Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/174

But	$p=q\ {\mbox{cosec}}\ \theta$
∴	$l={\frac {k\ q\ {\mbox{cosec}}\ \theta \ \sin \theta }{V}}={\frac {k\ q}{V}},$

which is constant. Therefore, if the particle, after cutting the tube at $G$ (Fig. 64) and continuing its course, recut the same tube at $H,$ the upward momentum communicated at $G$ will be equal to the downward momentum communicated at $H.$

But a particle of fluid traversing the field of force of the aeroplane may be regarded as passing through a series of regions bounded by adjacent lines of force, to each of which the foregoing result may be applied.
Fig. 66. Consequently the upward velocity acquired in traversing the ascending field to $C_{1}$ will be given up in traversing the descending field to the medial line $P\ Q$ (the line separating the front and rear portions of the field), and the downward velocity imparted to the particle in cutting the descending field to $C_{2}$ will be given up in traversing the corresponding ascending field, so that, in respect of the vertical component of motion, the final state of the fluid will be the same as its initial state.

Again, since the conditions determining the form of the field are symmetrical, the field itself must also be symmetrical about the plane of which the medial line $P\ Q$ (Figs. 64 and 66) is the trace.

Let $\ h_{1},h_{2},h_{3},h_{4},h_{5},$ etc. (Fig. 66), be points on the path of a particle of fluid cutting $P\ Q$ at $h_{4},$ corresponding to equal

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