Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/399

components of the terminal vortices, so that no error shall be introduced by an allowance for the “downthrow current,” as would be necessary were the tail plane situated in the ordinary position (Fig. 151).

Weight			.24 grams.
Area	1.125
+	1.030
+	.490
	———
=	2.645	sq. in.
Angle of aeroplane ${\tfrac {1}{20}}.$

Flight Data.

	Distance. Ft.	Time. Secs.
1 2 3 4	30 33 28 21	2 2 1.8 ?	${\begin{cases}{\mbox{Launched}}\\{\mbox{from 10 ft.}}\\{\mbox{altitude.}}\end{cases}}$ ${\begin{cases}{\mbox{Launched}}\\{\mbox{from 7 ft.}}\\{\mbox{altitude.}}\end{cases}}$

Whence,

Mean $V=15.5\ {\mbox{ft./sec.}}$
Mean $\tan \gamma ={\tfrac {1}{3}}$ or $\gamma =18{\tfrac {1}{2}}^{\circ }$ or,
Resistance $=\sin 18{\tfrac {1}{2}}^{\circ }\times .24$	$=.075$
Aerodynamic resistance $=W\beta =.24\times .05$	$=.012$
∴ Skin resistance	——— $=.063$

And area $=2.645$ sq. in. or skin resistance per sq. ft. area—

${\frac {.063\times 144}{2.645}}=3.43\ ({\mbox{grams}}).$

${\frac {3.43\times 32.2}{453.6}}=.243\ ({\mbox{poundals}}).$

But normal plane reaction at $15.5\ {\mbox{ft./sec.}}$

$=.7\times .078\times 15.5\times 15.5=13.1$

∴ $\quad \xi ={\frac {.243}{13.1}}=.0185$

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