The Sewerage of Sea Coast Towns Part 10

Chapter XIV

TRIGONOMETRICAL SURVEYING.

In the surveying work necessary to fix the positions of the various stations, and of the float, a few elementary trigonometrical problems are involved which can be advantageously explained by taking practical examples.

Having selected the main station A, as shown in Fig. 35, and measured the length of any line A B on a convenient piece of level ground, the next step will be to fix its position upon the plan. Two prominent landmarks, C and D, such as church steeples, flag-staffs, etc., the positions of which are shown upon the ordnance map, are selected and the angles read from each of the stations A and B. a.s.sume the line A B measures ll7 ft, and the angular measurements reading from zero on that line are, from A to point C, 29 23' and to point D 88 43', and from B to point C 212 43', and to point D 272 18' 30". The actual readings can be noted, and then the arrangement of the lines and angles sketched out as shown in Fig. 35, from which it will be necessary to find the lengths AC and AD. As the three angles of a triangle equal 180, the angle B C A = 180- 147 17'-29 23'= 3 20', the angle B D A = 180-87 41' 30"- 88 43'= 3 35' 30". In any triangle the sides are proportionate to the sines of the opposite angles, and vice versa; therefore,

A B : A C :: sin B C A : sin A B C, or sin B C A : A B :: sin ABC : A C, nr A C = (A B sin A B C) / (sin B C A) = (117 x sin 147 17') / (sin 3 20')

or log A C = log 117 + L sin 147 17' - L sin 3 20'.

The sine of an angle is equal to the sine of its supplement, so that sin 147 17' = sin 32 43', whence log A C = 2.0681859 + 9.7327837-8.7645111 = 3.0364585

Therefore A C = 1087.6 feet.

Similarly sin B D A: A B :: sin A B D: A D

A B sin A B D 117 x sin 87 41' 30"

therefore A D = --------------- = ----------------------- sin B D A sin 3 35' 30"

whence log A D = log ll7 + L sin 87 41' 30" - L sin 3 35' 30"

= 2.0681859 + 9.99964745 - 8.79688775 = 3.2709456

Therefore AD = 1866.15 feet.

The length of two of the sides and all three angles of each of the two triangles A C B and A D B are now known, so that the triangles can be drawn upon the base A B by setting off the sides at the known angles, and the draughtsmans.h.i.+p can be checked by measuring the other known side of each triangle. The points C and D will then represent the positions of the two landmarks to which the observations were taken, and if the triangles are drawn upon a piece of tracing paper, and then superimposed upon the ordnance map so that the points C and D correspond with the landmarks, the points A and B can be p.r.i.c.ked through on to the map, and the base line A B drawn in its correct position.

If it is desired to draw the base line on the map direct from the two known points, it will be necessary to ascertain the magnitude of the angle A D C. Now, in any triangle the tangent of half the difference of two angles is to the tangent of half their sum as the difference of the two opposite sides is to their sum; that is:--

Tan 1/2 (ACD - ADC): tan 1/2 (ACD + ADC):: AD - AC : AD + AC,

but ACD + ADC = l80 - CAD = 120 40', therefore, tan 1/2 (ACD - ADC): tan 1/2 (120 40'):: (1866.15 - 1087.6): (1866.15 + 1087.6),

778.55 tan 60 20'

therefore, tan 1/2 (ACD - ADC) = -------------------- 2953.75

or L tan 1/2 (ACD - ADC) = log 778.55 + L tan 60 20'

- log 2953.75 .

= 2.8912865 + 10.2444l54 - 3.4703738 = 9.6653281 .?. 1/2 (ACD - ADC) = 24 49' 53"

.?. ACD - ADC = 49 39' 46". Then algebraically

(ACD + ADC) - (ACD - ADC) ADC = --------------------------- 2

120 40' - 49 39' 46" 71 0' 14"

.?. ADC = ------------------------- = ------------ = 35 30' 7", 2 2

ACD = 180 - 35 30' 7" - 59 20' = 85 9' 53".

[Ill.u.s.tration: Fig. 35.--Arrangement of lines and Angles Showing Theodolite Readings and Dimensions.]

Now join up points C and D on the plan, and from point D set off the line D A, making an angle of 35 30' 7" with C D, and having a length of l866.15 ft, and from point C set off the angle A C D equal to 85 9' 53". Then the line A C should measure l087.6 ft long, and meet the line A D at the point A, making an angle of 59 20'. From point A draw a line A B, ll7 ft long, making an angle of 29 23' with the line A C; join B C, then the angle ABC should measure 147 17', and the angle B C A 3 20'. If the lines and angles are accurately drawn, which can be proved by checking as indicated, the line A B will represent the base line in its correct position on the plan.

The positions of the other stations can be calculated from the readings of the angles taken from such stations. Take stations E, F, G, and H as shown in Fig. 36*, the angles which are observed being marked with an arc.

It will be observed that two of the angles of each triangle are recorded, so that the third is always known. The full lines represent those sides, the lengths of which are calculated, so that the dimensions of two sides and the three angles of each triangle are known. Starting with station E,

Sin A E D: A D:: sin D A E: D E

A D sin D A E D E = -------------- sin A E D

or log D E = log A D + L sin D A E-L sin A E D.

From station F, E and G are visible, but the landmark D cannot be seen; therefore, as the latter can be seen from G, it will be necessary to fix the position of G first. Then,

sin E G D: D E :: sin E D G : E G,

D E sin E D G or EG= --------------- sin E G D

Now, sin E F G: E G :: sin F E G : F G

E G sin F E G F G = ------------- sin E F G

thus allowing the position of F to be fixed, and then

sin F H G : F G :: sin F G H : F H

F G sin F G H F H= ------------- sin F H G

[Ill.u.s.tration: FIG 36.--DIAGRAM ILl.u.s.tRATING TRIGONOMETRICAL SURVEY OF OBSERVATION STATIONS.]

In triangles such as E F G and F G H all three angles can be directly read, so that any inaccuracy in the readings is at once apparent. The station H and further stations along the coast being: out of sight of landmark D, it will be as well to connect the survey up with another landmark K, which can be utilised in the forward work; the line K H being equal to

F H sin K F H ------------- sin F K H

The distance between C and D in Fig. 35 is calculated in a similar manner, because sin A C D : A D:: sin CAD : CD,