Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/307

this is given in full by the expression—$\epsilon \ C\ \rho \ V^{2}\ A$ and $L^{2}=A,$ therefore $C_{3}=C_{\rho }.$

Substituting for $C_{1}$ and $C_{3}$ in Equation (6) we have—

$L^{4}={\frac {1-\epsilon }{V^{4}\times 2\ \rho ^{2}\ \kappa \ (1+\epsilon )\ \xi \ C}}\times$ $(W_{1}^{2}+\kappa \ W_{1}\ (2-q)\ L^{q}+\kappa ^{2}\ (1-q)\ L^{2q}).$

§ 195. A Numerical Example.—The employment of this equation may be illustrated by a numerical example. Let us take the case of an aerodrome of 1,000 poundals essential weight (31 lbs. approximately), to be designed for a velocity of 50 feet per second. The remaining data are as follows:—

	n taken as 12
therefore—	$C=$	.75
	$\epsilon =$	.75
	$\kappa =$	1.2.
It is found by trial design or by calculation that in the expression $W_{2}=k\ L^{q}$ that—
	$k=$	50
and	$q=$	1.5
We take	$\xi =$	.03
and	$\rho =$	.78.

Substituting values in Equation (7) we obtain—

$L^{4}=$ ${\frac {.25}{6,250,000\times 2.4\times .0061\times 1.75\times .03\times .75}}$ $\times (1,000,000+25,000\ L^{1.5}-1,250\ L^{3})$ $=70+1.75\ L^{1.5}-.0875\ L^{3}$ (approx.).

This form of equation can only be solved by plotting or by guessing^[1]; the solution gives $L=2.96,$ that is to say, the area $A\ (=L^{2})$ is $8.76.$

Now the total weight sustained by this area is $W_{1}+W_{2}$ and $W_{2}=50\ L^{1.5}=254$ (poundals), or $W_{1}+W_{2}=1,254,{\mbox{or}}\ 39$ lbs. almost exactly. We therefore have pressure per square foot

1. The particular case of q = 1.5 is an exception.

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