Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/279

field of force symmetrical about the plane $z\ z,M_{\alpha }$ will also represent the upward momentum communicated to the air in partially arresting its downward motion during recess.

And the air outside the surfaces $b\ b$ and $b_{1}\ b_{1}$ will be receiving upward momentum the whole time it is in the field. Let the sum of this momentum be denoted by $M_{1}.$

Now, since the total residuary momentum must be zero (§§ 5 and 117) the downward momentum remaining in the air between the surfaces $b\ b$ and $b\ b$ is also equal to $M_{1}$, and if $M_{\beta }$ be the downward momentum in the air when it quits the descending field we shall have: $M_{\beta }=M_{\alpha }+M_{1}.$

But according to the main hypothesis we may represent ${\frac {M_{\alpha }}{M_{\beta }}}$ by ${\frac {\alpha }{\beta }},$ that is

$\epsilon ={\frac {M_{\alpha }}{M_{\beta }}}={\frac {M_{\alpha }}{M_{\alpha }+M_{1}}}.$

It remains for us to assess the value of $M_{1}$ in terms of $M^{\beta }.$

Let us suppose that, in a manner analogous to the limitation of the sweep, the air external to the surfaces $b\ b\ b_{1}\ b_{1}$ be represented by the limited region cut off by two further surfaces $c\ c$ and $c_{1}\ c_{1};$ then it is evident that the distance separating these surfaces will be greater the greater the lateral extension of the aerofoil.

Calling the fore and aft dimension of the aerofoil unity, so that its lateral dimension will $=n,$ let us assume that the distance between $b\ b$ and $c\ c$ is proportional to $\kappa ,$ and let it be denoted by $a\ \kappa .$ We have no direct means of testing the accuracy of this assumption; we can only say that it is a reasonable assumption, since the conditions that influence the depth of the layer of air acted upon obviously affect the extent of the disturbance of the fluid in other directions.

Then the upward momentum received by the air at the time of its crossing the plane $z\ z$ is, from considerations of field symmetry, just half the total eventually imparted, that is $={\frac {M_{1}}{2}},$ but we