# Can the energy dissipated in a power-off descent be calculated and expressed in horsepower?

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.

### Background

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:

$$P = F*v$$

where

• $$P$$: 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.

• Most excellent! I had suspected that but wasn't sure whether or not I needed to first convert pounds to slug-furlongs per imperial fortnight, or something equally psychedelic. -NN Oct 23, 2021 at 19:19
• by the way, the sink rate has to be expressed in feet per second. -NN Oct 23, 2021 at 19:53
• @nielsnielsen slug-furlongs per imperial fortnight could be called "nielsens"? :) Oct 23, 2021 at 21:54
• I am honored! -NN Oct 23, 2021 at 22:48
• @RossMillikan aaaaand that, ladies and gentlemen, is why we must only use metric. :D Oct 24, 2021 at 7:52

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

• Not an advantage in this case, because you've got to convert the sink rate shown on your variometer to meters/sec, and the weight of your glider to kg. Oct 24, 2021 at 2:38
• Unless you know both of these things already. BTW weight is in Newtons, not kg.
– Frog
Oct 24, 2021 at 5:54
• @Frog: Maybe for the pedantic, but I assure you that if you visit Europe (or other metricified) place, and buy things that are sold by weight, the package is almost certainly going to be labeled in kg. I'm also willing to bet that if you look up the weight of European-built airplanes, it will be given in kg. Oct 25, 2021 at 1:00
• @jamesqf It's true, and may be due to the fact that the kilogram was, in the original definition, the unit of weight. In fact, and until the 1960s (more or less) many calculations were done with the MKS system, where the reference unit was the force, its unit being the kilogram-force or kilopond, (weight of the standard kilogram) and the mass was a derived unit, often called UTM (Unidad Técnica de Masa) or TME (Technische Masseneinheit), the amount of mass accelerated at 1m/s2 by 1 kilogram-force... Oct 25, 2021 at 6:40
• @xxavier: See any dictionary for the definition of pedantic :-) Oct 25, 2021 at 17:21