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. Assume 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 draughtsmanship 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
pricked 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".
[Illustration: 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
[Illustration: FIG 36.--DIAGRAM ILLUSTRATING 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,
AD sin CAD 1866.15 sin 59° 20'
or CD = ---------- = -------------------
sin SCD sin 85° 9' 53"
or log CD = log 1866.15 + L sin 59° 20' - L sin 85° 9' 53"
= 3.2709456 + 9.9345738 - 9.9984516
= 3.2070678. ' . CD = 1610.90 ft
The distance between any two positions of the float can be
obtained by calculation in a similar way to that in which the
length C D was obtained, but this is a lengthy process, and is
not necessary in practical work. It is desirable, of course,
that the positions of all the stations be fixed with the
greatest accuracy and plotted on the map, then the position of
the float can be located with sufficient correctness, if the
lines of sight obtained from the angles read with the
theodolites are plotted, and their point of intersection marked
on the plan. The distance between any two positions of the
float can be scaled from the plan.
The reason why close measurement is unnecessary in connection
with the positions of the float is that it represents a single
point, whereas the sewage escaping with considerable velocity
from the outfall sewer spreads itself over a wide expanse of
sea in front of the outlet, and thus has a tangible area. The
velocity of any current is greatest in the centre, and reduces
as the distance from the centre increases, until the edges of
the current are lost in comparative still water; so that
observations taken of the course of one particle, such as the
float represents, only approximately indicate the travel of the
sewage through the sea. Another point to bear in mind is that
the dilution of the sewage in the sea is so great that it is
generally only by reason of the unbroken fæcal, or other
matter, that it can be traced for any considerable distance
beyond the outfall. It is unlikely that such matters would
reach the outlet, except in a very finely divided state, when
they would be rapidly acted upon by the sea water, which is a
strong oxidising agent.