602 Transactions. — Chemistry and Physics. 



Art. LXIl.— ^ Graphic Method of Calculating Cubic Con- 

 tent of Excavation, as for Water-races on Uneven Grou7id. 



By George Hogben, M.A. 



[Bead before the Philosophical Institute of Canterbury, 2nd November, 



1898.] 



Plate LVII. 



An engineering friend lately asked me if I could find some 

 easy way of calculating the cubic content of excavation of 

 water-races on uneven ground, so as to avoid the heav)^ 

 arithmetical work involved in the use of the usual formulae. 



The problem set was as follows : Given the areas of two 

 cross-sections (a^ and 6^), and the perpendicular distance (x) 

 between them, to find the amount of excavation. 



The following is a graphic method of arriving at the re- 

 sult. The solution seems rather obvious ; but it is new to me, 

 and engineers to whom I have shown it do not seem to know 

 it ; so, as it would evidently save a large amount of labour, it 

 appears to be worth while to pubhsh it. a~, K-, and x are 

 found by the usual methods employed in surveying {a and h 

 not being found). 



The mass to be excavated can be treated as a frustum of a 

 pyramid, and can be calculated from the formula 



I {a" + ab + b'-). 



This involves multiplying a^ by 6^ finding the square root of 

 the product, and then adding a^, ab, and b^ in order to find 

 a^ -f a6 + b^, the area of the base of the equivalent pyramid. 

 Logarithms being out of the question, the process is a tedious 

 one. 



But a^ + ft6 -f b' may be easily found as follows (see 

 Plate LVIL). 



Draw two hues at right angles, AOB and OH. Along OA 

 mark off a distance OC on any convenient scale — say, 20 square 

 feet to 1 in. — to represent a^ ; and along OB mark off OD, to 

 represent b"^. 



Place a set square so that the right angle may be on OH, 

 and the sides containing the right angle may pass through C 

 and D ; let the right angle be at E. [If a pencil-point be 

 held at A, it is easy to slide one edge of the set square past 

 A until the other edge is at B and the right angle on OH. Or 

 we can, of course, describe a semicircle on CD instead; it 

 will cut OH at E ; then, without altering the compasses, we 

 mark off the radius twice, giving EF = CD.] 



