Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/445

system may distribute the momentum over a much greater mass of fluid than that coming within the sweep of the aerofoil, and so a far larger mass of the air than that coming within the sweep area will be "handled" in the sense of the present discussion. On this basis the area of the stratum from which energy may be drawn is given by the expression, ${\frac {1+\epsilon }{1-\epsilon }}\ \kappa \ A$ (comp. § 210).

In the following example the turbulence velocity is computed necessary to provide the requisite energy to a hypothetical albatros, whose data are:—

Weight		$=14\ {\mbox{lbs.}}$
Area		$=5\ {\mbox{square feet.}}$
${\mbox{n}}$		$=12,$ hence $\kappa =1.195$ and $\epsilon =.75.$
$\gamma$	taken	$=1/7$

The computation will be made both on the basis of sweep and that of peripteral area, and the figures will be given both for a simple harmonic motion and for circular motion, the assumption being in all cases that the whole of the available energy is utilised. As in all probability the bird can only utilise a comparatively moderate portion of the total available energy, the actual velocity of fluctuation will require to be very much greater than that stated in each case, in order that soaring should become possible.

Now resistance to flight = $W\ \gamma ,$ which from the foregoing data = 14 ÷ 7 = 2 pounds, or in absolute units = 64.4 poundals, or energy required per foot traversed = 64.4 ft. poundals.

Sweep = $\kappa \ A$ = 5 $\times$ 1.195 = 6 (approx.), and mass of air handled (on basis of sweep) per foot traversed = .078 = 6 = .47.

If $v$ be the velocity of mean square of turbulent motion, energy per foot traversed is

${\frac {.47\times v^{2}}{2}}$

∴	$.47\ \times v^{2}$		$=64.4\times 2$
	whence	$v^{2}$	$=274$
	or	$v$	$=16.5$

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