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Weigh your own head

...without removing it.
  (+22, -10)(+22, -10)
(+22, -10)
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A method for accurately finding out the weight of your own head, without taking the obvious - but terminal - route of removing it first.

This will find a true weight accurate to 1g of your head, from any chosen point on the neck.

It will not be an approximation using any displacement method.

Of course, the real point of this is for someone to suggest a method of actually doing this...

darsy, Aug 04 2000

Brain density changes during renal replacement http://www.sin-ital...12n3/ronco/ronc.htm
[AllenGrace, Aug 04 2000]

High acceleration and the human body http://www.distant-...98_feat_gershon.htm
[AllenGrace, Aug 04 2000]

Density of the human head http://www.ece.nus....elelilw/Handset.htm
Search for the above string of text to find the correct point in the article [wagster, Nov 10 2004]

[link]






       What, do you want us to do your homework for you?
StarChaser, Aug 05 2000
  

       I'm much more interested in the weight of thoughts and whether certain thoughts weigh more than others.
BMolly, Aug 05 2000
  

       I imagine this is best done by measuring the momentum of your head. Have a device clamp your shoulders still and vibrate your head back and forth. By measuring forces, velocities, and frictional losses, it should be pretty easy to determine the mass.   

       It doesn't sound very comfortable, but it surely beats decapitation.
egnor, Aug 05 2000
  

       Lie on a plank of wood. Your assistant puts a fulcrum underneath the plank, directly under your neck. Weights are put on the end of the plank above your head to make it balance. Then,

[added weights]*[distance of added weights from fulcrum] + [head weight]*[distance of CoG of head from fulcrum] = [body-head weight]*[distance of body-head's CoG from fulcrum]

We could assume that the Centre of Gravity of the head is halfway along, but we don't know the centre of gravity of the body-head, so this equation is insoluble.
A more sophisticated approach would be to use the same apparatus, but to balance the plank (by adding weights) at every point (for example, every 1cm) along the body. This would give you at every point the relative moments given by the masses of the top half and the bottom half. Correct application of some Clever Maths™ would then allow you to plot distance along body against mass of that 1cm chunk of body and from this taking the area under the curve for the lenght represented by the head would give you the mass of the head.
hippo, Aug 07 2000
  

       I'm afraid that would be rather innacurate, The Thing, as much of the weight of your head would still be supported by your body. You might get a better guess by suspending yourself from the ceiling by your feet, bringing your head up to your chest as far as possible, then lowering your self to the point of contact with the scales on the floor, and relaxing your neck muscles. There will still be a residual amount of your head weight supported by your body, though.
Lemon, Aug 11 2000
  

       Not really, the thing, but it could spark some obscure form of the "Size Matters" debate.
BigThor, Sep 06 2000
  

       I still think the original poster wanted us to do his homework for him.
StarChaser, Sep 07 2000
  

       Using NMR you could measure the proton density within each 1mmx1mmx1mm voxel of your head. Then do the same for carbon and calcium and whatever else is in there (can you do this?). Then calculate the mass in each voxel and sum to get the total head mass. NB. This would only work if you don't have any large metal objects in your head, for which decapitation is the preferred method. However this would occur as a by product of the NMR scan anyway because of the magnets so it's a quite general method.
lubbit, Mar 27 2001
  

       Take yourself to a zero gravity environment (not sure if this is necessary, but you might as well) and start rotating. Soon you'll be rotating about your centre of mass. Then remove a leg and recalculate the position of the centre of mass. Someone else can do the physics, but there must be an equation for this. Thus the cost of measuring your head's mass has been reduced from certain death to mutilation. You could do the other leg as well and then average the results. Of course, I would check that there really is an equation first.
lubbit, Mar 27 2001
  

       How about finding out your head displacment by putting it in a bucket of water and then using a sonogram/cat scan / x-ray to find out the average density of you head and simply multiply. I think the technology for finding out the density of stuff is already there 'cos I think they use it to test GOLD and other precious metals and to check cast steel girders for flaws.
CasaLoco, Mar 27 2001
  

       No displacement method? I was lower yourself, hanging from a scale, headfirst into a big graduated cylinder. Pause to measure the volume of the water in the cylinder, and the weight registered on the scale, right before one's head enters the water, and right after. You'll have to read up one your Archimedes for the rest of it.
nick_n_uit, Mar 28 2001
  

       Having just worked this out on a bit of paper with 9 variables, it turns out nearly everything cancels. The method is - Stick a bucket of water on some scales. Measure the weight. Stick your head in the bucket of water (don't touch the sides). Measure the weight (or get someone else to do it, I guess). Take first measurement from second. That's the weight of your head.   

       Is this intuitively obvious? I'm not sure.
Tomb, Mar 28 2001
  

       No, it's just wrong. I can't see what's wrong with the decapitation method myself.
DrBob, Mar 28 2001
  

       Tomb, I'm pretty sure that your method would just measure the *volume* of your head (divided by the density of water), not the *weight* of your head. Imagine if you did it with a heavy stone in your mouth --- it wouldn't change the result. nick_n_uit's solution has the same problem.   

       I'm pretty sure a sum-of-moments technique like Hippo's would work (it can also be done by spinning-in-zero-G a la lubbit's suggestion). First, you find your body's center of mass (usually near your waist). Then, add weights above your head to move your CoM to your neck. You end up with a system of three equations (the moments of your head, upper body, and legs around your waist; the moments of your head, upper body, legs, and the weight above your head; and your total body mass) and three unknowns (the masses of your head, torso, and legs). This can be solved using existing technology (er, algebra).
wiml, Mar 28 2001, last modified Mar 29 2001
  

       I liked Dewey's method from Malcolm in the Middle: Lay your head on a bathroom scale, and then look up really quick to see what it reads. It didn't work (fter repeated attempts), but its fun to watch.
nick_n_uit, Mar 29 2001
  

       Yeah, my bad. You'd just find out how much your head would weigh, if it was made of water. Not terribly useful.   

       I'm sure there must be an easy way, though.
Tomb, Mar 30 2001
  

       I've thought this one over before - without displacement or fulcrums, this is one way. Lie on a bench / bed etc with your head overhanging the end. Put a sling around your head with a rope to support it from the ceiling, with a spring balance / whatever attatched. tighted the rope till no slack is left. Now have muscle relaxants injected into your neck so it hangs free and the natural weight of your head should bear down on the sling. I think. *cross fingers*
Jharik, Mar 30 2001
  

       Locate someone with the same size head as your's and remove it. Then use combination of displacement and positron emission tomography, being careful to allow for energy/mass loss from emitted positrons. Or weigh it. Yeah, better: just weigh it.
Flooglehorn, Apr 02 2001
  

       If you are an ice skating champion, go down to the ice rink, and spin on the spot on the ice at a constant speed: then keeping the rest of your body still, tilt your head as flat as you can - this will change your angular velocity. (Cons. ang. mom., lalala).   

       Then: repeat the process, but this time lifting your left arm (from being vertically beside your body) up to the horizontal. This too will change your angular velocity.   

       From these three figures and a few simple anatomical measurements, you can then roughly deduce the ratio of the weight of your head to the weight of your left arm. Then it's simply a matter of cutting your arm off and weighing that to get an absolute figure.   

       Simple.
nickpelling, Jun 05 2001
  

       It may or may not be coincidence, but this very question appears in the 'Readers' questions' section of yesterday's Daily Mail (UK).
angel, Jun 06 2001
  

       Well, what's the answer?
jutta, Jun 08 2001
  

       No suitable answer was given.
angel, Jun 09 2001
  

       What if you were to surgicaly remove the bones + muscles from your neck but leave the nerves and blood vesels intact. Then you could weigh your head withought your body suporting it.
RobertKidney, Jun 10 2001
  

       Of course, you'd have to define pretty carefully where your "head" ends and "everything else" begins. Would you include the atlas bone? A vertebra or two? And the weight changes; your head would weigh more if you hadn't had a haircut in a year (or shaved your beard if you were male). It's like trying to determine the weight of your hand without removing it. (If you remove it, do you include the wrist and/or any untrimmed fingernails in the definition of "hand?"). Sorry, I just don't see a weigh.
whatsbruin, Aug 31 2001
  

       whatsbruin: the original idea specifies "from any chosen point on your neck".
wiml, Sep 01 2001
  

       Is there a Coroner in the house?

X Files Season 4: "Leonard Betts" SCENE 4
(Examination room. SCULLY places BETTS' head on a scale.
It reads 10.9 pounds. She carries it to an autopsy table and picks up a tape recorder.)
SCULLY: Case number 226897, Leonard Betts. As remains are incomplete all observations refer to a decapitated head. Weight: 10.9 pounds. Remains show no signs of rigor mortis or fixed lividity.
Nor do the corneas appear clouded which would seem inconsistent with the witnessed time of death now … (checks wall clock) 19 hours ago. I'll begin with the intermastoid incision and frontal craniotomy then make my examination of the brain.
(As SCULLY begins to make the incision, BETTS' eyes open and the mouth slowly opens, startling SCULLY.)
SCULLY: Oh, God!
(SCULLY is shocked and drops scalpel.)
SCULLY: Mmm.

thumbwax: How much did the rest of the guy weigh?

Skully: I dunno

FYI, I've emailed the experts at UC Davis to weigh in on this heady question and anxiously await their reply.
thumbwax, Sep 01 2001
  

       [thumbwax] you either have a script next you or know to much.....
kaz, Sep 01 2001
  

       Copy/Paste - of course - it was the last Copy/Paste I did on my old 4 Gig 'puter with 466 Celeron. This is the first annotation done with my new 80 GB hp pavilion 7865 with 1.2 Ghz Athlon, DVD, CD-RW and 256 MB of Ram. halfbakery looks so much prettier - even though I'm using the same monitor.
Why this particular idea to annotate first?
My head weighs more than it did yesterday.
thumbwax, Sep 02 2001
  

       With a good supply of a few thousand or so severed heads you should be able to get good measurements of volumes, densities and weights and their relationships, using the methods already outlined. Perhaps a formula could be generated from this data that would give us at least a fairly accurate estimate from measurements of a non-severed head's volume and density.   

       Why, though? Why?
Guy Fox, Sep 03 2001
  

       Because it's there
thumbwax, Sep 03 2001
  

       why weigh your head anyway?
LaLaLaLola, Sep 04 2001
  

       Because it's there. <-- (full stop)
thumbwax, Sep 04 2001
  

       There exist methods for measuring the average density of brain using a technique known as Computerized Tomography. Note that I know nothing about this except what you can read about in the link posted. [Brain density changes during renal replacement] I understand from this that you can get proportions of various elements within an area, which can be used to generate an average specific gravity for that area. Presumably this is applicable to other parts of the body than just brains. Further interesting reading to do with density of brains in [High acceleration and the human body].

So what's the point of this? You'd then be able to use the displacement method to find the volume of a head, as described above, and knowing the average specific gravity of your head would then *get you close* to knowing the weight.
AllenGrace, Sep 04 2001
  

       Yeah, but can you do it as a Bar Trick?
thumbwax, Sep 04 2001
  

       Tomography is what I do to the earth. I get a load of electrical parameters back, but no-one has yet managed to relate resitivity to hydraulic conductivity, permittivity, pore fluid density or quartz grain fracture networks, among many other things. (That's Chapter 1, the literature search. Bingo!). If the human head is anything like a sandy soil above Triassic mudstone (oh come on, can't you give a little?) then I doubt that tomography will give you what you're looking for, either. I vote for decapitation.

Having said that, couldn't we use a modification of those footprint pressure pads that sports scientists use to find where a runner's weight falls? I'm thinking of a flat surface that is made up of centimetre-square (or smaller) highly sensitive weighing scales like pixels. Headweighee lies on this multi-pixel surface and the weight per unit area is measured. I'm sure you would have to do something clever like tipping the surface slightly to find angular weights and so on. I just think that there is enough neck clearance for all the weight of the head to fall through to the pixels under the head. Unless you've got a fat neck.
lewisgirl, Sep 04 2001
  

       Bar Trick: try [Tomb]'s or [nick-n-uit]'s solution. Or maybe [Flooglehorn]'s, depending on your company.
jabbers, Sep 04 2001
  

       Wasn't there an old horror flick where a mad scientist kept his girlfriend's head alive until he could find a suitable donor for a body? If they can do it in an old black and White movie the technology must exist somewhere. I think it would be easier to just cut the head off and weigh it than to try to come up with a way to weigh it with out decapitation. Plus I think if the medical technology exists where you can sew a severed penis back on they have to be able to sew a head back on. My other solution is to make a special head weighing helmet and then stand in a room with a vacuum so that the rest of the body will be weightless... Does anybody have their Joke Detector on?
FractalAxle, Nov 09 2001
  

       Yes - not a flicker.   

       Wandering brieflly back on topic, I wonder if you could so something by vibrating the head an looking at the inertia. (Problem 1 - it qould give a false reading due to the fluid content. Problem 2 - You'd pass out due to the vibration of the brain on the skull.)   

       Might give another input into this knotty problem, though.
st3f, Nov 09 2001
  

       You could probably do this by passing a weight on a string/spring around the body and measuring how much the weight is attracted to the body at different places, and use that to plot the mass distribution of the body. You'd need the body to stay very still for this, I think, so it's probably not much better than method 2:   

       Method 2. Get a 1 kilogram weight. Cover it in glue, hang it from the ceiling by a rope, and get some system involving a camera that'll let you measure its position over time to a high degree of accuracy. Then let your neck go all floppy and crash into the weight. By measuring the speed of your head before impact, the speed of the weight, and the mass of the weight, you should be able to calculate the mass of your head by conservation of momentum. You should probably use muscle-relaxant drugs to avoid inadvertant reflex twitching at the point of impact.   

       But wait:   

       Method 3: Hang by your legs. Stretch your neck as far out as possible. Swing like a pendulum and calculate your period of swing. Now tuck your head in and do it again. Now calculate the movement of your centre of gravity, which will depend on the period of your swing. By measuring the movement of your head between the 2 positions, you'll be able to work out the mass of your head. Just make sure you don't move your shoulders.
pottedstu, Nov 09 2001
  

       There was an incredible scene in the film 'The Magic Christian' with Peter Sellers and Ringo Starr where ones head was made to shrink somewhat - or so it seemed - dialoguing about it the whole time - a very good illusion. That would be a good bar trick. I haven't seen that movie in upwards of 20 years.
thumbwax, Nov 09 2001
  

       Solution found yet?
thumbwax, Sep 19 2002
  

       14.76 grams, but I think I may have made a miscalculation somewhere due to the splitting headache I currently have.
PeterSilly, Sep 19 2002
  

       You could sit with bowed head over some scales on a table. Spit powerfully and calculated the head's weight by its proportional reaction upwards compared to the spit's force downward.   

       Stand on scales and nod the head back and forth between two points for an hour. From the weight lost, one gets the extra calories burned to accelerate and move the head over the total distance leading to its weight.
FarmerJohn, Sep 19 2002
  

       Arrange two trestle tables (or similar), a *head table* and a *body table*, with scales under all of the legs.   

       Place a weight of known mass (eg 100g) below the arbitrary *cut off* point (eg on your throat) and position your (suitably relaxed) body so that the added weight registers only on the *body table* scales (ie not at all on the *head table* scales).   

       If/when you finally get this to work, you will know that the *head table* is supporting the head and only the head, therefore the *head table* scales indicate the mass of your head, assuming your neck muscles are sufficiently relaxed as to be not supporting your head at all
three14's, Sep 22 2002
  

       //so that the added weight registers only on the *body table* scales//   

       How is this achieved? 100g plus how much?
FarmerJohn, Sep 23 2002
  

       I think that the answer can be found using one of those spring-loaded devices that fisherman use for weighing fish. If you pin the weighee's body to the floor, Gulliver style, but leave the head free of restraint and put the hook through the weighee's nose, then, at the moment when you have managed to raise their head off the floor, the reading on the device should give you the weight of their head. Resistance from the neck muscles should provide negligible distortion of the result because the pain that would be inflicted by the nasal hook as a result of any resistance and conversely, if they raised their head more than was justified by the upward force from the weigher then the spring would register no weight.
DrBob, Sep 23 2002
  

       << How is this achieved? >> I have absolutley no idea... The bit that went *If/when you finally get this to work...* will hopefully suffice as a get out clause ;)   

       Although, I *imagine* it would work... If a roll of carpet, instead of a body, was placed on two such tables, surely placing a weight on the carpet *precisely* above the point where the two tables abut one another, the weight would register equally on both sets of scales, whereas if the weight was moved, it would register only on one set of scales Finding that critical point is the key   

       Although in the case of the head, the critical point is determined beforehand depending only upon where (*officially* ) the body *stops* and the head *starts*   

       Position the body on the two tables, read the scales, add the weight (at the critical point) and re-read the scales If the weight does not register evenly on both sets of scales, move the subject and repeat the *read, add, re-read* steps until either the weight does register evenly or the subject decides to clear off to the pub
three14's, Sep 27 2002
  

       On the Web I saw and interview with a porn star who said each of her breasts weighed eight pounds. I wondered how she weighed them, especially since she said she was "a little drunk" when she did it.
think25, Sep 28 2002
  

       Weigh the person, then figure out and subtract the weight of the body (not including the head).   

       Why do we care how much our head weighs anyway? To prove how intelligent we are? To show how big our egos are?
"Ha Ha! My head's heavier than your head! Ner ner ner ner ner!"
NickTheGreat, Sep 28 2002
  

       OK. It's been a few years, but this should work:
Your neck muscles should be able to provide the same force regardless if you are on your back or stomach, so:
  

       1. Lie down on a table on your back, facing upwards, with your head sticking off the edge.
2. tie a belt around your head like a headband, and have a string attached to the weight, and have someone hang weights on the string until you "max out" (can't lift anymore.) That max-out facing up is say "X". You are using your neck muscles to push forward (up). Here you are lifting the weights, plus lifting the weight of your head.
3. Now, do the same thing, but facing down (on your stomach), and push forward (down) with your neck muscles. You'll need to have a the string attached to the belt to go around a pulley as you add weight. The max out in this position may be Y. Here you are lifting the weights MINUS the weight of your head (as the weight of your head is helping push down already).
4. So, the weight of your head is simply (Y-X)/2.
  

       You can't reliably & repeatibly "totally relax" your neck muscles, but you can more reliably "totally max" those muscles.   

       As far as "why", well I'm sure there's some fabulous prize. Where is it?
sophocles, Nov 10 2004
  

       You'd need some assistance to do this. Have a coffin like tank with a rubber "turtleneck" membrane where your head begins. The membrane divides the tank into "body" and "head" chambers. Now put your head through the turtleneck dividing the tank at your neck. Fill both sides of the tank with water. Add salt brine to the body chamber until the body is at neutral buoyancy while you hold your breath. Remember at this point to keep your neck relaxed. Now add brine to the head chamber until both head and body together are at neutral buoyancy. Obtain volume of head. Weight of equal volume of head chamber fluid is weight of your head.
hangingchad, Nov 10 2004
  

       [Sophocles]...yeah, and damage your neck muscles in the process.   

       How 'bout this... Use aforementioned plank of wood, with you lying down with your feet at the fulcrum and your head... well, obviously your head pointing away. Balance the plank using weights (or penguins, or little wizards, if you can stop them moving around... just make sure they're _calibrated_ little wizards).   

       Now lift your head, and get someone to measure the angle that your head has moved through. Raising your head has shifted the centre of gravity; by rebalancing the plank (moving little wizards around) you can work out precisely what that change was. Knowing your own weight, the original centre of gravity, the shifted centre of gravity and the angle you moved your head through, you should be able to work out how much... your... head.... weighed...   

       [sudden doubt]
moomintroll, Nov 10 2004
  

       [moomintroll] well, some would call that "exercise", not "muscle damage". Your angle-the-head method is a new & viable solution too...?except that as you raise your head up, you may also be adjusting your shoulders and would be touching the plank in a different spot. This shift in how you distribute your body weight on the plank may be too much signal to noise.
sophocles, Nov 10 2004
  

       Of course, weighing your head is just too easy using the Archimedes method. It would be more challenging—and more interesting, for sure—to weigh your own brain.
ldischler, Nov 10 2004
  

       The National University of Singapore tells us that the density of a human head is about 1050kg/m3 (see link), in case anyone wants to work out how much their head weighs using a bucket of water.   

       You can find out exactly how accurate their figure is by using the decapitation method and comparing results. You will of course require an assistant for this.
wagster, Nov 10 2004
  

       +1 always liked this one.
po, Nov 10 2004
  

       Exercise, muscle damage... why take the risk :-) [ldischler] take the point about moving the shoulders - maybe this calls for the addition of some sort of shoulder clamp. Roll out the clamping table... [sigh] the things I do for science.   

       Top mad scientist stuff.
moomintroll, Nov 11 2004
  

       I said it would be too easy using the Archimedes method, but now it occurs to me that that is wrong. The air in the various cavities would throw it off.
ldischler, Nov 11 2004
  

       I think an MRI would be useful here to virtually chop your head off. Localized densities and sizes are easily created with the data, and it's a simple matter of adding it all up. Now, for that million-dollar machine...
RayfordSteele, Aug 16 2006
  
      
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