Question

A fighter aircraft that has a landing weight of 23,000 lb uses a drogue (a small
parachute) to slow it down after landing. The aircraft touches down on the runway
at 135 knots with an idle power setting (i.e., no signi cant thrust), followed by the
immediate deployment of the drogue. Estimate the resulting stopping distance of the
aircraft. The drogue can be considered an open hemispherical shell of 3 m in diameter.
Assume that the braking action comes entirely from the drogue and it remains fully
deployed during the entire landing roll. Explicitly state any other assumptions that
you may make.

Answer 1

weight=23000 lb=102309.1 N

mass=weight/g=10439.7 kg

initial speed=135 knots=69.45 m/s

drogue is hemisphere with radius 3/2=1.5 m

drag force =0.5*rho*v^2*Cd*A

where rho=density of air=1.225 kg/m^2

v=speed

Cd=drag coefficient of hemisphere=0.42

A=cross sectional area=2*pi*radius^2=2*pi*1.5^2=14.137 m^2

then drag force=3.6367*v^2

==>mass*dv/dx=3.6367*v^2

==>m*dv/v^2=3.6367*dx

integrating,

m*(-1/v)=3.6367*x

using limits of v from 69.45 m/s to 0 m/s and limits of x from 0 to stopping distance d

m*(1/69.45)=3.6367*d

==>d=41.33 m

so stopping distance is 41.33 m