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

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.
Fig. 65.

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:—

${\frac {k\ p\ \sin \theta }{V}}$

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