Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/206

The expression then becomes:—

$P=C\ \rho \ V^{2},$

in which $C=32.2\ k/\rho ,$ and for air^[1] at 10 degrees C. and 760 mm. pressure we have $\rho =.078,$ whence,—

$C={\frac {32.2\times .0017}{.078}}=.7.$

The equation thus becomes:—

$P=.7\ \rho \ V^{2}$

The equation is identically the same in C.G.S. absolute units, and the constant is of the same value; that is to say, $P=$ dynes per square c.m., $V=$ c.m. per second and $\rho$ grammes per cubic c.m.

If we express $P$ in grammes per square c.m., and $V$ in metres per second, and substitute for $\rho$ for air at 10 degrees C., we obtain the equation in the form:—

$P=.009\ V^{2}.$

If the velocity is given in English miles per hour it is sometimes convenient to have the expression in the form:—

$P$ (pounds) $={\frac {V^{2}}{275}}.$

§ 135. Fluids other than Air.— If the whole physical properties of a fluid were represented by the symbols in the equation, or if, the equation being as it is, the fluid were incompressible and of zero viscosity, the constant $C$ would be the same for different fluids.

The experimental determination in the case of sea-water has been made by Captain Beaufoy, and independently by R. E. Froude, the results being in close agreement. In absolute units we have:—

$P=.55\ \rho \ V^{2},$

that is to say, the value of the constant is ${\frac {55}{70}}$ or approximately four-fifths of that in the case of air.

This difference is undoubtedly due to the lower kinematic viscosity of water, which is less than air in the ratio of 1 : 14. The nature of the relationship connecting the function kinematic viscosity and the changes in the value of the constant, is not very

1. From the determination of Regnault.

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