How big of an electric motor do you need to power a six foot (1.8m) Piper Cub model? Finding the answer is easy!

#### Article Series

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:

### Part 2: A Power System for our Piper Cub

### Part 3: Flying and Landing Speeds for our Piper Cub

#### 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.

#### Flying Weight

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.

#### Motor 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?

Good website Carlos. Thanks for keeping us up to date on the hobby! Is KQEO still on the air??? Loved that station when I lived there…

Thank you for visitng the website. It doesn’t sound like KQEO is still around (https://en.wikipedia.org/wiki/KQEO).

Oh well…should have looked it up before I asked. That was our hoppin’ and boppin’ station when I lived there during High School. Thanks again…

Thank you. This is very well explained – it is not often you get this much of the math in one place, and all explained in a way that makes sense.

I am assuming that the difference between ‘similarly’ and ‘exactly’ has to do with something about scale (the same math for an 18″ span plane has different results than a full sized cub) as well as little details like airfoil, etc. but this certainly helps me in looking at a plane I know the handling of and relating it to an idea or plane that I am thinking of.

Thanks again.

Why, thank you! Spread the word! LOL

Air molecules are small, but not small enough that the effects of their size can be ignored. When you get down to the size of a model airplane, air just behaves differently than a full-size airplane.

Thank you for the information, but I am still in the dark since your calculations are specific only to motor weight. Now tell me how much “POWER” is needed. Can this be expressed as watts / #? or watts / sq ft (cm) of wing area?

Tks

Matching the power requirements of an airplane to motor weight is a lot easier than matching it to a given power output. That’s the beauty of my formula. It’s easier to remember and easier to work with. Of course, for a better estimate, use my calculator.

As I explain in my Model Airplane Design book, my formula is roughly equivalent to 75 watts per pound.

[…] a previous article, we figured that our Piper Cub model needed to weigh about 4.8 pounds (2.2 kg) to match the flying […]

[…] Part 1: Quick and Easy Model Airplane Sizing […]

Propeller, electric outrunner, speed control needed

I have a requirement for at least four systems where I want a 22 inch pusher style fan blade to run in a stationery fan ring that it about 10 inches deep. I need to move the maximum number of CFM in free air or with some static differential across the fan ring or about 75 Pascals. I need the props, motors and controls that will allow me to run the fans from near zero to maximum speed using a control knob and/or a control signal. I want to move in excess of 5000 cfm and would prefer to go up as high as I can go with a limitation of 2000 Watts of input power on the upper end. The batteries do not have to be super light since they will not be airborne but should be in some kind of pack that would allow them to be easily carried around. The system should also run with each single fan connected to 120 or 240 V AC maximum 2000 Watts by means of some kind of rectifier/charger arrangement.

I realize that the application is a bit out of the ordinary but believe that all the components are currently available and it’s just a matter of selecting the correct ones for my application. The weights are not so important although the motor should be less than 7 lb.

On batteries, the system should run at full speed for at least 20 minutes and while plugged into mains power should run at full speed for an hour.

Colin Genge

Since this is not directly related to the article, I have moved the comment and my response to the forums:

http://forums.rcadvisor.com/index.php?p=/discussion/16/propeller-electric-outrunner-speed-control-needed#Item_1

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