[Physics] How to measure the mass and balance point of a human forearm

home-experimentmassMeasurements

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? .

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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.

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).
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.
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.
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.

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[Physics] What truly is mass, and is there a direct way to measure it

You ask, "How do we describe mass to the aliens, who don't know about our (g)?" This is an example of a class of questions referred to by Martin Gardner as "Ozma problems." The classic Ozma problem is how we describe to aliens the distinction between right and left; the answer is that we do it by describing the weak nuclear force.

Your statement of your Ozma problem seems a little ambiguous to me. Essentially you're asking how we describe to the aliens a unit of gravitational mass. (You don't say so explicitly, but it seems clear from context that you don't mean inertial mass.) Futhermore, there is a distinction bewteen active gravitational mass (the ability to create spacetime curvature) and passive gravitational mass (what we measure with a balance). Not only that, but your question could be interpreted as asking whether we can compare with the aliens and see whether the value of the gravitational constant $G$ is the same in their region of spacetime as it is in ours.

We can easily establish 1 g as a unit of inertial mass. For example, we can say that it's the inertia of a certain number of carbon-12 atoms.

The equivalence principle holds for us, so presumably it holds in experiments done by the aliens as well. This establishes that our 1 g unit of inertial mass can also be used as a unit for the passive gravitational mass of test particles.

You didn't ask about active gravitational mass, but the equivalence of active and passive gravitational mass is required by conservation of momentum, and has also been verified empirically in Kreuzer 1968. Cf. Will 1976 and Bartlett 1986.

The other issue is whether $G$ is the same for the aliens as for us. Duff 2002 has an explanation of the fact that it is impossible to test whether unitful constants vary between one region of spacetime and another. However, there are various unitless constants that involve $G$, such as the ratio of the mass of the electron to the Planck mass.

A more fundamental difficulty in the fundamental definition of mass is that general relativity doesn't seem to offer any way of defining a conserved, global, scalar measure of mass-energy. See, e.g., MTW, p. 457

Bartlett, Phys. Rev. Lett. 57 (1986) 21.

Duff, 2002, "Comment on time-variation of fundamental constants," http://arxiv.org/abs/hep-th/0208093

Kreuzer, Phys. Rev. 169 (1968) 1007

MTW: Misner, Thorne, and Wheeler, Gravitation, 1973.

Will, “Active mass in relativistic gravity: Theoretical interpretation of the Kreuzer experiment,” Ap. J. 204 (1976) 234, available online at adsabs. harvard.edu.

[Physics] How to you accurately determine your own mass

I'm not sure that it makes sense to try to measure your body weight to a precision of 100 g. For example I was just thirsty and drank a 20 ounce bottle of water, which transferred about 600 g extra mass to my stomach. Even just breathing changes your mass: if you take ten half-liter breaths per minute and your exhalations contain 5% carbon dioxide by volume, that's a mass loss of tens of grams of carbon per hour. (Moisture is probably a bigger effect there, too.) To measure a 50–100 kg mass to a precision of 0.1 kg is a fractional uncertainty of $10^{-4}$, which is about two orders of magnitude better precision than most college-course laboratory experiments. Furthermore you would expect to see changes of several hundreds of grams over the course of the day (which would be interesting, which is maybe why you're asking).

You won't in general find $10^{-4}$ precision in cheap consumer electronics. I'd expect a bathroom scale to have an absolute precision of 1%–5%, or one to seven pounds for a 150 pound person, with poorer quality loosely associated with cheaper scales.

If what you want is a well-calibrated absolute weight with three or four significant figures, the best setup for you is going to be a cantilever system with well-calibrated reference weights. That's what's at your doctor's office — sorry. If it weren't the most cost-effective way to get a reasonably accurate weight, then doctors would buy something else.

If, on the other hand, you're interested in seeing kilogram-level changes in your weight with sub-kilogram precision, you might not need the absolute weight after all. If you can convince yourself that your scale is linear for small deviations from your weight, then maybe you can take your scale's last two digits, the kilogram and decigram digits, at face value. Here's one way you could test that:

Get several similar-but-different sized weights, about the size of the mass differences you're hoping to measure. Brick pieces might work. Label them somehow: A, B, C, etc.
Put a base load on the scale so that it reads somewhere near the value that you're interested in. For instance, if you're weighing yourself, you could stand on the scale and have someone help you with the next steps.
One at a time, add your test weights to the scale. Each one will increment the reading on the scale by some amount. You'll make a data table like
```
reading  load
-------  -----
80.0 kg  just you
81.2 kg  you + A
82.2 kg  you + A + B
```
and so on. From this you can find the mass of each little weight. (This is how veterinarians weigh stubborn cats, but they do it one cat at a time).

Now repeat the measurements with the same base, but with the other weights in a different order:

reading  load
-------  ----
95.2 kg  you plus all your weights
80.2 kg  just you
81.1 kg  you + B
82.3 kg  you + B + A

There are a couple of things that you might learn from this procedure:

One is the random error inherent in each scale. For instance, I made the two "just you" weights different in the last digit. That's not unreasonable: essentially all digital readouts have what's called a "Schmidt trigger" that puts some hysteresis in the last digit, so that it doesn't flicker between adjacent values; however that means that the uncertainty in the last digit of a digital readout is at least $\pm1$ in the final digit.
You might also find that the same brick fragment C reliably takes the scale from 80.0 kg to 82.0 kg with one base load, but from 95.0 kg to 97.2 kg with another base load. That would mean that your scale is "nonlinear," since the same increase in signal gives a different increase in output starting from a different place. You'd have to decide how much this bothers you, if you find it.

This technique doesn't address the question of stability: presumably you're interested in measuring your weight over many days. I'd suggest essentially the same test for measuring the stability of your scale(s): find an inert weight that's close enough to your body weight that you expect the scale to be linear for nearby values, and compare your weight on the scale to that rather than simply to the reading of the scale. Depending on the precision you're interested in, you may still have some strange stuff happen. For instance some electronics will respond differently in humid weather than in dry weather; also some weights will absorb moisture from the air and have different masses in humid weather than in dry weather.

As the saying goes: quick, cheap, or correct, pick any two. You're not going to be able to get the precision that you want without some expenditure of money or time, but you can perhaps get the result that you want a little easier.

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[Physics] What truly is mass, and is there a direct way to measure it

[Physics] How to you accurately determine your own mass

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