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Prof. Templeton
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I am shocked that you braindenizens don't know the answer. The solution's been determined and there's only one person who can do that. No that's not Chuck Norris.(He could do that though. :P )

Not guessed him yet? :P

Spoiler for He is a gentleman....:

Still no answer? :P

Spoiler for He is a cannibal...:

YEAH! He is the one and only Dr. Hannibal Lecter. :D

He could slice your head, so neatly that a drop of blood won't be wasted. BTW, he loves to have fried-to-a-crisp Cerebral Lobes for dinner. :P

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uh oh, maybe he's a burn victim and he's very sensitive about it. perhaps we should drop the question.

Actually... yes i am :(

No scars, just no hair... eyebrows and all that stuff i have, just no hair hair! :unsure:

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don't be a sissie! :D

Hey, if anyone is able to solve this with a good deal of accuracy, then I would really like to know how to calculate the weight of my breasts (my poor shoulders are aching all the time! :) )

cut em of, put em on a scale, and have a professional put them back on to a size you prefer :D

Voila! Problem solved...

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how about a trip or search for a cryogenic lab where they store whole boddies and some heads too - they must have ooodle of info

edit: btw - prof, we said get ahead not a head - you dont really have to do this

Edited by Lost in space
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If I remember right, the average human head weighs about 15 lbs... so maybe PG's head is full of air and not so dense after all, lol :rolleyes: ... just kidding PG

The 15 lbs came from when my brother was in traction (he's a quadroplegic *spelling?*), they had to counter-weight 15 lbs to keep him from passing out when he was turned in that circlectric bed.

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If I remember right, the average human head weighs about 15 lbs... so maybe PG's head is full of air and not so dense after all, lol :rolleyes: ... just kidding PG

The 15 lbs came from when my brother was in traction (he's a quadroplegic *spelling?*), they had to counter-weight 15 lbs to keep him from passing out when he was turned in that circlectric bed.

hmm and your brother is not dense, does i run in the family? jk
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Weight yourself on ground first.

Wp=Wh + Wb (weight person = weight head + weight body).

We have Wp now.

The human body is mostly water, so it floats in water, but such that only the tip of the head is barely above water. Hence, create a deep enough pool such that when you put a scale at the bottom and stand on it, only your chin is above water. Read the scale (it will be non-zero since your chin is above water; in other words water is not carrying all your weight yet).

this number is Wb' (bouyant weight of body).

Wb' = Wb - density of water*Vb

We can find Vb (volume of body upto chin) by dunking it in water (already mentioned).

Now we can solve for Wb and then Wh.

Forgot one thing: I think we may have to account for the bouyant force acting on the scale as well.

Edit: clarification

Edited by Suicide
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Edit: clarification

Weight yourself on ground first.

Wp=Wh + Wb (weight person = weight head + weight body).

We have Wp now.

The human body is mostly water, so it floats in water, but such that only the tip of the head is barely above water. Hence, create a deep enough pool such that when you put a scale at the bottom and stand on it, only your chin is above water. Read the scale (it will be non-zero since your chin is above water; in other words water is not carrying all your weight yet).

this number is Wb' (bouyant weight of body).

Wb' = Wb - density of water*Vb

We can find Vb (volume of body upto chin) by dunking it in water (already mentioned).

Now we can solve for Wb and then Wh.

Right! i have no idea what you saying... :wacko:

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Weight yourself on ground first.

Wp=Wh + Wb (weight person = weight head + weight body) (Eq. 1).

We have Wp now.

The human body is mostly water, so it floats in water, but such that only the tip of the head is barely above water. Hence, create a deep enough pool such that when you put a scale at the bottom and stand on it, only your chin is above water. Read the scale (it will be non-zero since your chin is above water; in other words water is not carrying all your weight yet). Let's call this number is Wp'.

Wp' = Wh+Wb - density of water*Vb (Eq. 2)

We can find Vb (volume of body upto chin) by dunking it in water (already mentioned).

This leads to a redundant system of equations. I guess the key is to differentiate the coefficients in eq 2.

Am I even on the right path?

Forgot one thing: I think we may have to account for the bouyant force acting on the scale as well.

Edit: Small typo & grammar

Edited by Suicide
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I thought you were onto something for a minute there Suicide but dunkings in water don't really give us a way to differentiate how much of your weight is down to head or body. I reckon there's probably some way of doing it by looking at how angular momentum varies as you spin someone who is lying down and move the center of rotation about, since that depends on weight distribution. But I'm not going near those equations.

If you just knock someone unconscious and place their body lying down level with a set of scales you could get a rough idea since they wouldn't be exerting much supporting force.

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I thought you were onto something for a minute there Suicide but dunkings in water don't really give us a way to differentiate how much of your weight is down to head or body. I reckon there's probably some way of doing it by looking at how angular momentum varies as you spin someone who is lying down and move the center of rotation about, since that depends on weight distribution. But I'm not going near those equations.

If you just knock someone unconscious and place their body lying down level with a set of scales you could get a rough idea since they wouldn't be exerting much supporting force.

Alright everybody! New assignment, go knock out some random person on the street and weight their head :)

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I just saw a clip from Jerry Maguire last night...the human head supposedly weighs 8lbs

I just tried to weigh mine....

either 5.5 or 8.5.

I laid on the ground and rested my head on the scale.

had to use a mirror to read the numbers.

so could that have been 5.5 or 5.8???

LOL

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I thought you were onto something for a minute there Suicide but dunkings in water don't really give us a way to differentiate how much of your weight is down to head or body. I reckon there's probably some way of doing it by looking at how angular momentum varies as you spin someone who is lying down and move the center of rotation about, since that depends on weight distribution. But I'm not going near those equations.

If you just knock someone unconscious and place their body lying down level with a set of scales you could get a rough idea since they wouldn't be exerting much supporting force.

can you take them in space and spin them remove a leg (keeping them alive), check the effect and then the opposite arm, no I'm going random..

It must be inertia check - like crash test dumbies

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First you find out how much times your head can go into the rest of your body. Now you find out your body weight. If it fits in fifteen times, divide your body weight by 15 and so on. (Am I missing something?)

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First you find out how much times your head can go into the rest of your body. Now you find out your body weight. If it fits in fifteen times, divide your body weight by 15 and so on. (Am I missing something?)
Hmm tricky. I could get my head into the rest of my body one time but it wasn't easy. I think I could do that as many times as I wanted to now. Only I don't want to any more.
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Hmm tricky. I could get my head into the rest of my body one time but it wasn't easy. I think I could do that as many times as I wanted to now. Only I don't want to any more.

Wow... this is a really old thread. Hmmm... so how do you weigh your head????

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