Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/271

is this probable if the angle $\beta$ is considerable. Apart from the practical considerations introduced by the necessity for thickness in the aerofoil, which probably imposes a minimum limit on its fore and aft dimension, there are reasons (which will be discussed hereafter) for supposing that perfect continuity of motion is not possible under the conditions of finite lateral extent. Whatever defect in the theory may he introduced by considerations of this nature may be legitimately ignored at the present stage.^[1]

§ 174. Grliding Angle.—Let $\gamma$ represent, in circular measure, the gliding angle—that is, the angle of flight path at which the force to overcome the resistance is exactly provided for by the component of gravity in the path of flight. It will be assumed that $\gamma$ comes within the definition of a small angle, i.e., $\gamma =\sin \gamma =\tan \gamma$ with sufficient approximation. Then—

$\gamma ={\frac {\mbox{Resistance}}{\mbox{Weigh}}}.$

Now weight $(W)=\rho \ \kappa \ A\ V^{2}\ (\alpha +\beta )$ and resistance comprises—

(1) Aerodynamic resistance $={\frac {\mbox{Energy per sec.}}{\mbox{Velocity}}}$

$={\frac {\rho \ \kappa \ A\ V^{2}\ (\beta ^{2}-\alpha ^{2})}{2}}$

and

(2) Skin frictional $=\xi \ C\ \rho \ A\ V^{2},$ and

(3) Body resistance $C\ \rho \ \alpha \ V^{2}$ where $\alpha$ is a normal plane area to which the body resistance is equivalent.

Now (3) is a superadded resistance with which for the moment we will not concern ourselves, so that we have—

$\gamma ={\frac {\beta ^{2}-\alpha ^{2}}{2\ (\alpha +\beta )}}+{\frac {\xi \ C}{\kappa \ (\alpha +\beta )}}$ $={\frac {(1-\epsilon )}{2}}\ \beta +{\frac {\xi \ C}{(\epsilon +1)}}.$

But we know that for Least Resistance these terms are equal, consequently under the conditions of Least Gliding Angle ($=\gamma _{1}$) we have $\gamma _{1}=(1-\epsilon )\ \beta _{1},$ that is to say, the least gliding angle will be

$\gamma _{1}=(1-\epsilon )\ {\sqrt {\frac {2\ \xi \ C}{\kappa \ (1-\epsilon ^{2})}}}.$

1. Compare § 189.

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