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

Rh 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—

§ 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:—

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 ; 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