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Volume 31, 1898
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Art. LXII.—A Graphic Method of Calculating Cubic Content of Excavation, as for Water-races on Uneven Ground.

[Read 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 heavy arithmetical work involved in the use of the usual formulæ.

The problem set was as follows : Given the areas of two cross-sections (a2 and b2), and the perpendicular distance (x) between them, to find the amount of excavation.

The following is a graphic method of arriving at the result. 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, ii appears to be worth while to publish it. a2, b2, and x are found by the usual methods employed in surveying (a and b not being found).

The mass to be excavated can be treated as a frustum of a pyramid, and can be calculated from the formula

[The section below cannot be correctly rendered as it contains complex formatting. See the image of the page for a more accurate rendering.]

x/3 (a2 + ab + b2).

This involves multiplying a2.by b2, finding the square root of the product, and then adding a2, ab, and b2 in order to find a2 + ab + b2, the area of the base of the equivalent pyramid Logarithms being out of the question, the process is a tedious one.

But a2 + ab + b2 may be easily found as follows (see Plate LVII.).

Draw two lines 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 a2; and along OB mark off OD, to represent b2.

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; ii will cut OH at E; then, without altering the compasses, we mark off the radius twice, giving EF = CD.]

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OE represents the product ab in the formula (Euclid, vi., 8).

Mark off EF = CD.

Then OF represents a2 + ab + b2.


We can read this off and multiply by x/3; or


Use tables giving the content of pyramid with base a2 + ab + b2 and height x; or


If the distance between our sections is always the same—say, 162 ft.—we can graduate OH to represent cubic yards—1 in. = 40 cubic yards—and read off the cubic content at once without calculation.

If the distance between the faces, is not always the same, we may place the scale at the right-hand side, and have a straight edge or rigid steel wire radius movable about a pivot at S. Move, this edge or radius about S until its edge is over F. Run the eye along the edge or radius until you reach, say at F, the vertical line corresponding to the perpendicular distance between the faces, given at the top MHN. The reading on the same horizontal line as this gives the cubic content. (This is easily seen to follow by similar triangles.)

No calculation whatever is needed, and the whole operation may be as readily performed as any ordinary consultation of tables for rectangular areas and volumes.

To avoid the use of a rigid wire or edge for radius—(a) A silk thread may be attached at S, and stretched so as to pass over F, and the reading taken as before; or (b), if only one or two distances between the faces are commonly used, scales of cubic yards may be constructed to correspond, and the proper scale may be placed along OH, and the cubic content read off thereon.

The scales I have chosen are rather large for practical needs; but it would be easy to put various scales along AOB, MHN, and at the line of volumes at the right, to suit different kinds of work.

Example.—The figure shows the method of calculation for faces of 50 square feet and 70 square feet. With horizontal distance 162ft. the content is 358 cubic yards; 216ft., 478 cubic yards, &c.

The method, it will be noted, is an exact one, not a mere approximation; the accuracy depends chiefly on the accuracy of the original measurements, and on the accuracy with which the scale can be read.*

[Footnote] * The scale referred to above can be read much more nearly than probably the areas of the faces can be estimated, for an error of ½ a square foot in each face would give an error of about 3 cubic yards in the result.