Page:Lowell Hydraulic Experiments, 4th edition.djvu/93

 ness of the sill. It is essential, however, that the corners of the sill and sides of the weir presented to the stream, should be full and sharp, and not rounded or bevelled in any degree.

121. Two modes present themselves for studying, experimentally, the laws governing the discharge of water over weirs. First, that which has been uniformly adopted heretofore, namely, to obtain by direct measurement the quantity of water discharged in a given time, through an aperture of known dimensions; this is evidently the only mode of resolving the question completely. To perform the experiments, however, upon a scale of magnitude corresponding to the ordinary practical applications, usually requires an apparatus of great cost, and such as is beyond the reach of most experimenters. The great difficulty is, to obtain a suitable basin, in which to make the direct measurement of the quantity discharged by the weir.

The second mode dispenses with a direct measurement of the quantity. If we have two weirs of the same form, but of different lengths, and we know that the quantities of water discharged by them, in certain circumstances, are equal; knowing also the depth upon the sill of each weir, we have the data for an equation by which one unknown quantity may be determined. Neither the coefficient of contraction, nor the absolute discharge can, however, be obtained by such an equation.

122. The discharge over weirs is commonly assumed to vary as the square root of the third power of the depth; let us suppose it to be unknown, and equal to $$a$$.

Suppose also $$l$$ the length, and $$h$$ the depth, on one of the weirs; and $$l'$$ and $$h'$$ the corresponding dimensions for the other weir; $$C$$, a constant coefficient; $$Q$$, the quantity which, by hypothesis, is the same for both weirs. Assuming, according to the common formula, that the quantity is proportional to the length of the weir, we have $$Q = C l h^a ;$$ $$Q = C l' h'^a ;$$ consequently, $$Clh^a = Cl' h'^a ;$$ $$\left ( \frac{h}{h'} \right )^a = {{l'} \over {l}};$$ taking the logarithms, we have $$a(\log\ h - \log\ h') = \log\ l' - \log\ l$$