If the weight in pounds of an aircraft is known along with its sink rate in feet per minute at best power-off glide conditions, can the horsepower being dissipated by drag as the plane descends via gravity be calculated?
Horsepower = weight(lbs) * sink_rate(fpm)/(550*60)
It really is that simple.
At a constant airspeed glide, kinetic energy isn't changing, only potential energy. So we know that all the energy going into drag is coming from the descent.
A basic law of physics is:
$P = F*v$
- $P$: the power
- $F$: the force, in our case the force due to gravity
- $v$: the velocity, in our case the sink rate
The only trick is that we have to make sure that units align. In the case of horse-power, which is 550ft-lbs/s, we have to convert fpm to fps (divide by 60) and then scale by 550 to get the answer in HP.
Even better is when we don't use non-standard units. Then there is no conversion factor, because power in Watts is identically Nm/s.
That power is sink rate x glider weight. If you use m/s for sink rate and newton for weight, you'll get the power in watt. Advantages of SI...
It's easy to understand why it is so. Within a uniform gravitational field with acceleration g, a mass m is pulled down with a force m·g. The potential gravitational energy E at a given height h is E = m·g·h
Now, for an infinitesimal variation of height dh, the corresponding infinitesimal variation of energy is dE = m·g·dh
And if that variation dE takes place in an infinitesimal time dt, the power implied is W = dE/dt = m·g·dh/dt. Since, in our case, dh/dt is the sink speed w, we have that power W = m·g·w