Energy Loss Due to Drag per distance

[SoarNeck]

Hi folks,

Looking for a formula for energy loss per distance due to drag.

Drag is a force, so units are (mass)(distance/time^2). Energy is (mass)(distance/time)^2. I'm guessing drag*distance, but I'm not sure if there's a 1/2 constant in there or the like.

Thanks,
Adam

[Ollie]

Energy is simply force times distance, where they are in the same direction no matter how fast or slow. Power is force times distance divided by time. Kenetic energy is one half of mass times velocity squared.

Of course the zero lift aerodynamic drag force is equal to half of the mass density of air, times area, times coefficient of drag, times velocity squared. If lift is involved then the induced drag must be added to the zero lift drag.

[Sparky Paul]

Work = force (drag) * distance (whatever)
Newtons times meters=Joules
1 Newton-meter=1 Joule=.738 ft-lb
.
Power= work done/time
1 Watt=1 Joule/sec=1.34X10^-3 hp
1 hp= 550 ft-lb/s=746 W
.
for a constant force (drag) at a contstant velociy, power=forece*velocity
.
Efficiency (this can be the good one:) )=Pout/Pin

[JMP_blackfoot]

If you are into gliders, the power spent by the drag force is equal to the power supplied by gravity.
Drag x Forward Speed = Weight x Sinking speed.
It is relatively easy to ascertain Forward Speed, Sinking Speed and Weight, then have an idea of the drag force.