Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/72

Kinematic viscosity, which we will denote by the symbol $\nu ={\frac {\mu }{\rho }},$ and and is consequently of the dimensions ${\frac {L^{2}}{T}}.$

Writing the law of viscous resistance in its kinematic form we have $R=A\ \nu \ {\frac {V}{l}}$ where $A$ is the area of the surface; it will be noted that this expression is dimensional.

If we similarly write the law of skin-friction $R=A\ \nu \ {\frac {V^{1.5}}{l}},$ we find that the dimensions do not harmonise.

Let us examine this expression in a general form, where

$R=c\times A^{p}\times \nu ^{q}\times V^{r};$

dimensionally:—

	${\frac {L^{4}}{T^{2}}}=c\times L^{2p}\times {\frac {L^{2q}}{T^{q}}}\times {\frac {L^{r}}{T^{r}}};$
∴	$2p+2q+r=4,$
and	$q+r=2,$
∴	$q=2\ {\boldsymbol {-}}\ r,$
∴	$2p+4\ {\boldsymbol {-}}\ 2r+r=4,$
∴	$2p\ {\boldsymbol {-}}\ r=0,$
	$p={\frac {r}{2}}.$

The general expression therefore becomes:—

$R=c\ \nu ^{q}\ A^{\frac {r}{2}}\ V^{r}$

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in which

$q+r=2.$

This is the general equation to the kinematic resistance of bodies in viscous fluids, and correlates the variations in respect of viscosity, area, and velocity; the application extends to both normal and inclined planes and bodies of the most diverse form.^[1]

It may be illustrated here in its relation to the law of skin-friction; we have, $R$ varies as $V^{1.5}$ and the full kinematic expression therefore becomes—

$R=c\ \nu ^{\frac {1}{2}}\ A^{.75}\ V^{1.5},$

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1. The method of dimensions presumes geometrical similarity of figure.

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