Hawksley = 40 in
Neville = 37 in
Santo Crimp = 36 in
Saph and Schoder = 34-1/2 in
The circumstances of the case would probably be met by constructing the outfall 36 in in diameter.
It is very rarely desirable to fix a flap-valve at the end of a sea outfall pipe, as it forms a serious obstruction to the flow of the sewage, amounting, in one case the writer investigated, to a loss of eight-ninths of the available head; the head was exceptionally small, and the flap valve practically absorbed it all. The only advantage in using a flap valve occurs when the pipe is directly connected with a tank sewer below the level of high water, in which case, if the sea water were allowed to enter, it would not only occupy space required for storing sewage, but it would act on the sewage and speedily start decomposition, with the consequent emission of objectionable odours. If there is any probability of sand drifting over the mouth of the outfall pipe, the latter will keep free much better if there is no valve. Schemes have been suggested in which it was proposed to utilise a flap valve on the outlet so as to render the discharge of the sewage automatic. That is to say, the sewage was proposed to be collected in a reservoir at the head of, and directly connected to, the outfall pipe, at the outlet end of which a flap valve was to be fixed. During high water the mouth of the outfall would be closed, so that sewage would collect in the pipes, and in the reservoir beyond; then when the tide had fallen such a distance that its level was below the level of the sewage, the flap valve would open, and the sewage flow out until the tide rose and closed the valve. There are several objections to this arrangement. First of all, a flap valve under such conditions would not remain watertight, unless it were attended to almost every day, which is, of course, impracticable when the outlet is below water. As the valve would open when the sea fell to a certain level and remain open during the time it was below that level, the period of discharge would vary from, say, two hours at neap tides to about four hours at springs; and if the two hours were sufficient, the four hours would be unnecessary. Then the sewage would not only be running out and hanging about during dead water at low tide, but before that time it would be carried in one direction, and after that time in the other direction; so that it would be spread out in all quarters around the outfall, instead of being carried direct out to sea beyond chance of return, as would be the case in a well- designed scheme.
When opening the valve in the reservoir, or other chamber, to allow the sewage to flow through the outfall pipe, care should be taken to open it at a slow rate so as to prevent damage by concussion when the escaping sewage meets the sea water standing in the lower portion of the pipes. When there is considerable difference of level between the reservoir and the sea, and the valve is opened somewhat quickly, the sewage as it enters the sea will create a "water-spout," which may reach to a considerable height, and which draws undesirable attention to the fact that the sewage is then being turned into the sea.
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. 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'.