But let’s say I divided the head up into a bunch of small cubes (e.g. 1 cm x 1 cm x 1 cm). And I knew the density of each cube. Could I then calculate the mass of the head after the overall volume is measured?
If so, then all you have to do is “divide” the head up into a bunch of small cubes and measure the density of each cube. Is there a technology that will do this? X-Ray? MRI?
One still needs assumptions about the isotopic composition of the head. X-rays, for example, only measure the proton (or, equivalently, electron) density. A block of carbon-12 and a block of carbon-13 look the same to an X-ray, but one is 8% more dense. Even assuming typical isotopic ratios for each element, one still needs to know the elemental composition since the proton-to-neutron ratio is not the same for all elements.
Muon radiography was another thing I toyed with here, but it has the same proton/neutron problem, as muon scattering depends only on the density of the protons.
One thing that is immune to this problem is neutron scattering. Neutrons are agnostic to whether they bounce off a proton or a neutron. However, it’s pretty tough to measure the outgoing neutrons, and one would need to average over much more than a lethal dose’s worth of data to get anything competitive.
Here’s a good source from 1969, although I think it’s been superceded:
" Charles E. Clauser " # “and” #
" John T. McConville " # "and" #
" John W. Young "
,
title = "{
Weight, volume, and center of mass of segments of the human body
}",
institution = " Aerospace Medical Research Laboratory (AMRL) ",
address = " US-OH ",
type = " Technical Report ",
number = " AMRL-TR-69-70 ",
year = " 1969 ",
keys = " biomechanics, antropometry, body model, data "Heck, if you divided the head up into one head, that’d work, too, if you knew the density.
And on further thought, I’m not sure that even measuring the gravitational field would be enough, unless you had some means of probing the field inside the head as well as out (gravitational lensing of some sort of penetrating radiation might work). Suppose you took a person of uniform density. Now take a sphere of that material in the neck, with center just below the cut plane. Shrink that sphere down to a smaller uniform sphere with the same mass. You’ve just removed some mass from the head, without changing the gravitational field anywhere outside the original sphere. Nor the moment of inertia, nor buoyancy, for that matter.
If you perform a tracheotomy on the subject, you don’t even need to rig up a complicated breathing mechanism. Hey, maybe Anthony Federoff has a useful career ahead of him after all!
All right, I’ll admit it. I don’t get this. A song lyric? Internet subculture in-joke?
There’s a distintion between calculate or extrapulate the weight of my head, and actualy weighing my head.
Psst… post #58. A little bit of Canadiana for you.
Ooops - thanks! I’d overlooked that.
Find someone of the same gender and approximately same age as you who has no bone wasting diseases (unless you do) and who has a head slightly larger than yours (to make sure we don’t come up short) and hair similar to yours (in thickness) but longer than yours (I’m assuming you want to measure the hair since it is part of your head). Make sure that they don’t have any glass eyes or other prosthetics (unless you do). Make sure that they have the same number of teeth as you. Cut their hair to the same length and style as yours.
Now make a plaster mold of your head. After the mold dries, weigh it. Call this weight y. Now remove the head of the person you found. Without letting it drain, put it into some sort of powerful blender smasher thing which will render a completely homogenized pulp which is of equal density throughout. Fill the plaster mold with the pulp. There should be a little left over since we started with a slightly larger head. Weigh the pulp filled mold. Call this weight x.
x - y should be a good approximation of the weight of your head.
Actually, since your method creates a mechanical average of head-density, if you had trouble finding a head slightly larger you could just use two heads. You would just end up with more leftovers.
True, but now that I think about , I’m not sure that this method would work too well any way. It doesn’t account for air pockets such as in the sinuses, mouth, nose, ears, etc. Those would have to be accounted for somehow. The only way I can think of would be to fill your mouth, sinuses ears, nose, etc. with some of the pulp then remove it, weigh it, and substract that weight from your final x - y value.
Couldn’t you probably get a pretty accurate measurement by putting your head in a bucket of blended head to see how much it displaces, then weighing the displaced blended head?
What about using rotational inertia, spinning the body along different axis’?
Now this I rully, rully like, if only for the notion of blended head. Just make sure it’s only blended and not puréed so it retains the proper proportion of air pockets and voids. Should have a nice heady aroma, though.
Possibly, but this still doesn’t account for the air pockets.
Hi guys…
Really neat physics problem here.
I THINK I came up with a good way, but I will have to double check my math first.
This is physics at its best.
Simple method, probably as accurate as anything else you could do, and MUCH easier than all those fancy, smancy high tech methods. And you dont even have to measure how much water is displaced.
I’ll double check my math and hopefully post the answer in a day or two
take care
Blll
Just don’t let it go to your head.
Imagine you’re suspended from your scale. You are holding a balloon full of air in your hands. So the downward force is equal to your weight + the balloon’s weight. Now you hold the balloon over your head and lower yourself, balloon first, into water until the balloon is completely submerged and pushing up against your hands. The balloon displaces water and you are bouyed up by the weight of the water displaced. Your scale now reads your weight + the weight of the balloon - the weight of the water displaced by the balloon. The water displaced by the balloon weighs far more than the balloon, so the net effect is that your scale reads significantly less than it did. A delta of much more than the weight of the balloon.
Enjoy,
Steven