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Model Airplane Design Made Easy
Many years ago I read a series of articles in Radio Control Modeler Magazine by Chuck Cunningham. These inspired me and my brother into a very rewarding hobby of building and designing model aircraft. We couldn’t believe that with just a few simple formulas we could design and build our own models. I have kept this series of articles over the years and still follow these simple principles in model airplane design today.
Let’s talk about a basic design for a model airplane. When first approaching a concept design, you need to think about a few questions about how you want your model to fly. Will the role model be a trainer, a sportsman, an intermediate or any number of combinations? How big of a model do you want to design and build? What kind of energy are you going to use? Will the model be gas or electric? I will use very standard assumptions to arrive at a model that, if built and balanced correctly, will fly with almost no trim changes on first flight – just the basic design.
Let’s start with a basic gas-powered model with a 60″ wingspan and 12″ cord, powered by a .60 cu. in. 2 stroke engine. The numbers will be almost the same for an electrically powered aircraft or an aircraft with a smaller or larger engine and wingspan. So get out your calculator and follow.
Wing area and aspect ratio
The area of the wing is nothing more than the length of the wing X the chord. It will be a constant chord kite 60″ long by 12″ wide or 60″X12″ = 720 in². Then the aspect ratio or wingspan squared divided by the wing area (AR=B2/S) will give a basic idea of the model’s flight characteristics. Higher or lower will determine if the model is a float or a brick. It also helps determine how much power the model needs to fly. Using the formula and values so far:
B2/S=AR or 3600/720=5 or an aspect ratio or 5 to 1 (5:1)
Most sports models have an aspect ratio between 4:1 and 7:1. Below 4:1 and you become a NASA test pilot and above 7:1 you get a glider type model. Using the chart in Figure 1, an aspect ratio of 5:1 results in a model with good overall maneuverability and smoothness. So, with the assumptions above, a 60″ wingspan with a 12″ chord, our total wing area is 720 square inches and the aspect ratio is 5:1. With that in mind we will continue with the basic fuselage and use the assumptions to arrive at the basic overall dimensions of the fuselage.
Basic fuselage design
With our assumptions from above, we are ready to lay out the basic fuselage design. To keep this as simple as possible in the design of a simple sport plane model, we have established the model’s wingspan and drawstring. We’ll assume the fuselage will be 75% of the model’s wingspan and our formula will be 75% of 60″ or .75 X 60=45, so our overall fuselage length will be 45″. If we look at the side view of our model, we know that the fuselage is basically made up of two different components, the nose and the tail with the wing somewhere in the middle. In our example, we’ll use a 20% nose length, 11″ or the distance from the trailing edge of the propeller to the leading edge of the wing. At this point, we won’t worry about the C/G as we will discuss later in the model design For the tail moment we will double the nose moment or 40%, 18″. This length is the distance between the trailing edge of the wing and the leading edge of the horizontal stab. Yes, I know to be pure in design, the length should be from Propeller Aft to Wing C/G and Tail Moment should be from Wing C/G to Tail C/G . This would involve more calculations to arrive at the desired results. I just try to keep it simple.
Over the years I have assumed that the horizontal stabilizer is in the range of 20% to 30% of the wing area. I generally use 22-23% in my designs. Please note that deltas and fly wings are different designs and require different considerations. With our assumptions above, we will use 22% of the wing area. So 22% of 720 is about 158 square inches, and we’ll assume that’s 3 times the cord or 3C. We’ll round these numbers up and use just a little math, so our chord will be 158/3=52 and the square root of 52 equals our 7″ chord. The stabilizer span will be 3C or 7″ XC = 21 “. Our stabilizer now has the decreases of 21″ X 7”.
The vertical drift
Again over the year I assumed the vertical aileron was within a range of about 1/3 of the horizontal stab area. I usually assume it goes from the top of the horizontal stab to the top of the vertical fin. So again, with just a little math we can come up with some basic designs. The horizontal cutter has an area of 158 square inches. So 1/3 of 158 equals approximately 52 square inches. Using the square root or 52 we arrive at a vertical fin height of 7″ and a bead of approximately 7.25″. Kind of an ugly looking plane, so just adjust the height and drawstring to come up with a set of numbers that will keep approximately the same area for the vertical fin. A dorsal fin could be added to increase surface area and decrease the height and width of the fin profile.
In the next article, we will look at the rest of the design dimensions, as there are still other assumptions to consider. So far we have our basic design with a 60″ span and 12″ cord. Overall, the fuselage length is 45″ and the nose moment is 11″ while the tail moment is 18″. The horizontal stabilizer measures 21″ X 7″ and the vertical stabilizer measures 7″ long. high. We still need to consider the elevator, aileron and rudder area. Also the thrust lines as well as the incidence and of course the center of gravity. Later I will discuss how to present the design in CAD and how to design a good simple and flying airplane model.
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