Given the dimensions and gross weight of an RC model aircraft, how can I find the center of gravity? Can I calculate it, or perhaps there's some way to approximate it?

  • $\begingroup$ It's impossible just using dimensions because it doesn't give any information as to what is inside the structure, for example the location and size of tanks, the APU, seating, cargo, etc. $\endgroup$
    – Ron Beyer
    Commented Jan 3, 2017 at 15:09
  • $\begingroup$ Sorry I was asking about CG calculation for a RC model aircraft can we make assumption from its wing characteristics and its gross weight and engine avionic location to find cg $\endgroup$
    – kcihtrak
    Commented Jan 3, 2017 at 15:15
  • $\begingroup$ You can, of course, approximate it by simply finding the point where the aircraft balances. If your model is small enough, put a broomstick under the plane, parallel to the wings, and figure out how far forward/back to move it to make the plane balance. $\endgroup$
    – abelenky
    Commented Jan 3, 2017 at 15:39
  • $\begingroup$ What If I don't have the plane just data $\endgroup$
    – kcihtrak
    Commented Jan 3, 2017 at 15:41
  • 3
    $\begingroup$ Do you mean "find where the CG is now" or "find where the CG ought to be"? $\endgroup$
    – Steve
    Commented Jan 3, 2017 at 21:38

3 Answers 3


You cannot.

Assuming that by 'dimensions' you mean the values for length, width and height, there is no way to calculate CofG from it. Two planes with exactly the same dimensions, but one with a thicker wings or tail, will have different C of G.

Even if you know the precise shape of the plane, the C of G cannot be calculated. Replacing a wooden tail with a heavier metal one with the same shape would shift the C of G. So would replacing the engine with a heavier one.

You need either the plane itself, or exact plans showing all the components and their weights (and a lot of maths).


The question you have asked is competently answered by DJClayworth.

But I guess you want to ask something different, that is where the center of gravity must be in order to have a well-behaved aircraft. The aerodynamicist's answer would be: Slightly ahead of the neutral point.

First, calculate the areas of both the wing and the horizontal tail. Interpolate between the wing roots to include the section which is covered by the fuselage. These areas are needed for scaling the influence of the wing and the tail.

Next, pick a reference point. The nose of the wing root would be a good choice. Then measure the distance between this reference point and the quarter chord point of both the wing and the horizontal tail. For unswept wings, the quarter chord point of the wing root is sufficient; however, with swept wings you need to find the spanwise station where half of the wing area is inboard.

Use the sketch below for a graphical method: The grey area is the swept wing's planform. First, add one tip chord length at each end of the root chord and vice versa with one root chord length at each end of the tip chord (blue lines). Then connect the ends of the resulting lines (red lines). The spanwise station where they cross is your mean chord (green line). Use the quarter point of this chord (black circle) and carry it over to the root to measure the distance to your reference point.

Graphical mean chord interpolation

Do the same for the horizontal tail. If it is a V-tail, use the projected planform in the horizontal plane.

Now all that is left to do is to interpolate the two distances $x_{NP}$ with the respective areas $A$: $$x_{NP} = \frac{x_{NP_{wing}}\cdot A_{wing} + x_{NP_{horiz. tail}}\cdot A_{horiz. tail}}{A_{wing} + A_{horiz. tail}}$$

Now that you know the neutral point, place the center of gravity at 20% of the mean chord (remember the green line?) ahead of this point. In a normal RC aircraft with unswept wings this is approximately at one third of the chord length from the nose of the wing root chord.


If the model aircraft can actually fly you can narrow down where the center of mass is.

The center of mass will be along the centerline of the aircraft (usually by virtue of symmetry).

When the aircraft is sitting on the ground the CoG will be above the triangle described by the gear. If it has a nose wheel then the CoG will be more towards the rear (so the aircraft can pitch up on takeoff). If it is a tail dragger it'll be more towards the front.

Beyond that the center of mass will most likely be near or just in front of the wing root. That way loss of control will let it remain stable instead.

  • $\begingroup$ The type of landing gear a plane has would have no bearing on its CG. The CG would need to remain the same if you switched a plane from tricycle to taildragger gear. $\endgroup$ Commented Jan 4, 2017 at 12:07
  • $\begingroup$ But it helps narrow down where it is. Not to mention that switching gear types requires moving either the main gear or the CoG so the plane can actually sit on the ground correctly. $\endgroup$ Commented Jan 4, 2017 at 12:16
  • $\begingroup$ You can't safely move the cg to accommodate landing gear. That's backward thinking. The cg is a critical component to the flight characteristics of the plane. $\endgroup$ Commented Jan 4, 2017 at 14:05

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