How do I weigh my head?
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Assume I want gram level accuracy, also, the head has to remain intact and attached at the end of the experiment. Methods that rely on inference from volume * density are ruled out.
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Answer:
Here is an idea that may or may not work (this is related to the see-saw idea, but maybe less horribly inaccurate): You stand in the center (on the axis) of a smoothly and freely rotating platform. Your body should be completely immobilized except for your neck and head. Using an array of cameras, the 3D position of your head and neck are monitored continuously. You then start moving your head around in any way that you can. As you do so, your moment of inertia changes, and therefore your rotational velocity changes by conservation of angular momentum (like a spinning ice-skater extending or withdrawing their arms). You can accurately measure the rotational velocity of the platform, and then correlate the changes in the rotation with the captured motion of your head. It would probably be possible to take into account slowing of the rotation due to friction by comparing readings in the same head position at different times. Then the tricky part: The moment of inertia is given by the integral of density x r^2 over all space, denoted Int[mr^2]. What we want is the total mass of the head Int[m]. m is of course a complicated function of r: the density of your head is not uniform. The question is whether you can back out the function m(r) from the large dataset collected with your head in different positions. It might be helpful to also rotate your body about different axes and collect more data that way. If it is possible to extract m(r), then you can easily calculate the mass of your head, simply by integrating this function over the desired region. (You will have the density as a function of position for the entire head/neck region, so you can choose where the cutoff line is between the head and neck).
Jesse Berezovsky at Quora Visit the source
Other answers
From the Guardian: http://www.guardian.co.uk/notesandqueries/query/0,,-1244,00.html But for a less dramatic method, take a seesaw, a ruler, and a large inanimate object. Balance flat on your back on the seesaw, then bend your head forward on to your chest. The seesaw will tilt: shuffle along to restore the balance. Measure the distance you have to shuffle and divide this by the amount by which you moved (the centre of) your head. This gives the weight of your head as a fraction of your whole body weight. Peter Green, Department of Mathematics, University of Bristol.
Azeem Azhar
You can't. Not to gram accuracy and any such measurement would be wrong by the time you had written it down. This is because the mass of the head changes all the time as hydration changes, body fluids are exuded and swallowed or evaporated, hair is grown, etc. A gram is well within the margins of change. (Added after question revised to gram accuracy.) For a practical ball park measurement: Stick it in a full bucket of water and weigh the displaced water. You do this by catching it in a tray, or weighing the bucket of water before and after, and doing the math. (The human body weighs about the same as water because it is mostly water and the remainder balances out as some parts are denser and others are less dense.) I did add a second solution which a couple of comments relate to but it turned out it was rubbish, so I've removed it. Third solution. This will work. ;) Weight yourself, Weight your body only (i.e excluding head). Subtract weight of body from total weight.
Matt Langley
According to scientist Ramaar Norhway, and proven studies of a recent 2012 Human Anatomy Skull Structure, the only way to actually weigh your head properly is to take measurements (circumference of your head) and apply the same measurements to a pumpkin through carving it. Also add exact measurement of your facial features. Than, add at least two pounds or 0.907185kg precisely of left over carvings, to equal the amount your bones and brain weigh in your head. Good luck, Charlie Winston
Anonymous
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