California Girl
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- Oct 8, 2009
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- #1
weigh your own head?
Without using medical equipment.
Without using medical equipment.
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weigh your own head?
Without using medical equipment.
weigh your own head?
Without using medical equipment.
You can only do it once and it requires an ax and a friend. It hurts less if you weigh your friend's head.
Trust me. I'm an engineer and we get paid to know these things.
What if the density of all tissues is not knownWater displacement is the only way I can think of.
Water displacement is the only way I can think of.
Water displacement is the only way I can think of.
What if the density of all tissues is not knownWater displacement is the only way I can think of.
You'd have to assume that the density of the head is, on average, the same as the density of the rest of the body.
I have a rock and a balloon that displace the same amount of water...Water displacement is the only way I can think of.
^^^^^ That.
If you fill a bucket with water (or, in Radio's case, beer) and weigh it, then stick your head in it, and weigh it again... that gives you the weight of your head. I did not know that till today.
Water displacement is the only way I can think of.
^^^^^ That.
If you fill a bucket with water (or, in Radio's case, beer) and weigh it, then stick your head in it, and weigh it again... that gives you the weight of your head. I did not know that till today.
What if the density of all tissues is not known
You'd have to assume that the density of the head is, on average, the same as the density of the rest of the body.
I'm not sure whether that's an accurate assumption.
Also, some bodies are more adipose than others.
I think my method would be more accurate- especially in the cases of lardasses whose excess adipose tissue causes disproportionate displacement compared to persons with greater muscle mass and lower body fat percentageYou would have to assume that your head is the same approximate density as the rest of your body though.
Water displacement is the only way I can think of.
^^^^^ That.
If you fill a bucket with water (or, in Radio's case, beer) and weigh it, then stick your head in it, and weigh it again... that gives you the weight of your head. I did not know that till today.
Actually, I was thinking of filling a tub to the tip-top, getting in and totally immersing yourself. Measure the water that spills out. Then do it again, and this time leave your head above water. Measure the spillage again. Use your weight to calculate the volume to weight ratio, and then the difference in spillage each times could give you your approximate head weight.
You would have to assume that your head is the same approximate density as the rest of your body though.
I think my method would be more accurate- especially in the cases of lardasses whose excess adipose tissue causes disproportionate displacement compared to persons with greater muscle mass and lower body fat percentageYou would have to assume that your head is the same approximate density as the rest of your body though.
^^^^^ That.
If you fill a bucket with water (or, in Radio's case, beer) and weigh it, then stick your head in it, and weigh it again... that gives you the weight of your head. I did not know that till today.
Actually, I was thinking of filling a tub to the tip-top, getting in and totally immersing yourself. Measure the water that spills out. Then do it again, and this time leave your head above water. Measure the spillage again. Use your weight to calculate the volume to weight ratio, and then the difference in spillage each times could give you your approximate head weight.
You would have to assume that your head is the same approximate density as the rest of your body though.
Be expensive to fill a tub with beer.