I've been trying to find this out for about a month now. I'm usually met with humorous but not really helpful comments about chopping my own arm off. I'm preferably looking to do this on a budget using common household items, but I'm willing to pay good money if required.

I've been told elsewhere that you can't use a kitchen scale. Explanation is " If you consider your arm as a rigid rod with a center of mass located at distance x from the elbow, then the torque about the elbow is mgx. If the scale is placed at a distance y from the elbow, it provides a torque equal and in the opposite direction and the weight measured by the scale is mgx/y. So, naturally as you decrease y, the force measured by the scale increases. But this doesn't tell you anything about the distribution of weight along the arm. The measurements will be the same for any distribution that has the center of mass at the same location."

There are a few solutions from a similar question. I lack the rep to comment on that, hence this question. - [Link][1] Can I use either of these methods to measure mass distribution and centre of mass a limb, by measuring the limb in segments? I'm not sure what the margin of error is for henne's or emitabsorb's solutions. Would either method likely be more accurate then just using the average mass data, as given in most biomechanics textbooks as a percentage of total body mass?

Edit. Just further question, based on the comments.

I have a decent estimate of the centre of mass of my forearm using the archimedes volume measurements, in 5mm segments. Just not the mass of forearm, which is obviously the crucial component. I could then estimate the mass distribution, using historical data on the density of forearm muscle/bone mass.

I'm just having issues with steps 1 and 3, if anyone can help. This is beyond my basic comprehension. .

For step 1, I couldn't find anything about lying on the plank, so I wasn't sure of the equation for the measurements I took. I lied on a plank, balanced by a scale and books on either end, and measured my forearm and hand when vertical. The readings on the digital scale jumped a lot, so I took an average from ten measurements. The weight rose by 0.05kg for the hand, and 0.52kg for the forearm. I know that the length of my forearm is 26.3cm and the centre of mass is 10.2 cm from the elbow.

For step 3, I think I understand the formula for step 3. It's just Mr^2. With R being the rotation of axis/gyration. Do I have to subtract the MOI of the arm from the MOI of the entire body?

I'm also wondering how to estimate the Moment of inertia about the centre of mass, using the measurements I've taken for the mass distribution in 1cm segments. Is it (M1L1^2+M2L2^2...)(MtotalH^2) Sorry for the layout. Having issues with Latex at the minute. M being mass, L=Length and H= Balance point.

[1]: Practical method to weigh human limbs with common household items? .

  • $\begingroup$ What about this article? people.unica.it/brunoleban/files/2012/04/… and this one to check if your numbers are of the magnitude? dtic.mil/dtic/tr/fulltext/u2/710622.pdf $\endgroup$
    – Farcher
    May 22, 2016 at 20:27
  • $\begingroup$ Thanks very much. Yep, I'd read both of them. The only methods in the first study that haven't been mentioned here, are only possible on human cadaviers. The figures in the second link are for cadaviers again. I'm going to compare my results from all the suggestions below, then against the human data from this link. exrx.net/Kinesiology/Segments.html . (de Leva, 1996) Though, I agree with Han-Kwang's reasoning for paying less attention to these figures. Also, most studies I've checked, cadavier or human, are smallish sample sizes of one population group. $\endgroup$
    – Kief
    May 24, 2016 at 19:14
  • $\begingroup$ Someone's been hitting the gym. 😏 $\endgroup$ Jun 2, 2018 at 16:39
  • $\begingroup$ I wish, tendinitis in my elbow has kept me from hitting the gym for 3 years now. $\endgroup$
    – Kief
    Jun 3, 2018 at 12:33

1 Answer 1


The method with measuring the instantaneous weight while swinging the arm with a fixed angular velocity sounds rather impractical. You might try to combine data from multiple sources. I assumed that you are talking about the bit of limb from elbow up to and including the hand, with the hand stretched, but it turns out that you wish to exclude the hand. The steps below apply to the forearm including the hand; you'd have to repeat it for the hand, with the wrist as a pivot point, and make subtractions accordingly.

  1. Measure the product WL of weight and distance from elbow to center of gravity, by using method #1 in the answer that you referred to. (Put a plank on a pivot on one side, on a scale on the other side, lie down, compare the weight distribution with the arm flat versus the forearm vertical).

  2. Measure the volume distribution of your forearm using the Archimedes method: fill a deep bucket of water, placing the bucket on a scale, and measuring the increase in weight as you insert more and more of your arm into the water. This will give you the cross-sectional area A(x) of your arm as a function of distance x from the elbow.

  3. Measure the moment of inertia of your arm by sitting on a frictionless swivel chair (alternative: plate on marbles on a hard floor) and see how your body rotates in the opposite direction if you rotate your forearm in the horizontal plane. This will also require that you measure the moment of inertia of your entire body, but that is a relatively simple problem. The moment of inertia is $I=\int A(x) \rho x^2\,dx$, where $\rho$ is the density of your arm.

  4. Combine the data from the previous three points. To use the volume distribution from #2, you will have to make the assumption that the density of your forearm is constant over its length, but that's a better approximation than assuming that your forearm has the same density as the rest of your body from biomechanics text books, including lungs and intestines. You can combine the previous results in different ways to find your answer; the difference between the different methods will be an indication for the accuracy.

  • $\begingroup$ Updated answer (intro paragraph and step 4). $\endgroup$ May 24, 2016 at 6:33
  • $\begingroup$ good answer, I immediately thought of Archimedes principle, but did not think of segmenting the measurements to get a measure of density. The segmentation should also give the balance point. $\endgroup$
    – anna v
    May 24, 2016 at 7:24

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