# How do I calculate the centre of gravity for the below question?

Textbook Answer for above question: BEM of 25,450kg & the CG is 57 cm in front of the main wheels.

I'm having trouble on finding the CG position. Is there any ways to find the datum of the aircraft in order to calculate arm, moment and centre of gravity position?

• If the CG were above the main gear then the loading would be 24725 on the main and zero on the nose. If the CG were directly over the nose wheel then there would be 24725 on the nose and zero on the main. So the ratio is the loadings gives you the ratio of the distances between each wheel and the CG, in this case 24000:725. By inspection we can see that the CG is closer to the mains than the nose wheel. Is that enough to point you in the right direction?
– Frog
Aug 22, 2022 at 9:40
• At first, I don't quite get it, to be honest. But after reading it a couple of times, I think I get a brief idea. So based on my understanding, we can roughly assume the CG position by using the ratio method to determine where it is likely to be closer to either the nose wheel or mains wheel. Do I understand your explanation correctly? Thanks for your explanation, by the way. Aug 22, 2022 at 13:34
• having a formula is good but I’ve always preferred the idea of having an intuitive understanding of the problem, visualising what’s happening and then applying the numbers. In this case, most of the weight is on the main wheels and a little on the nose wheel so it follows that the CG must be a little in front of the mains. If you get a numerical answer that doesn’t agree then you know there’s a problem.
– Frog
Aug 22, 2022 at 17:56
• Putting the numbers in, the ratio 24000:1450 (sorry I counted the nose wheel loading incorrectly before) applied along the 10m distance gives 9.43:0.57metres so the CG is 57cm from the mains and 943cm from the nose wheel.
– Frog
Aug 22, 2022 at 17:56
• I agree. Well explained, and thanks again for your great help and clarification. Aug 24, 2022 at 13:22

To find the position of the centre of gravity, you need to find the point where the sum of the moments of the weights on the wheels is 0.

$$\sum_{i=0}^n \left[W_i \times \left(x_i-x_{c.g.} \right)\right] = 0$$

You can choose the datum point yourself; the centre of gravity will be found with respect to that datum point.

It often helps to put the datum at one of the axles, because it simplifies the equations.

If we put the datum at the main wheel axle, the equation becomes:

$$\left( 2\times 725 \right) \times \left(1000 - x_{c.g.} \right) - \left( 4\times 6000 \right) \times x_{c.g.} =0$$

Solving for $$x_{c.g.}$$ is left up to the reader, but it should give the correct answer.

• Hi DeltaLima, the formula does work. Thanks for your big help. I have been stucking on this question for quite a while. Aug 22, 2022 at 13:45
• Hi again, I'm not very familiar with the formula that you used because I'm quite new to the topic of the centre of gravity, so I don't have a strong foundation. Is there any recommended place where I can better understand how the formula works and practice? Aug 22, 2022 at 15:08
• @ChuaRenyu - In English, the formula says "the sum of all moments is zero" and "moment is weight times lever" and "lever is distance from c.g." Don't memorize the formula, learn the meaning. Aug 22, 2022 at 21:44