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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]

[Voice, Aug 31 2023]

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,

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

 — 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 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

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

 — jutta, Jun 08 2001

 — 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

 — 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:

 — 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).

 — 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?
 — 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:

 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

 [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.

 — 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

 // The air in the various cavities would throw it off. // That would only throw it off by the mass of the air inside the mouth and sinus cavities. Taking a wild guess of 4 cubic inches of air space there, adds 0.0002 pounds.

I'd say that the bigger problem is that the definition of the "cut off" between head and neck must be a flat plane that does not intersect the chin in order to use the displacement method.
 — scad mientist, Aug 16 2023

IPhone app would do this pretty easily. Take a picture of from the front, from the side, add height and weight. The app does the math using the percentage of the body calculation from the pictures. 300 pounds, head is 10% of the area calculated from the pictures, that's a 30 pound head. Then you cut your head off and weight it. Not necessarily in that order.
 — doctorremulac3, Aug 16 2023

[doctorremulac3]; that method assumes that the density is uniform; which may be the case, but I would think it most-likely isn't (higher proportion of bone in the head, for a start).

 OK here is a proposal. In my opinion the key requirement in [darsy]'s original specification from 23 years ago is that the measurement be //accurate to 1g//. This demands that the method proposed be able to accurately calculate the error or uncertainty on the final presented weight, and further, that said calculated error or uncertainty is <±1g

 So here goes. First, take any stones out of your mouth or nose.

 Note that to fulfil the criteria demanded, of ±1g, the weighing scale has to be more accurate than that. Recommended is to use a ±0.1g scale.

 Iteration 1: h+b is your total body weight, and h is an approximation of the weight of your head. But we cannot calculate the accuracy of this measurement, and so it is useless.

 Now chug a load of muscle relaxant drugs and try again. This will lead h to approach closer to the actual weight of your head. However we still are not quantifying the accuracy, and we are also not defining the //chosen point on the neck//.

 So first of all, take a black ballpoint pen and draw a dashed line around the back of your neck, to define the plane where you would slice if you wished to remove your head to weigh it. You can write x8 at the end of the dashed line if you wish.

 Now we need a third weighing scale [N] which produces a reading n. This third weighing scale has a kind of cradle which is shaped like a shallow U, made perhaps from 2mm thick metal with a safe rounded edge. This thin curved cradle is inserted under the neck, on a height-adjustable support, so that the edge of the U shaped cradle presses on to the dotted line. The support is cranked up so that weighting machine [N] is taking the weight of your neck. Now the total body weight is h+n+b, and the weight of the head can be approximated as h+(n/2)

 Of course this is still an unquantifiable approximation because the weight recorded by [N] is not exactly equally from your head and from your body. So what we need to do is simply raise or lower the head cradle to a position where the weight n is composed exactly 50% of the head weight and 50% of the body weight.

 The way to do this is to duplicate [N]. Add a second U shaped thin cradle beside and very close to [N]. This second one is called [M] and so now you are being supported by four supports, each resting on a weighing machine. Your head is resting on [H], your upper neck is on [N], your lower neck is on [M] and your body is on [B]. The halfway point between the cradle contact points of [N] and [M] is marked by a dotted line drawn on to the skin of your neck in black ball-point pen.

 Your body weight is given by h+n+m+b.

 Now comes the clever bit. Adjust the height of the head support until n=m .

 Raise and lower the combined support positions of [N] and [M] and graph the difference between n and m, and also graph the value of h+n for different vertical displacements.

 This can be repeated with the thin curved cradles [N] and [M] closer together (ensuring that they remain equidistant from the dotted line). Again graph the results.

The graphs showing the variation of h+n for different spacings and heights gives the accuracy to which h+n approximates the weight of your head. If this experiment can be run with a deviation of less than 1g then it counts as a success and you have succeeded. The calculated accuracy of the result can be given as ± the maximum or statitstically fudged variation in h+n for the different spacings and heights of [M] and [N].
 — pocmloc, Aug 18 2023

Probably better to have an array of parallel neck cradles, arranged so the central ones either side of the dotted line are slightly higher, to give a kind of bell curve distribution of weights on these neck weighing machines. Then as you raise and lower the entire array the bell curve will get higher but may also shift side to side. You want to adjust the head height to the position where the bell curve remains symmetrical through a reasonable range of heights. The sum all the recorded weights to the left side of the dotted line and calculate the uncertainty as before.
 — pocmloc, Aug 18 2023

This shows how much the halfbakery has evolved as this post is actually devoid of a halfbaked solution and should be deleted. All it does is present a problem - which anyone can do - eg: when you accidentally put too much milk in your coffee, wouldn't it be great if someone else would generate the halfbaked solution for getting it out? (the head problem is easily solved with a water volume displacement calculation - it's primary school physics)
 — xenzag, Aug 18 2023

[xen] read the spec: //It will not be an approximation using any displacement method.//
 — pocmloc, Aug 19 2023

 Simple.

3. Calculate the difference - that's the weight of your head. Easy.
 — tatterdemalion, Aug 21 2023

 The original spec ruled out decapitation. Pity, as it would improve many people.

Current company excepted, of course.
 — a1, Aug 21 2023

CT scan. Results are density based. Cut where you like, weigh the voxels.
 — mylodon, Aug 27 2023

That method would fail where the subject's skull had been replaced with a 2mm thickness of lead.
 — pertinax, Aug 27 2023

On the plus side - while the subject may not know his or her weight of head, they will be free of 5G contamination.
 — mylodon, Aug 29 2023

 //Current company excepted, of course.//

Sp: accepted.
 — 21 Quest, Aug 31 2023

Mondegreen.
 — a1, Aug 31 2023

 Hmm, sounds kind of fiddly, do we need to add any weights [Hippo]?

 Balance a plank non centrally on a fulcrum and place a scale under the long end of the plank and record the weight applied to the scale.

 Weigh the plank normally.

 Cut the plank in two at the point it rested on the fulcrum, weigh both pieces and compare results.

What's the relationship between the first measurement and that of the cut long 'half' of the plank.
 — Skewed, Sep 25 2023

[skewed] the relationship requires a longitudinal density curve in order to calculate meaningfully.
 — pocmloc, Sep 25 2023

Meh! now I'm going to have to go away and find out what the words mean damn you ;p
 — Skewed, Sep 25 2023

I'd just like to compliment [Skewed] on their 23 years of careful thought and contemplation before replying to my comment
 — hippo, Sep 25 2023

 [Big grin] Well I like to take my time don't I, not on everything of course.

Any rumours to the effect that I'd never seen this before and took less than five minutes to throw a comment out without fully reading your comment, far less progressing beyond it to others, are obviously baseless slanders that I'd like to take this opportunity to strongly deny, to any who might point to my first appearance fourteen years after this post as in any way persuasive that I couldn't have seen it before then I say "Poppycock!" .. I was clearly just lurking on another account for a long time.
 — Skewed, Sep 25 2023

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