Page:Aerial Flight - Volume 1 - Aerodynamics - Frederick Lanchester - 1906.djvu/266

Let	$x$	= direct resistance, of which—
	$x_{1}$	is that due to body resistance, and
	$x_{2}$	is that due to aerofoil area.
„	$a_{1}$	= a normal plane area whose resistance is the equivalent of $x_{1},$ and
„	$a_{2}$	= a normal plane area whose resistance is the equivalent of $x_{2}.$
„	$V$	= velocity of flight.
„	$\xi$	= coefficient of skin friction.

$k$ and $q$ are constants, as in preceding section, and $C_{1}\ C_{2}\ C_{3}$ are further constants.

It can be shown that—

$y=C_{1}\ {\frac {W^{2}}{L^{2}\ V^{2}}},$  and $\ x=C_{2}\ a_{1}\ V^{2}+C_{3}\ \xi \ V^{2}\ L^{2}.$

We require to know the minimum value of $x+y;$ or, we require to solve for minimum value the expression—

$C_{1}\ {\frac {W^{2}}{L^{2}\ V^{2}}}$$\ +C_{2}\ a_{1}\ V^{2}+C_{3}\ \xi \ V^{2}\ L^{2}.$

Now $W=W_{1}+W_{2},$  and $\ W_{2}=k\ L^{q},$ so that expression becomes

$C_{1}\ {\frac {(W_{1}+k\ L_{q})^{2}}{L^{2}\ V^{2}}}$$\ +C_{2}\ a_{1}\ V^{2}+C_{3}\ \xi \ V^{2}\ L^{2}.$

or

$C_{1}\ W_{1}^{2}{\frac {1}{L^{2}\ V^{2}}}+C_{1}\ W_{1}\ 2\ k\ {\frac {1}{L^{2-q}\ V^{2}}}$$\ +C_{1}\ k^{2}\ {\frac {1}{L^{2-2q}\ V^{2}}}$$\ +C_{2}\ a_{1}\ V^{2}+C_{3}\ \xi \ V^{2}\ L^{2}.$

where $L$ and $V$ are variables.

Making a temporary substitution of constants in order to abbreviate, we have—

${\frac {a}{L^{2}\ V^{2}}}+{\frac {b}{L^{2-q}\ V^{2}}}$$\ +{\frac {c}{L^{2-2q}\ V^{2}}}$$\ +c\ V^{2}+f\ V^{2}\ L^{2}.$

Differentiating in respect of $L$ and $V$ gives simultaneous equations as follows:—

${\boldsymbol {-}}2\ {\frac {a}{L^{3}\ V^{2}}}\ {\boldsymbol {-}}\ (2\ {\boldsymbol {-}}\ q)\ {\frac {b}{L^{3-q}\ V^{2}}}$$\ {\boldsymbol {-}}\ (2\ {\boldsymbol {-}}\ 2q)\ {\frac {c}{L^{3-2q}\ V^{2}}}$$\ +2\ f\ L\ V^{2}=0.$

(1)

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