How big of an electric motor do you need to power a six foot (1.8m) Piper Cub model? Finding the answer is easy!
In this series I use a six foot (2 m) scale model of a Piper Cub as an example for performing various simple calculations useful for model airplanes. This is the first article in the series. Here’s a link to the other articles:
Six Foot Cub
Let’s say we decide to design and build a Piper Cub because the world cannot have enough Cubs. We want to make the wing span exactly six feet (1.8 m) because that is the size of our building table. How big of an electric motor do we need to power it? Whoa! Did your mind just go blank? You say you have no idea? It’s not as hard as it looks. Let us take it one simple step at a time until we get to the answer.
Full Size Cub
Look up the specifications for the Piper Cub. For the sake of argument, let’s say that it weighs 1,000 pounds (454 kg). The wing span is 35.25 feet (10.75 m) and the wing area is 178.5 feet squared (16.6 m^2).
The formula for the aspect ratio of a wing is (wing span * wing span) / (wing area). That doesn’t look too hard, does it? The units are not important as long as they are consistent. Plug in the numbers from the specifications and you should come up with the value 7. I know that it’s really a little bit less than that. Let’s just round it out. For the sake of this discussion, the aspect ratio of a Piper Cub is 7.
Given the wing span and aspect ratio, let’s compute the wing area of a Piper Cub as a check on the numbers. The wing area of a wing is (wing span * wing span) / (aspect ratio). Plugging in the numbers, you should get a value close to 178 feet squared (16.5 m^2). Again, we are still talking about the full-size Piper Cub.
Similarly, we can plug in the numbers we have to compute the wing area of our six foot Cub. I get 5.1 feet squared (0.47 m^2).
Airplane Type Constant
There’s a very simple formula that relates the weight, wing span, and wing area of an airplane. The value that you get is an indicator of how that airplane will handle in the air. The formula is (weight) / (wing span * wing area). The units are not important as long as they are consistent. In this example, we will use ounces as the unit of weight and feet (or feet^2) as the units of length and area.
Let us plug in the numbers for a full-size Piper Cub to find its airplane type constant. I get a value of about 2.5. So if we come up with a six foot model of the Piper Cub that has an airplane type constant of 2.5, it will behave in the air similarly to the full-size Cub.
Note that I said similarly, not exactly. The point of this article is to quickly come up with some numbers that can be used as the starting point of a detailed airplane design.
The airplane type constant formula may look very simple, but it is extremely handy. In fact, this is the formula from my first book that I use the most.
Let us figure out what the flying weight of our model Cub needs to be to match the airplane type constant of the full size Cub. Rearranging the airplane type constant formula, we can compute the target weight of an airplane using (airplane type constant * wing span * wing area). For our model Cub, I get a value of about 4.8 pounds (2.2 kg). That is the target flying weight.
We are in the home stretch! My electric model airplane power rule says that for a sport model the motor should be about 10% of the flying weight of the model airplane. I get a value of about 8 ounces (227 g). Since this is a Cub, we could get away with using less.
That’s the final answer: the electric motor weight of a six-foot Piper Cub model airplane is about half a pound.
That was not as hard as it looked at first, was it?