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How to measure the mass and balance point of a human limb?


davekm

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Preferably the forearm. I can't spend big money on x-rays, nor can I chop my arm off :D  I've asked this elsewhere - LINK

I know the volume of my forearm, using the Archimedes method, for 10 equal segments of my forearm.

I've also looked at x-rays to work out roughly tissue distribution. Only issue, in terms of precision, is there is quite a big difference in density between bone, fat and muscle. On average, bone is 1.7500 g/cm and muscle is 1.0599 g/cm3.

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Isn't there also going to be a big difference between people's forearms? Or is this strictly to measure only your forearm?

Also, please describe what you mean by "balance point". What is the forearm being balanced on the point of?

Finally, why do you want to know this? Is there a practical application you had in mind, to help focus responses?

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Pretty hard without chopping your limb off, however:

Using the principles of a "Lamboley test"(it has been almost 30 years since I was involved with one) which was devised to find the c.g. and moment of inertia of small sailing dinghys. (to control weight distribution and therefore costs and durability of the boats, as lightness in the ends was preferred in wave conditions)

So if you had a pendulum apparatus you sat "fixed in" and testing at two different height settings...Knowing the mass distribution of the apparatus you should have enough information to know your mass, the height of the centre of gravity of your mass in that position, and the radius of gyration/moment of inertia with respect to the axis of rotation.

Then, rotating only your one limb to a different position on the plain of rotation do the two tests again, and find the difference. That should give you almost enough to tell you the c.g. and moment of inertia  of your limb with a few assumptions and calculations. You might need extra data points as you don't know the mass of the limb as I would when moving a fitting or lead corrector weight to make the boat "class legal".

Less accurately (maybe more accurately if you can't sit rigidly during the swing test?) you could do a static test on a board just to check the change in c.g. of your "system", but with two unknowns, mass and position, you need more assumptions.

Edited by J.C.MacSwell
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It is for a biomechanical computer simulation. I only need to measure my forearm. I only need the measurement for the forearm solely. Not forearm+hand or upper arm. Sorry, I should of said center of mass, that's what I mean't by balance point. @J.C.MacSwell I will take a look later, thanks!

Edited by davekm
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2 hours ago, davekm said:

It is for a biomechanical computer simulation. I only need to measure my forearm. I only need the measurement for the forearm solely. Not forearm+hand or upper arm. Sorry, I should of said center of mass, that's what I mean't by balance point. @J.C.MacSwell I will take a look later, thanks!

In practice what I am suggesting may not work as experimental errors may be higher than simply estimating from what you have already done...but in theory you could get the info for forearm plus hand...and then get it for just hand and subtract the difference...this all assuming your interests are from joint to joint....so get results for body...rotate at elbow and so get difference for forearm plus hand...then rotate at wrist to get results for hand,

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13 hours ago, J.C.MacSwell said:

Pretty hard without chopping your limb off, however:

Using the principles of a "Lamboley test"(it has been almost 30 years since I was involved with one) which was devised to find the c.g. and moment of inertia of small sailing dinghys. (to control weight distribution and therefore costs and durability of the boats, as lightness in the ends was preferred in wave conditions)

So if you had a pendulum apparatus you sat "fixed in" and testing at two different height settings...Knowing the mass distribution of the apparatus you should have enough information to know your mass, the height of the centre of gravity of your mass in that position, and the radius of gyration/moment of inertia with respect to the axis of rotation.

Then, rotating only your one limb to a different position on the plain of rotation do the two tests again, and find the difference. That should give you almost enough to tell you the c.g. and moment of inertia  of your limb with a few assumptions and calculations. You might need extra data points as you don't know the mass of the limb as I would when moving a fitting or lead corrector weight to make the boat "class legal".

Less accurately (maybe more accurately if you can't sit rigidly during the swing test?) you could do a static test on a board just to check the change in c.g. of your "system", but with two unknowns, mass and position, you need more assumptions.

As far as I can tell, that's the experiment you would do to determine the mass, and balance point of an arm...

after you remove it from the body.

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2 hours ago, John Cuthber said:

As far as I can tell, that's the experiment you would do to determine the mass, and balance point of an arm...

after you remove it from the body.

right...and if you attach your removed arm to the boat and rotate it to different positions and make a few tests and assumptions you can glean that information as well...and then start to wonder why you chopped off your arm...

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On 6/14/2018 at 10:15 PM, J.C.MacSwell said:

right...and if you attach your removed arm to the boat and rotate it to different positions and make a few tests and assumptions you can glean that information as well...and then start to wonder why you chopped off your arm...

True, but dull.

OK, for the benefit of the OP, imagine I'm lying on a bed which is mounted on an axle perpendicular to my "height"- the line between my head + my feet.

We know my mass is 70 Kg and I'm 170 cm high.

I can shuffle up + down the bed until I'm balanced with (for example) my arms by my sides.

And if I raise my arm above my head I make that end of the bed heavier so the bed tilts.

And I guess we can put a mass on the other end of the bed to bring it back to balance.

Or, I could stick my arm out to the side and so on...

If I'm prepared to do silly exercises like this how do we answer the OP's question?

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13 hours ago, RjMaan said:

Actually there is not a direct way for the measurement of mass but there is an indirect way by using the gravity. You have to fill the small bucket with water and then put your limb in it that you want to measure. Some or most of the water will pour out because of putting your limb in to the bucket. If you know the exact weight of the bucket will full water and after putting your limb in bucket, by subtracting them will give you the weight of your limb. You can also check the more appropriate way from any authentic source like Wikipedia, Qanda or any other authentic source.

No.

Imagine I made a statue of my arm from lead.

When I put it in the bucket, it will displace just the same weight of water as my real arm does.

So the weight of water will be the same, but a lead copy of my arm is obviously a lot heavier than the real arm.

 

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4 hours ago, John Cuthber said:

No.

Imagine I made a statue of my arm from lead.

When I put it in the bucket, it will displace just the same weight of water as my real arm does.

So the weight of water will be the same, but a lead copy of my arm is obviously a lot heavier than the real arm.

 

Need more info...

Stick yourself underwater on a scale, adding weight if necessary to hold you in place and get a reading. Then stick the part of your arm out of the water that you wish to weigh. The difference will be the weight of that part of the arm, plus the weight of the displaced water. So then deduct the weight of the displaced water and you have the weight of that part of the arm.

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On 25/6/2018 at 12:52 AM, J.C.MacSwell said:

Need more info...

Stick yourself underwater on a scale, adding weight if necessary to hold you in place and get a reading. Then stick the part of your arm out of the water that you wish to weigh. The difference will be the weight of that part of the arm, plus the weight of the displaced water. So then deduct the weight of the displaced water and you have the weight of that part of the arm.

How would you differentiate between a lead or a flesh arm? Both will continue to weigh down on the scale and the Archimedes force only depends on the displaced water, which is equal in both situations.

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32 minutes ago, Bender said:

How would you differentiate between a lead or a flesh arm? Both will continue to weigh down on the scale and the Archimedes force only depends on the displaced water, which is equal in both situations.

It made sense when I wrote it...I better try again +1

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