Sometime back, I asked the question about how to calculate the cruise power for level flight if I know the L/D, weight, glide speed, sink rate, and battery parameters.

I have lost the answers in my awesome filing system. Would one (or two) of you be so kind as to fill me in on this again?

From : Don Stackhouse

Let's walk through this step-by-step.

If we know the weight and L/D and the airplane is in level flight, then lift is equal to weight and therefore the drag is equal to the weight divided by the L/D.

Work is equal to force times distance, and power is the work per unit time. If we multiply the Drag (the force in this case) times the airspeed, we get power. At that point all we have to do is convert the units of the result to get watts.

For example, let's say the weight is 10 ounces and the L/D is 8:1, at an airspeed of 20 feet per second. The drag is therefore 10/8 = 1.25 ounces. Dividing by 16 ounces per pound, we find that this is equal to .0781 pounds of drag.

Multiplying this by the speed of 20 fps, we get 1.56 foot-pounds/second of power consumed by the drag.

Since 1 horsepower = 550 foot-pounds/second, this is equal to .0028 horsepower.

Since 1 watt = .7376 foot-pounds/second, our airplane is using 2.11 watts to overcome drag.

Now, let's assume the prop is about 70% efficient, and the cheap can motor is only about 55% efficient, for an overall powerplant efficiency of 38.5%. This means that we need 2.11/0.385, or 5.5 watts coming from the battery to get 2.11 watts worth of power actually delivered to the airframe.

So if we combine all of this, we get the formula:

Power req'd.(watts) = Weight(pounds) * L/D * speed(fps) / (.7376 * powerplant efficiency)

The only catch here is that if we're using glide tests to measure the glide ratio and then assume that glide ratio is equal to the L/D, that method is only valid if weight is equal to lift. In descending flight, this is only true if the glide ratio is better than about 5:1. At steeper glide angles, the lift (which is measured perpendicular to the flight path) is not really approximately equal to the weight anymore, since some of the plane's weight is being supported by the upward component of drag. An extreme example of this is a parachute in vertical descent, where lift = zero, and drag is equal to weight.

In this regard, sink rate can be a better measure of power required. The weight of the plane times the sink rate (its vertical speed) is a direct measure of the power being consumed by the plane in a glide. So, if our model descended at 2.5 fps, the power represented is 10/16 = .625 pounds times 2.5 fps, or 1.56 foot-pounds per second. This approach works regardless of L/D. Divide by the powerplant efficiency and convert to watts, and you have your answer. The resulting formula is:

Power req'd.(watts) = Weight(pounds) * sink rate(fps) / (.7376 * powerplant efficiency)

The only other possible major pitfall in these methods is ground effect. Make sure that your measurements are taken when the model is at least more than about one wingspan above the ground. Ground effect comes into play below that altitude, and will give you aircraft efficiency measurements that are substantially greater than in normal out-of-ground-effect flight.

Don Stackhouse
DJ Aerotech