° 21' 20" W.
= 64-62 miles (104-0 km)
45° was observed to the top
of elevation of
of a tower. 42 metres nearer to the tower a further angle of elevation of
60° was observed. Find (a) the height of the tower, (b) the distance the observer
Fig.
2.
is
from the foot of the tower.
16
InF]ig.2.16,
AC H i.e.
also i.e.
=
cot
AC = H
BC H BC AC BC
A
cot A
=
cotS
=
H cot B. AB = ft (cot ,4
- cotB)
AB
H cot
A -
cot
B
SURVEYING TRIGONOMETRY
71
42 cot 45° -cot 60°
42
1-0-577 4 —
BC = H
AC
Check
99 38ln "
0-4226 cot
-
B
=
99-38 cot 60°
=
99-38 x 0-577 4
=
DC BC
=
=
99-38
=
AC
=
99-38 - 42
-
57-38
m
AB 57-38
=
m
Exercises 2(a) 1. A flagstaff 90 ft high is held up by ropes, each being attached to the top of the flagstaff and to a peg in the ground and inclined at 30° to the vertical; find the lengths of the ropes and the distances of the pegs from the foot of the flagstaff.
(Ans. 103-92
ft,
51-96
ft)
From the top of a mast of a ship 75 ft high the angle of depression of an object is 20°. Find the distance of the object from the ship.
2.
(Ans.
206 -06ft)
A
tower has an elevation 60° from a point due north of it and 45° If the two points are 200 metres apart, find the height of the tower and its distance from each point of observation. (Ans. 126-8 m, 73-2 m, 126-8 m) 3.
from a point due south.
A
boat is 1500 ft from the foot of a vertical cliff. To the top of the and the top of a building standing on the edge of the cliff, angles of elevation were observed as 30° and 33° respectively. Find the 4.
cliff
height of the building to the nearest foot.
(Ans.
A
5.
What
vertical stick 3
m
108
ft)
long casts a shadow from the sun of 1-75 m.
is the elevation of the
sun ? (Ans.
59° 45')
X and Y start walking in directions N17°W and N73°E; find their distance apart after three hours and the direction of the line joining them. X walks at 3 km an hour and Y at 4 km an hour.
6.
(Ans. 7.
A, B, and
C
are three places.
B
is
30 km
15
N67H°E
km S70°08'E) of A, and
C
SURVEYING PROBLEMS AND SOLUTIONS
72 is
40 km S221/2 °E of B. Find the distance
Circular Measure
2.2
The circumference
and bearing of C from A. (Ans. 50 km, S59°22'E)
of a circle
= 2rrr where
77
= 3*1416 approx.
The radian
2.21
The angle subtended at the centre of a known as a radian.
circle by an arc equal in
length to the radius is
Thus
2
77
radians
=
360°
radian
=
M
1
/.
277
,
=
57°17 45" approx.
=
206 265 seconds.
This last constant factor is of vital importance to small angle calculations and for conversion of degrees to radians.
Example
Convert 64° 11' 33" to radians.
2.5.
64° 11' 33" = .-.
231093 seconds.
231093
=
no. of radians
=
1-120 37 rad.
206 265
Tables of radian measure are available for 0°-90° and, as the radian measure is directly proportional to the angle, any combination of values produces the same answer for any angular amount.
By
tables,
64°
= 11'
64°
11'
1-11701
= 0-003 20 33"
=
0-000 16
33"
—
1-120 37
It now follows that the length of an arc of a circle of radius r and subtending d radians at the centre of the circle can be written as
=
arc
This
is generally superior to the
* When 6
N.B.
To
is written
it
-
r.6 rad
(2.14)
use of the formula
2m *
m
(21S)
implies 6 radians.
find the area of a circle.
A
regular polygon
ABC
...
A
is
drawn inside a
circle, Fig. 2.17.
SURVEYING TRIGONOMETRY Draw
OX
+
BC +
tOX (perimeter When the number
=
...)
of polygon)
of sides of the
increased to infinity
is
OX becomes
(oo),
=
of polygon
±OX(AB
polygon
AB
perpendicular to
Then area
73
the radius,
the perimeter becomes the cir-
cumference, and the polygon becomes the circle .*.
area of circle =
=
The area of
V2.r .2ttt
Fig.
2. 17
TFT 2
the sector
ari=>a
OAB. area of sector
e
area of circle
2tt
of sftrtnr
rrr
^
" _
\r*Q
(2.16)
2tt
2.22
Small angles and approximations
For any angle < 90°
(i.e.
<
77/2 radians) tantf
>
> sim
A
Fig.
Let angle
AOC
2.
18
= 6
OA = OC and
AB be a tangent to the arc AC AD perpendicular (1) to OB.
let
Draw
= at
r
A
to cut
OC
produced at B.
SURVEYING PROBLEMS AND SOLUTIONS
74
Then
OAB
area of triangle
=
±0A AB
=
h.rtand =
.
2
Now
area of sector
OAC
area of triangle
OAC
triangle
OAB
±r 2 tan0 2
=
±OC.AD
=
Ir.r sin0
=
|r 2
=
|r 2 sin0
> sector (X4C > triangle OAC.
Ar 2 tan0 > |r 2
> |r 2 sin0
tan0 > 9 > sin0 This
is obviously true for all
Take
to
< it/2.
values of
be very small.
Divide each term by sin0, *
then
>
cos0 It
known
is
that as
Thus
~
cos0 .*.
cos0
then
-*
,
3
>
1
sin0 -*
1.
when
1
is small
must also be nearly
..
1.
sin0
The
result
shows
that sin0
may be replaced by
0.
Similarly, dividing each term by tan0, 1
1
i.e.
>
>
tan0
> s * n fl tan0
"
> cos0
A
tan0
Tan0 may It
also be replaced by 6.
can thus be shown that
tan0
The following values
for very small =2..
^
angles
sin0.
are taken from H.M. Nautical
Almanac Office. (These
Five-figure Tables of Natural Trigonometrical Functions. tables are very suitable for most machine calculations.)
it can be seen that $ may be substituted for sin 6 or tan0 to 5 figures up to 2°, whilst 6 may be substituted for sin0 up to 5° and for tantf up to 4° to 4 figures, thus allowing approximations to be made when angles are less than 4°
Example 2.6. If the distance from the earth to the moon be 250000 miles (402000 km) and the angle subtended 0° 30', find the diameter of the moon.
250 OOO miles
Fig.
The diameter
Example
2.7.
2.
19
~
arc
2L
250000 x 30'rad
-
250000 v 30 x 60 206265
ABC
=2
2180 miles (3510 km)
Find as exactly as possible from Chambers Mathemati-
cal Tables the logarithmic sines of the following angles:
A = 00° 02' 42"
and
C = 00° 11' 30"
Use these values to find the lengths of the sides AB and AC in a angle ABC when BC = 12-736 ft. Thereafter check your answer by
tri-
another method, avoiding as far as possible using the tables at the same places as in the first method.
The lengths
are to be stated to three places of decimals.
(M.Q.B/S)
SURVEYING PROBLEMS AND SOLUTIONS
76
As the sines and tangents of small angles change so rapidly, special methods are necessary. Method 1 Chambers Mathematical Tables give the following method of finding the logarithmic sine of a small arc:
To
the logarithm of the arc reduced to seconds, add 4-685 574 9 and sum subtract 1/3 of its logarithmic secant, the index of
from the
the latter logarithm being previously diminished by 10.
00°02'42" = 162"
logl62
2-2095150
constant
4-6855749
6-8950899
-|(logsec02'42" -
10) =
| x 0-000 000 2
= 0-0000001
6-8950898 00° 11" 30" = 690"
log 690
2-8388491
constant
4-685 5749
7-5244240 -!(log sec
11'
30" - 10) =
I x 0-0000024
= 0-0000008
7-524423 2
Method 2
As
the sines and tangents of small angles approximate to the
value,
radian value of
logO-000785397 4 radian value of
B
00° 11' 30"
Summary
(1)
690 „ 206 265 „
=
=
0-000785 3974
6-8950895
= =
logO-003 3452112
To
* 6^ 206 265
A 00° 02' 42" =
= 0-003 345 2112
7-524423 5
(2)
Vegas tables
(to l")
00° 02' 42"
6-895089 8
6-895089 5
6-8950898
00° 11' 30"
7-524 423 2
7-524 4235
7-524 4231
AB
find the length of sides
AB
=
BC
sin
sin
A
C
=
and
AC when BC
= 12-736
12-736 sin (02' 42" sin 02' 42"
+
:
11' 30") (see 2.51
SURVEYING TRIGONOMETRY
77
12-736
=
Logs 1-1050331
S/14' 12"
=
2-929 2793
4-6855749 8-719887 3
-|(sec-10)
0-0000035
8-7198838
AB
=
66-803
AC
=
BC
S/02' 42"
6-8950898
AB
1-8247940
12-736
1-1050331
ft
sin
12-736 sin 11' 30" sin 02' 42"
B
sin^4
S/ll' 30"
7-524423 2 8-629 4563
AC
=
2.3
54-246
S/02' 42"
6-895 0898
AC
1-7343665
ft
Trigonometrical Ratios of the Sums and Differences of two angles (Fig. 2.20)
To
prove:
sin (A
+ B) = sin A cos B +
+ cos A sin B (2.17)
B -
cos (A + B) = cos A cos
- sin A sinS (2.18)
Let the line OX trace out the angle A and then the angle B. Take a point P on the final line 0X 2 Draw PS and PQ perpendicular to OX and OX, respectively. Through Q draw QK parallel to OX to meet .
Fig. 2.20
PS
at R.
Draw
pendicular to OX.
ffPQ sin(A + B)
=
fl OP
=
= = =
KQO
=
A
RS + PR
OP RS OQ OQ 'OP
RS PR OP OP PR PQ PQ' OP
sin^4 cos
6 + cos A sinB
,
QT
per-
SURVEYING PROBLEMS AND SOLUTIONS
78
cosC4 +
fi)
=
OT-ST
OP OP OP OT 0£_ ST P£ OQ' OP PQ' OP
angle
B
is
m
cos 4 cosB - sin 4 sinB
= If
OT_ST
OS =
OP
now considered -ve,
4 cos (-B) + cos 4 sin (-B) = sin 4 cosB - cos 4 sinB
sin(4-B) =
sin
(2.19)
Similarly,
cos(4-B) =
cos 4 cps(-B) - sin 4 sin(-B)
= cos A cos B + sin A sin B tanG4'+B)
=
s * n (A + ff) _ cos (4 + B)
(-by cos4 cosB) =
sin A
B be
cosB + cos A sinB
B - sin4
cos A cos
sin
4 + tanB - tan A tan B
tan 1
Similarly, letting
(2.20)
B (2 2 1) .
-ve,
tanC4-B) =
tan 4 + l-tan4
- tanB
tan A 1
tan(-B) tan(-B)
+ tan4tanB
(2.22)
A cosB + cos 4 sinB
If
sin(4 + B)
=
sin
and
sin (A - B)
=
sin A
then
sin (A + B) + sin(4 - B)
=
cosB
(2.23)
and
sin(4 + B) - sin (A - B)
= 2 cos A sinB
(2.24)
cosB - cos A sinB 2 sin A
Similarly,
as
cos (4 + B)
=
cos 4 cosB - sin 4 sinB
and
cos(4 - B)
=•
cos 4 cos B + sin 4 sin B
If
A«
cos(4 + B) + cos(4 - B)
=
cos(4 + B) - cos(4 - B)
=
(2.25)
-2
sin 4
sinB
(2.26)
B, then
cosA
+ 4)
=
sin
2A
=
2 sin4
cos(4 + 4)
=
cos 24
=
cos 2 4 - sin 2 4
=
1
sin (4
or
cosB
2 cos 4
- 2 sin 2 4
=2 cos
2
4 -
1
(2.27)
(2.28)
(2.29)
79
SURVEYING TRIGONOMETRY
A
2 tan
= 1
(2.30)
- tan 2 A
Transformation of Products and Sums
2.4
From
tan 2 A
=
tan04 + 4)
cosB
sin(4 + B) + sin(A - B)
=
2 sin A
sin 04 + B) - sinC4 - B)
=
2 sinB cos
A + B = C A - B = D
if
and
A =
then
C-±D
sin
B = £^_D
and
""
2
C + sinD =
"
2
2 sin
C + D cos C ~ D
2 sin
6 ~
Similarly,
sinC
sinD
-
"
D
cos
C + D
2
From
=
2 cos ,4
cos04 + B) - cos(A - B)
=
-2 sinA
cos
C +
cos D
=
2 cos
cosC - cosD = -2
and
sin
(2.32)
2
cos(A - B)
cos(v4 + B) +
(2.31)
c + D
cos
C + D
sin
cosB sinfi
C - D
C ~ P 2__
2
(2.33)
(2.34)
These relationships may thus be tabulated sin (A ± B)
=
sin A cos
cos 04 ± B)
=
cos
tan 04
sin04 + B) + sin 04 - B)
=
2 sin A
=
2 cos
cos(A - B)
=
2 cos
cos(i4 + B) +
cos(/4 + B) - cos(/4
=
2 sin ,4 cos
cos 2A
=
cos 2
24
=
tan
1
sin
B
± tanB T tan 4 tan B
sin04 + B) - sin (4 - B)
A -
sin
cos B + sin A sin B
tan>l
±B)
1
sin 2,4
/4
B ± cos A
cosB
A sinB 4 cos B
-B) = -2sini4sinB
/}
sin 2 A
=
1
- 2sin 2 4
=
2 cos 2 ,4 -
2tan ^ - tanM
A + sinB
2 sin
4 +
B„^A cos
-
B
1
SURVEYING PROBLEMS AND SOLUTIONS
80
A -
sin
B =
sin
2 sin'
4
~ B cos^4 + B 2
cos
4 + cos £ =
2 cos
2
^ + B cos ^ ~ g 2
B = -2
cos ,4 - cos
2
^ + B
sin
sin
^ ~ B
2
The
2.5
2
Solution of Triangles
The following important formulae are now proved: Sine rule
a
b
=
.
sin
=
sinS
A
_£_
=
sinC
2R
(2.35)
Cosine rule cz
a 2 + b z - 2ab
=
cosC
sinC = 2l^\s(s -a){s - b){s -
(2.36) c)\
(2.37)
ab
Area of triangle = lab sinC =
(2.38)
^s(s-a)(s-b)(s-c)
(2.39)
Half-angle formulae f (s-b)(s-c)
sini =
V
2
A
A(s-a) W be
=
2 tan
(2.41)
As-b)(s-c)
A =
V
2
(2 .40)
be
s (s
_
(2.42)
a)
Napier's tangent rule
B -C tan*^ an —— -l
=
An£tan£±^
( 2>43)
b + c 2.51
Sine rule (Figs. 2.21 and 2.22)
Let triangle /4SC be drawn with circumscribing circle. Let 4J3, be a diameter through /I (angle ABC = angle
AC
In Fig. 2.21,
In Fig. 2.22,
j£- = sinB
i£
AB
=
sin(180-B)
=
sinB
|
JL = sinB 2R
4^0.
SURVEYING TRIGONOMETRY
Fig. 2.21
Fig. 2.22
b
=
sinB
2R
b
Similarly
sin A
2.52
81
sinB
sinC
=
2R
(2.35)
Cosine rule (Fig. 2.23)
Obtuse Fig.
AB 2
=
AD.22 AD 2
=
b 2 sin 2
=
AD 2
=
6
=
b 2 sin 2
=
.'.
2
+
„„ 2 BD
2.
23
Acute'
The cosine
rule
(Pythagoras)
+ (BC - CD) 2
C + {BC -
b cos
with
Cf
C
acute
+ (BC + CD) 2
sin 2 (180 -
AB 2
O
C + (BC 2
+ {BC + b cos (180 - C)\ 2 b cos
=
b sin z C +
(BC-bcosC) 2
=
b 2 sin 2 C + a z - 2ab
=
a
=
a 2 + b 2 - lab cos
2
}
with
C
obtuse
C) 2 in either
case
cosC + b C. 2 + b (sin 2 C + cos 2 C) - lab cosC
C
z
cos z
(2.36)
SURVEYING PROBLEMS AND SOLUTIONS
82
Area of a triangle
2.53
2
=
cos C
=
From
c
2 a 2 + b - lab cosC z z ~2 a + bu - c~
lab sin 2 C
=
1
- cos 2 C
=
2 Z 2 2 _( a +b -c \
1
lab
\
=
(l
+ a
z
-c%
(a + b)
2
- c
2
x
c
2
- (a -
2ab
-c 2 )
lab
J
2
b)
2afe
(a + b + c)(- c
+ a + bXc - a + (2ab)
4s (s -
/ 2
a + b
A
lab
V
_
+b
2
2
a) (s
- b)(s -
b)(c
+ a -
fc)
2
c)
a2bz
where 2s = a + b +
c.
sinC =
i_ /s(s-a)(s-b)(s-c)
(2.37)
ab*j In Fig. 2.23,
Area of triangle
=
\AD EC
=
lab
.
sin
C
(2.38)
= ^Lab -4 V s ( s = 2.54
a)(s
- b)(s -
c)
^s(s-a)(s-b)(s-c)
(2.39)
Half-angle formulae
From Eq.(2.28), tint* A
-%1-co.A)
.
^(l-»Lt£L-L) q
2
_
(fc
_ C)
2
4bC (a - b + c)(a + b - c)
Abe =
(s-b)(s-c) be
sini = 2
Ks-bKs-e) fee V
(2 40)
SURVEYING TRIGONOMETRY
83
Similarly, 2
cosz±A =
2
= l(l + b +c 2\ 2bc
I(l + cos4)
2
2
2
(b + c)
=
-a 2 \ *
- a2
4bc (b + c + a)(b + c
=
-
a)
Abe s(s be
_
/sjs-a)
=
cosil
V
2
(241)
be
A
.
sin
tan_d
a)
-
= cos
/
=
(s-bXs-c)
V
A
(2>4 2)
s(s - a)
2
The last formula is preferred as (s - b) + (s - c) + (s - a) = provides an arithmetical check. 2.55
s,
which
Napier' s tangent rule
From the sine
rule,
b
sinB sinC
_
c b - c
then
sinS - sinC sinS + sinC
b + c
2cos S + C
sin
2 r sin
2
B + C ™c 2
tan
B
~
B-C
B-C
(By Eqs. 2.31/2.32)
2
C
2
B+ C tan
•*.
tan
B ~ c = b-c 2
2.56
tQV
b + c
B+ C 2
Problems involving the solution of triangles All problems
come within the following
four cases:
(2.43)
SURVEYING PROBLEMS AND SOLUTIONS
84 (1)
Given two sides and one angle (not included)
to find the other
angles. Solution
Sine rule solution ambiguous as illustrated in Fig.
:
2.24.
Given AB and AC with AC may cut line BC at
angle B, C, or
C2
Fig. 2.24
The ambiguous case
of the
sine rule (2)
Given
all the
Solution: (3)
angles and one side to find all the other sides. Sine rule
Given two sides and the included angle Solution: Either cos rule to find remaining side Napier's tangent rule (this is generally pre-
or
ferred using logs) (4)
Given the three sides cos rule
either
Solution:
half-angle formula (this is generally preferred
or
using logs.)
Example 2.8
(Problem
1)
Let
c a
m = 190-5 m =
466-0
A = 22° Using sine sin
c
C
15'
rule,
sin
A
a
sinC = csinA a
= 466-0 sin 22°
15'
190-5
466-0 x 0-378 65 190-5
= 0-926 25
Fig. 2.25
3
,
SURVEYING TRIGONOMETRY
C2 =
Angle
or
180
190-5 sin [180 -(67° 51' 30"
+22°
C,
To
67° 51' 30" 112 08' 30"
85
=
now Problem
find side b (this is
2)
a sin
B fo
sin
_
A
a sinB sin
i.e.
,4
15' 00")]
sin 22° 15' 00"
190-5 sin89°53'30" sin 22° 15' 00"
=
190 ' 5 x 1-0
=
503-10
0-37865 5,
=
m
190-5 sin[(180 -(112° 08' 30"+ 22° 15' 00")] sin 22° 15 '00" 190-5 sin 45° 36'
N.B. The notation 9-578 24 is preferred to 1-578 24 - this is the form used in Chambers, Vegas, and Shortredes Tables. Every characteristic is increased by 10 so that subtraction is simplified
—
the ringed figures are not usually entered.
Also "I.B.
e.
logfc
=
log a + log sin
B +
log cosec,4
Addition using log cosec4 is preferable to subtracting log sin A. logfc,
N.B. A gap is left between the third and fourth figures of the logarithms to help in the addition process, or it is still better to use squared paper.
Example 2.9 (Problem
3)
Let
a
= 636
c
=
818
m m
B = 97° 30' To find b, A and By cosine rule b
2
=
C.
cosB
a* + c 2 - 2ac
=
636
2
+ 818
2
- 2 x 636 x 818 x cos
= 404496 + 669124 + 135815-94
b
=
1209435-94
.
1099-74
m
sin A
oinA SmA
B
sin
a
b
=
=
636 sin 97° 30' 1099-74
=
0-57337
asinB b
A = 34° 59'
B = The
first part of the
figures get large suitable.
it
calculation is essentially simple but as the
becomes more
The following approach
As c>
a,
C>
10"
47° 30' 50"
and logs are not recommended.
difficult to apply is therefore
A. Then, by Eq.(2.43),
tan%4 2
,
£^JltanC±J. c + a
818 - 636 818 + 636
2 .
182 tan 41° 15'
1454 = 0-10977
(180
-97° 30') 2
SURVEYING TRIGONOMETRY
\{C-A) l
-(C + A)
6° 15' 50"
=
41° 15' 00"
c = 47° 30' 50" A = 34° 59' 10"
By adding By
=
87
subtracting
Now, by sine
=
b
rule,
a sin
6 cosec A
= 636 sin 97° =
1099-74
30'
cosec 34° 59' 10"
m
N.B. This solution is fully logarithmic and thus generally preferred. Also it does not require the extraction of a square root and is therefore superior for machine calculation.
Example 2.10 (Problem
c = 932
b = 719
a = 381
Let
To
4)
find the angles.
From then
a2
cos A
b 2 + c 2 - 2bc cos
b
2
2
+ c - a 2bc
719
2
+ 932
A
2
2
- 381
2
2 x 719 x 932
516 961 + 868 624 - 145 161 340 216
1
0-925 54
A = 22° cosfi
a
2
15'
00"
2 „2 + c - b 2ac t
145161 + 868 624 - 516961 2 x 381 x 932
496824
= 0-69957
710 184
B = 45° 36' cosC =
a
2
30 2
+ b - c lab
2
145161 + 516961 - 868624 2 x 381 x 719 -206 502 = -0-37691 547 878
SURVEYING PROBLEMS AND SOLUTIONS
88
180 - 67° 51' 30"
C =
112° 08' 30"
=
Check
22° 15' 00" + 45° 36' 30" + 112° 08' 30"
=
180° 00' 00"
Alternative
By
tan
half-angle formula,
=
S.
- b)(s - c) s(s - a)
a
=
381
s - a
=
635
b
=
719
s - b
=
297
c
=
932
s - c
=
84
2s
=2032
s
= 1016
s
= 1016
tanA - / 297 * 84
then
V
2
This
Ks
V
2
1016 x 635
is best solved by logs
log
tani = I[(log297 + log84) 2
2
7'
(log 1016 + log635)]
30"
=
11°
=
22° 15' 00"
297
2-47276
84
1-924 28
4-39704 1016
3-00689
635
2-80277 5-80966 2) 18-58738
tany1/2 9-29369
tan^ = 2
-^ =
/
635
><
84
635
2-80277
84
1-924 28
V1016 x 297 22° 48' 15"
2
4-727 05
1016
3-00689
297
2-47276
5-47965
B = 45° 36' 30"
2)
tan
19-24740
B/2 9-623 70
SURVEYING TRIGONOMETRY The sides
Example 2.11
AB
AC
= 36ft Olin., lo
ABC
of a triangle
= 30 ft if in.
and
4
89
measure as follows:
BC
= 6ft0lin. 4
(a) Calculate to the nearest
20 seconds, the angle BAC. (b) Assuming that the probable error in measuring any of the sides is ± 1/32 in. give an estimate of the probable error in the angle A.
(M.Q.B/S)
Fig.
AB
=
AC BC
c
=
=
b
=
=
a
=
0*
2.
26
=
36-036 ft
30ftlf-in. 4
=
30-146
ft
S
6ft0±in.
=
6-021
ft
s
=
36
ft
in.
c
=
0-0655
- C
=
5-9555
s - a
=
30-0805
s
=
36-1015
16
4
2 |72-203
36-1015
Using Eq. (2.42) tan-d 2
=
/Cs-bHs-c)
V
s(s-a) 5-955
5x0-065 5
J,36*1015x30-0805 4-
=
1°05'10"
2
A = tan# =
2
o
/,
/(s-o)(s-"c)
V
2
,
10 20
s(s -
6)
30-0805x0-065 5
y36-1015x5-9555 ^
=
5° 28' 05"
2
g = tan£ = 2
10° 56 '10"
Rs-a)(s-b)
V
J
s(s-c) 30-0805x5-9555 36-1015x0-0655
SURVEYING PROBLEMS AND SOLUTIONS
90
83° 26' 45"
£. = 2
C = 166° 53' A + B + C = 180°
Check (b)
30"
The probable error of ±1/32 in. = ±0*003 ft. The effect on the angle A of varying the three sides
culated by varying each
is best cal-
of the sides in turn whilst the remaining
sides are held constant.
To
two
carry out this process, the equation must
be successively differentiated and a better equation
for this
purpose
cosine rule.
is the
cos A
Thus
b
=
z
2
+ c - a 2bc
2
Differentiating with respect to a,
2a 8a 2bc
-sin A8A a = =
8A„
a 8a be sin A
Differentiating with respect to b,
-sin A8A h
=
(2bcx2b)-(b 2 +C 2-a*)(2c) 4b*c'-
X
2b 2c
c
- c2
a2 + b 2
2b c cos
but
C
a
+ b - c 2ab
cos :osC C 8b = - 8A cos C 8A b6 _ = -a a *
be sin A
Similarly, from the
(as 8a = 8b)
symmetry of the function:
8A C = _ « cos g g c = be sin
A
SAa cos B
(as 8a = 8c)
Substituting values into the equations gives:
8A b =
6-021 x ±0-003 x 206265 30-146 x 36-036 x sin 2° 10' 20" = S4 cos 166° 53' 30"
find the height of an object having a vertical face
The ground may be
(a) level or (b)
sloping up or down from the
observer, (a)
Level ground (Fig. 2.27)
V.D.
Fig. 2.27
The observer from the object.
away The
of height h is a horizontal distance (H.D.)
The
vertical angle (V.A.) = 6 is measured.
vertical difference
V.D.
=
(2.44)
H.D. tanfl
Height of the object above the ground = V.D. + h
V.D.
V.D.
\B
Horizontal line ._o< Depression"
H.D
Fig. 2-28
(b) Sloping
ground (Fig. 2.28) is measured as
The ground slope
V.D.
=
a
H.D. (tan0 ± tana)
Height of object above the ground = V.D. +
(2.45)
ft
N.B. This assumes that the horizontal distance can be measured.
;
SURVEYING PROBLEMS AND SOLUTIONS
92
To
2.62
A
find the height of an object
when
its
base
is
inaccessible
base line must be measured and angles are measured from
its
extremities. (a)
Base
line
AB
level and in line with object. (Fig. 2.29)
a and
Vertical angles
are measured.
/8
Fig.
Thus
A,C,
=
EC, cot a
B,C,
=
EC, cot/8
A,B,
=
A,C, - B,C,
EC,
=
a -
j8
Height of object above ground at
A
=
D
Base
line
AB
EC, +
ft,
above ground at A
= (b)
EC, (cot a -cot 0) (2.46)
cot
at
=
AB cot
Height of ground
2 29
EC, +
h,
— h2
level but not in line with object (Fig. 2.30)
Angles measured
at
A
horizontal angle
at
B
horizontal angle
vertical angle
/8
vertical angle
In triangle
a
ABC AC
=
AB
sin<£ cosec(0
+ 0)
(sine rule)
SURVEYING TRIGONOMETRY
93
Fig. 2.30
BC
and
Then Also
C,£
C,£
=
AB sin 6 4Ctana
=
,46 sin
=
BC 4B
=
=
cosec(0 +
tana
(2.47)
sin0 cosec(0 + 0) tan£
(2.48)
cosec(0 +
=
A
C,£ +
A,
Height of ground (D) above ground at
A
(c)
<£)
tan/8
Height of object (E) above ground at
=
<£)
C,E +
/i,
-
Sase Zme 4B on sloping ground and
Fig. 2.31
/i
2
in line with object (Fig. 2.31)
SURVEYING PROBLEMS AND SOLUTIONS
94
Angles measured
a
A: vertical angle
at
(to object)
vertical angle 8 (slope of ground)
Angle measured
Then
B
at
A^EB
In triangle
vertical angle
:
=
a - 8
B,
=
180 - (£ - 8)
E
=
p - a
A,E = A,B,
AB
sin {180
sin(/8
-
-(0-S)} cosec(/8 - a) cosecQ8 - a)
8)
Height of object (£) above ground at
EC
=
EC, + h
=
A^E
=
AB
sin
}
a +
sin(fi
/i,
- 8) cosec(/8 -
= EC, +
AB
(2.49)
A
Height of ground (D) above ground at
(d) itase line
(to object)
X ,
4,
=
j8
a)
sina +
/i,
(2.50)
A
hy
— h2
.
on sloping ground and not in line with object (Fig.2.32)
Fig. 2.32
Angles measured
at
A
horizontal angle 6
:
vertical angles
a (to
object)
8 (slope of ground) at
B: horizontal angle vertical angle
A B 2 = AB cos y
/S
SURVEYING TRIGONOMETRY Then
4,0,
= A
X
B2
sin<£cosec(0 +
95
)
EC, = 4,C, tana
— AB cos 8 sin0
cosec(fl +
E,C 2 = B2 C, =
EC 2
EC +
=
=
X
EC 2 + fi,B 2 + = 4g cos 5 sing =
y4,B 2 sin0
B,C2
cosec(0 +
cosec(fl +
2.63
To
)
j4
tan/6
+
AB
sing +
ft,
hi
— hz
EC 2 +
h,
- h2 +
AB
sinS
find the height of an object above the ground
when
its
and top are visible but not accessible (a)
Base
line
AB, horizontal and
Vertical angles measured at at
A
in line with object (Fig. 2.23) :
B:
a, and /S,
Fig. 2.33
,~ co\
4
= EC\ + =
<£)
tan/3
h,
Height of ground (D) above ground at
or
(2.51)
hi
Similarly, height of object (E) above ground at
EC
tana
A
Height of object (E) above ground at
EC
and
a2 Bz
base
SURVEYING PROBLEMS AND SOLUTIONS
96
From Eq.(2.46)
C,E =
AB cot a, - cot/3,
AB
C,D =
Also
ED
Then
C,E - C,D
=
cot
a 2 - cotB z
=
H
H = AB\
[cota, - cot^3,
(b)
Base
line
AB
(2.53)
cota 2 - cotB z
horizontal but not in line with object (Fig. 2.34)
Angles measured
at
A
:
at
B
:
horizontal angle 6 vertical angles a, and horizontal angle
a2
and
Bz
vertical angles
jS,
I
Fig.
AC ED .'.
2.
MHEWM^WAW/SigWig
34
H
=
AB sin0 cosec(# + AC (tan a, - tan a 2 ) AB sin cosec(0 + ^>)(tana,
H
=
AB
= 4,C, = =
J
H
=
)
- tana 2 )
(2.54)
Similarly,
sin 6 cosec (0 +
<£)
(tan
B ,
tan /8 2 )
(2 55) .
SURVEYING TRIGONOMETRY
97
line AB on sloping ground and in line with object (Fig. 2.35) Vertical angles measured at A a,, a 2 and 8
Base
(c)
:
at
B:
/8,
and
/8 2
Fig. 2.35
From Eq.(2.50)
and
Then
EC,
=
/IB sin(/3, - 8) cosec(/9, - a,) sin a,
DC,
=
4B
ED H
sin(/3 2 - 5)
=
EC, - DC,
=
A8[sin(/6, - 5) cosec(/3, - a,) sin a, -
- sin(/8 2 (d)
Base
line
cosecQ3 2 - a 2 ) sin a 2
AB
8) cosec(j8 2
on sloping ground and not
- a 2 ) sin a 2 ]
(2.56)
in line with object
(Fig. 2.36)
Angles measured
at A:
horizontal angle 6 vertical angles a, and a 2 8 (slope of ground)
at B:
horizontal angle
cf>
vertical angles
/3,
and
£2
SURVEYING PROBLEMS AND SOLUTIONS
98
Fig.
2.
36
From Eq.(2.51),
ECi
=
DC ED
=
AB cosS AB cos 8
=
EC, - DC,
=
AB AB
X
Then
ED
Also
To
2.64
=
sin<£ cosec(0 +
)
tana,
sin<£ cosec(<9 +
)
tana 2
cos 5 sin0 cosec(0 + <£)(tana, - tana 2 )
(2.57)
cos 5 sin 6 cosec(<9 + <£)(tan)6, - tan B z )
(2.58)
find the length of an inclined object (e.g. an inclined
flagstaff) on the top of a building (Fig. 2.37)
Base
line
AB
measured and
is
if
on sloping ground reduced to
horizontal.
Angles measured
at A:
horizontal angles 0, and dz to top and bottom of pole
a, and
vertical angles
at B:
a2
to top
and
horizontal angles
<£,
bottom of pole and 2 to top and bottom of pole
vertical angles
)6,
and
Bz
to top
and
bottom of pole. In plan the length
ED
In elevation the length
is projected
ED
as E^D^ (=
is projected as
E ZD).
EE 2
,
i.e.
height
Then
AE,
=
Also
AD,
=
AB AB
sin>,
cosec(0, +
<£,)
sin
cosec(0 2 +
>
2
2)
the difference in
a
:
SURVEYING TRIGONOMETRY
e
99
d
Fig. 2.37
The length E,D = L
is
now best calculated using co-ordinates
(see Chapter 3)
Assuming bearing
AB
= 180° 00'
Ae = AE, 4d = 4D, Then
ed
and similarly, E,D 2 In the triangle
sin(90 -6,)
=
AE, cos0,
sin(90 -
=
4D, cos0 2
=
D 2 D, =
=
E,e _
E,D,D2
The bearing
D 2 e = 4£,
.
Length E,D, find difference in height
=
To
E,D2 (sec bearing)
=
A = AE, tana, A = AD, tan a 2
AE, tana, - AD, tan
find length of pole In triangle
EDE2
i.e.
,
ED 2 ED
D2 D,
E E2
Height of base above
EE 2
_ tan -i
AB)
E,D 2
Height of top above
Length
sin0, - AD, sin0 2
of the direction of
inclination (relative to
To
2)
- 4e = AD, cos62 - 4E, cos0, /Id
EE\ + E2 D Z
-VEEl + E,D
2
2
SURVEYING PROBLEMS AND SOLUTIONS
100 2.65
To
find the height of an object from three angles of elevation
(Fig. 2.38)
only
Fig. 2.38
Solving triangles
ADB h
cos* =
and
2
cot
2
ADC
by the cosine
a + xz
rule,
-h 2 cot z p
2hx cot a 2 fc
cot 2 a + (x + y)
2
2
- h cot 2 (2.59)
2h(x + y) cot a 2
(x+y)[fc (cot
•••
i.e.
2
a -
cot
2 j8)
2 + x ]
=
2 2 x[h\cot a - cot 0) + (x + y)
2 ]
2 2 2 2 2 h [(x + y)(cot a - cot j9) - x(cot a - cot 0)]
=
x(x + y)
2
- x 2 (x + y)
2
i.e.
h
2
=
(x + y)[x(x + y) - x ] 2 2 2 2 (x + y)(cot a - cot /S) - x(cot a - cot 0)
2 (x + y)(cot a
[
[x(cot 2
H
(x + y)(xy) - cot 2 j8) - x(cot 2 a - cot 2 0)
xy(x + y) ] - cot 2 /8) + y(cot 2 a - cot 2 j8)J
(2.60)
x = y,
h
y/2x
=
2 [cot 6 - 2 cot 2 /3 + cot a]* 2
(2.61)
1
SURVEYING TRIGONOMETRY
101
Example 2.12 A,B and C are stations on a straight level line of bearing 126° 03' 34". The distance AB is 523-54 ft and BC is 420-97 ft. With an instrument of constant height 4'- 3" vertical angles were successively measured to an inaccessible up-station
At
B
7° 14' 00" 10° 15' 20"
C
13° 12' 30"
A
Calculate (a) the height of station
D
(b) the bearing of the line (c) the horizontal length
h\cot a - cot fi) + x 2 * 7T2hx cot a s 2 2 2 175-16 (cot 7°14'00 - cot 10° 15' 20") + 523-54 2 x 175-16 x 523-54 x cot 7° 14' 00"
(b)
(c)
=
0-85909
=
30° 47' 10"
Thus bearing of
Length of line
2.66
AD AD
y
The broken base Where a base line
the following system
=
126° 03' 34' - 30° 47' 10"
=
095°
16' 24"
=
h cot
a
=
175-16 cot 7° 14'
=
1380-07
line
problem
AD
ft
cannot be measured due to some obstacle
may be adopted,
Fig. 2.40.
Broken base
Fig. 2.40
line
Lengths x and z are measured. Angles
To
a,
Method
and 6 are measured at station E.
j8
calculate
BC = y
:
1
In triangle
AEB
EB =
x sin EAB sin
In triangle
AEC
EC =
a
+ y) sin EAC sin (a + /S)
(x
SURVEYING TRIGONOMETRY
S?
Then Also
in triangle
* sin ( a +
ft)
(x + y) sin
a
=
EC
EDB EB
+
Cv
=
EDC
EC
ft)
sinEDC
z
=
(2.62)
EDB
z) sin
sin(0 + in triangle
103
sin#
ER EC
Then
+ z> sin(^ zsin(0+ft)
fr
=
(2.63)
Equating Eqs (2.62) and (2.63) (y
+
(x
i.e.
x sin (a +
ft)
ft)
(x + y) sin
a
z)
*z sin (a +
_
z) sinfl
z sin (0 +
+ yXy +
=
+
ft) sin(fl
a sin# - sin ( a + ft)
ft)
sin
yz + y( x + z)
Then
+
xz 1
sin
This
is a quadratic
y
=
=
ft)
(2.64)
equation in y. Thus
_x+ z + 2
±£+ 2
sin ((9 +
a sin#
//*_+_z\ \
/J
2
V
2
_
^[l
J
Z /(* - )
V
2
_ sin(q +
I
2
+ r7
ft) sin(fl
sin a sin0
Si "( a
+
ft) 1
J
^ Sin (0 7ft)
sin
/
+
(265)
a sin#
Method 2 Area of triangle 4 BE
=
Axfc
=
I^E.EB
sin
a
(1)
BCE
=
iyfc
=
IfiE.
EC
sin
ft
(2)
CUE
=
±zh =
ADE
=
|(jc
=
±AE ED
Dividing (1) by (2)
y Dividing JV(4), 6 (3) by
Z ~
x +y + z
-
sin (9
(3)
+ y + z)h .
AE EC
x
ICE. ED
sin(a +
ft
+
0)
(4)
sin a (5)
sin ft
CE
Sin *
AB f AEsin(a + 8 +
_ 6)
(6)
SURVEYING PROBLEMS AND SOLUTIONS
104
Multiplying (5) by (6),
sin/3 sin(a +
z)
Then
y
-(*±±)
=
\
2
/
j8
+ d)
- xz sin/3sin(a + /3 + sin a sin#
y* + y(x + z)
i.e.
a sin0
sin
xz
y(x + y +
/fc±£V
+
sj\
2
fl)
=
sin P sin < a + Z3 + ^> + xz
sina
/
sinfl
(2 66) v '
Method 3 (Macaw's Method) In order to provide a logarithmic solution an auxiliary angle is
used.
From the quadratic equation previously formed, y
i.e.
Now
{y
+
2
+ y(x +
jfc,
+ z)P tan'M
let
z)-
=
xz
sin/3sin(a + sina sin#
J(, +
zY +
y + |(x + z)}
2
we
fl)
=
(2 67)
" Sin fins^°, +/ + g) (?)
(2-68)
(2 69)
get
=
I(x +
=
i(jc
+
2
z) (l 2
z)
+
tan 2 M)
sec 2 iW
secM
y + I(x + z)
=
|(x + z)
y
=
I(x + z)(secM -
y
+
4«z si„ fl gin (a + p + (x + z) sin a sin
Substituting this in Eq. (2.68), {
/3
1)
=
(x + z)
sec/Wkl - cosM)
=
(x + z)
secM
sin 2 iiW
(2.70)
Example 2.13 The measurement of a base line AD is interrupted by an obstacle. To overcome this difficulty two points B and C were established on the line AD and observations made to them from a station
E
as follows:
AEB BEC CED Length
AB
= 527-43
ft
and
=
20° 18' 20"
= 45° =
CD
19' 40"
33° 24' 20"
= 685-29
ft.
Calculate the length of the line AD. (R.I. C.S.)
1
SURVEYING TRIGONOMETRY
a = 20°
Here
p =
x
18' 20" \
=
45°19'40"
a+p a+p+
=
33° 24' 20"
p
527-43
j
= d
65° 38' 00"
= 99°02'20" =
+ d
105
78° 44' 00"
\{x + z)
=
1(1212-72)
=
606-36
|(jc^z)
=
§(157-86)
=
78-93
|
z
By method y
=
685-29
=
J
1
As .93'
-606-36 +
,
527-43 x 685-29 sin 65° 38' sin 78° 44'
V
sin20°18'20"sin33°24'20"
=
-606-36 + V6230 + 1690057 -606-36 + 1302-415
=
696-055
=
ft
AD
Then
=
1212-72 + 696-055
=
1908 -775
ft
By method 2
,
y = -606-36 + yfa)6-36 + 527 43 x 685 29 sin 45°19'40" sin99°02'20" ^ sin 20° 18' 20" sin 33° 24' 20" '
Let (1) lines /4B, and B,C be inclined to the horizontal plane by a and /S respectively. (2) Horizontal (3) (4)
AB BC
In triangle
= =
h cot
h cot/8
AB 2 h
2
2
+
cot
Similarly in triangle
=
a
=
=
AB,
=
h
coseca
B,C = fccosec/3
ABC,
4C 2
AC
= 6
Angle in inclined plane AB^C = B,B = h
Then
2
ABC
angle
h cosec
2
BC 2
a+
- 2 AB.BC cos (9
h
2
cot 2 fi - 2 A 2 cot
a
cot
/8
cos
AB,C, 2
a +
2 /z
cosec 2 /3 -
2fc
2
coseca cosec/S cos<£
SURVEYING TRIGONOMETRY
107
Then z
h cot
2
2
a+
/i
cot
2 ft
2
2
-2fc cotacotft cos0 =
/i
2
cosec 2 a + h cosec 2 B
- 2h z coseca cosecft cos<£ cos*0 = (cosec
2
a -
2
cot a) + (cosec
2
cosec 2 a - cot 2 a =
as
Then
ft
- cot 2 ft) + 2 cot a cot ft cosfl
coseca cosecft
cosec 2 ft - cot
2(1
=
cos<£
2
2 ft
=
1
+ cot a cot ft cosfl)
coseca cosecft = sin a sin/8 + cos a cos ft cos# 2
or
cos
=
cos 6
(2.71)
- sin a sin ft cos a cos ft
,n y
From a station 4 observations were made to stations with a sextant and an abney level.
Example 2.14
B
and
C
With sextant
-
angle
With abney level
-
BAC =
84° 30'
angle of depression (AB) 8° 20' angle of elevation
(AC) 10° 40'
Calculate the horizontal angle BA^ C which would have been if a theodolite had been used
measured
(R.I.C.S./M)
From equation
(2.72),
cos0 =
cos
- sin a sin ft cos a cos ft
=
cos 84° 30' - sin (-8° 20') sin 10° 40' cos (-8° 20') cos 10° 40'
cos 84° 30' + sin8°20'sinl0°40' cos 8° 20' cos 10° 40'
0-09585 + 0-14493 x 0-18509 0-989 44 x 0-98272
0-12268 0-97234
d
Example
2.
ABC.
AB
BC
If
15
A
=
0-12617
=
82° 45' 10"
pipe-line is to be laid along a bend in a mine roadway
2 in a direction 036° 27', whilst rises due South at 1 in 3-5, calculate the angle of bend in the pipe. falls at a gradient of 1 in
(R.I.C.S.)
SURVEYING PROBLEMS AND SOLUTIONS
108
3.5
Fig. 2.42
Plan
From equation (2.71),
cos0 =
sin
a
sinjS + cos
a
cos/3 cos#
where
a =
£
1
cot" 2 1
= cot" 3-5
26° 33'
= =
15° 57'
6 = 036° 27' - 00° .-.
cos0 = cf>
sin 26° 33' sin 15°
= 35° 26' 40"
36° 27'
=
57'+ cos 26°
i.e.
33' cos 15° 57' cos 36° 27'
35° 27'
Exercises 2(b)
Show that for small angles of slope the difference between horiz zontal and sloping lengths is h /2l (where h is the difference of vertical height of the two ends of a line of sloping length I) If errors in chaining are not to exceed 1 part in 1000, what is the
8.
greatest slope that can be ignored ?
[L.U/E Ans. 2°
34']
The height of an electricity pylon relative to two stations A and B (at the same level) is to be calculated from the data given below.
9.
Find the height from the two stations the theodolite axis is 5'— 0".
if
at both stations the height of
SURVEYING TRIGONOMETRY Pylon
Data:
AB
109
= 200
ft
Horizontal angle at Vertical angle
A
"
'
x fl
10.
X,
XY
= 56
Y and Z ft
and
B =
40° 20'
at
A
at
B
43° 12' 32° 13'
= =
mean height 139-8 ft)
Ans.
(R.I.C.S.
A = 75° 10'
at
are three points on a straight survey line such that
YZ =
80
ft.
From X, a normal offset was measured to a point A and X/4 was found to be 42ft. From Y and Z respectively, a pair of oblique offsets were measured to a point B, and these distances were as follows:
YB =
96 ft,
ZB
= 88
ft
Calculate the distance 4fi, and check your answer by plotting to some suitable scale, and state the scale used.
(E.M.E.U. Ans. 112-7 ft) 11.
From the top
point
A
B
of a tower 120
ft high, the angle of depression of a and of another point B is 11°. The bearings of A and from the tower are 205° and 137° respectively. If A and B lie in a
is
15°,
horizontal plane through the base of the tower, calculate the distance
AB. (R.I.C.S.
A, 6, C,
12
D
Ans. 612
ft)
are four successive milestones on a straight horizon-
tal road.
From a point due W of A, the direction of B is 84°, and of D The milestone C cannot be seen from 0, owing to trees. If
is 77°.
the direction in which the road runs from and the distance of from the road.
A
to
D
is 0, calculate 0,
(R.I.C.S. Ans. = 60°06' 50", OA = 3-8738 miles) At a point A, a man observes the elevation of the top of a tower B to be 42° 15'. He walks 200 yards up a uniform slope of elevation 12° directly towards the tower, and then finds that the 13.
elevation of
has increased by 23° 09'. Calculate the height of
B above
B
the level
of A. (R.I.C.S. Ans. 823-82 ft) At two points, 500 yards apart on a horizontal plane, observations of the bearing and elevation of an aeroplane are taken simultaneously. At one point the bearing is 041° and the elevation is 24°, and 14.
at an-
other point the bearing is 032° and the elevation is 16°. the height of the aeroplane above the plane. (R.I.C.S.
Calculate
Ans.
1139ft)
SURVEYING PROBLEMS AND SOLUTIONS
110
Three survey stations X, Y and Z lie in one straight line on plane. A series of angles of elevation is taken to the top of a colliery chimney, which lies to one side of the line XYZ. The angles measured at X, Y and Z were: 15.
the
same
at
X, 14° 02';
Y, 26°
at
The lengths XY and YZ are 400
34'; ft
at Z,
and 240
ft
18° 26' respectively.
Calculate the height of the chimney above station X. (E.M.E.U. Ans.
112-0
ft)
The altitude of a mountain, observed at the end A of a base line of 2992-5 m, was 19° 42' and the horizontal angles at A and B were 127° 54' and 33° 09' respectively. 16.
AB
Find the height of the mountain. (Ans.
1804 m)
between two inaccessible points A and B by observations from two stations C and D, 1000 m apart. The angular measurements give ACB = 47°, BCD = 58°, BDA = 49°; ADC = 59°. 17.
It
is required to determine the distance
Calculate the distance
AB (Ans.
18.
B
An aeroplane same
at the
the aeroplane is
Show
is
level,
observed simultaneously from two points A and a distance (c) due north of B. From A
A being
S^°E
and from
B N
E.
that the height of the aeroplane is
c tan
a
sin
sin(0 +
and find
2907 -4 m)
its
)
elevation from B.
/L.U.
Ans.
B =
tan-'
sin
V
/
base line ABCD is sited such that a portion of BC cannot be measured directly. If AB is 575-64 ft and CD is 728-56 ft to one side of ABCD are and the angles measured from station 19.
It is proposed to lay a line of pipes of large diameter along a roadway of which the gradient changes from a rise of 30° to a fall of 10° coincident with a bend in the roadway from a bearing of N 22° W to N 25° E.
20.
Calculate the angle of bend in the pipe. (Ans.
11M on » N ny ,J?n( ou )
SURVEYING TRIGONOMETRY 21.
At a point
A
111
at the bottom of a hill, the elevation of the top of
a tower on the hill is 51°
18'.
At a point
B
on the side of the
same vertical plane as A and the tower, the elevation AB makes an angle 20° with the horizontal and the distance in the
feet.
hill,
and
is 71°40'.
AB = 52 Determine the height of the top of the tower above A. (L.U. Ans. 91-5 ft)
Two points, A, B on a straight horizontal road are at a distance 400 feet apart. A vertical flag-pole, 100 feet high, is at equal distances from A and B. The angle subtended by AB at the foot C of the pole (which is in the same horizontal plane as the road) is 80°. Find (i) the distance from the road to the foot of the pole, 22.
(ii)
the angle subtended by
(L.U.
AB
at the top of the pole.
Ans.
(i)
258-5
ft,
(ii)
75° 28')
Bibliography
USILL, G.w. and HEARN, G., Practical Surveying (Technical Press). HADDOCK, M.H., The Basis of Mine Surveying (Chapman and Hall). LONEY, s.c, Plane Trigonometry (Cambridge University Press). blakey, J., Intermediate Pure Mathematics (Macmillan).
3 CO-ORDINATES A
point in a plane
may be defined by two systems:
(1) Polar co-ordinates. (2)
Rectangular or Cartesian co-ordinates. 3.1
Polar Co-ordinates
This system involves angular and linear values,
i.e.
bearing and
length, the former being plotted by protractor as an angle from the
meridian.
B
Fig.
3. 1
(s,8)
Polar co-ordinates
A normal 6 inch protractor allows plotting to the nearest 1/4°; a cardboard protractor with parallel rule to 1/8°; whilst the special Booking protractor enables 01' to be plotted.
The displacement
of the point being plotted depends on the physi-
cal length of the line on the plan, which in turn depends on the horizontal projection of the ground length and the scale of the plotting.
s
A Fig. 3.2 If
a
is the
Displacement due
to angular error
angular error, then the displacement
112
CO-ORDINATES
BB a
and as
= s tana
X
is small
~
BB, s = 300
If
ft
and
a=
=
300 x 60 x 12 inches 206 265
~
1 inch
DD
1
sa
01' 00",
fiS'
i.e. 1
113
.
,
minute of arc subtends 1 inch in 100 yards,
second of arc subtends
Similarly, on the metric system,
dd = BB,
i. then
6000 yards,
1 inch in
i.e.
3 x/a miles.
s = 100 metres and
if
100x60
a = 01
'
00",
.
metres
206 265
= 0-0291 m, Thus,
29 mm
i.e.
1
minute of arc subtends approximately 30
1
second of arc subtends approximately
1
1cm
or
A i.e.
point plotted on a plan
0'Olin. in 1 yard (0*25 0*1 in. in 1 yard (25
mm
mm
may be assumed in 1 metre)
in 1 metre)
—*
—
>
1
mm mm
to
in
100m,
in
200 m
in 2 km.
be 0*01 in., (0*25mm),
minute of arc,
10 minutes of arc.
As this represents a possible plotting error on every line, seen how the error may accumulate, particularly as each point ent on the preceding point.
it
is
can be depend-
Plotting to scale
3.11
The length
some
of the plotted line is
definite fraction of the ground
length, the 'scale' chosen depending on the purpose of the plan and the
size of the area.
Scales may be expressed in various ways: (1) (2)
(3)
3.12
As inches (in plan) per mile, e.g. 6 in. to 1 mile. As feet, or chains, per inch, e.g. 10ft to 1 inch. As a representative fraction 1 in n, i.e. 1/n, e.g. 1/2500.
Conversion of the scales
40 inches to
1
mile
—
insult 1 inch I
lin.
ranracanfc represents
= 132 ft = 44 yd = 2 chn
— — .-—
.-
-rp:
40
feet
SURVEYING PROBLEMS AND SOLUTIONS
114
1
inch to 132 ft
-
1
inch represents 132 x 12 inches
lin.
Thus the representative
= 1584 in. fraction is 1/1584.
Scales in common use
3.13
Ordinance Survey Maps and Plans:
Large scale:
Medium
1/500, 1/1250, 1/2500.
scale: 6 in. to 1 mile (1/10 560), 2y2 in. to 1 mile (1/25000).
Small scale:
1/625000, 1/1250000.
2,l, 1/2 , 1/4 in. to 1 mile;
Engineering and Construction Surveys: 1/500, 1/2500, 10-50 ft to 1 inch, (See Appendix, p. 169)
3.14
1/4,
V8, Vl6in. to
1ft.
Plotting accuracy
Considering 0*01 in. (0*25 mm) as the size of a plotted point, the following table shows the representative value at the typical scales.
suggested measurement
O.S. Scales
precision limit 3 in. (76 mm)
1/500
0-01 x 500
=
1/1250
0-01 x 1250
= 12-5 in.
1ft (0 -3 m)
1/2500
0-01 x 2500
= 25-0 in.
2 ft (0-6 m)
1/10 560
0*01 x 10 560 =
105-6
in.
5
ft
(1-5 m)
1/25000
0*01 x 25000 = 250-0
in.
10
ft
(3-0 m)
5 in.
Engineering Scales
0-01 x 600
= l-2in. = 6-0 in.
6
0-01 x 792
= 7-92 in.
6 in. or l/2 link
0-01 x 1584
= 15-84 in.
1 ft or 1 link.
ft
0-01 x 120
lin. to 50ft
1 in. to
lin. to lin. to
10
lchn 2chn
lin. in.
Incorrect scale problems
3.15
If a scale of 1/2500 is used on a plan plotted what conversion factor is required to
to
scale 1/1584
(a) the scaled lengths, (b) the area (a) 1 in.
On
computed from the scaled length?
the plan 1
in.
= 1584 in. whereas the scaled value shows
= 2500 in. All scaled values must be converted by a factor 1584/2500
= 0-6336.
CO-ORDINATES (b) All the
115
computed areas must be multiplied by (0*6336jf
0-4014
Bearings
3.2
Four meridians may be used, Fig. 3.3: 1.
True or geographical north.
2.
Magnetic north.
3.
Grid north.
4.
Arbitrary north.
GN.
M.N
T. N.
M.N.
T.N.
6.N.
4
Convergence
of
the meridian
Magnetic declination
Fig.
3.
3 Meridians
True north
3.21
The meridian can only be obtained precisely by astronomical The difference between true bearings at A and B is the
observation.
convergence of the meridians to a point, i.e. the north pole. For small surveys the discrepancy is small and can be neglected but where necessary a correction may be computed and applied. 3.22
Magnetic north
There is no fixed point and thus the meridian is unstable and subjected to a number of variations (Fig. 3.4), viz.:
— the annual change in the magnetic declinabetween magnetic and true north. At present the magnetic meridian in Britain is to the west of true north but moving towards it at the approximate rate of lOmin per annum. (Values of declination and (a)
Secular variation
tion or angle
SURVEYING PROBLEMS AND SOLUTIONS
116
the annual change are shown on certain O.S. sheets.) (b) Diurnal variation
the
mean value
minima
—a
daily sinusoidal oscillation effect, with
and 6-7 p.m., and maxima and and 1p.m.
at approximately 10a.m.
at approximately 8a.m.
(c) Irregular variation
— periodic magnetic
fluctuations thought to
be related to sun spots.
Secular variation
13001 25* 20
15
10
5
5
10
15*
East
West
Diurnal variation
0.00
24.00 Hours G.M.T.
East
Fig. 3.4 Approximate secular
and diurnal variations
in
magnetic
declination in the London area (Abinger Observatory)
Grid north (see section 3.7).
3.23
O.S. sheets are based on a modified Transverse Mercator projection which, within narrow limits, allows: (a)
Constant bearings related to a parallel
(b)
A
grid.
scale factor for conversion of ground distance to grid dis-
tance solely dependent on the easterly co-ordinates of the measurement site. (See page 39).
3.24
Arbitrary north for absolute reference and often the leg of the traverse is assumed to be 0°00 '.
This may not be necessary first
CO-ORDINATES Example 3.1 If
True north is 0°37'E of Grid North. Magnetic declination in June 1955 was 10°27' W.
the annual variation
grid bearing of line
line
AB
in
117
AB
was
10'
082°32'
,
per annum towards North and the what will be the magnetic bearing of
January 1966? G.N.
MM'66
MK'55
»•
Fig. 3.5
Grid bearing Correction
082°32'
True bearing
081°55'
Mag
-0°37
/
declination
June 1955
10°27'
Mag. bearing
June 1955
092°22'
Variation for
January 1966
3.25
-lO^xlO'
-1°45'
Mag. bearing January 1966
090°37'
Types of bearing There are two types in general use:
(a) Whole circle bearings (W.C.B.), which are measured clockwise from north or 0°-360°.
SURVEYING PROBLEMS AND SOLUTIONS
118 (b)
Quadrant bearings (Q.B.), which are angles measured to the
east or west of the
N/S meridian.
For comparison of bearings, see Fig.
3.6.
N
Case
(i)
Case
(ii)
W Case
(Hi)
Case
(iv)
270'-
W 270Fig. 3.6
Comparison
of
bearings
119
CO-ORDINATES Case
(i)
Whole circle bearing in the W.C.B. of Q.B. of
Case
(ii)
AB AB
= a°
AC AC
= a°2
=
first
quadrant
— 90°
N a?E
90°- 180° W.C.B. of Q.B. of
= S
0°E
= S(180-a 2)°E
Case
180°- 270°
(Hi)
W.C.B. of Q.B. of
AD AD
= a°3 =
S0°W
= S(03-180)°W
270°- 360°
Case(iv)
W.C.B. of Q.B. of
AE AE
= a°4 = N
W
= N (360
-a^W
Example 3.2 072° =
N
148° =
S32°E S 16° W N30°W
196° =
330° = N.B. fix is
72°
E 180-148 = 32° 196-180 = 16° 360-330 = 30°
i.e. i.e. i.e.
Quadrant bearings are never from the E/W always N or S.
line,
so that the pre-
It is preferable to use whole circle bearings for most purposes, the only advantage of quadrant bearings being that they agree with the values required for trigonometrical functions 0-90° as given in many mathematical tables (see Chapter 2), e.g.:
(Fig. 3.7a)
sin 30° = 0-5
cos 30° = 0-8660 tan 30° = 0-5774 (Fig. 3.7b)
sin 150° =
sin (180 -150)
= sin 30° cos 150° = -cos (180 -150) = -cos 30° tan
150°=
Sinl5 °
cos 150 = -tan 30°
=
+sin3° -cos 30
SURVEYING PROBLEMS AND SOLUTIONS
120
*E Cos 30°
Bearing N E
(a)
+ +
Bearing S E
-Cos 30°
*E (b)
210°
-Cos 30* Bearing S
W
Bearing N W + ~ + Cos 30°
330" Fig. 3.7
(Fig. 3.7c)
- sin (210 - 180) = -sin 30°
sin 210° =
cps210° = -cos (210- 180) = -cos 30°
CO-ORDINATES tan 210° =
- sin 30
sin 210 cos 210
= +tan30 (Fig. 3.7d)
121
-cos 30 c
sin 330° = - sin (360 - 330)
= - sin 30° cos 330° = cos (360 -330) = + cos 30° - sin 30 + cos 30
sin 330 cos 330 = -tan 30°
tan 330° =
Conversion of horizontal angles into bearings. (Fig. 3.8)
3.26
Fig. 3.8 Conversion of horizontal angles into bearings
Forward Bearing
AB
Back Bearing
BA = a BC = a
Forward Bearing If
the
=
sum exceeds 360° then 360
Bearing BC(j8) =
a
± 180 +
a°
± 180° ± 180 +
is subtracted, i.e.
- 360 = a + Q ± 180
This basic process may always be used but the following rules simplify the process. (1)
To
the forward bearing add the clockwise angle.
sum is less than 180° add 180°. sum is more than 180° subtract 180°. some cases the sum may be more than 540°, then subtract
(2) If the If the (In
540°) N.B.
If
the angles measured are anticlockwise they must be sub-
tracted.
122
SURVEYING PROBLEMS AND SOLUTIONS
Example 3.3
Fig. 3.9
bearing
+ angle
AB ABC
= 030°
N
30°
E
= 210° 240°
bearing
+ angle
-
180°
BC BCD
= 060° = 56°
N60°E
116°
bearing
+ angle
+
180°
CD CDE
= 296° = 332°
N64°W
628°
bearing
-
540°
DE
= 088°
N88°E
CO-ORDINATES Check
AB
bearing
123
= 030° = 210°
angles
56°
332°
628°
-«xl80°,
i.e.
-3x180° -
bearing
DE
540°
= 088°
The final bearing is checked by adding the bearing of the first line sum of the clockwise angles, and then subtracting some multiple
to the
of 180°.
Example 3.4 to
The clockwise angles
of a closed polygon are observed
be as follows:
A B C
223°46'
241°17' 257°02'
D E F
250°21'
242° 19' 225°15'
If the true bearings of BC and CD are 123° 14' and 200° 16' respectively, and the magnetic bearing of EF is 333°2l', calculate the magnetic declination.
(N.R.C.T.)
From the size of the angles it may be initially assumed that these are external to the polygon and should sum to (2n + 4)90°", i.e.
l(2x6) + 4}90 = 16x90 = 1440° 223°46'
241° 17' 257°02'
250° 21' 242° 19' 225°15'
Check 1440°00'
To
obtain the bearings,
Line
BC
bearing
123° 14'
BCD
257°02'
+ angle
380° 16'
-
180°
SURVEYING PROBLEMS AND SOLUTIONS
124
bearing
+ angle
CD CDE
200° 16' (this checks with given 250°21' value) 450°37'
bearing
+ angle
DE DEF
180° 270°37'
242° 19' 512°56' 180°
bearing
EF
332°56'
+ angle
EFA
225° 15'
558° 11' 540° bearing
FA
018°11'
+ angle
FAB
223°46' 241°57'
180° bearing
+ angle
AB ABC
061°57' 241°17'
303° 14' 180° bearing
BC
123° 14'
Magnetic bearing
EF EF
333°21'
True
bearing
3.27
332°56' 0°25'
Magnetic declination
Check
W
Deflection angles (Fig. 3.10)
In isolated cases, deflection angles are measured and here the normal notation will be taken as:
Right angle deflection— positive. Left angle deflection— negative.
Taking the Example
3.3,
AB
Deflection right
030° + 30°
Deflection
left
-124°
Deflection right
+ 152°
Bearing
CO-ORDINATES
125
•152°(R)
Fig.
3.
10 Deflection angles
Bearing
AB
030° + 30°
Bearing
BC
060°
+ 360° 420°
-124° Bearing
CD
296°
+ 152° 448°
-360° Bearing
DE
088°
Check
AB
030° + 30°
+ 152° + 212°
-124°
DE
088°
;
SURVEYING PROBLEMS AND SOLUTIONS
126
Exercises 3(a) 1.
Convert the following whole circle bearings into quadrant bearings: 214°30'
311°45' (Ans.
027°15'
;
218°30'
;
S34°30' W; S 22°30' E;
N N
;
;
287°45'
078°45'
;
157°30';
;
244°14'
;
278°04.'
N
72°15' W; 38°30' S W;
W
S 86°30' E;
N
78°45' E;
)
Convert the following quadrant bearings into whole circle bearings:
N
S 34°15' E; 82°45' W;
25°30' E;
N
S 18°15' W; (Ans.
3.
093°30'
27°15' E; 48°15' W;
S64°14' W; N81°56' 2.
;
S 42°45' W; S 64°14' E;
025°30'
;
145°45'
;
222°45'
277°15'
;
115°46'
;
214°30')
;
N
79°30' W;
S 34°30' W.
280°30'
The following clockwise angles were measured
;
198°15'
in a
;
closed
tra-
verse. What is the angular closing error?
163°27'36";
324°18'22";
62°39' 27"
181°09'15";
305°58'16";
188°02'03";
;
330° 19' 18"
292°53'02";
131°12'50" 4.
(Ans. 09")
Measurement of the
interior anticlockwise angles of a closed tra-
ABC
verse D E have been made with a vernier theodolite reading to 20 seconds of arc. Adjust the measurements and compute the bearings of the sides if the bearing of the line AB is N 43° 10' 20" E.
Angle
5.
EAB ABC BCD CDE DEA
135°20'40"
(R.I.C.S. Ans.
60°21'20" 142°36'20"
89°51'40"
AB N 43°10' 20" E BC S 17°10'52"E CD S 20°12'56" W DE N 69°38' 36" W EA N01°29'08" W)
111°50'40"
From the theodolite readings given below, determine the angles ABCDE. Having obtained the angles, correct them to
of a traverse
the nearest 10 seconds of arc and then determine the bearing of the bearing of AB is 45° 20' 40".
Back
BC
Readings Forward Station
Theodolite
Forward
Station
Station
E
A
B
0°00'00"
264°49'40"
A B C D
B
C
264°49'40"
164° 29' 10"
C
D E
164° 29' 10" 43°58' 30"
314° 18 '20"
A
314°18' 20"
179°59' 10"
Station
D E
if
Back Station
43°58'30"
(R.I.C.S. Ans. 125°00' 20")
CO-ORDINATES Rectangular Co-ordinates
3.3
A
127
may be fixed in a plane by linear values measured xy axes. The x values are known as Departures or Eastings whilst the y values are known as Latitudes or Northings. The following sign convention is used: point
parallel to the normal
Direction
East
+x
+ departure
+ Easting (+E)
West
-x
- departure
-Easting (-E)
North
+y
+ latitude
+ Northing (+N)
South
_y
- latitude
-Northing (-N)
N.W. quadrant (270*-360*) N
N.E.
quadrant (0*-90')
Latitude
Departure (eastings)
270* W-+
*90*E
(+4,-2)
(-2,-4) S.W. quadrant (180'- 270*)
quadrant OC-ieO*)
Rectangular co-ordinates
Fig. 3.11
N.B.
S.E.
0°-90°
-+
NE
i.e.
90°- 180°
—
S
E
i.e.
180°- 270°
-»
S
W
i.e.
-N -E
-lat -dep.
270°- 360°
-»
NW
i.e.
+ N-E
+ lat -dep.
+E -N +E +N
or
+ lat + dep. -lat
+ dep.
This gives a mathematical basis for the determination of a point with no need for graphical representation and is more satisfactory for the following reasons: (1)
Each station can be plotted independently.
(2) In plotting, the point is not
dependent on any angular measur-
ing device. (3)
Distances and bearings between points can be computed.
SURVEYING PROBLEMS AND SOLUTIONS
128
Rectangular co-ordinates are sub-divided into: (1) Partial Co-ordinates, (2)
3.31
which relate
Total Co-ordinates, which Partial co-ordinates,
AE,
to a line.
relate to a point.
AN
(Fig. 3.12)
These relate one end of a line to the other end. They represent the distance travelled East (+)/West
(-)
and
North (+)/South (-) for a single line or join between any two points.
Fig. 3.12 Partial co-ordinates
Given a line of bearing d and length
AE
Partial departure =
A E AB
N.B.
always compute
difference in Eastings
= s sin Q
AN
Partial latitude =
AN^g
i.e.
s,
difference in Northings
scosd
= in
i.e.
(3.1)
(3.2)
bearings not angles and preferably quadrant
bearings.
3.
32
Total co-ordinates (Fig.
These relate any point The following notation Total "
Total «
Total "
Easting of Northing of
Easting of Northing of
Easting of Northing of
3.
to the
13)
axes of the co-ordinate
system used.
is used:
A = EA A = N^
B = E A + A E AB B = N A + AN AB C = E B + A EBC = EA + A EAB + A E BC C = N B + &NBC = N^ + AN^ + AN BC
129
CO-ORDINATES
Fig. 3.13 Total co-ordinates
Thus
in general terms
Total
Easting of any point =
E^+SAE
(3.3)
= Total easting of the first point + the sum of the partial eastings up to that point.
Total Northing of any point = N^ +
2 AN
(3.4)
= Total northing of the first point + the sum of the partial northings up to
that point.
N.B.
If
a traverse is closed polygonally then
SAE SAN i.e.
the
sum
Example
(3.5)
=
(3.6)
of the partial co-ordinates should equal zero.
3.5
AB BC CD
m m 210° 100 m co-ordinates of A E 50 m N
Given: (Fig. 3.14)
Total
Line
=
AB
045° =
Partial departure
N
AE^
45°
E
045°
100
120°
150
40 m
100 m
= 100 sin 45° = 100 x 0-707
= + 70*7 m
Total departure (E^)
A= +
Total departure (E fl )
B=
50*0
m
+ 120-7 m
130
SURVEYING PROBLEMS AND SOLUTIONS
-50* (200-6.-50-9)
Fig. 3.14
Partial
Line
latitude
BC
A N^
= 100 cos 45°= 100x0*707
120° = S 60°
AE BC
Partial departure
= + 70-7 m
Total latitude (N^)
A = +
Total latitude (N5 )
B=
40*0 m
+ 110*7 m
E 150 m
= 150 sin 60° = 150 x 0*866
= + 129*9 m
B = + 120*7 m (EC )C = + 250*6 m
Total departure (EB ) Total departure Partial
Line
latitude
AN^C
CD 210°= S30°W
Partial departure
Partial
= 150 cos 60° = 150 x 0*5
latitude
Check ED = EA +
AE CD
AN CD
= - 75*0
Total latitude
(NB )fi = +110*7 m
Total latitude
(Nc )
C = +
35*7
m
100 m
= 100 sin 30° = 100 x 0*5
= - 50*0
Total departure (E^)
C
= + 250*6 m
Total departure (E^)
D
= + 200*6 m
= 100 cos 30° = 100 x 0*866 Total
latitude (Nc )
Total
latitude (N^)
= - 86*6m
C = +
m D = - 50*9 m
&E AB + AE BC + AECZ? = 50*0 + 70*7 + 129*9 - 50*0 = + 200*6 m
35*7
CO-ORDINATES
131
N^= KA + Mi AB + ANSC + ANqd = - 50-9 m
= 40-0 + 70-7 - 75-0 - 86'6 Exercises 3(b) (Plotting) Plot the following traverse to a scale of
6.
= 100
1 in
thereafter obtain the length and bearing of the line in square yards of the
AB
links, and and the area
enclosed figure.
N N
21°
W
120 links from
28°
E
100 links
N60°E
117 links
E 15° E 40° W
105 links
N S S
(Ans. From scaling
32°
A
200 links
75 links
N62°45'E 340
to
B
links; approx. area
3906 sq yd,)
The following table shows angles and distances measured
7.
theodolite traverse from a line length 110 ft.
AB
in a bearing due South and of horizontal
Angle
Angle value
Inclination
ABC BCD CDE DEF
192°00'
+15°
92°15'
0°
93°30'
-13°
170°30'
0°
Inclined distance
BC CD
DE EF
(ft)
150
200
230 150
Compute the whole
circle bearing of each line, plot the survey to a scale of 1 in. = 100 ft and measure the horizontal length and bear-
ing of the closing line.
(M.Q.B./M. Ans. 260 ft; 076°30')
The following notes reter to an underground traverse 8. the mouth, A, of a surface drift. Line
Bearing
AB BC CD DE
038°
325
111°
208
level
006°
363
level
308°
234
made from
Distance (links) dipping at 1 in 2-4
rising at 1 in
3-2
Plot the survey to a scale of 1 chain to 1 inch. Taking A as the origin, measure from your plan, the co-ordinates of E.
What is the difference in level between
A and E
to the nearest
foot?
(M.Q.B./UM Ans. E, E 233 links N 688 78
ft)
links; diff. in level
AE
SURVEYING PROBLEMS AND SOLUTIONS
132 9.
Plot the following notes of an underground traverse to a scale of
lin
**
100
ft.
Line
Distance
Bearing N 28° W
AB
354 ft dipping at
W S 83° W N 8°E
BC CD DE EF
N
83°
S 89°
133
ft
1 in
7
level
253ft level
E
219
ft
rising at 1 in 4
100
ft
level
Points A,B,C and D are in workings of a lower seam and points and F are in the upper seam, DE being a cross measure drift between the two seams. It is proposed to drive a drift from A to F. Find the bearing, length, and gradient of this drift. (M.Q.B./UM Ans. N 40° W; 655 ft; +1 in 212)
£
10.
The co-ordinates
in feet, relative to a
common
point of origin
A, are as follows:
Departure
A B C
E E 360 E
D Plot the figure
Latitude
A BCD
275
237
552
230
N N
174 S
to a scale of 1 inch to 100
ft
and from the
co-ordinates calculate the bearing and distance of the line AC. (M.Q.B./UM Ans. N 67°24' E; 598ft) 11.
An area
ABC
in the form of a triangle
co-ordinates of the points
A B and C
has been defined by the as
in relation to the origin 0,
follows:
A B C
South 2460
ft
East 3410
North 2280
ft
North 1210
ft
East 4600 ft East 1210 ft
Plot the positions of the points to a scale of
ft
1 in. to
find the area, in acres, enclosed by the lines joining
1000
AB,
and and
ft,
BC
CA. (M.Q.B./M Ans. 169-826 acres) There is reason to suspect a gross angular error in a five-legged 12. closed traverse in which the recorded information was as follows: Interior angles:
Sides:
A
B
110°;
AB EA
180
ft;
245
ft
150°;
BC
C
70°;
420 ft;
CD
D
110°;
350 ft;
E 110° DE 410 ft;
CO-ORDINATES Plot the traverse to a scale of 100
ft
133
to lin.
and locate the gross
angular error*, stating its amount.
(L.U./E)
A
13.
rough compass traverse of a closed figure led to the following
field record:
Line
Length
AB EC
422
Bearing 57°
405
316°
CD
348
284°
DE
489
207°
EA
514
109°
Plot the figure (scale lin = 50ft) and adjust it to close using a graphical method. Letter your plan and add a north point (magnetic declination 10° W).
(L.U./E) 3.4
Computation Processes
As tables of trigonometrical functions are generally tabulated only terms of angles 0°-90°, it is convenient to convert the whole circle bearings into reduced or quadrant bearings. in
The signs of the partial co-ordinates are then related to the symbols of the quadrant bearings, Fig. 3.15.
E + W -
1 >
N
r^
Departures
+
J
::)
NW + -
Latitudes
Alternatively, the whole circle
bearings are used and the sign of the value of the partial co-ordinate is derived from the sign of the
N E + +
W
+
sw
SE +
trigonometrical function.
— s Fig.
The process may be (i)
(ii)
3.
15
either
by logarithms or by machine (using natural trigonometrical functions).
See Chapter 6 on location of errors.
m
SURVEYING PROBLEMS AND SOLUTIONS
134
Computation by logarithms
3.41
AB
Let
= 243° 27'
423-62
m
(.4
2063-16 m
E 5138 -42 m N)
(243° 27' = S63°27' W)
Logs partial departure
(AE) 2-578 579
AB
1-951 602
\
distance
2-626977
1\
cos bearing
1
-650 287
)
partial latitude
2-277 264
sin bearing
(&NAB )
The log distance
N.B.
is written
EA
2063-16
t^AB EB
-378-95
"a
down once
1684-21
m m
5138-42
ANAB
-189-35
NB
4949-07
m
only, being added to the
log sin bearing above and the log cos bearing below, to give the partial
departure and latitude respectively. It is
often considered good computing practice to separate the log
figures for convenience of adding though the use of squared paper
would obviate
this.
Computation by machine
3.42
partial departure
AR._
sin bearing
0-894 545
EA AB.o EB
-378-95
"a
5138-42
2063-16
1684-21
J
A 23-62
distance
0-446 979
cos bearing partial latitude
AN ^MB
J
\
1
> *
WAB "b
Using a normal
digital machine, the distance (being
189-35
4949-07
common)
in
is
set once in the machine and then separately multiplied by the appropri-
ate trigonometrical function. In the case of the twin-banked Brunsviga, the processes are simul-
taneous.
N.B.
For both natural and logarithmic trigonometrical functions the
following tables are recommended:
Degrees only Degrees and minutes
5 figure tables
Degrees, minutes and seconds
6 figure tables
4 figure tables
Degrees, minutes, seconds and decimals of seconds 7 figure tables
135
CO-ORDINATES Tabulation process (Fig. 3.16)
3.43
Nottingham Regional College of Technology
Traverse computation sheet
Compiled by
_^±y_-
Date
_?j_r„/ii.7 : .
Line
Bearing
AB
N*
Traverse
1
4E =
length x sin B'g
Length
sin/cos S£r
4N= length x cos AE
AH
243* 27'00" 423-62
S63*27'W
s 0-894
A to
Z
Computed by .^/i.jGreen) Checked by _!??•„.
(Red)
B'g
E
A
494907
B
-189-35 1684-21
042* 32' 00" 221-38 N42*32' E
Ftoint
545 -378 95
C0446 979
BC
N
2063-16 5138-42
SO676 019 H49-66 cO-736 884
+163-13
£
-229-29 -26-22
1833 87 5112-20
Check
C
2063-16 5138-42
1833-87 5112-20
Fig.
3.
Tabulated computation
16
Example 3.6
Calculate the total co-ordinates, in feet, of a point the bearing of AB is 119° 45' and the distance is 850 links on a slope of 15° from the horizontal.
B
if
The co-ordinates
E
264-5
A
of
relative to a local origin are
N
5356*7
ft
ft.
(M.Q.B./UM)
To find horizontal length
(Fig. 3.17)
Fig.
=
AB, but 100 links = 66
be multiplied by
ft;
K
logs,
17
cos 15° links
therefore to convert links to feet the length must
= 0*66,
AB,
By
AB
3.
i.e.
=
K.AB
=
0-66 x 850 x cos 15°
cos 15°
0-66
1-81954
850
2-92942
cos 15°
T-984 94
AB,
2-73390
SURVEYING PROBLEMS AND SOLUTIONS
136
To find
Fig.
119° 45'
By
AB
partial co-ordinates of line
3.
(Fig. 3.18)
18
S60°15' E
=
+ 264-5
logs, partial departure
2*672 52
EB
+734-9
1-938 62
AB,
2-73390 (see above)
cos 60° 15'
1-695 67
partial latitude
2-429 57
To
N
5356-7
5087-8
ft
AN'AB
-268-9
ND
5087-8
E 734-9 ft
obtain the bearing and distance between two points given
their co-ordinates
Let the co-ordinates
Then tan bearing
N.B.
+ 470-4
sin 60° 15'
Co-ordinates of B, 3.44
A£'AB
(0)
(Fig. 3. 19) of
A and B
=
For convenience this Bearing
F
—2
be
E^N^
and
E s Nfl
respectively.
_ F
± N nA ™B - N
(3.7)
AE•AB ANAB
(3.8)
is frequently written
AB
=
tan-'AE/AN
(3.9)
(the sign of the differences will indicate the quadrant bearing).
Length
AB
=
>/(AE*
+
AN*
)
(3.10)
137
CO-ORDINATES This
N.B.
is not
a very good solution for computation purposes and
the trigonometrical solution below is preferred.
AB
=
(3.11)
cos bearing (0)
AN^g sec# =
AB
or
(3.12)
AEAB
(3.13)
sin bearing (0)
AE^cosectf
=
(3.14)
If both of these determinations are used, their agreement provides a check on the determination of 0, but no check on the subtraction of the Eastings or Northings.
AZab N„ 'B 1
>B
,
^/* ^a
*"ab
.
•
N/,
A
EA
To
Fig. 3.19
Example
E
find the length
and bearing between two points
3.7
N
E A
632-16
B
925-48
m m
+ 293-32
m
AE
Bearing
AB
AN
tan"'
m
421-74
m
- 528-14
+ 293 32 -528-14 '
=
949-88
(E) (S)
B
SURVEYING PROBLEMS AND SOLUTIONS
138
By
logs,
293-32
2-467 34
528-14
2-72275
tan (0)
1-744 59
—
»
i.e.
Length
AB = AN
sec 6
E
150^57'
AE
or
= 528-14 sec 29° 03'
By
S 29° 03'
cosec 6
293-32 cosec 29° 03'
logs,
528-14
2-72275
sec 29° 03'
0-058 39 2-781 14
or
293-32
2-46734
cosec 29° 03'
0-31375 2-78109
The
first
solution is better as
solution is obtained
if
-*
604-14
m
->
604-07
m
AN > AE,
but a more compatible
the bearing is more accurately determined, using
7 figure logs,
AE AN
2-7227491
tan ((9)
9-7445926
Bearing (0)
AN sec#
2-4673417
=
29°02'50"
2-7227491 10-058 3786
2-7811277
AE
or
cosec
- 604-13 m
2-467 3417
10-3137836
2-7811253
604- 12 m
The following horizontal angle readings were recorded Example 3.8 during a counter-clockwise traverse ABCD. If the line AD is taken as an arbitrary meridian, find the quadrantal bearings of the remaining lines.
Find also the latitudes and departures of the line length is 893-6 m.
AB = S 22° 42' 40" E BC = N36°18'10"E CD = N26°20'00" E line CD AE = -396 -4m AN
= +800*9
m
Example 3.9 In order to continue a base line AC to G, beyond a building which obstructed the sight, it was necessary to make a traverse round the building as follows, the angles being treated as deflection angles
when traversing
ACD CDE DEF Calculate EF
=
in the direction
92° 24' to the
ABCDEFG.
left
= 90° 21' to the
right
CD
=
56-2
ft
89° 43' to the
right
DE
=
123-5
ft
= for
F
to be
on
AC
produced and find
EFG
and CF. (L.U.)
IG i I
89°43'
£
^_J
$__».-r i i
i
I i i i i
I I
i
X. 1i i
I I I I
I I i I I
I
90°21y 1
92°24', -o-
Fig. 3.20
Assuming the bearing
AC =
0°00'
WW CO-ORDINATES i.e.
AC ACD (left)
-angle
360° 00'
=
Bearing
92° 24'
CD CDE (right)
267° 36'
DE DBF (right)
357° 57'
Bearing
+ angle
i.e.
S 87° 36'
i.e.
N 02° 03'
i.e.
N 87° 40' E
o
90 2l'
Bearing
+ angle
141
89° 43'
447° 40'
-360° 00'
EF
Bearing
087° 40'
FG EFG
Thus, to obtain the bearing of the deflection angle
= bearing AC, = 87° 40'
left.
Check on deflection angles +
To
90° 21'
92° 24'
89° 43'
87° 40'
180° 04'
180° 04'
obtain the co-ordinates of
Line Distance (ft)
AC CD
56-2
DE
123-5
Bearing
E
sin Bearing
Ae
cos Bearing
An
o°oo'
87°36'w
0-999 12
0-04188
-56-15
-
N02°03'w
0-035 77
0-999 36
-
+ 123*42
S
4-42
2«35
+ 123.42 -
2.35
E F - 60.57 N^ +121-07 Thus F must be +60*57
east of E. The line
ft
EF
has a bearing
087° 40' Length
EF
AE^
=
sin bearing
=
To
60-62
60 57 sin 87° 40' '
=
,
ft
find the co-ordinates of F,
&NEF
=
60-62 cos 87° 40'
=
+
N£
=
+121-07
N/r
=
+123-54
2-47
SURVEYING PROBLEMS AND SOLUTIONS
142
.-.
F is CF =
ft ft.
EF EFG CF
Ans.
Example and
above C on the bearing due N.
123*54
123-54
B
= 60-6
ft
=
87° 40' deflection
=
123-5
ft
The co-ordinates (metres)
3.10,
left
of the base line stations
A
are
E
A
26 543-36
B
26895-48 E
35432-31 35983-37
N N
The following clockwise angles were measured as traverse:
part of a closed
ABCDEA ABC BCD CDE DEA EAB
183° 21'
86° 45'
329° 17' 354° 36' 306° 06'
Determine the adjusted quadrant bearings of each of the lines relative to the meridian on which the co-ordinates were based.
26 543-36
B
26895-48 m
AE
352-12
tan bearing
m 35 983-37 m AN 551-06 m
m
A
35 432-31
m
AB =
352-12 551-06
bearing
AB
=
032° 34'
corr.
2
Angles
Corrected Angle 183° 20'
183° 21'
-01'
86° 45' 329° 17'
-01'
-01'
329°
354°
36'
-01'
354° 35'
306° 06'
-01'
306° 05'
Angles should equal i.e. .*.
16'
1260° 00'
1260° 05'
£
86° 44'
(2n + 4) 90
(2x5 +
4)90
=
1260<
error is 05' distributed as 01' per angle.
CO-ORDINATES
143
Calculation of bearings
Bearing
Angle
AB ABC
032° 34'
-
>
N
32° 34' F
-
>
N
^5° 54' F
*
N57°22'W
183° 20'
215° 54' -
Bearing
Angle
BC BCD
180°
035° 54' 86° 44' 122° 38'
+ 180° Bearing
Angle
CD CDE
302° 38' 329° 16'
631° 54'
-540° Bearing
Angle
DE DEF
091° 54'
-
i
266° 29' -
»
^ ftR°nfi'
f
354° 35' 446° 29'
-180° Bearing
Angle
EA EAB
S
86°29'W
306° 05' 572° 34'
-540° Bearing
032° 34'
AB
Check
A disused colliery shaft C, situated in a flooded area, surrounded by a circular wall and observations are taken from two points A and 6 of which the co-ordinates, in feet, relative to a local Example 3.11
is
origin, are as follows:
C
is
Station
Eastings
Northings
A
3608-1
915-1
B
957-6
1808-8
approximately N.W. of A.
Angles measured are
at
A
to the tangential points 1 and 2 of the walls
BAC, = 25° 55' and BAC Z = 26° 35'.
Angles measured at B to the tangential points 3 and 4 of the wall C3 BA = 40° 29' and CA BA = 39° 31'. Determine the co-ordinates of the centre of the shaft in feet relative to the origin, to one place of decimals and calculate the diameter are
SURVEYING PROBLEMS AND SOLUTIONS
144
formed by the outside of the wall.
of the circle
(M.Q.B./S)
Fig. 3.21
N
E A
3608-1
915-1
B
957-6
1808-8
AE'AB
AN„ D + VU3
_ 2650-5
893-7
In Fig. 3.21,
Bearing of
AB
=
tan
-iAE
AN
= N71°22'W Length
In triangle
AB
=
AN
-2650-5
_ =
tan"
i.e.
+ 893-7 288° 38'
sec bearing or
=
893-7 sec71°
=
2797>1
22' or
AE
cosec bearing
2650-5 cosec 71° 22' 2797-1
ABC
Angle A
=
I {25° 55'+ 26° 35'}
=
26° 15'
Angle
fi
=
I 40° 29' +39° 31'}
=
40° 00'
Angle
C = 180° -
=
113° 45'
{
(26° 15' + 40° 00')
180° 00'
.
CO-ORDINATES
By
145
the sine rule,
BC
AC
=
AB
=
2797-1 sin 26° 15' cosec 113° 45' =
=
BC
=
1351-6 sin 40° 00' cosec 26° 15'
sin
sin
A cosec C 1351-6
B cosec A
Bearing
+ Angle Bearing
Bearing
- Angle Bearing
=
AB BAC AC
314° 53'
BA
108° 38'
1964-3ft
288° 38' 26° 15'
CBA BC
40° 00'
068° 38'
To find co-ordinates of C Line
BC
068° 38'
AE^ Ec
AN5C Nc
N68°38'E
i.e.
1351-6
1351-6 sin 68° 38'
=
ft
=
+1258-7
=
Es + ^EBC
=
957-6 + 1258-7
=
+2216-3
=
1351-6 cos 68° 38' =
+ 492-4
=
NB + ANSC
=
1808-8 + 492-4
= +2301-2
Check Line
AC
314° 53'
AE^ EC
To
N45°07'W
i.e.
=
1964-3
1964-3 sin 45° 07'
= EA + & EAC = 3608-1 - 1391-8
ft
=
-1391-8
=
2216-3
+1386-1
AN^ C
=
1964-3 cos 45° 07' =
Nc
=
NA + AKAC
=
915-1 + 1386-1
=
2301-2
find the diameter of the wall
Referring to Fig.
a =
3.
21
o |(40 29'-39°31')
R = BCsinO°29' =
ft
cz
1351-6 x 29 x 60
206265
=
BC =
0°29' x 0°29'(rad) 11.40
ft
SURVEYING PROBLEMS AND SOLUTIONS
146
Check
p = 1(26° 35' -25° 55') = 0°20'
R = AC
x 0°20'(rad)
1964-3 x 20 x 60
n^ft
=
206 265 22-8
Diameter of wall
To Find
3.5
3.51
ft
the Co-ordinates of the Intersection of
Two Lines
Given their bearings from two known co-ordinate stations
As an
alternative to solving the triangle and then computing the
may be
co-ordinates the following process
Given
bearings
applied:
A (EA NA ) B (E 5 Nfl ) a and jQ C (E N ) C C
Fig.
3.
22
From Fig. 3.22,
E c = EA = EA = EB = EB .-.
Then the
Nc (tana
-tanjS)
(N^-N^tana
+
AN^
(3.15)
tana
+ (Nc -NB )tan/8 + ANsc tan0
(3.16)
EB - EA + N^ tana - NB tan0
C E g - E^ + HA tana - NB
total northing of
Nc =~ To
=
+
tanft
tana - tanjS
obtain the partial co-ordinates from equation (3.17)
(3 1?)
147
CO-ORDINATES Partial Northing 1,C
Nc
-
NA =
MiAC
Nc
=
- NA
tan a ~ Ng ta " ^ - N^ tana - tan/8
Eg ~ Ea + Na
E 5 - E^ + N^tan a - N^tanft - N^ tana + NA tanft tana -
tanft
_ (Es-E.0 -(NB -K4) tanft tan a - tan ft A E^, -/IS - A IWa. =
AN^
Then
tana -
(3 lg)
tan/3
Similarly, from equation (3.17),
AUBC = N^-N*
=
Eg-E^ + N^tana-Nstanft tana -
WBC
Then The following
=
& E AB - &NAB
tanft
tan a
"T tana - f.
(3-19)
tan/8
alternative process
may be used:
Nc = N4 +(E c -Ex)cota = N^ + AE^ = N B + (E c -E5 )cotft = NB + AE 5C As
a
(3.20)
cotft
(3.21)
cot
before, the total and partial co-ordinates are given as:
E C = Nb ~ Na + Ea cot a ~ EB cot a - cot ft
cot
n 2?
.
ft
^
~ AE^cotft AE^ m*B cot a - cot
and
'
(3.23)
ft
AE
AN4 g - AE^ cot a cot a - cot ft
=
(3.24)
N.B.
Theoretically, if Scot > Stan, then it is preferable to use the cot values, though in practice only one form would be used.
Example 3.12
Let the co-ordinates be the bearings be
tana
=
1-7321
tan ft
=
-0-5774
Using the tan values
cot
A = E4, N6 a = 060°
8 = E13, N4 ft
a =
0-5774
=
-1-7321
cot ft
= 330°
;
from equation (3.17)
Nc =
1-732 1) + (4 x 0-577 4) (13 - 4) + (6 x -• ' -
_
=
y-397 g
4 + (9-397-6) x 1*7321 =
9-884
-
.
1-7321 + 0-5774
from equation (3. 15)
Ec =
SURVEYING PROBLEMS AND SOLUTIONS
148
or equation (3. 16)
Ec =
13 + (9-397 -4) x -0-5774=
9-884
Using the cot values, from equation (3.22),
E„ C
From equation
4 - 6 + 4 x 0-5774+ 13 x 1-7 321 0-577 4+1-7321
^ Q .oo d
=
(3.20),
Nc =
6 + (9-884 - 4) x 0-577 4
Nc
4 +
=
9-397
or equation (3.21)
=
(9-884-13) x -1-7321 = 9-397
If the formulae using partial values are employed the individual equation computation becomes simplified.
Using the previous values (Ex.
3.5),
from equation (3.18)
WAe AC Then
Nc
.
-
+ 3-397
AN^
6 + 3-397
this value is
From equation
0,S774 -?-<1-?JL: 1-7321 + 0-577 4
= N^ + =
When
(13
= 9-397
known, equation (3.15) may be used as before.
(3.23);
AEUc = Ec
(4-6) - (13-4) x -1-7321
=
+5-884
0-577 4 + 1-7321
=
ea +
=
4 + 5-884
^AC = 9-884
When this value is known, equation (3.20) may be used as before. The above process is prefered and this can now be given in a tabulated form.
Example 3.13
(1)
AN'AC
_ —
AB ——a-——AN^tan/8 —^————I— —&E—tan - tan /S
(2)
AE^
(3)
AN5e = AE^ B
=
tOAAC
tana -
b&A B tana
tana -
(4)
AEBC
=
ANflC
tan/3
tan^S
149
CO-ORDINATES Igloo
Oriented diagram
E base
E
Stations
_E_B_ase_
N
Bearings
^AX + 13 486-85 m + 12 759-21 m
13' 57"
+ 10 327-36
182° 27' 44"
+ 13 142-72
a 278°
_Iglpo__(B)
m m
tana -6-911745 2
&AB
-727-64
tanjS +
a -
tan
AN^tan/8
&&AC
W Base Ec
0-0430004
ANAB
+ 2815-36
l
tan
+ 121-06 -848-70
-J-
-843-44
*
-6-9547456"
—
x tana
+ 122-03
m
12 643-41
10 449-39
m
Check
& EAB
-727-64
Afy^tana -19459-05 tan
+
+ 18 731-41
&BC W Base E c 3.52
+ 12643-40
tan /8
-6-9547456"
—
115-81
a -
«
—
x tan/3
2693-33
m
10 449-39
Given the length and bearing of a line and C, Fig 3. 23
AB
and
all the
AN,BC l
m angles
A,B Given:
(a)
Length and Bearing of AB,
Ec - E^ = =
b cos(A +6) b(cosA cos 6 c sin
sin
(b)
Angles
A sin#)
B cos A cos^ -
c
sinB sin 4 sin 5
sinC but
E« -
E\d
=
c cos Q
j4B sin
A,B and
bearing^
C.
SURVEYING PROBLEMS AND SOLUTIONS C
Fig.
AB
N, - N4
=
c sintf
C
=
180 - 04 + B)
sinC
=
sin A
.
and
=
3.
23
cos bearing^
cosB + cos A
sin 6
Then E,7
— E.
=
(EB - E4) sinB cos 4 - (NB - N^) sing sin ,4 sin A cos B + cos A sin S
,
E4 sin A cos g + E^ cos A sin B + Eg sin B cos 4 - E^ sin g cos 4 - NB sin B sinA + Ha sin g sin A sin 4 cos g + cos A sin B
EA cotB + EB cot 4 + (N.4 - NB ) cot A + cotB E.4
cotg + Efl cot /I - AN^s cot 4 + cotB
Similarly,
N^ =
N,4
(3.25)
cotB + NB cot ,4 + &EAB cot A + cotB
(3.26)
Check
E^cotB -
1)
E fl (cot4 + 1) + N^cot B + 1) + + N^cotA - 1) - (E c + Nc )(cot4 + +
N.B. All these methods are mathematically sound but the first has the advantages that (1) no formulae are required beyond the solution of triangles, (2) additional information is derived which might be required in setting-out processes.
Example 3.15. Equalisation of a boundary line. The following survey notes refer to a boundary traverse and stations A and E are situated on the boundary. Line
Bearing
AB BC CD DE
Horizontal length
N
83° 14'
E
253-2
S
46°30'E
426-4
N 36°13'E
543-8
S 23°54'E
1260-2
(ft)
proposed to replace the boundary ABCDE by a boundary AXE is a straight line and X is situated on the line DE. Calculate the distance EX which will give equalisation of areas on each side of the new boundary. (M.Q.B./S) It is
where
AX
Computation of co-ordinates with A as the origin, Fig. 3.25 Line
AB N
83° 14' E
253-2
ft
Logs
Line
AE
2-400 42
sin#
9-99696
length
2-40346
cos 6
9-071 24
AN
1-47460
BC S46°30'E
E,
0-0
+ 251-44
EB +
NA
251-44
0-0
+
29-83
No +
29-83
EB +
251-44
426-4
AE
2-490 38
sin0
9-86056
-»
+ 309-30
155
CO-ORDINATES length 2-62982
cos 6
9-83781
AN
2-46763
N5
+
- 293-51
Nc CD N
Line
36° 13'
E
29-83
- 263-68
543-8
Ec +
AE
2-50691
sin0
9-77147
560-74
+ 321-30
->
ED +
882-04
length 2-73544
cos#
9-90676
AN
2-642 20
Nc
- 263-68
_+ 438-74
HD
+ 175-06
-200 Fig. 3.25
-400
-600
800-
-1000
200
By
400
800
C
ED BDC
800
1000
1200
1400
construction,
Join BD.
Draw
line parallel to
BD
through
to cut
Area of triangle BDC X = area of triangle between same parallels).
at C, (triangles on
Join
C
Draw
line parallel to C,/4 through B to cut ED at 8,. of triangle AB^C = area of triangle ABC^.
X
Area Line
•'•
same base and
A.
AB
X
(X) equalises the irregular boundary in such a triangle
ABP + Length
triangle
EX =
QDB^ =
977.84
ft
triangle
(calc.)
way
that
PQC
= 978
ft
(scaled).
156
Line
SURVEYING PROBLEMS AND SOLUTIONS
DE
E
S 23° 54'
1260-2
ED +
882-04
AE
2-70805
+
510-57
sin0
9-60761
EE +
1392-61
N^ +
175-06
length 3-100 44 costf
9-96107
AN
3-06151
- 1152-15
N^ Checks
AE
AN +
977-09
+
251-44
29-83
- 293-51
+
309-30
+ 438-74
-1152-15
+
321-30
+ 468-57
+
510-57
2AE + 1392-61 ABCDEA (see
Area of figure
-1445-66
+ 468-57
SAN
- 977-09
Chapter 11)
ft)
(2)
(3)
(4)
N
E
F.dep.
B.dep.
+ 251-44
+1392-61
A B C
+ 29-83 - 263-68
+ 251-44 + 560-74
+ 560-74 + 882-04
D E
+ 175-06 - 977-09
+ 882-04
+ L392-61
+ 1392-61
0-0
Double areas
0-0
(1)
0-0
+ 251-44 + 560-74
0-0
+ 882-04
(5) (3)
+ 560-74 + 630-60 + 831-87 - 882-04
x (5)
A
B C D E
167 26-8
166 276-0
145627-0
861832-0 + 1024185-8 - 166 276-0 2;
857 909-8 428 954-4 sq
ft
From co-ordinates Bearing
EA
- 1 92 ' 61
=
tan~1
=
305° 03' 16"
^
=
(4)
-1141-17
N 54° 56' 44" W
157
CO-ORDINATES A A 1392-61 cosee 54° 56' 44' -
Length
EA = =
ED AED
1701-2
ft
I'
'
(x)
23° 54' 00"
W
336° 06' 00"
Bearing
=
N
Angle
=
336° 06' 00" - 305° 03' 16"
=
31° 02' 44"
To find length EX
=
(a) such that the area of triangle
AXE
equal
is
to
428 954-4 sq.ft.
Area of triangle
AXE
AED
=
^ax
-
2 area triangle
sin
AXE
x sin AED 2 x 428 954-4 1701-2 sin31°02' 44"
977-84
=
(length EX).
ft
Exercises 3(c) (Boundaries) 14. The undernoted bearings and measurements define an irregular boundary line on a mine plan between two points A and B, the latter being a point on a straight line XY bearing from South to North. Plot the bearings and measurements to a scale of 1/2 in. = 100 ft, and thereafter lay down a straight line from A to a point on XBY so that the areas to the North and South respectively of that line will be ,
equal.
From A
N 63° 30' W S45°00'W S80°45'W N 55° 15' W S 60° 30' W
Check your answer by calculation
185
ft
245
ft
175
ft
250
ft
300
ft
to
B
of the respective areas.
(M.Q.B./M)
The undernoted traverse was taken along an irregular boundary 15. between two properties:
AB BC CD DE EF A
N32°45'E N71°30'E S61°15'E N71°30'E S40°30'E
lies on a straight boundary fence
XY
464
ft
308
ft
528
ft
212
ft
248
ft
which bears
N
7° 30' W.
,
158
SURVEYING PROBLEMS AND SOLUTIONS
Plot the traverse to a scale 1/2400 and thereafter set out a F to a point G on the fence XAY so that the areas North and South of the line are equal. straight line boundary from
What length of fencing will be required
How
far is
G
from
?
A? (N.R.C.T.
3.6
Ans.
1480
ft;
AG
515
ft)
Transposition of Grid
New
Fig.
3.
axis
Transposition of grid
26
Let the line AB (Fig. 3.26) based upon an existing co-ordinate system have a bearing 6 and length s.
AE^ ^AB
Then
the
The co-ordinate system new system is 0. The co-ordinates of the
is
=
s sin (9
=
s
now
COS0
to be
changed so that the origin of
position of 'slew' or rotation
A =
E^
and the axes are rotated clockwise through an angle
+a
to give a
new
bearing of AB. i.e.
or
/8
- a
=
a =
6 -
The new co-ordinates
EB =
i.e.
/8
of
Ejj
Old bearing -
New
bearing
(3.28)
B may now be computed: + s sin/S
=
EA +
Efl
=
^A + &&AB cos a - &^AB sin a
Nfl
=
NJ^
s sin 6 cos
Similarly,
+ s cos/8
a -
s cos 6 sin
a (3.29)
CO-ORDINATES
N'b If
=
NA +
s cos 6 cos
=
N^ +
AN^
159
a +
cos a +
s sin 6 sin a
AE^sina
(3.30)
a scale factor k is required (e.g. to convert feet into metres),
then,
E 'a + AE^ = E^ + ZctAE^gCos a -
E* =
K5
and
i
= =
AN^
sin a]
(3.31)
l^ + iiiMfl ANJS Ni N^ + /ctAN^ cos a + AE^sina]
(3.32)
From the above,
-
AN^ nAN^
+
nAEAB
AE^ B = fctAE^ cos a =
AN^ S = where If
m=
k cos
mAE^ mAN^
a and n = k
sin
sin a] (3.33)
(3.34)
a
the angle of rotation (a) is very small, the equations are simcosa-»0 and sin a -> a radians.
plified as
E5 =
VA
+ k[AEAB
-,
NB = N^ + tfAN^ + Example 3.16
Transposition of grid
Fig.
3.
27
fiNAB a]
(3.35)
AE^
(3.36)
a]
SURVEYING PROBLEMS AND SOLUTIONS
160
In Fig. 3.27,
OA OB OC OD
Let
=
AE
045°
i.e.
AN
N45°E
400
+ 282-84
+ 282-84
=
120°
+ 303-10
- 175-00
210°
S60°E S30°W
350
=
350
-175-00
-303-10
=
330°
N30°W
400
-200-00
+ 346-40
the axes are now rotated through -15 creased by +15°. If
OA' =
060°
OB' =
135°
OC' =
225°
N60°E S45°E S45°W N15°W
OD' = 345°
c'
the bearings will be in
AE'
AN'
400
+ 346-40
350
+ 247-49
-247-49
350
+ 247-49
-247-49
400
-103-52
+ 386-36
+ 200-00
Applying the transposition of the grid formulae; equation (3.29)
AE'
=
AEcosa- AN
equation(3.30)
AN'
=
AN
AE
Ae
An
cosa
AN
cos a + sina
OA + 282*84 + 282-84 +273-20
-73-20
OB + 303.10 -175-00 + 292-77 OC - 175-00 -303.10 -169-04 OD -200-00 + 346-40 -193-19
+ 45-29
N.B.
sina
AE sin a
An cosa Ae
sina
Ae'
An'
+ 273-20 - 169-04
-73-20 + 346-40 + 200-00 -78-45 + 247-48 -247-49
+ 78-45 -292-77 -89-68 + 334-60
+ 51-76 -103-51 + 386-36
(1)
cos (-15°) - +0-965 93
(2)
sin (-15°) = -0-258 82
+45-29 -247-49 -247-48
had a co-ordinate value (EqNq) based on the new axes, these values would be added to the par-
(3) If the point of rotation (slew)
tial values,
AE'AN'
3.7
to give the
new
co-ordinate values.
The National Grid Reference System
Based on the Davidson Committee's recommendations, all British Ordnance Survey Maps will, on complete revision, be based on the National Grid Reference System with the metre as the unit. The origin of the 'Modified Transverse Mercator Projection' for the British Isles is
Latitude
Longitude
49° 2°
N
W
To provide positive co-ordinates for the reference system a 'False was produced by moving the origin 100 km North and 400 km
Origin'
West
.
The basic
grid is founded
upon a 100 km square; commencing from
the false origin which lies to the S.W. of the British Isles, and all
squares are referenced by relation to this corner of the square.
CO-ORDINATES
/0
100
200
300
400
161
500
600
False origin
_1_ 49° N
True
-^T
origin
2*
Fig. 3.28
W
latitude
long'itucle
Old O.S. grid reference system
700km
E
162
SURVEYING PROBLEMS AND SOLUTIONS 'Eastings are always quoted Originally the 100
first.'
km squares were given a
reference based on the
number of 100 kilometres East and North from the origin (see Fig. 3.28). Subsequently, 500 km squares were given prefix letters of S, N and H, and then each square was given a letter of the alphabet (neglecting I). To the right of the large squares the next letter in the alphabet gives the appropriate prefixes, T, O and J (see Fig. 3.29). 1
km
M
L
H- Hebrides
Central meridian
o
N
P
M
L 1
R
U
Q
Y
z
V
W
C
D
E
A
B
H
J
K
F
G
O
P
L
Q
R
S
V
W
X
A
B
F
G
L
M
T
J
800
N- North
SJ
M L-*-
T
U
Q
X
Y
z
V
w
B
c
D
E
A
B
G
H
J
K
F
Q
R
s
V
W
A F
G
T
-ir
300
S- South
M
L
-N- \
O
P
L
U
a
z
V
II
Q
.'
V
W
S
/ False origin
True origin
Fig. 3.29
New
Y
X
u R
.
-?#— 49* N latitude -^"^ .
-^
<
2
W longitude
O.S. grid reference system
Square 32 becomes SO 43 becomes SK 17 becomes
NM
The basic reference map
is to the scale
1/25000
(i.e.
approxi-
mately 2y2 inches to 1 mile), Fig. 3.30. Each map is prefixed by the reference letters followed by two digits representing the reference numbers of the SW corner of the
SK54. This shows the relationmaps and the manner in which each
sheet. See example (Fig. 3.30), i.e.
ship between the various scaled
sheet is referenced.
CO-ORDINATES
163
Nottingham Regional College of Technology has the E 457 076-32 m, N 340 224-19 m. Its full 'Grid Reference' to the nearest metre is written as SK/5740/076 224 and the sheets on which it will appear are:
A
point
P
in
grid co-ordinates
Reference
Grid size
Sheet size
Scale
SK 54
1/25 000
( ^ 2*/2 in.
SK 54 SE
1/10 560
(
SK 57 40
1/2 500
(
SK 57 40 SW
1/1 250
( 2;
6
50
mile)
in. to 1
^ 25 in.
1km 1km 100 m 100 m
km 5 km
10
to 1 mile)
1km m
to 1 mile)
500
in. to 1 mile)
10km 350 000 SK 54 (1/25000)
49
48
46
5km 45
SK 54 SE (6" to imile)
SK 5740
SK 5740
SW
(1/2500)
(1/1250)
340000 $5
o 8 o
-*
"»
u>A Fig. 3.30
oi
noo» o
O)
O.S. sheet sizes
o»
o ° 8
Exercises 3(d) (Co-ordinates) 16.
The co-ordinates
of stations
A and B
Latitude
Departure
A
+8257 m
S
+7542 m
+ 1321m - 146 m
Calculate the length and bearing of 17.
are as follows:
The co-ordinates
of two points
A
N
B
S 495-4
188 -6
m m
AB
(Ans.
A and B
E E
922-4 58-6
244°
01';
are given as:
m m
1632 m)
SURVEYING PROBLEMS AND SOLUTIONS
164
P midway between A
Calculate the co-ordinates of a point
S 153-4 m,
(Ans.
and B.
E 490-5 m)
18. The bearings of a traverse have been referred to the magnetic meridian at the initial station A and the total co-ordinates of B, rela-
tive to A, are found to be 368
m
W, 796
m
S.
Calculate (a) the length and magnetic bearing of AB, declination is 13° 10'
metres of two points, X and Y, are as 582-47 m N 1279-80 m
W E
X
follows:
Y
m
1191-85
S
755-18
Calculate the length and bearing of XY. (Ans. Survey station X has a Northing 424-4 a height above Ordnance datum of 260*8 ft.
20.
m
2699-92 m; ft,
138°54'50")
Easting 213*7
ft
and
Station Y has a Northing 1728-6 ft, Easting 9263-4 ft and a depth below Ordnance datum 763*2 ft. Find the length, bearing and inclination of a line joining XY (Ans. 9143-3 ft; 081° 47' 54"; 1 is 8-92)
The co-ordinates
21.
of four survey stations are given below:
Station
North
A B C
822 164 210
D BD.
East (ft) 90 469 614
(ft)
718
81
Calculate the co-ordinates of the intersection of the lines AC and (L.U. Ans. N 520, E 277)
22.
Readings of lengths and whole circle bearings from a traverse
carried out by a chain and theodolite reading to 1 minute of arc were as
follows, after adjusting the angles:
AB
Line
BC
0°00'
35° 40'
46° 15y2
Length (ft)
487-2
538*6
448-9
W.C.B.
Length
(ft)
'
156° 13' 295-4
FA
AG
GH
HD
180° 00'
270° 00'
64° 58'
346° 25'
37° 40'
963-9
756-2
459-3
590-7
589-0
EF
Line
DE
CD
W.C.B.
CO-ORDINATES
AB
Taking the direction parture of each line.
A
165
as north, calculate the latitude and de-
and the mean co-ordinates as obtained by the three routes are taken as correct, find the coordinates of the other points by correcting along each line in proporof
If
is taken as origin
D
tion to chainage (answers are required correct to the nearest 0*1
(L.U. 23.
Ans.
D
= 1234-7
Angle observed 198° 06' 30"
284° 01 '30" 200° 12' 30" 271° 33' 30"
268° 01 '30"
AB BC CD DE EF FG
245
310 480
709 430
607
N
61° 27' 40" W)
From the following notes, calculate the length and bearing DA: Line Bearing Length
AB BC
015° 30'
630 m
103° 45'
540
CD
270° 00'
227
The notes
of the
m m
(Ans. 25.
E)
Length (metres)
Calculate the length and bearing of the line GA. (Ans. 220-6 m; line
ft
traverse:
14° 48' 00"
AB
ABC BCD CDE DEF EFG
ft)
N, 637-6
The following notes were taken during a theodolite Bearing of line
24.
ft
668 m; S 44° 13' W)
of an underground traverse in a level
seam are as
follows:
Line
Azimuth
AB BC CD
30° 42'
Distance
86° 24'
150-6
32° 30'
168-3
(ft)
315° 06' DE 45-0 The roadway DE is to be continued on its present bearing to a point F such that F is on the same line as AB produced. Calculate the lengths of EF and FB. (M.Q.B./M Ans. EF 88-9 ft; FB 286-2 ft)
A shaft is sunk to a certain seam in which the workings to the dip have reached a level DE. It is proposed to deepen the shaft and 26.
connect the point E in the dip workings to a point X by a crossmeasures drift, dipping at 1 in 200 towards X. The point X is to be
SURVEYING PROBLEMS AND SOLUTIONS
166
134 ft from the centre of the shaft A and due East from
AX
it,
being
level.
The following
made
ate the notes of a traverse
the centre of the shaft
A
in the
seam from
to the point E.
Vertical Angle
Line
Azimuth
Distance
AB BC
270° 00'
127
Level
184° 30'
550
Dipping 21°
CD
159° 15'
730
Dipping 18y2 °
DE
90° 00'
83
Level
Calculate (a) the azimuth and horizontal length of the drift
and (b) the amount by which
it
is
EX
necessary to deepen the
shaft.
(M.Q.B./M
Ans.
(a)
358° 40'
(b) 434-5
1159-6
ft
ft)
The notes of a traverse between two points A and £ in a certain seam are as follows: Angle of Inclination Distance (ft) Azimuth L,ine
27.
AB BC CD DE It is
+6°
89°
600
170°
450
181°
550
level
280°
355
level
-30°
proposed to drive a cross-measures
another point
F
exactly midway between
from a point
drift
A and
E
to
B.
Calculate the azimuth and length EF.
(M.Q.B./M
Ans. 359° 33'; 867
888-3
ft,
ft
inclined)
Undernoted are details of a short traverse between the faces of two advancing headings, BA and DE, which are to be driven forward 28.
until they meet:
Azimuth
Line
AB BC CD DE
Distance
80°
270-6
ft
180°
488-0
ft
240°
377-0
ft
350°
318-0
ft
still to be driven in each heading. (M.Q.B./M Ans. BA + 168-4 ft; DE + 291-5 ft)
Calculate the distance
29.
In order to set out the curve connecting
two straights of a road
to be constructed, the co-ordinates on the National Grid of /, the point of intersection of the centre lines of the straights produced, are re-
quired.
167
CO-ORDINATES a point on the centre line of one straight, the bearing Al being 72° 00' 00", and B is a point on the centre line of the other straight, the bearing IB being 49° 26' 00"
A
is
.
Using the following data, calculate with
full
checks the co-ordinates
of/.
Eastings
Northings
(ft)
(ft)
A
+43758-32
+52202-50
B
+45165-97
+52874-50
The length AB
is
and the bearing 64° 28' 50". (N.U. Ans. E + 45 309-72 N + 52706-58)
1559-83
ft
proposed to sink a vertical shaft to connect X on a roadway upper horizon with a roadway GH in the lower horizon which passes under CD. From surveys in the two horizons the following data
30.
It
CD
is
in the
are compiled:
Upper horizon Station
Horizontal Angle
Inclined
Inclination
Remarks
Length
A
co-ordinates of
+
1 in
200
854-37
400
943-21
276° 15' 45"
B
N +1
in
1356-24
Bearing
88° 19' 10"
C
A
E 6549-10 ft
N
ft
AB
30° 14' 00" E.
736-21
Level
D Lower horizon Station
Horizontal Angle
Inclined
Inclination
Remarks
Length
Co-ordinates of
+ lin50
326-17
E 7704-08 ft N 1210-88 ft
+1
278-66
Bearing
193° 46' 45"
G
in 20
83° 03' 10"
N
EF
54° 59' 10" E.
626-10
level
H Calculate the co-ordinates of 31
.
The surface levels
X
(Ans.
of two shafts
E X
8005*54 ft,
and
Y and
respectively as follows:
Surface Level
Depth
X
820-5
ft
A.O.D.
200yd
Y
535*5
ft
A.O.D.
150 yd
N
1918*79 ft)
their depth are
£
SURVEYING PROBLEMS AND SOLUTIONS
168
The co-ordinates of the centre of the two shafts in fact, are respectively as follows:
E
N
X
-778-45
+2195-43
Y
+821*55
+ 359-13
Calculate the length and gradient of a cross-measures nect the bottom of the shaft. (Ans. 32.
2439-3 ft(incl.), 2435-6 ft(hor.); 3°
The co-ordinates
S44°ll' E
are
M
i.e. 1 in 18)
N25m E13m. From ,4 a line AB runs AB an equilateral triangle ABC is
C
to the north of
AB.
Calculate the co-ordinates of B and C. (Ans. B E + 94-5m, N-55-9m. C 33.
10',
con-
117m. On the line
for
set out with
A
of
drift to
E+
(a) Calculate the gradient (as a percentage)
N+
126-4 m,
53-7m)
between two points,
and N, which have been co-ordinated and heighted as given below:
p
Co-ordinates E(ft) N(ft)
•
t
M # (b)
(ft)
6206-5
3465-2
212-4
5103-2
2146-8
196-6
6002-5
2961-4
Determine the length (in centimetres) of the line 500 (assume 1 ft = 0-3048 m).
plotted at a scale 1 (c)
Height
Calculate the bearing of the line MO (R.I.C.S. Ans. 0-92%;
104-8cm;202°03')
From an underground traverse between 2 shaft wires the following partial co-ordinates in feet were obtained: 34.
AB BC CD
MN when
:
E 150-632 ft, E 528-314 ft,
N
82-115
ft
E
N
428-862
ft
26-075
ft,
S 327 -958
A and D
ft
Transform the above partials to give the total Grid co-ordinates of B given that the Grid co-ordinates of A and D were: N 432 182-684 metres A E 520 163-462 metres,
station
D
E 520378-827
(Aide memoire)
X
metres,
N 432238-359
=
x,
+ K(x -yd)
Y =
y,
+ K(y + xd)
metres
(N.R.C.T.) Bibliography
MINISTRY OF DEFENCE, Textbook of Topographical Surveying, 4th ed. (H.M.S.O.)
169
COORDINATES GLENDENNING, J., Principles of Surveying (Blackie) CLARK, D., Plane and Geodetic Surveying, Vol. 1 (Constable) HOLLAND, J.L., WARDELL, K. and WEBSTER, A. C, Surveying, Coal-
mining Series, (Virtue) Brief Description of the National Grid and Reference System. O.S. Booklet No. 1/45 (H.M.S.O.) middleton and CHADWICK, Treatise on Surveying, Vol. 1 (Spon). salmon, V.G., Practical Surveying and Field Work (Griffin) BRINKER, R.C. and TAYLOR, W.C., Elementary Surveying, 4th ed. (International Textbook Co.)
A
Appendix Comparison of scales Scales in
common
use with the
metric system Scales in
Recommended
Other Alternative
byBSI
Scales
i
:
I
:
1
:
1
:
1
:
ioooooo 500000 200000 100000 50000 20000 10000
1
:
1
:
1
:
1
:
1
:
5000 2000 1000
1
:
500
1
:
1
:
1
200 100
:so 20
1
:
1
:
1
:5
10
\ :
1
:
1
:
1
:
1
:
1
:
1
:
625000 250000 125000 62500 25000
2500 1250
use with the
A
1
:
1
:
ioooooo 625000 250000 126720 63360 25000 10560
1
:
2500
1 in to
1
:
1
1
:
1250 500
1
:384
1
1
common
foot/inch system
1
:
1
:
1
:
1
:
1
:
1
:
1
:
192
1
96 48 :24
1
:
1
=
in to 1 mile approx.
mile approx.
Id in to 1
i in to \ in to
1
mile approx.
1
mile
1 in to 1
mile
2 J in to 1 mile approx. 6 in to 1 mile
208.33
ft
in to 104.17 ft
1 in to
A
41.6
in to r
ft
ft
is in to 1 ft
J in to 1 ft i in to 1 ft \ in to 1 ft
12
1
4
3 in to
in to 1 ft 1 ft
From Chartered Surveyor, March, 1968.
.
Laws
4.11
The
(1)
4
INSTRUMENTAL OPTICS
4.1
Reflection at Plane Surfaces
of reflection incident ray, the reflected ray, and the normal to the
mirror at the point of incidence all lie in the
The angle
(2)
same plane.
= the angle of reflection
of incidence (i)
...
A
(r).
8
C Fig. 4.1
AO, Fig. 4. 1, is inclined at a (glancing angle the minor MN. Since i = r, angle BON = MO A = a. If AO duced to C, Angle MO A = NOC = BON = a
The
ray
Thus the deviation
of the ray
AO
is 2a.
MO A)
to
is pro-
Therefore the deviation
angle is twice the glancing angle, i.e.
D
= 2a
(4.1)
Deviation by successive reflections on two inclined mirrors
4.12
(Fig. 4.2)
Ray
AB
is incident
on mirror
A^/V, at
a glancing angle a.
thus deflected by reflection + 2a The reflected ray BC incident upon mirror
angle
j8
is deflected
It
is
M2 NZ at a glancing by reflection - 2j8 (here clockwise is assumed
+ve).
170
INSTRUMENTAL OPTICS
171
\
Fig. 4.2
The
total deflection
In triangle
D
BCX,
(2a- 2/3) = 2(a-j8).
is thus )8
=
a+
=0-a D =
i.e.
is
2(a-j8) = 20
(4.2)
As is constant, f/ie deflection after two successive reflections constant and equal to twice the angle between the mirrors.
The
4.13
optical square (Fig. 4.3)
This instrument, used above principle.
By
for setting out right angles,
employs the
Eq. (4.2), the deviation of any ray from 2 incident on mirror a to the normal = 20, i.e. 2 x 45° = 90°.
M2
at an angle
4.14
Deviation by rotating the mirror (Fig. 4.4) Let the incident ray AO be constant, with a glancing angle a. The mirror M, N, is then rotated by an anticlockwise angle /3 to MgN^.
When the glancing angle
is
a
the deviation angle is 2a.
After rotation the glancing angle is (a+ is therefore
/3)
and the deviation angle
2(a+ /S )
Thus the
reflected ray is rotated by
= 2(a+£)-2a =
2/8
(4.3)
SURVEYING PROBLEMS AND SOLUTIONS
172
// the incident ray remains constant the reflected ray deviates by twice
the angular rotation of the mirror.
Fig. 4.3 Optical square
Fig. 4.4
4.15
The sextant
Principles of the sextant (Figs. 4.5 and 4.6) Mirror M,
,
silvered, is connected to a pointer P.
As
M,
is rotated
the pointer moves along the graduated arc. Mirror M2 is only half-silvered and is fixed. When the reading at P is zero, Fig. 4.5, the image K, reflected from both mirrors, should be seen simultaneously with K through the With a suitable object K as the horizon, the plain glass part of M 2
.
mirrors should be parallel.
INSTRUMENTAL OPTICS
173
Fig. 4.5 Zero setting on the sextant
When observing an elevated object S, Fig. 4.6, above the horizon K, the mirror M, is rotated through angle until S is simultaneously observed with K. The angle being measured is therefore 6. From triangle
M EMz y
2a =
,
2/3
from triangle
M^QM2
a = 90 + p +
90 +
,
a =
4 i.e. f/ie rotation
+ 6
2(a-0)
6 =
j8
=
+
a-j8 = £0
(4.4)
of the mirror is half the angle of elevation.
with the index pointer MP at When the mirrors are parallel, = zero on the graduated arm. When the angle 6 is being observed, the mirror is turned through angle 0, but the recorded value MP2 = 6, i.e.
2.
measured if the objects are at the same height relative to the observer, which means that in most cases the angle measured is in an inclined plane. The horizontal angle may be computed from the equation (see page 107):
N.B.
(1) Horizontal angles are only
cos H.A. =
cos
- cos a, cos a. sina2 i
sin a,
where 6 = the measured angle = vertical angles. (2) Vertical angles artificial horizon.
in the inclined
plane and a, and
must be measured relative
a2
to a true or an
SURVEYING PROBLEMS AND SOLUTIONS
174
s
Recorded value 8
Fig. 4.6 Principles of the sextant
4.16
Use
of the true horizon
(a) As the angle of deviation, after two successive reflections, is independent of the angle of incidence on the first mirror, the object will continue to be seen on the horizon no matter how much the observer moves. Once the mirror W, has been set, the angle between
the mirrors is set, and the observed angle recorded.
This
is the
main advantage of the sextant as a hand instrument,
particularly in marine and aerial navigation where the observer's position is unstable. (b) If the observer is well
above the horizon, a correction 50
required for the dip of the horizon, Fig. 4.8.
is
INSTRUMENTAL OPTICS
175
\ Fig. 4. 7
Box sextant
The Nautical Almanac contains tables
for the correction factor
due to the dip of the horizon based on the equation:
86 = -0-97
where h = height in feet above sea or
where
H
86 = -1-756
yjh
minutes
86
(4.5)
level,
V^
minutes
= height in metres above sea level.
Fig. 4.8 Dip of the horizon
4.17
Artificial horizon (Fig. 4.9)
On land, no true horizon is possible, so an 'artificial horizon' is employed. This consists essentially of a trough of mercury, the surface of which assumes a horizontal plane forming a mirror.
SURVEYING PROBLEMS AND SOLUTIONS
176
£-|>
Mercury plane
^^J
Fig. 4.9 Artificial horizon
The
between the object S and the
vertical angle observed
tion of the
image
S,
reflec-
in the mercury is twice the angle of altitude (a)
required.
Observed angle = S,ES
= 2a
True altitude = MS, S = a
Rays SE and
SS^ are
assumed
parallel due to the distance of S from
the instrument.
4.18
Images in plane mirrors
Image
Fig. 4. 10 Images in plane mirrors in front of the mirror is
Object situated at
/,
From the glancing angles a and (a) triangles
(b) triangles
Thus the point mirror as
seen
at
E
as though
it
were
Fig. 4.10.
/
OFC OFD
/S
it
can be seen that
and ICF are congruent, and IDF are congruent.
(image) is the same perpendicular distance from the
(object), i.e.
OF =
FI.
INSTRUMENTAL OPTICS
177
Virtual and real images
4.19
As above, the rays reflected from the mirror appear to pass through the image thus being unreal or virtual. For the image to be real, the object would have to be virtual.
/,
The it
can
real test is
be—
it
whether the image can be received on a screen:
is real, if
4.2
not—
it is
if
virtual.
Refraction at Plane Surfaces
Normal
/= Angle
of incidence
r» Angle
of
Boundary
of
refraction
media
Refracted ray
Fig. 4.11
The e.g.
air
incident ray AO, meeting the boundary between two media, and glass, is refracted to B, Fig. 4.11.
Laws
4.21 (1)
The
of refraction
incident ray, the refracted ray, and the normal to the bound-
ary plane between the two media at the point of incidence all lie in the
same plane. (2)
For any two given media the
ratio
— sin
.
is a constant
known as
r
the refractive index (the light assumed to be monochromatic).
Refractive Index =
Thus
4.22
sin
i
sin
r
(4.6)
Total internal reflection (Fig. 4.12) If
a ray
AO
is incident
on a glass/ air boundary the ray may be
refracted or reflected according to the angle of incidence.
SURVEYING PROBLEMS AND SOLUTIONS
178
(b) angle
(C)
Reflected
Critical
Fig. 4.12
When the angle
of refraction is 90°, the critical angle of incidence
is reached, i.e.
sin c
rMa g'
1-5
For crown glass the refractive index a fi g
sinc-^g
•'
c If
/a n\ (4.7)
.
= TTTono = sin c
^
41° 30'
the angle of incidence (glass/ air)
internally reflected, and this principle is
> 41° 30' the ray will be employed in optical prisms
i
,
:
within such surveying instruments as optical squares, reflecting prisms in binoculars, telescopes and optical scale-reading theodolites. Total internal reflection can only occur when light travels from N.B.
one medium to an optically less dense medium, 4.23
e.g.
glass/ air.
Relationships between refractive indices (Fig. 4.13)
index from air to glass is a fj, g fractive index from glass to air is g fMa (a) // the refractive
Therefore e.g., if a /J. g
e
= 1-5 (taking
8 (b)
fjL
a
as
air
=
u„a ^
= 1),
—
Given parallel boundaries of i
/**
then
= 0-66
-Ma = gfia sin ia
_ air,
glass,
air,
then
= constant =
then the re-
(4.8)
1-5
sin
,
(4.9)
sin i a sin
i
sin
i
sin
i,
8
aPi
= a fi g sin ig = constant
INSTRUMENTAL OPTICS
179
Fig. 4.13
The emergent ray is parallel to the incident ray when returning same medium although there is relative displacement.
(c) to the
This factor
is
used in the parallel plate micrometer.
Refraction through triangular prisms
4.24
When the two refractive surfaces are not parallel the ray may be bent twice in the same direction, thus deviating from its former direction
by an angle D. can be seen from Fig. 4.14 that
It
and
A =
ft + ft
D
(a,
=
- ft) +
(4.10) (cl,-
(4.11)
ft)
= (a 1+ a 2 )-(ft + ft) i.e.
D =
(a,
+ a2 ) -
A
Thus the minimum deviation occurs when If
A
(4.12) (4.13) a,
+ Og = A
(4.14)
is small, then
a = and
D
/x/3
= /ift + /xft -A = Mft+ft)-^ = AQi-1)
(4.15)
SURVEYING PROBLEMS AND SOLUTIONS
180
Fig. 4.14
Instruments using refraction through prisms
4.25
The
Refraction through a triangular prism
line ranger (Fig. 4.15) 0^
Prism P 2
Prism
P^
Field of view
Fig. 4.15
The
line ranger
a+)8 = 90° .'.
Thus
0,
C0 2
2(a+
is a straight line.
/3)
= 180°
INSTRUMENTAL OPTICS The prism square (Fig.
181
4.16)
^-~*
Fig. 4.16
This
is precisely the
The prism square
same mathematically as the
optical square
(Fig. 4.3), but light is internally reflected, the incident ray being
greater than the critical angle of the glass.
The double prismatic square (Fig. 4.17) combines the advantages of both the above hand instruments.
Fig. 4.17
SURVEYING PROBLEMS AND SOLUTIONS
182
Images 2 and 3 are reflected through the prisms. 0, is seen above and below the prisms.
The
parallel plate micrometer (Fig. 4.18) t
Fig. 4.18
A t
The
parallel plate micrometer
parallel- sided disc of glass of refractive index /a and thickness
is rotated through
an angle
Light is refracted to produce displace-
6.
ment of the line of sight by an amount x =
DB
=
AB
=
——
sin(0 -
cos
~
:
x.
)
x sin(0 -
f(sinfl cos
d>)
- cosfl
sin
cos<£
= f(sin# - cos0
tan>)
but refractive index fj.
=
sin0 =
sin# sin
4>
sinfl
and
2 cos<£ = y/\l - sin
vV
2
<£}
-sin 2 0}
J
INSTRUMENTAL OPTICS -
=
t
sin0
1-
=
t
sin 6
2
VOtx
- sin2
*
x
.-.
1
0).
cos0 2 VO? -sin 0)
sin0~0 rad and
is small, then
If
cos0 sin#
-<[ sin0
183
sin
2
may be
neglected.
ifl(l-i)
(4.17)
A parallel plate micrometer attached to a level is to Example 4.1 show a displacement of 0*0 1 when rotated through 15° on either side of the vertical.
Calculate the thickness of glass required
if its
refractive index is
1-6.
State also the staff reading to the nearest thousandth of a foot
when
the micrometer is brought to division 7 in sighting the next lower readto 20 with 10 for the normal posiing of 4*24, the divisions running tion. (L.U.)
Using the formula x =
t
t
sin
'Hfc£h)\
= siae
h
.
// *J
[
(l-s^)(l
\(jjl
-
+
si,^ 1
sin 0)Qi + sin &))}
0-01 x 12 in.
sin icoN _ L
sinl5) )] If (l-sinl5)(l + a/ 1(1-6 -sin 15) (1-6+ sin 15)/
0-01 x 12
0-25882 x 0-388 24
= 1-1940 N.B.
If
in
'
in.
the approximation formula is used,
The micrometer
is
t
= 1-222
in.
geared to the parallel plate and must be corre-
lated. Precise levelling staves are usually graduated in feet
and
so the micrometer is also divided into 20 parts, each representing 0*001 ft. (The metric staff requires a metric microfiftieths of a foot,
meter).
To avoid confusion, the micrometer should be set to zero before each sight is taken and the micrometer reading is then added to the staff reading as the parallel plate refracts the line of sight to the next lower reading.
SURVEYING PROBLEMS AND SOLUTIONS
184
Z*
a
3
ZP
!R -
•
-» - s -* \
Reading 4-24
Staff
0-0070 Micrometer 4-2470
Fig. 4.19
Use
of the parallel plate
micrometer in precise levelling
Exercises 4(a) 1.
Describe the parallel plate micrometer and show how
precise work when attached If
it
is
used in
to a level.
an attachment of this type
is to
give a difference of 0-01 of a thickness of glass
foot for a rotation of 20°, calculate the required
when
the refractive index is 1*6.
Describe how the instrument may be graduated to read to 0-001 of mean. a foot for displacements of 0-01 of a foot above and below the (L.U. Ans. 0-88 in.) Describe the method of operation of a parallel plate micrometer in glass is 1*6 and precise levelling. If the index of refraction from air to rotation angular the calculate the parallel plate prism is 0*6 in. thick, 0*001 ft. of image of the of the prism to give a vertical displacement 2.
(L.U. Ans. 3° 03' 36") 4.3
4.31
Spherical Mirrors
Concave or converging mirrors (Fig. 4.20)
A narrow beam of light produces a real principal focus F. P is called the pole of the mirror and C is the centre of curvature. PF is the focal length of the mirror. A ray AB, parallel to the axis, will be reflected to F.
normal to the curve at B, so that
Angle
As
AB
is parallel to
ABC
= angle
CBF
=
= angle
ABC
= 6
6.
PC,
Angle
PCB BF
= FC.
BC
will
be
INSTRUMENTAL OPTICS
Fig. 4.20
As
Concave
185
mirror
the beam is assumed narrow
:.
If
the
beam
PF ^ BF ~ FC PC ^ 2PF = 2/ ^
r.
(4.18)
of light is wide a cusp surface is produced with the
apex at the principal focus. The parabolic mirror overcomes this anomaly and is used as a reflector for car headlights, fires, etc, with the light or heat source at the focus.
Fig. 4.21 Wide
beam on a
circular mirror
SURVEYING PROBLEMS AND SOLUTIONS
186
Fig. 4.22 Parabolic mirror
Convex
4.32
or diverging mirrors (Fig. 4.23)
C^"
Fig. 4.23
A
Convex mirror
narrow beam of light produces a virtual principal point F,
being reflected away from the axis. The angular principles are the same as for a concave mirror and
e
r
The
4.33
2/
relationship between object and image in curved
mirrors
Assuming a narrow beam, the following rays are considered
in all
cases (Fig. 4.24). (a)
Ray OA,
parallel to the principal axis, is reflected to pass
through the focus F. (b)
Ray OB, passing through the focus F,
is then reflected
parallel to the axis. (c)
Ray OD, passing through the centre
of curvature C, and thus
a line normal to the curve.
N.B.
In graphical solutions,
it
is advantageous to exaggerate the
vertical scale, the position of the
As
image remaining in the true position. it should be represented as a
the amount of curvature is distorted,
INSTRUMENTAL OPTICS
187
straight line perpendicular to the axis.
Any two
at their intersection,
above rays produce,
of the
tion of the image
the posi-
/.
\
\^i*
^
'!
/
/C
^^r\.
1
Oy
/*\
l
!
%
Fig. 4.24
relationships between object and image for concave mirrors
The are:
(a)
When the object
is at infinity, the
image is small,
real,
and
inverted. (b)
When the object is at the centre of curvature C, the image is same size and inverted. When the object is between C and F, the image is real, en-
also at C, real, of the (c)
larged and inverted. (d) (e)
When the object When the object
image
is at F, the is
between
F
is at infinity.
and P, the image is virtual, en-
larged and erect.
For convex mirrors, in and erect, Fig. 4.25.
all
cases the image is
virtual,
diminished
Fig. 4.25
4.34
Sign convention
There are several sign conventions but here the convention Realmany advantages provided the work
is-positive is adopted. This has is not too advanced.
SURVEYING PROBLEMS AND SOLUTIONS
188
All real distances are treated as positive values whilst virtual
distances are treated as negative values— in
all
cases distances are
measured from the pole.
shown as solid lines whilst
In the diagrams real distances are
N.B.
distances are dotted.
virtual
Derivation of Formulae
4.35
Concave mirror (image
4.26
real), Fig.
„
Object
c
r
v
b
r
1
f
u
|,
1
/
Fig. 4.26
To
prove:
where
1
2
/
r
=
1 -1 -+
U
"
V
/
= the focal length of the mirror
r
= the radius of curvature
u = the distance of the object from the pole v = the distance of the image from the pole
The
OA
ray
is reflected at
to
AI making an equal angle
a on
AC.
either side of the normal
From Fig.
A
P P
4.26,
a+
6 =
and
(f>
a = 6-13
.'.
fi
= 2a +
j8
= 209-j3) + 8 i
2B-P
=
<£+j8=
i.e.
As the angles
26.
a, /3,
6 and
are
^ small, B
is closely
to P. rad
P rad 6 rad .
h
1P
+
~
Sin<£ =
£!
Sin
-
sin0
=
2h
h
0P
j8
~
CP
=
I is real
so IP is +ve.
Jp
OP h
CP
is real
so
OP
is
+ve.
adjacent to
189
INSTRUMENTAL OPTICS 1 i.e.
v
+
1
=
u
2
< as
J
r
(4.19)
f=T )
virtual), Fig. 4.27
Concave mirror (image
Fig. 4.27
From Fig.
4.27,
£-a
= and
= 2a -
-
j8
= -20 +
As
j6
208-0-0
=
/.
a =
•••
/8
j8-0
20 =
before,
2fc
_ft
_
CP~ OP
_^_ /P
i.e.
v
u but the image is virtual, therefore v
11
2 r
Convex
is negative.
~ f
1
(4.19)
v
u
mirror, Fig. 4.28
Fig. 4.28
= a+
2a = .-.
i.e.
2(<£-0)
=
>
=
2(9
•••
+
/S
+
j3
<£-j8 = 20
+
/S
a =
- 6
190
As
SURVEYING PROBLEMS AND SOLUTIONS before,
h
trad
~
Siti
Prad
~
sin/3 ~
^
sin0
.
rad
h
(0
is real
OP
(C
is virtual
h
~ -PC
Thus
h
_
IP 1
^-
i.e.
.*.
(/
OP
-
-v
h
_
-PC
1 — u
-r
1
u
is +ve).
PC
JL = _L r
f
Therefore, using the sign convention, the formula is both types of mirror in all cases.
4.36
is -ve).
2_
— +— 1
•••
is -ve.)
2h
OP
v
IP
is virtual
~ -IP
(4.19)
common
to
Magnification in spherical mirrors (Fig. 4.29) u
Fig. 4.29 Magnification in spherical mirrors
IB
is the
image of OA.
In the right-angled triangles OPO^ and /P/, the angle a is common, being the angles of incidence and of reflection, and therefore the triangles are similar. /7 (image size) /, P (v) ... A Thus, magnification = 0, P (w) 00^ (object size) i
.
m
=
—
neglecting signs
(4.20)
INSTRUMENTAL OPTICS
191
Example 4.2 An object 1 in. high is placed on the principal axis 20 in. from a concave mirror which has a radius of curvature of 15 in. Find the position, size and nature of the image.
If a thin lens is assumed to be split into a series of small prisms, any ray incident on the face will be refracted and will deviate by an angle (Eq.4.15) D = A(/x -1)
Fig. 4.32
N.B. The deviation angle
D
Formation of images is also related to the height h
and the
focal length /, i.e.
D =
h/f
(4.21)
INSTRUMENTAL OPTICS The
4.43
193
relationship between object and image in a thin lens
The position
of the image can be
drawn using three rays, Fig. 4.33.
Fig. 4. 33
N.B.
Two
(a)
Ray OA
OB
Ray
F2
exist.
,
parallel to the principal axis is refracted to pass
through principal focus (b)
and
principal foci, F,
F2
.
passes through the principal focus
F,
and
is then
refracted parallel to the principal axis. (c)
Ray OPI passes from object
to
image through the pole
P
without refraction.
Convex lens (a) When the object focus
F2
(b)
,
real
is at infinity, the
image
is at the principal
and inverted.
When the object
is
between
infinity
and
F,
,
the image is real
and inverted. (c) When the object is between F, and P, the image is virtual, magnified, and erect, i.e. a simple magnifying glass.
Concave lens The image 4.44
is
always
virtual, erect
and diminished.
Derivation of formulae
The real-is-positive sign convention is again adopted, but for convex lenses the real distances and focal lengths are considered positive, whilst for concave lenses the virtual distances and focal lengths are considered negative.
As with
mirrors, thin lens formulae
depend on small angle approxi-
mations.
Convex lens (a)
Image
real, Fig.
4.34
D
=
a +
(3
SURVEYING PROBLEMS AND SOLUTIONS
194
By Eq.
(4.21),
D
=
1
1
=
1+ V
U ....
=± + L
i
V
U (b)
Image
virtual, i.e. object
between
Fig.
4.
D = a h.
i.e.
=
i
.e.
/3
A_A U
/
but v is virtual,
35
V
negative.
4/
-iu + iv
D
=
Concave lens (Fig. 4.36) ft
- a
F
and P, Fig. 4.35.
INSTRUMENTAL OPTICS
195
i.e.
V
f
u
but v and / are negative, being virtual distances
Fig.
4.45
v
u
/
4.
36
Magnification in thin lenses (Fig. 4.37)
Fig.
As with spherical triangles with angle
4.
37
OPO^ and
mirrors,
/P/,
are similar right-angled
a commpn.
magnification
m
=
_L //i
(image size)
OOi
(object size)
UP P
(object distance u)
y
—
(image distance
as before
v)
SURVEYING PROBLEMS AND SOLUTIONS
196
N.B. This should not be confused with angular magnification or magnifying power (M), which is defined as
the angle subtended at the eye by the image the angle subtended at the eye by the object
For the astronomical telescope, with the image
M
=
focal length of objective
_
focal length of eyepiece
4.5
4.51
at infinity,
jo
(4.22) fe
Telescopes
Kepler's astronomical telescope (Fig. 4.38)
Objective
Eyepiece
Diaphragm
Fig. 4.38
The telescope
is
Kepler's atronomical telescope
designed to increase the angle subtending dis-
tant objects and thus apparently to bring them nearer.
The objective
lens, converging and of long focal length, produces
an image FX, inverted but real, of the object at infinity. The eyepiece lens, converging but of short focal length, is placed
F so as to produce from the real object magnified but similarly inverted.
close to
IY
,
4.52
FX
a virtual image
Galileo's telescope (Fig. 4.39)
The eyepiece
is
concave and produces a
objective.
As the
is unsuitable for
virtual, magnified, but
XF produced by the image lies outside the telescope eyepiece, it surveying purposes where cross hairs are required.
erect image IY of the original inverted image latter
INSTRUMENTAL OPTICS
197
Objective
Galileo's telescope
Fig. 4.39
4.53
Eyepieces Ideally, the eyepieces should reduce chromatic
and spherical
aberration.
Lenses of the same material are achromatic is equal to the
average of their focal lengths,
If their
distance apart
i(/,+/2 )
=
d
if their
i.e.
(4-23)
distance apart is equal to the differences between their
focal lengths, spherical aberration is reduced, i.e.
d
=
/,
-
U
(4.24)
For surveying purposes the diaphragm must be between the eyeThe most suitable is Ramsden* s eyepiece,
piece and the objective. Fig. 4.40.
Diaphragm
Fig. 4.40
The
Ramsden's eyepiece
focal length of each lens is the same, namely
/.
Neither of the
conditions (4.23) or (4.24) is satisfied.
Chromatic Spherical
|-(/i+/2 ) /, - / 2
= =
/
compared with 2/3/
compared with 2/3
/
Huyghen's eyepiece, Fig. 4.41, satisfies the conditions but the It is used in the Galileo telescope.
focal plane lies between the lenses.
SURVEYING PROBLEMS AND SOLUTIONS
198
Diaphragm
Huyghen's eyepiece
Fig. 4.41
Chromatic condition
-j (3/
= 2/ =
d
=
d
=
2/
An astronomical telescope consists
Example 4.3 in. apart.
f)
3/ - /
Spherical condition
24
+
If
of
two thin lenses
the magnifying power is xl2, what are the focal lengths
of the two lenses ?
Fig. 4.42
magnifying power
= 12/.
But
+ fe
=
26
in.
12/e + fe
=
26
in.
fe
=
=-r
=
12/e
f /.
fo
4.54
is
The
26
-
2
in.
24
=
12
=
h
eyepiece lens in.
objective lens
internal focussing telescope (Fig. 4.43)
The eyepiece and objective are fixed and an internal concave lens used for focussing. For the convex lens, by Eq. (4.19)
1 /,
1
+
1
199
INSTRUMENTAL OPTICS Internal focussing lens
Objective
Diaphragm
Fig. 4.43
Internal focussing telescope
1
-
i.e.
1
1
1
1
1
or
u,
u,
For the concave lens,
= -(f,-<0
«2
.1+1V
1
"2
/a
1
=
u
+
- d
v,
1
1
l-d
1
1
-d
v,
/a
An
2
1
/
internal focussing telescope has a length
to the diaphragm.
The respective
internal focussing lens are
/,
(4.25)
-d /
from the objective
focal lengths of the objective and the
and /2
.
To find the distance d of the focussing lens from the objective when the object focussed is u, from the objective, Fig. 4.43. For the objective,
11
\
For the focussing lens,
-JU /a
1
i.e.
_
1 u.
v,
-
(v, i.e.
d)(/ - d) =
d
2
-
d
2
-
-d
/a(/
-
/
d) -
/2 (v,
-
d)
-/2 (/-v,)} =
dtf
+
v,)
+
{
dC/
+
v,)
+
W/ + /
/ Vl
-d
a
)
-f2 l\ =
(4.26)
SURVEYING PROBLEMS AND SOLUTIONS
200
This
is a quadratic
equation in d and
its
value will vary according to
the distance «, of the object from the instrument.
Example 4.4
Describe, with the aid of a sketch, the function of an
internal focussing lens in a surveyors' telescope and state the advant-
ages and disadvantages of internal focussing as compared with external focussing. In a telescope, the object glass of focal length 7 in. is located 9
away from the diaphragm. The focussing lens is midway between these when the staff 60 ft away is focussed. Determine the focal length
in.
of the focussing lens.
(L.U.) Internal focussing lens
Diaphragm
Fig. 4.44
For the convex objective lens, /,
=
7
m,
=
60 x 12
in. =
720
in.
Then, by Eq. (4.19), 1
i
-
JL -
=
1- JL 720
7
720 - 7 720 x
713 5040
For the focussing lens, u2
v,
- 4-5
=
7-068
4-5
=
2-568
4-5 1
1
u
u.
+
2-568
4-5
-4-5 + 2-568 11-556
u
-
-5-98
in.
(i.e.
the lens is concave)
201
INSTRUMENTAL OPTICS
In an internally focussing telescope, Fig. 4.43, the objective of focal length 5 in. is 7*5 in. from the diaphragm. If the internal focussing lens is of focal length 10 in., find its distance from
Example 4.5
the diaphragm
when focussed to infinity. /, = 5 in. and thus the position
For the objective, 5
in.
of F,
will be
from C,.
C 2 F,
5 - d
=
For the internal focussing lens,
1
u2
-(5-d)
v,
=
7-5 - d
± + JU
=
f2
-
.
5 _ d
-(5 - <0(7-5 - d)
d
7-5 - d
+ 10(5 - d)
-10(7-5 -
)
=
-75 + lOd + 50 - lOd = -25
- 12-5d + 12-5
=
+ d
z
d
= 4-235
v9 = i.e.
'
=
(37-5 - 12-5d 2
X
l
10
i.e.
*2
2
1
i.e.
-10
= =
/a
d)
in.
7-5 - 4-235
the internal focussing lens will be 3-265
phragm when focussed to
4.55.
= in.
3-265
in.
away from the
dia-
infinity.
The tacheometric telescope
(external focussing) (Fig. 4.45) Staff
\fertical axis
Diaphragm
Fig. 4.45
The tacheometric telescope (external focussing)
Let a, b and c represent the three horizontal cross hairs of the diaphragm, ac being a distance i apart and b midway between a and
202
SURVEYING PROBLEMS AND SOLUTIONS With the telescope in focus, these lines will coincide with the
image of the staff observed at A, B and C respectively; the distance AC = s is known as the staff intercept. The line bOB represents the line of collimation of the telescope, with bO and OB conjugate focal lengths of the lens, v and u, respectively. The principal focal length of the lens is
FO
(/),
whilst the vertical axis is a distance k from
the principal focus F.
ACO
Because the triangles acO and
AC
QB
=
are similar,
±
or
ob
ac
=
i
u_
v
Using the lens formula, Eq.(4.19),
v
u
f
and multiplying both sides by uf gives,
V Substituting the value of u/v from Eq.(4.27),
=
u
s- +
/
i
Thus the distance from the vertical axis
to the staff is given as
D = sl+ (/+d)
(4.28)
This is the formula which is applied for normal stadia observations with the telescope horizontal and the staff vertical.
The
ratio f/i
-
M
is
given a convenient value
sionally 50), whilst the additive constant (/ + d) = ing upon the instrument.
of,
K
say, 100 (occawill vary depend-
The formula may thus be simplified as
D
=
M.s + K
(4.29)
The constants M and K for a certain instrument were 100 and 1*5 respectively. Readings taken on to the vertical staff were Example 4.6
3*15, 4*26 and 5*37
ft respectively, the telescope being horizontal. Calculate the horizontal distance from the instrument to the staff.
The stadia
intercept s
Horizontal distance
=
5-37-3-15 =
D = =
If
the instrument
trunnion axis at 4-83
was ft,
set at 103-62
2-22
ft
100 x 2-22 + 1-5 223-5 ft (68-1 m) ft
A.O.D. and the height to the
203
INSTRUMENTAL OPTICS then the reduced level of the staff station
=
103-62 + 4-83 - 4-26
=
104-19
ft
A.O.D. (31-757 m)
N.B. 4-26 - 3-15 If
5-37 - 4-26
=
=
= Is
1-11
the readings taken on to a metre staff were 0*960, 1*298, 1*636
respectively, then the horizontal distance = 100 x (1*636-0*960)
= 67*6m + 0*5 If
the instrument
was
m
set at 31*583
=
m
68*1
m
A.O.D. and the height of the
trunnion axis at 1*472 m,
then the reduced level of the staff station
4.56.
The
=
31*583 + 1*472 - 1-298
=
31*757
m
anallatic lens (Fig. 4.46) Vertical axis
Staff Anallatic
Diaphragm
•*>*
Fig. 4.46
The
anallatic lens
In the equation D = s(f/i) + (/ + d), the additive factor (/ + d) can be eliminated by introducing a convex lens between the objective and the diaphragm.
The basic principles can be seen in Fig. 4.46. The rays from the Ad and Ce will for a given distance D always form a constant angle d intersecting at G. If this fixed point G is made to fall on the
staff
vertical axis of the instrument the additive term will be eliminated. i.e.
Consider the object lens with the object neglecting the anallatic lens.
By Eq.(4.19)
and the image
1= 1+1 U u
v
a, c,,
(4.30)
V
f
and by Eq. (4.27)
AC
_
s (4.31) a,c.
Consider the anallatic lens with the object as a^c
x
and the image
204
SURVEYING PROBLEMS AND SOLUTIONS
as ac. Thus the object distance
=
-x
v,
and
the image distance = v - x Applying the previous equations to this lens,
i U
*
~
_
*
v - *
(4.32)
v - x
i
v ~ * = J^L v - x a c
and
(4.33)
'
x
N.B. The object distance
is
t
assumed positive but the image distance
is negative.
An expression a^c^
D
for
can now be found by eliminating
from these four equations. From Eq. (4.32) v,
- x
*< v ~ *> + v - *
=
/,
From Eq.(4.33) a,Ci
ac(v - x)
=
v,
- x
Combining these gives a "fr
1
—
v-x)
Qc (/i + 7
but from Eq. (4.30)
v =
-Hi u - f
Substituting in the above
a c i
t
Uf
+
ac[f,
u-f
=
- «)
U but from Eq. (4.31) a,c,
=
sv
sf
u
u - f
giving
+
ac(fi
sf
i.e.
s//,
Writing ac as
i,
sfU
u
=
aciu -
O
(/,
+
uf
u-f
- ')
£- -
*)
the distance apart of the stadia lines
=
HfM-f)
=
*[«(/,+/-*) + /(*-/,)]
=
+ uf - *(«-/)]
s //i
»(/+/,-*)
/(* ~
/
+
/,
/t)
- x
v,
v,
and
.
205
INSTRUMENTAL OPTICS
D =
but
u + d
+ /,-*)
t(/
^ x ~ f'>
= Ms and
d
if
f
+
U-x
+ d
^-^
=
(4-34)
U-x
+
f
D = Ms
M
where
(4.35) //i
= i
(4.36)
+ /,-*)
(/
a constant factor usually 100. therefore choose the lenses where the focal < x
The manufacturer can
length / t is such that
considered anallatic as
The
ally considered to
ample
it
is
movable.
variation of the focal length of the objective system is gener-
be negligible
4.7), manufacturers
for
most practical purposes (see Ex-
aiming at a low value for K, and in many
cases the telescopes are so designed that when focussed at infinity midway between the objective and the diaphragm. This allows accuracies for horizontal sights of up to 1/1000 for most the focussing lens is
distances required in this type of work.
Example 4.7 6
in.
An
focal length.
axis 4
anallatic telescope is fitted with an object lens of If
the stadia lines are 0*06 in. apart and the vertical
from the object lens, calculate the focal length of the anallatic lens and its position relative to the vertical axis if the multiplying in.
constant is 100. From Eq. (4.34) the distance between objective and axis
f+U-x When
/
= 6
d
in.
_
6 (*-/i) 6 +
M
Also, from Eq.(4.36),
/,
= 4
- x
&
~
!(/+/, -X) Therefore when
M
= 100,
M
-
i
= 0-06, /=6, 6/,
0-06(6 +
100
U-x)
206
SURVEYING PROBLEMS AND SOLUTIONS
Combining these equations,
4(6+/,-*) = 6(*-/,) 10* = 24 + = 6/,
and
10/,
100 x 0-06(6 + /,-*)
6*
=
36+0
*
=
6 in.
/,
=
3*6 in.
and
Thus the focal length of the anallatic lens (6-4) = 2 in. from the vertical axis.
is 3*6 in.
and
its posi-
tion is
Example 4.7a An anallatic tacheometer in use on a remote survey was damaged and it was decided to use a glass diaphragm not originally designed for the instrument. The spacing of the outer lines of the new diaphragm was 0*05 in., focal lengths of the object glass and the anallatic lens 3 in., fixed distance between object glass and trunnion in., and the anallatic lens could be moved by an adjusting screw between its limiting positions 3 in. and 4 in. from the object glass. In order to make the multiplier 100 it was decided to adjust the
axis 3
if this proved inadequate to graduate a special staff for use with the instrument. Make calculations to de-
position of the anallatic lens, or
termine which course was necessary, and
if a special staff is required, determine the correct calibration and the additive constant (if any). What is the obvious disadvantage to the use of such a special staff?
(L.U.)
From Eq. (4.36),
M
,,
=
ILL
/
=
+ u
u -JL Mi
3
100 x 0-05
=
6 - 1-8
=
4-2 in.
the anallatic lens should be 4*2 in. from the objective. As this is not possible, the lens is set as near as possible to this value, i.e. 4 in. ^-*-3Then = = 90 0-05(3 + 3 - 4) i.e.
—
M
The additive
factor
=
K
from Eq.(4.34)
/(*-/,)
/+/,-*
=
3(4-3) 3 +
3-4
=
V5in
207
INSTRUMENTAL OPTICS
the multiplying factor is to be 100, then the staff must be graduated in such a way that in reading 1 foot the actual length on the staff is
If
12 x
15 in.
i.e.
13!
9
4.57.
in.
3
The tacheometric telescope
(internal focussing) (Fig. 4.47) Mechanical axis
Diaphragm
H43£? u2
Fig. 4.47
To tor
M
for the
Tacheometric telescope (internal focussing)
find the spacing of the stadia lines to give a multiplying fac-
for a given sight distance: if
i
=
stadia interval,
and
s
=
stadia intercept;
convex lens (objective) i.e.
s
= £Tl
x
=
m ,s
u,
where m, is the magnifying power, for the concave lens (focussing) *
_
v2
=
i.e.
Vfir
but
distance -
i
=
xYl = m2 x
i
=
m^TTizS
(4.37)
D = Ms
'-$ M
(4.38)
Example 4.8 An internally focussing telescope has an objective 6 in. from the diaphragm. The respective focal lengths of the objective and the internal focussing lens are 5 in and 10 in. Find the distance apart the stadia lines should be to have a multiplying factor of 100 for an
observed distance of 500
ft.
SURVEYING PROBLEMS AND SOLUTIONS
208 At 500
the object will be 500
ft
i.e.
ft
- 6/2
«,
=
500 x 12 - 3
v
=
X ? 5997 - 5
i
mi
=
5997
5-004 2
=
from the objective,
in.
in
in.
From Eq.(4.26), d
2
- d(/ + d
i.e.
d
v,)
+
- d(6 +
- /2 /}
+/2 )
{v,(/
=
+ {16v, - 60} =
v,)
2
- 64 v, + 240}]
=
| [(6 + v) ±
VK6
=
|[(6 + v) +
V(6-v )(46-v )].
+
v,)
1
1
= i[ 11-004 2 ± VCO'995 8 x 40-995 8)]
d
i.e.
z
=
2-308
v
=
/
u2
=
v,-d
in.
_ d =
6
=
- 2-308
5-004
=
3-692 in
2-2-308 -
2-696 in
From Eq.(4.38), _
.
Dm m 2
Dv^V2
_
}
Mu u 2 M 500xl2x 5-0042x3-692 x
100 x 5997 x 2-696
=
0-068 56
in.
What errors will be introduced if the previous instrument Example 4.9 used for distances varying from 50 to 500 ft ?
is
At 50
ft
rrt
50 x 12 - 3
=
«,
597x5
=
597
2985
=
597 - 5
=
in.
5-0422 in.
592
Then, from Eq.(4.26),
d
|[(6 + v,) +
=
|[ll-042 2 - VC0-9578 x 40-9578)]
=
2-389
v2 = "2
4(6-^X46-^)]
=
in.
6 - 2-389
=
= 5-042 - 2-389
3-611
=
in.
2-653
in.
INSTRUMENTAL OPTICS The
0-068 56 x
=
stadia intercept (s)
209 " "2 1
v,v 2
0-06856 x 597 x 2-653 5-042 x 3-611 =
5-9641
-
0-497 Oft
in.
The value should be 0-500 =
error
representing
At
An
Example 4.10
0-0030
ft
Q-30
in 100
ft
100
ft
error
=
0-27
ft
200
ft
error
=
0-20
ft
300
ft
error
400
ft
error
= 0-09 ft = 0-01 ft
500
ft
error
= 0-00
ft
ft
internal focussing telescope has an object glass
The distance between the object glass and the diaphragm is 10 in. When the telescope is at infinity focus, the internal focussing lens is exactly midway between the objective and the diaphragm. Determine the focal length of the focussing lens. of 8 in. focal length.
At infinity focus the optical centre of the focussing lens lies on the line joining the optical centre of the objective and the cross-hairs, but deviates laterally 0*001 in. from at
25
ft.
it when the telescope is focussed Calculate the angular error in seconds due to this cause.
(L.U.) With the telescope focussed at infinity, For the focussing lens, 1
1
1
u
«2
V2 1
= v,
u With focus at 25
ft
=
1
l-d
1
-5
7-5 in.
(assuming 25
u,
=
25 x 12
v,
=
J£jL «,
-
/,
1
2
10-5
15
Focal length of focussing lens
ft
= =
/,
1
1
8
=
l-d
- d
U-d =
v,
from object lens.)
300
30 °x 8 300 - 8
= 8-2192
in.
SURVEYING PROBLEMS AND SOLUTIONS
210
From Eq.
(4.26), 2
dO
d d
i.e.
2
+
+ IV1O +
Vt)
/2) -/a/I
- 18-219 2d + (143-836 - 75)
Solving for d,
d
=
= 5-348
/-if,
\\i
x '
H-* oooi
x J X\
M
T Fig. 4.48
With focus at 25
the image would appear at x, neglecting the
ft
OX
internal focussing lens, i.e. off line, the line of sight is
=
With the focussing lens moving all images produced by the
v,.
now EXJZ and
objective appear as on this line.
The
line of sight through the objective is thus displaced
the length
To
v,.
calculate XX,.
XE UE
XX, /,/2
x
i.e.
0-001 x
=
(/
-
v,)
l-d 0-001 x (10 - 8*219) 10 - 5-348
0-001781
=
=
o-OOO 391
in.
4-552
To
calculate the angular error (5), tan 8
=
8
=
4.6
4.61
Ml
_
X
OX 206
265x0-000 391 Q.010
9 . 8 nccondc
Instrumental Errors in the Theodolite
Eccentricity of the horizontal circle
In Fig. 4.49, let
0,
= vertical axis
XX,
in
INSTRUMENTAL OPTICS 2
=
Graduated circle axis
0^0 2
=
e
Z A,
~
0,4 2 =
=
211
eccentricity r
Fig. 4.49
the graduated circle (Centre 2 ) is not concentric with the veraxis (centre 0,) containing the readers A and B, the recorded value 6 will be in error by the angle a. As the instrument is rotated, the readers will successively occupy If
tical
positions 4,6,,
AZ B 2 A 3 B 3 ,
a,
=
.
0i-<£i tan-
M
A ZE
e sin
tan r
a,
~
tan
If
(4.40) r
and as a
is small,
a rad ~ tan a
e sin
a2 = a.
(4.39)
e sin>
Since e is small compared with Similarly,
- e cos
r
+ e cos0
e
sin0
(4.41)
(4.40)
the readers are 180° apart, 4,0,8, is a straight line and the true value of the angle .
mean of the recorded values 6 give the
212
SURVEYING PROBLEMS AND SOLUTIONS i.e.
=
0,
-
a,
2cf>
=
0,
+
2
4>
=
§(0, +
•"'
=
Z
+ a2
~ a2
as a,
(4.42)
2)
N.B. (1) On the line a = 0. 2 (2) At 90° to this line, a = maximum. ,
X
(3) If the instrument has only one reader, the angle should be repeated by transitting the telescope and rotating anticlockwise, thus giving recorded values 180° from original values This is of particular
importance with glass arc theodolites in which the graduated circle is of small radius.
To determine
the amount of eccentricity and index error on the
horizontal circle: (1) Set
index
A
to 0°
B
to 0°
and read displacement of index
B
from
180°, i.e. 5,.
and read displacement of index A from 82 Repeat these operations at a constant interval around the
(2) Set
180°, i.e. (3)
index .
plate, i.e. zeros at multiples of 10°. If the readers A and B are diametrically opposed, placement of reader B, from 180°, Fig. 4.50. Index A at 0°. Index B, at 180° - 8,, i.e. 180 - (2a + A).
let 5,
y
Reading
fli
180-6 Index error \
Fig. 4.50
Let §2 = displacement of reader Az from 180°, Fig. 4.51, Index B2 at 0. Index A 2 at 180 - 8 2 i.e. 180 - (2a + A). ,
= dis-
INSTRUMENTAL OPTICS
213
Index error X
Fig. 4.51 If
there is no eccentricity and
S2 = If
A and B
are 180° apart, then 5, =
0.
there is eccentricity and
A and B
are 180° apart, then
5,
= 8Z =
a constant If
there is no eccentricity and
+ 5, = -S 2
,
i.e.
A and B
are not 180° apart, then
equal, but opposite in sign.
there is eccentricity and A and B are not 180° apart, then 5, and 8 2 will vary in magnitude as the zero setting is consecutively changed around the circle of centre but their difference will remain conIf
2f
stant.
A plotting of the values using a different zero for each pair of dex settings will give the results shown in Fig. 4.52.
in-
360
+x and -x are 180" apart Fig. 4.52
4.62.
The
line of collimation not perpendicular to the trunnion axis
Let the line of sight make an angle of 90° + € with the trunnion axis inclined at an angle a, Fig. 4.53.
SURVEYING PROBLEMS AND SOLUTIONS
214
Vfcrtical
Fig.
4.
axis
Line of collimation not perpendicular
53
to the trunnion axis
The angular error Q in the horizontal plane due to the error e may be found by reference to Fig. 4.53. =
*I YZ
tane =
*X TY
tan0
TY = YZ sec a
But
tan0
.-.
=
X]f
i.e.
YZ
i.e.
TYt * ne
=
TY
=
=
tan e
TY cos a
tan € sec
a
(4.43)
TY cos a and e are small,
If
then If
tions
6
(4.44)
observations are made on the same face to two stations of eleva-
a and a 2 ,
,
then the error in the horizontal angle will be
2 )
=
±tan-1 (tan £ sec a,) - tan"' (tan e sec
±(0,-0 2 )
2:
±e(seca, -seca 2 )
±(#i
-
On changing sign.
€ sec a
=
-
a2 )
(4.45) (4.46)
face, the error will be of equal value but opposite in
Thus the mean
of face left and face right eliminates the error
to collimation in azimuth.
The sign
due
of the angle, i.e. elevation of de-
pression, is ignored in the equation.
The extension of a straight line, Fig. 4.54. If this instrument is used to extend a straight line by transitting the telescope, the following conditions prevail
*
With the axis on the line TQ the line of sight will be 0A V To observe A, the instrument must be rotated through the angle e to give pointing (1)
T,Qr On
— the
axis will be rotated through the
same angle E
transitting the telescope the line of sight will be (180°
A,OB, =
A0B 2
.
B2
is thus fixed
-pointing
On changing face the process
is
to
- 2a)
(2).
repeated — pointing (3) — and then
INSTRUMENTAL OPTICS
215
Fig. 4.54
pointing (4) will give position S4 The angle B2 0BA = 4e, but the rect extension of the line AO. .
mean position B
will be the cor-
The method of adjustment follows the above process, measured on a horizontal scale.
The collimation graticule to read on
The
4.63
error
B3
,
B 2 B4
being
may be corrected by moving the telescope
i.e.
aB^.
trunnion axis not perpendicular to the vertical axis
(Fig. 4.55)
The trunnion
(horizontal or transit) axis should be at right angles
to the vertical axis;
if
the plate bubbles are centralised, the trunnion if a trunnion axis error occurs. Thus the
axis will not be horizontal
on transitting, will sweep out a plane inclined to the verby an angle equal to the tilt of the trunnion axis.
line of sight, tical
Fig. 4.55
Trunnion axis not perpendicular to the vertical axis
the instrument is in correct adjustment, the line of sight sweeps out the vertical plane ABCD, Fig. 4.55. If
If the trunnion axis is tilted
by an angle
e,
the line of sight sweeps
SURVEYING PROBLEMS AND SOLUTIONS
216
out the inclined plane ABEF. In Fig. 4.55, the line of sight is
assumed to be AE. To correct
for
necessary to rotate the horizontal bearing of the line of sight by an angle 6, to bring it back to its correct position. the
of the plane
tilt
Thus
sin#
sin#
i.e.
and
it is
=
EC ED
=
AD ED
tane
BC tan e ED tana tane
tana tane
(4.47)
6 and e are small, then
if
6
where
6 e
= =
e tan
a
(4.48)
correction to the horizontal bearing
trunnion axis error
a = angle On
=
of inclination of sight
transitting the telescope, the inclination of the trunnion axis
will be in the opposite direction but of equal magnitude.
mean
of face left
and face
Thus
the
right eliminates the error.
Method of adjustment (Fig. 4.56) (1) Observe a highly elevated target A,
e.g. a church spire.
(2) With horizontal plates clamped, depress the telescope to observe a horizontal scale 8. (3)
Change face and re-observe A.
(4)
As
before, depress the telescope to observe the scale at C.
(5) Rotate horizontally to
D midway between B
and C.
(6)
Elevate the telescope to the altitude of A.
(7)
Adjust the trunnion axis until
A
is observed.
(8)
On depressing
D
should now be observed.
the telescope,
Fig. 4.56 Field test of trunnion axis error
INSTRUMENTAL OPTICS
217
Vertical axis not truly vertical (Fig. 4.57)
4.64
If the instrument is in correct adjustment but the vertical axis is not truly vertical by an angle E, then the horizontal axis will not be
by the same angle E. due to this will be
truly horizontal
Thus the
error in bearing
E tana This
is
(4.49)
a variable error dependent on the direction of pointing tilt of the vertical axis, and its effect is
relative to the direction of
not eliminated on change of face, as the vertical axis does not change in position or inclination.
e
Face
1/
t£^
Horizontal line
left
i
f
/
Face
r ight
Trunnion axis not perpendicular to the vertical axis
Face
Horizontal
left
lin e
S
\
V—.
Fac e right
Vertical axis
unchanged on change of face Vertical axis not truly vertical
Fig. 4.57
In Fig. 4.58, the true horizontal angle (0)
A£)B 2 -( c i)Ei tana, + Thus the
(c 2 )
Ez
tan
a2
error in pointing (0) is
A OB = angle x
y
(>)
.
dependent on (1) the
tilt
of the
axis E, which itself is dependent on the direction of pointing, vary-
maximum (£ ) when on the line of tilt of the vertical axis to zero when at 90° to this line, and (2) the angle of inclination of the ing from
line of sight.
To measure the value 4.59.
of e and
E
a striding level is used, Fig.
SURVEYING PROBLEMS AND SOLUTIONS
218
Fig. 4.58
Bubble axis
Trunnion axis
Face right
Vertical axis
Vertical axis
Fig. 4.59
Let the vertical axis be inclined at an angle
E
Let the trunnion axis be inclined at an angle e
Let the bubble be out of adjustment by an angle
Then face
j3.
left tilt
of the trunnion axis
=
tilt
of the bubble axis
=
tilt
of the trunnion axis
=
tilt
of the bubble a^is
E E -
e j8
face right
E + = E +
e jS
INSTRUMENTAL OPTICS Mean
tilt
of the trunnion axis
= -|[(E -
e)
Mean
tilt
of the bubble axis
= -±[(E -
j8)
219
+ (E + + (E +
e)]
j8)]
=
E
(4.50)
=
E
(4.51)
Therefore the mean correction taking all factors into account is
Etana N.B. The value of
E
(4.52)
is related to the direction of observation and its
effective value will vary from
maximum
to nil. Tilting level readings
should be taken for each pointing. If
E
is the
maximum
of the axis in a given direction, then
tilt
E,
where
is the
E
=
cos 6
angle between pointing and direction of
maximum
tilt.
does not strictly need to be in adjustment, nor is it necessary to change it end for end as some authors suggest, the mean of face left and face right giving the true
Then the bubble recording the
tilt
value. If
the striding level is graduated from the centre outwards for n
pointings and 2n readings of the bubble, then the correction to the
mean observed
direction is given by
=
c
A (XL -1R) tana
(4.53)
2n
where c = the correction in seconds d = the value of one division of the bubble in seconds
XL =
the
2/? = the
a =
sum sum
of the readings of the left-hand end of the bubble of the readings of the right-hand end of the bubble
the angle of inclination of sight
n = the number of pointings.
The sign
of the correction is positive as stated.
pends upon the sign of
XL
N.B. The greater the change in the value of on the horizontal angle. 4.65
Any changes de-
- 2/2 and that of a.
a
the greater the effect
Vertical circle index error (Fig. 4.60)
When the telescope
is horizontal,
the altitude bubble should be
central and the circle index reading zero (90° or 270° on whole circle
reading instruments). If
the true angle of altitude the recorded angles of altitude the vertical collimation error
and the circle index error
= = = =
a a, and
6,
a2
{
SURVEYING PROBLEMS AND SOLUTIONS
220
Recorded value (F.L.)
=
(F.R.)
=
a
=
-1(0.,
a =
a, -
a = a2 + + a2 )
- 6
cf>
+ (4.54)
Collimation error
Objective end
Bubble axis Index error
Eyepiece end
Face
left
S ^\^X^" / Collimation error & Ssf VC -''A
^fe.~r* ble
axis—
\
«2 \ -.-4
Index error
^J
Mf
^f^-
\^s^
Face right
Fig. 4.60
Thus, provided the altitude bubble is centralised for each reading, of face left and face right will give the true angle of altitude. If the bubble is not centralised then bubble error will occur, and, depending on the recorded displacement of the bubble at the objective and eyepiece ends, the sensitivity will indicate the angular error. As the bubble is rotated, the index is also rotated. Thus 6 will be subjected to an error of ± i"(0 - E)8", where 8" = the
mean
the angular sensitivity of the bubble. If the objective end of the bubble is higher than the eyepiece end on face left, i.e. L > EL then 6 will be decreased by ±-(0 - B )8",