Page:Aerial Flight - Volume 2 - Aerodonetics - Frederick Lanchester - 1908.djvu/391

And by Equation 16, 28,

$\cos \Theta _{i}={\frac {56}{192}}+{\frac {4\cdot 35}{\sqrt {56}}}=\cdot 873$

or

$\Theta _{i}\quad 29^{\circ }10'.$

Curve No. 5. (Inflected type.)

$H_{n}=64.\qquad H_{1}=16.\qquad \cos \Theta _{1}=1.$

Vide § 28.

Curve No. 6.

$H_{n}=64.\qquad H_{1}=8.\qquad \cos \Theta _{1}=1.$

$C={\sqrt {8}}\times \left(1-{\frac {8}{192}}\right)=2\cdot 71$.

Lowest point on curve,

${\frac {H}{192}}+{\frac {2\cdot 71}{\sqrt {H}}}=1$

whence

$H\ {\text{(max.)}}\ =148\cdot 7.$

Point of inflection

$H_{i}=\left({\frac {192\times 2\cdot 71}{2}}\right)^{\tfrac {2}{3}}=40\cdot 75$

and

$\cos \Theta _{i}={\frac {40\cdot 75}{192}}+{\frac {2\cdot 71}{\sqrt {40\cdot 75}}}=\cdot 636$

or

$\Theta _{i}\quad 50^{\circ }30'.$

Curve No. 7. Special case. Semicircle.

Curve No. 8. (Tumbler.)

$H_{n}=64.\qquad H_{1}=8.\qquad \cos \Theta _{1}=-1.$

$C={\sqrt {8}}\times \left(-1-{\frac {8}{192}}\right)=-2\cdot 946$.

Lowest point on curve,^[1]

${\frac {H}{192}}+{\frac {-2\cdot 946}{\sqrt {H}}}=1$

whence

$H=229$.

1. The lowest point on the tumbler curve, i.e., when cos θ = + 1, is calculated as a check on the plotting.

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