Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/173

Rh to take place in the direction of the lines of force. Let this displacement at $$G$$ and $$H$$ be equal to $$s_1$$ and $$s_2$$ respectively; then, since the flux is everywhere equal, $$\frac{s_1}{s_2} = \frac{q_2}{q_1} .$$

But the acceleration of the particles is proportional to the rate of displacement, and therefore to the displacement itself.

Hence $$\frac{\mathit{Acceleration\ at}\ G}{\mathit{Acceleration\ at}\ H} = \frac{s_1}{s_2} = \frac{q_2}{q_1} ,$$ that is, the intensity of the field is inversely proportional to the distance between the boundary lines of force.

Taking the velocity of the fluid through the field as $$V,$$ let $$k$$ be the intensity of the field (Figs. 64 and 65), where $$q$$ is the normal distance between two adjacent lines of force, $$J_1, J_2$$ (so that $$kq$$ is constant), and let $$p$$ be the distance in the line of relative motion, and $$\theta$$ be the angle at which the path of the particle cuts the lines of force. Then the time taken by the particle to traverse the “tube of force” $$J_1, J_2 = \frac{p}{V} ,$$ the momentum imparted in the direction of the lines of force $$= \frac{k\ p}{V} ,$$ of which the vertical component $$l$$ is:—