# How can you weigh your own head in an accurate way?

I read some methods but they're not accurate. They use the Archimedes principle and they assume uniform body density which of course is far from true. Others are silly like this one:

Place it on some scale

I'm looking for some ingenious way of doing this accurately without having to lie on your back and put your head on a scale which isn't a good idea.

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Pity I don't have the rep to post an answer, as I think I have a more accurate solution than the "relax your neck muscles" and similar ones: go to the morgue. Cut off the head of a recently deceased person with a similar body type. Measure the ratio of the weights of the head and body. Measure your weight. Even more accurate: measure the density of the head you just cut off. Now measure the volume of your own head by immersing it into water. The volume of the head does not vary as much as of the body, where muscle, fat, etc. play a role. For better accuracy, cut off many heads for an average. –  vsz Jul 12 '13 at 20:59
This question appears to be off-topic because the actual physics content is minimal. It exists merely because of the joke. Nobody is learning anything that they couldn't have invented themselves after a couple of minutes of thought. The post is taken up space on the site front page, and thereby drowning more serious posts. –  Qmechanic Jul 12 '13 at 22:50
Voting to reopen because it patently is about physics. –  Nathaniel Jul 13 '13 at 2:58
I see answers appearing addressing tomography and Compton scattering. I challenge the remark "nobody is learning anything that they couldn't have invented themselves after a couple of minutes of thought". –  Johannes Jul 13 '13 at 4:03
I respectfully disagree with my colleagues on this (I wasn't around during the first moderator conference on the matter), and am going to cast the last re-open vote here. However, I want to emphasize that this I'm taking this step because there is physics here (good physics that does come up in an experimental context at times) not because the question is framed in an amusing way. Wannabe imitators whose questions don't have the physics content will be disappointed. –  dmckee Jul 14 '13 at 13:22

Get someone to relax their neck as much as possible, stabilize their torso, then punch them in the head with a calibrated fist and measure the initial acceleration. Apply $\vec F=m \vec a$.

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This ignores any reaction force from the neck. What you will get is the effective mass of a pivoted rigid body, $m_{eff} = \frac{I+m c^2}{L^2}$ where $c$ is pivot to center of mass, and $L$ is pivot to force distance. –  ja72 Jul 12 '13 at 13:45
It does, but assuming the neck was elastic to some extent (i.e. more like a spring than a rigid pivot), then by approximating an ideal impulse (making the spatial duration short and the applied force large), I was hoping the reaction force from the neck could be made negligible. Of course there are practical problems - the shorter the distance over which the observation is made, the greater the error bar on the acceleration. –  twistor59 Jul 12 '13 at 13:55
A similar method has been used, though I don't think the goal was to measure the weight of the participants' noggins. If only they took a few measurements during the procedure... –  Phil Jul 12 '13 at 15:01
This technique fails for all the obvious reasons, but it's too funny not to upvote –  Mike Pennington Aug 17 '13 at 14:22
@MikePennington OK I did introduce the callibrated fist for its comedy value, but I still think something along these lines could be made to work: attach accelerometers to your head, and apply a known impulsive push over a very short distance (a few millimeters?), short enough such that the reaction force from the neck was negligible when the head was moved through this distance. If the head was relaxed/floppy at the time the impulse was applied there may be some possibility of a reasonable estimate. –  twistor59 Aug 18 '13 at 11:44

Step 1: Take a trip to deep space (space suit recommended; means of transportation left as an exercise for the reader). It is important to compute the Hill sphere of your body to make sure it is large enough at this stage. Really it's a Hill-roughly-person-shaped-spheroid-blob, but feel free to assume a spherical you to simplify the calculation.

Step 2: Deploy a constellation of micro-satellites in orbit around yourself (see Fig. 1).

Step 3: Carefully track the motion of every satellite.

Step 4: Process telemetry data to measure the mass distribution of your body, head included.

Step 6: Profit!

Figure 1: Micro-satellites deployed in gravimetric tracking mode. Careful monitoring of telemetry is required to determine the mass distribution of the central body.

(composited from source images here and here)

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I think your comment on babou's answer with some equations could be a good answer. –  whatever Jul 12 '13 at 13:10
@metacompactness I was quoting his suggestion which he has now edited out of his answer. –  Michael Brown Jul 12 '13 at 13:14
This experiment isn't very practical but at least step 5 is optional. –  whatever Jul 13 '13 at 7:23
Won't the equipment and space suit skew the results? –  Thomas Jul 14 '13 at 23:39
@Thomas with enough data you could account for that. It would take some advanced numerical computation to do it accurately, but in principle you could measure the complete mass distribution this way. There won't be any simple analytical formula in this case. By the way, I said the space suit was "recommended" - if you don't want to wear one that's your choice. ;) –  Michael Brown Jul 15 '13 at 0:13

Petroleum engineers would all provide you with the same answer "use Compton scattering", as this is how the mass density of rock formations gets measured deep in oil wells.

A more complete answer is: Compton scattering can provide you with a measurement of the bulk density of your head. Combine this with a volumetric measurement (dipping your head in a tank and measuring the rise in level), and you're done. You also need a (once off) calibration of the density measurement, as explained below. The big advantage of such a Compton measurement is that it can be repeated and automated at will. Based on accuracies obtained in boreholes, I would expect Compton measurement capable of yielding head weights with a sub-percentage accuracy.

The Compton scattering measurement is done by scanning your head with a gamma ray beam, and measuring the beam attenuation. The gamma ray source should create photons with high enough energies to interact by Compton scattering, but low enough to avoid pair production. 137Cs provides a good source. Although Compton scattering depends on electron density and not on mass density, the measurements can be calibrated to give the correct bulk density for human heads using the ratio of the electron density to the nucleon density. This is based on the observation that the electron density is equal to the bulk density multiplied by Z/A where Z is the average atomic number and A the average atomic weight of the atoms in a human head. In practice, this calibration is best done using heads decapitated from dead bodies.

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"Based on accuracies obtained in boreholes" You had me up to boreholes. –  Chris Jul 12 '13 at 16:09
I thought you were about to suggest another technique used by petroleum engineers, which is to blow an explosive in one ear and measure the wave propagation with sensors over the rest of the head, to determine the internal structure. –  babou Jul 12 '13 at 16:48
"The Compton scattering measurement is done by scanning your head with a gamma ray beam" sounds safe enough, you try first. –  Thomas Jul 14 '13 at 21:08
"Scanning your head with a Gamma ray beam" is only slightly less destructive than cutting you head off and putting it on a scale. –  WillMcLeod Aug 1 '14 at 17:21

Here are some methods I came up with:

Newton's Method:

1. Measure the whole body's mass, let's call it $M$
2. Now detach the head, and put it a distance $d$ apart from the body
3. Measure the gravitational attraction of the two parts of the body(let's call it F)
4. We have a system of equations to solve: $$m_1+m_2=M \\ \frac{G m_1 m_2}{d^2}=F \\ \Rightarrow \frac{G}{d^2}m(M-m)=F$$ which is a quadratic equation and has two real roots. Head's mass is probably the smaller one!

Einstein's Method 1:

1. Shine a light ray, close to the person's head
2. Measure the deflection angle $\delta \phi$
3. The mass can be calculated from this equation: $$M=\frac{c^2 b \delta \phi}{4 G}$$ where $b$ is the distance of closest approach, and $c$ is the speed of light.

Einstein's Method 2:

1. Measure the wavelength of an out-coming ray(an absorption line for instance) from your head.
2. Calculate the redshift, and the rest is easy peasy .

Heisenberg's Method:

1. Measure the whole body's weight, using a scale.
2. Continuously measure the rest of the body without looking at the head(Now this step is crucial, no one should look).
4. Now that the head is effectively detached, you can weigh the rest of the body and calculate the difference.

Fermi's Method:

1. By all means you can consider the head to be a sphere with radius $r=0.1$ meters.
2. The density of head is more than water($10^3 \ \frac{kg}{m^3}$) and less than stone($5.5 \times 10^3 \ \frac{kg}{m^3}$); we'll take the geometrical mean $\rho = 2.3 \times 10^3 \ \frac{kg}{m^3}$
3. Now the mass will be:$m=\frac{4}{3}\pi r^3 \rho = 10 kg$("$\approx$" is not necessary :)

Hooke's Method:

The neck can be modeled as a spring, let's assume its stiffness is $k$.

1. lie on the ground and measure the neck's rest length($l_0$).
2. Hang from a tree upside down, and measure the neck's length($l_h$); now if we know $k$, the head's weight will be: $$W_h=(l_h-l_0)k \tag{1}$$
3. We still don't know $k$, and we have to find it. While you are still upside-down, hang some known weights to your head and measure the neck's length respectively.
4. Plot the weights with respect to spring's length, the slope will be its stiffness constant. Now using formula $(1)$ we can calculate head's weight.

Mansfield's Method:

Take MRI pictures of the head, which, using my technique will give the 3D density plot of head within few seconds. Now the rest of the problem is just banana cake.

Zeno's Method:

While this method might not be practically efficient, it has incredible philosophical advantages!

1. Cut half of what has remained from the head,
2. weigh it,
3. Go to step (1)

Pair Annihilation:

Find your dual(which presumably has opposite quantum numbers than yours), slowly bang your heads together(be careful with the rest of the body). We will measure the released energy(call it $E$), your head's mass will be: $$m=\frac{E}{2c^2}$$

To be completed ...

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My head is an observa-Bell. –  Soham Chowdhury Jul 12 '13 at 19:52
@Ali If you think (some of) downvoters (on that closed question) now want to take their votes back, you must edit the question to make it possible for them. (Only if you think so; it may have inverse consequences too!) –  Mostafa Jul 13 '13 at 9:19
"No need for the $\approx$ sign"... hahaha. +1 for that. But I think Hooke's Method is actually possible, if you consider the skin and neck muscles loose enough, and can somehow find out an average value of stiffness of the human spine. –  udiboy1209 Jul 13 '13 at 11:22
@dimension care to avoid rollbacking yourself so much? :) You're bumping the post.. –  Manishearth Aug 17 '13 at 11:02
@udiboy1209 I thought that too at first. But that's only the case for an ideal spring which the neck is far from. I suppose one might be able to do some destructive tests on others' necks and arrive at a neck-spring model, but at that point you might as well weigh their heads and compare circumference. –  WillMcLeod Aug 1 '14 at 17:31

Thanks a lot for your votes for the "answer" below. Unfortunately I think now the solution does not work. It is great for two slices, but that is the end.

There is another solution that should give 3 slices, which is still a bit short. And I am afraid I do not see how to use the two available steps to start building a recursion :-)

Why ? It is clearly possible to cut the rod in $n$ "homogenous" slices, and then do the $n$ measurements, so as to get a set of $n$ equations with $n$ unknown, the weights of each slice. So we feel the problem conquered. The hitch is that these equations are linearly dependent and have an infinity of solutions, because balancing the rod depends only on the center of mass and the total weight.

Equations are a tricky business, especially when we forget to solve them.

Too bad, Especially today: I am in France right now, and the country has its national holidays : july 14th, aka Bastille day.

So I was hoping that this year, cutting heads could be declared unnecessary.

Let's try to do better for next year.
And do not lose hope ... mistakes are interesting too, in their own way, and very educational.

My first reaction was to measure volume with archimedes principle. But that does not work since the body density is variable. I was realizing that as @Manishearth made the comment.

An alternative, is to put yourself on a scale composed of a plank with a central pivot so that only your head is on one side. And you balance the difference between head and body with weights. This does not work a priori because you do not know the mass repartition of your body, and not all the mass has the same distance to the pivot.

Then what you can do is to do this measurement many times, while moving progressively the body from one side of the plank to the other side, measuring each time the difference in torque due to the weight.

I have not yet worked out the mathematics (and I am not sure I can still do that, probably yes with some effort) but I think that should provide data to determine the longitudinal weight repartition of the body. This reminds me a bit of the way medical scanners work, though this weight problem is probably simpler.

To be convinced that the approach may work, one can first approximate the body as a rod, such that the two halves are each homogenous but have different weights. Two measurements should allow finding the respective weights of the two halves.

You then get the following equations:

$m_1\ l/2\ =\ m_2\ l/2\ +\ m_3l$

$m_1\ 3l/4\ +\ m_2\ l/4\ =\ m_4l$

The weigths $m_3$ and $m_4$ are known, and the length $l$ is of course measured and known. Thus the equations can be solved to get $m_1$ and $m_2$.

This can be generalized to any number of slices.

In my above quip about being able to do the mathematics is not with respect to resolving a set of linear equations. Rather, I was wondering is there is some kind of transform to express a continuous description of this approach. As I said, I was somewhat inspired by the way medical scanners are working (or maybe the analogy hit me afterwards ... hard to know).

This can be generalized to many homogenous segments approximating the body density repartition.

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"Now there is an alternative that avoids this difficulty but is somewhat unpleasant. You hang yourself by the feet to the scale and measure your weight. Then you do the same with your head only immerged in water." If anyone does this please put it on youtube. –  Michael Brown Jul 12 '13 at 9:58
Uh, no, this assumes that the avg density of the head is same as that of the body. This is false, the head is significantly denser. The difference of measurements is $\rho_{water}\Delta V$, where $\Delta V$ is the volume of the head. –  Manishearth Jul 12 '13 at 10:01
@Manishearth Thanks. I was just realizing that when you wrote the comment. Too bad I had to remove the text, as it amused me (and apparently others). Here is another solution, without all the math worked out, but I am convince it is workable, with arbitrary precision, up to restlessness of the measured subject. But theoretical slicing of people still makes me uneasy. I prefer to hang them in water. –  babou Jul 12 '13 at 10:40
It's OK, I've made the exact same mistake on a different question here. Re:slicing: I was actually in the process of typing an answer to that effect, see below :) –  Manishearth Jul 12 '13 at 10:51
Hmm, can't remove the downvote unless you edit it again. Oh well :/ –  Manishearth Jul 12 '13 at 11:17

I would do it like this:

The muscles of the neck have to be as relaxed as possible so that it approximates a flexible linkage, such that at some point along this linkage (which we can identify as the division between the head and the body, and thus the mass of the head includes a portion of the neck), if we separate the body into two free-body diagrams, there is no net vertical force being exerted between the two.

An assistant can read the scale.

The real difficulty is introduced if you define "head" as that portion of the skeleton, and ajeacent tissue, which excludes all vertebrae of the spine and everything below.

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you get +1 for the alien as a person picture. Also +1 because if he is not to picky this quite possibly would do what he want's to do. –  ryan Jul 12 '13 at 14:49
You don't need an assisstant, if you are ok with attaching some mirrors. –  Dimensio1n0 Jul 15 '13 at 18:34
"which isn't a good idea" according to the question! –  Ali Jul 17 '13 at 8:03
@Ali I didn't even notice that, and it's in the question's very first edit. Be that as it may, it's not a reasonable constraint. The question doesn't justify why it's not a good idea. –  Kaz Jul 17 '13 at 21:04

1 Person A lays down on a carousel with its neck over the edge. Person B knocks A out. Person C spins the carousel. B clocks the time. Person D takes pictures of A's head as it goes around.

2 Measure the angle of the head by looking at the pictures.

3 Do the vector math.

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What about accounting for the neck's material resistance to, and limited degree of, flexion? –  Matt Ball Jul 12 '13 at 18:47
This can be done with asymmetric objects. The angle must be measured from the head and necks shared gravitational centre. The mass of the neck will therefore be part of the final answer. It's not a pendulum, more like a hinge. Flexion is fine in the range of potential angles (face down). –  user2574264 Jul 12 '13 at 22:20
I think you broke his neck.. interesting approach though +1 –  Thomas Jul 14 '13 at 21:10
The angle would be independent of the mass, this would just tell you the rotational velocity. –  Rick Sep 28 at 13:59

Assume the body to be made of N sections $S_i$ of uniform linear density $\lambda_i$ and length $l_i$ and center of mass $r_i$ (from one end). Increase N as much as you want to get a more accurate reading. The sections need not be of the same length. Try to slice your body into approximately homogenous sections (homogenous with respect to mass-per unit length, not volume density).

Now, set up a seesaw-like system with a plank on a log or other pivot. Find a friend or grad student.

Now, sleep on the plank with the boundary between section $S_j$ and section $S_{j+1}$ right above the plank. This will be at a distance $L_j=\sum\limits_{i=1}^{j} l_i$ from one end. Have a friend attempt to balance the system using weights. He will get that some weight $M_j$ will balance the plank when placed at $R_j$.

Repeat this measurement for $j= 1 \to N-1$. Now, write torque equations for each system, assuming that each section has a linear density $\lambda_i$. These equations will be of the form:

$$M_j R_j=\sum\limits_{i=1}^N \int\limits_{L_j}^{L_j+l_i} \lambda_i xdx$$

Solving the integral, we get:

$$M_j R_j=\sum\limits_{i=1}^N \frac{\lambda_i}{2}(l_i^2 +2l_iL_j)$$

You now have $N-1$ equations in the $N$ variables $\lambda_i$. Your $N$th equation will be $\sum l_i\lambda_i =M_{you}$

Now, simply calculate the head-mass for the idealized system, and that will be approximately your head mass. Accuracy increases as $N$ does, though you can also increase accuracy by smartly partitioning the system.

The simpler method is decapitation and measurement.

Reattaching the head is, of course, left as an exercise to the reader.

Alternate solution #2:

(Note: In all seriousness, this probably won't work because LIGO isn't tuned to detect the gravitational waves of head wiggles)

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A brilliant book by Tom Cutler: 211 Things a Bright Boy Can Do has a section which shows you how to weigh your head accurately.
This method is easy to understand, and quite practical. It doesn't even involve harming yourself as most of the other methods listed above do. You can actually do it.

Here's the excerpt:

There can’t be many problems more vexing than that of weighing your head. For a start, how do you decide where your head ends and your neck begins? And once you’ve decided this and marked the junction with an indelible pen, what next? It’s a riddle that beat even Archimedes. So here is a method that will produce a good estimate of the weight of your head the next time you find yourself at a loose end.

Required:
· A plastic kiddie pool
· A plastic rainwater barrel
· A plastic bucket
· A plastic measuring jug
· A plastic chair
· Some bathroom scales

Method

1. On a nice day, inflate your kiddie pool in the yard. Divert any bathers from wading into it.

2. Put your barrel into the pool and fill it (the barrel) with warm water, right up to the top. Do not allow it to overflow.

3. Shave all your hair off. You’ll get a more accurate measurement this way. Because you are bald no water can be removed by capillary action.

4. Stand on your chair, hold your breath, and lower your head slowly into the water until it is submerged up to your Adam’s apple. The displaced water will flow down the sides and collect in the bottom of the pool. Slowly take your head out again.

5. Pour the collected water (from the pool) into your bucket. This is the most annoying part of the experiment because the pool is full of water and you need to move it without spilling any. You can, if you are ambitious, begin the experiment by digging a deep hole in the lawn, into which you now crawl with your bucket, so as to siphon the water from the pool.

6. Measure, and jot down, the volume of water in your bucket by pouring it into your measuring jug. If the jug’s too small, do it in steps.

7. Empty the bucket and the jar for the next measurement.

8. Put the barrel back and fill it to the brim again. Take all your clothes off and weigh yourself. Note down this weight.

9. Get on the chair and climb carefully into the barrel, completely submerging yourself. After a moment climb out and pour the displaced water into the bucket and then into the measuring jug, keeping a note of the final amount.

10. Multiply your body weight by the ratio between the two figures. This will give you the weight of your head.

11. Put your clothes back on.

$P.S:$ I've added a bit to the text because some parts were unclear. :)

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+1 for funniness but it assumes that your bodyie's density is the same as your head's density, isn't it. ? –  Dimensio1n0 Jul 15 '13 at 9:59
@Dimension10 Aah, I know, I realised that soon after I posted the answer. :/ –  mikhailcazi Jul 15 '13 at 10:16
Does it all have to be plastic? It might be easier for some people to use glass or metal. ;) –  AJMansfield Jul 24 '13 at 23:53
Hahha I'm sure it will. But since this has been taken from a book, I didn't wanna change it. –  mikhailcazi Jul 25 '13 at 9:49
@AJMansfield I don't think you get glass/metal inflatable kiddie pools though :/ Neither glass buckets/barrels. :P –  mikhailcazi Jul 25 '13 at 9:53

## Method 1: The Chemist's method

Firstly, take pictures of your head from at least $400^3=64000000$ different equally spaced out angles. Samples:

Then load them onto the computer, and use some 3D visualising software (maybe Scientfic Computing.SE can help you with that. ) to intersect the images at different angles, to get good, but of course, approximate enough (you don't want to crash your computer. !) description of your face. Load the data onto Wolfram Mathematica (or something, I don't know if Mathematica works actually, ) and then get a good parameterisation of your head's surface.

$$x=0.1\cos\theta\sin\varphi$$

$$y=0.1\sin\theta\cos\varphi$$

$$z=r\cos\varphi$$

The above are the parameterisations you would get if you're head's surface was a perfect sphere.

Then, take some random samples of the flesh of your head using a knife and an injection-giving machine. Now, your head will be perforated, but that's ok, (After all, what's that in comparison to the great gain for physics! It will be so useful in phenemonology, right?!)

Perforated head. Just try to ensure that your head doesn't crumble or that will be a a loss to physics.

Not very painful.

Now, measure the densities of these samples, and plot them as a scalar field across an image containing it. .

Now, do a simple volume integral (assuming that your brain is still functioning, which it shoouldn't be , because then the experiment, would be unfair. ) .

$$M=\iiint_V \rho\left(x,y,z\right) \mbox{ d}V$$

## Method 2: The Murderer's method

1. Commit suicide.

2. Be reborn with an identical twin (PSEUDOSCIENCE ALERT!!!).

3. Enssure that your identical twin goes through the exact same conditions as you.

Step 1 to 3 are strongly encouraged if you already have an identical twin (to ensure accuracy, due to 3.) but compulsory if you don't.

1. After you're as old as you were when you were reading this question, kill him.

## Method 3: The Teacher's method

Now, attach the head to a spring balance, and measure the force required to drag it at a constant velocity. . .

Then, just apply:

$$F_k=\mu_kgm$$

## Method 4: A Quantum Mechanic's method

. Pluck out your head. Give NASA/USSR/ISRO/some random space organisation ssome money to launch it into a place with very low potential, and then keep observing your head.

You may also try throwing your head as hard as possible. It may work, with some small probability.

Apply Schrdoinger's equation. (Time-independent, potentialless, non-relativistic)

$$i\frac{\hbar}{2m}\nabla^2\Psi=\frac{\partial\Psi}{\partial t}$$

Now, determine $|\Psi|$ experimentally through your observations. Apply the above equation.

## Method 5: Another Quantum Mechanic's method

. Similar to Method 4 but instead use path integrals. Measure the phase intensityr and apply : :

$$A=\sqrt{\frac{2i\pi\hbar }{mN}}$$

## Method 6: A standard-modelist's method

Somehow Use the Standard Model $\mathcal L$

$$\mathcal L = - \frac{1}{4}{H^{\mu \nu \rho }}{H_{\mu \nu \rho }} + i\hbar {c_0}\bar \psi \not \nabla \psi + {c_0}\bar \psi \phi \psi + h.c. + {\left\| {\not \nabla \phi } \right\|^2} - U\left( \phi \right){\text{ }}$$

 .

To compute the mass of your head . ! . ! .

## Method 6: A general relativist's method

Pluck out your head. Throw it into outer space. Calculate the metric tensor near your head . . Compare it with the Kerr metric tensor.

$$c_0^{2} d\tau^{2} = c_0^{2}\left( 1 - \frac{r_{s} r}{\rho^{2}} \right) \mbox{d}t^{2} - \frac{\rho^{2}}{\Delta} \mbox{d}r^{2} - \rho^{2} \mbox{d}\theta^{2} - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \mbox{d} \phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} c_0 \mbox{d}t \mbox{d}\phi$$

## Method 7 (or was it 8? Or 6? ). : . The Special Relativist's method.

Settle down completely to rest. Now, let someone come by, and somehow (if it's even possible) convert your head's rest energy into the Kinetic Energy of a donkey with a known mass. Now, using the velocity of the donkey, apply

$$m=\frac12m_\mathrm{donkey}\frac{v_\mathrm{donkey}^2}{c_0^2}$$

## Method 8: The Weird Person's method*

Change youur neck to a spring balance and hang upside down.

But I guess the best method is still @user2574624's /.

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+1 for the effort, even if it's a little comical in tone –  Larry Harson Jul 15 '13 at 12:05
Method 3 is obviously not reliable as the sliding of the head on the table will entail more parts of the head coming in contact with the table, giving a variable value of $\mu_k$. Not to mention that the head + neck system is a continuous mass distribution, so $m$ is variable too. –  shortstheory Oct 3 '13 at 14:07

You'll need something like this. The green things are pulleys(consider massless, frictionless)

Also, get some ropes, tie them around your feet and gently lower yourself until your head is under the liquid.

Measure the mass of the displaced water $m_d$ .

At this point, a buoyant force $F_b$ is acting vertically upwards. Weight $W$ is acting downwards.

Case 1(weight exceeds buoyant force)

If weight is more(try and feel with your neck muscles, in which direction the force is acting), make the apparatus as follows.

The the strings to your head(the ears or around under your chin is a good spot to tie) Experiment with different weights and obtain a mass $m_e$ such that you feel no force on your neck muscles.

Now $$W=m_e g+F_b$$ $$mg=m_e g+m_d g$$

Case 2(Buoyant force exceeds weight.)

Set the apparatus as follows.

Obtain a mass $m_e$ in the same way such that you feel no external force.

Now $$W+m_e g=F_b$$ $$W=F_b-m_e g$$ $$W=m_dg -m_eg$$

Now about calculating that force you feel in your neck muscles, you can attach a modified (extremely sensitive)spring balance like instrument which rests between your chin and shoulders, calibrated to indicate "zero" force in space.

One drawback is the slight compressibility of the body(around your neck) which might absorb some of the buoyant force.

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This also assumes that the head is uniform in density . –  Dimensio1n0 Aug 5 '13 at 10:44
How come? I'm balancing the forces. Buoyant force depends on total volume of the medium displaced. The head doesn't change much in volume does it? –  user80551 Aug 5 '13 at 11:47
Oh, I see you're not measuring displaced water . However, the fact that your body is there will make a difference . Probably need to hook your body to something , or just chop itj off . –  Dimensio1n0 Aug 5 '13 at 12:31
What kind of a difference? –  user80551 Aug 5 '13 at 15:36
From the weight of your body . The pulley s are at the centre of the head, no special reason that I see for why the body's weight doesn't affect . –  Dimensio1n0 Aug 5 '13 at 15:37

A safe (albeit not entirely comfortable) method is the following:

Ask a partner to hang you upside down using a rope around your ankles. Tell him to swing you with a small amplitude. You relax your body and ask him to measure the swing period. Now bend your head, keep it in fixed position, and again ask your partner to measure your swing period.

Knowing the local gravitational acceleration $g$, and using the expression relating the swing period to the distance the center-of-mass is away from the pivot point: $$t_{swing}^2 = \frac{4\pi^2}{g} \frac{\int\limits_{0}^{L} m(z) zdz}{\int\limits_{0}^{L} m(z) dz}$$ (with $z$ measuring the distance away from the pivot point), from the change in period you can work out the change in the position of your center of mass.

Provided you weigh yourself afterwards (or better yet: yourself including the rope) and provided you have an accurate estimate of the change in height of the center-of-mass of your head (tip: record the scene with a digital video camera, and ensure you have access to reliable information on the center-of-mass of human heads), you get a good estimate of the mass of your head.

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It feels like 'reliable information on the center-of-mass of human heads' is cheating - if you know the center of mass of your own head, you probably know its mass. –  jwg Jul 15 '13 at 7:34
Human heads vary in size and weight, but little in position of center-of-mass. You need to know that on average human heads have their centre-of-mass right between the ears. No other a-priori info is needed, and you certainly don't start of with any knowledge on the mass of your own head. –  Johannes Jul 15 '13 at 13:35

The solution has to be static. Any motion through the neck (relative DOF from body) is going to require very detailed modeling of the bones and muscles to achieve the proper result.

I propose a straight up weight of the head, with careful attention in keeping the neck as loose as possible and as horizontal as possible. See below for an example. You want no muscle forces on the neck (apply anesthetic if needed). Now move the head slowly up and down until you get a linear relationship between elevation and head force, and so you need to average the force around where you think the natural position of the neck is.

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This is highly dependent on the linear mass distribution of the body, as an $r^2$ term comes into the picture. –  Manishearth Jul 12 '13 at 15:11
If there is no force through the neck, then only measuring $F_C$ is sufficient. –  ja72 Jul 12 '13 at 17:57
Ah, those are separate scales. Well, the issue here is that there are a lot of forces on your head from the neck. I don't think this would give an accurate answer :/ –  Manishearth Jul 12 '13 at 18:01
@Manishearth I agree and that is where the anesthetic comes in! –  ja72 Jul 12 '13 at 18:14
@ja72: Alternatively try shooting yourself so that you'll be completely stationary. But you just wouldn't know the mass of your head. Still, a lot of use to phenemonologists! +1. ! –  Dimensio1n0 Jul 15 '13 at 12:12

Here is a way to measure it using the person's moment of inertia. Also, compared to many of the other answers to this question, this method is not only feasible, but actually reasonably safe for the subject.

1. Measure the mass $m$ of the person; the length $l_b$ and average cross-sectional width (from the front) $w_b$ of the person's body (excluding the head); and the length $l_h$ and average cross-sectional width $w_h$ of the head of the person whose head we wish to know the mass of.
2. Obtain a rigid stretcher that includes restraint straps, like this one:
3. Securely mount the person whose head is to be measured onto the stretcher, using the included restraint straps, so that the 'front' of the person is facing upward.
4. Measure the moment of inertia $I_{gross}$ of the person in the stretcher about an axis extending upward through their neckn.
5. Measure the moment of inertia $I_{tare}$ of the stretcher (without the person in it) about the same axis.

Now compute the net moment of inertia: $$I = I_{net} = I_{gross} - I_{tare}$$

For simplicity, I will model the head and body of the subject as if they were thin rectangular plates rotated about the axis at one end.

Let $m_h$ be the mass of the head of the person, and $m_b$ be the mass of their body. We know, then that $m_h + m_b = m$, or in other words, that the masses of their head and body together add up to the mass of the entire person.

$$I = \frac{m_bh_b^2}{3} + \frac{m_bw_b^2}{12} + \frac{m_hl_h^2}{3} + \frac{m_hw_h^2}{12}$$

Solving this equation yields:

$$m_h = \frac{4 l_h^2 m+w_h^2 m+12 I}{4 h_b^2+w_b^2+4 l_h^2+w_h^2}$$

Going further:

If you wished, you could use a more accurate model of the person, resulting in a more complicated equation for their moment of inertia. Also, if one wished to obtain a more accurate measure, one could measure the person's moment of inertia about several different axes, in order to compensate for nonuniformities in the density of the body as a whole, rather than just differences in the density of the head and the rest of the body.

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You can't separate the $I$ of the head and the body without modeling it. You've modeled it as thin plates of uniform density. It doesn't take into account varying densities within the "plates" (ie bones versus brain). It would be nice to basically define a plane that "bisects" the neck into "head" and "body" and come up with a measurement technique that requires no modeling of the tissue. –  Mark Lakata Jul 12 '13 at 20:59
@MarkLakata The extension to the solution that I discussed at the end for this problem only requires modeling of the volumetric properties of the body, so a model is much easier to obtain, because you can directly measure it. Basically, the more sections you subdivide the body into, and the more points you measure the moment around, the better the answer you can get. Getting this volumetric model is what the first step is all about, you can make it more complex as desired. –  AJMansfield Jul 12 '13 at 23:44
You are still assuming that density is known. Moments of inertia are calculated as the integral of mass density times distance. Your "thin rectangular" plate model, for example, assumes the mass density of the rectangle is the constant. –  Mark Lakata Jul 14 '13 at 6:03
@MarkLakata What I am assuming is that as the size of each subsection approaches zero, its density becomes constant. As you subdivide the body into smaller and smaller sections, the density of any one section gets more and more constant. While it is perhaps true that these subsections will never actually have completely constant density, you can just keep subdividing into smaller sections (with consequently less variation between them) until the variations within each section are low enough to allow us to get our target number of significant digits. –  AJMansfield Jul 14 '13 at 17:56
Yes, that would work. The key is that you will need to make at least N measurements at different axes to determine the N unknowns in your model, and then solve all the equations at once. –  Mark Lakata Jul 15 '13 at 16:40

A more accurate solution than the "relax your neck muscles" and similar ones:

Go to the morgue. Cut off the head of a recently deceased person with a similar body type. Measure the ratio of the weights of the head and body. Measure your weight.

Even more accurate: measure the density of the head you just cut off. Now measure the volume of your own head by immersing it into water. The volume of the head does not vary as much as of the body, where muscle, fat, etc. play a role. For better accuracy, cut off many heads for an average.

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What about MRI? I don't know the specifics, but a single google search reveals this link (http://link.springer.com/article/10.1007%2Fs002230050015#page-1) which suggests one can measure tissue density using it, and it's rather obvious one could also use MRI to reconstruct a 3D image of the head.

Alternatively, measure the relevant tissue densities of a cadaver then use MRI to evaluate the actual volume occupied by each tissue type which could give an estimate.

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I have mentioned that in my answer(which I just made community wiki) –  Ali Jul 16 '13 at 9:34

Let us take the matrix of tensometers as mattress (about 450x150 points 5mm matrix distance). Let us put examine man/woman on this matrix. Then the total weight is sum of partial forces being measured by tensometers multiplied by gravity. Weight of head is the sum of forces measured by tensometers under head multiplied by gravity. To eliminate muscles force of neck, you may apply anesthetic for examined object.

The sophisticated procedure to eliminate forces of neck´s muscles will come soon.

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On a turning roundabout, if you pull your body in to the middle, it will speed up due to conservation of angular momentum. If you lean out, it will slow down, for the same reason.

I speculate that, at some expense, it would be possible to build a mechanism that could:

• Continuously rotate you while
• Moving your body on the rotating platform and
• Carefully measure the change of angular momentum resulting at each translation.

You'd be stuck on it, turning around, for a while. But it has the up-side of not being fatal.

Using computation, and a mathematical process called deconvolution, I believe it would be possible to build up a volumetric mass density map of your whole body. By simply weighing yourself, you can turn densities in to actual masses.

Now, you can decide which part of you is your head, and ask the software to measure its mass.

To get a 3D volume model, the mechanism would have to also roll you over, as well as translating you in x & y (while you gently turn). To obtain a 2D map, which would be sufficient for the head measuring use case, just x & y translation would be sufficient.

I'm not certain this approach works. It may be the case that the measurements are not linearly independent of each other, and that all you will discover is the centre of mass. I have asked another question here on Phy.S.E. about this.

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OK, here is my take:

In a large bath immerse the body in salt water so that it floats. Get a photo , the plate perpendicular to the horizontal plane along the body, and measure the height out of water.

Get the volume of body and head by measuring the weight of the body up to the neck and then fully immersed, knowing the density of water.

Suppose for ease of algebra it is two balls

a large ball with density $\rho_1$, volume $V_1$, and a small with density $\rho_2$, volume $V_2$ and $\rho_1 < \rho_2$. The mass of $V_2$ + mass of $V_1$ is $M$

Let the ratio of the height out of water of $V_1$ to the height of $V_2$ be $a$ .

We have :

$\rho_1V_1+\rho_2V_2=M$ and $\frac{\rho_1}{\rho_2}=a$

solving

$\rho_1=a\rho_2$

$a\rho_2V_1+\rho_2V_2=M$

Hence $\rho_2=\frac{M}{V_2+aV_1}$

Now with a computer numerical integration one can do an integral over heights in the photo for a real body with irregular features and work with averages.

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It is incorrect to say that $\frac{\rho_1}{\rho_2} = a$. The height out of the water of a floating body does not bear any relation to the distribution of mass, only to the total mass and the distribution of volume by height. –  jwg Jul 12 '13 at 15:03
This assumes that the average density of the head is the same as that of the body, which is untrue (The head is much denser). See physics.stackexchange.com/questions/70839/… (That answer has since been edited, but my comment there applies here too) –  Manishearth Jul 12 '13 at 15:12
@Manishearth I have two densities, one for V1 ( body) and one for V2 ( head). The denser the body the deeper it sinks. –  anna v Jul 12 '13 at 16:35
@annav Ah, oops. Still, it's pretty inaccurate to assume that the body has uniform density. –  Manishearth Jul 12 '13 at 16:40
@jwg en.wikipedia.org/wiki/… " Any floating object displaces its own weight of fluid.". You could critisize my assumption of the vertical displacement being proportional to the volume. –  anna v Jul 12 '13 at 16:44
1. put on a tight-fitting helmet of known mass.
2. attach a known spring between the helmet and a solid wall.
3. have someone give the helmet+head a knock, and measure the frequency of vibration.

That gives the mass of head + helmet. Then subtract mass of helmet.

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If you can find a way to calculate the density of your head by looking at the amount of the head that is - for example - bone, brain or space, and take the density of each of them, then you could get an average density for the whole head. Then simply measure the volume of the head by possibly submerging it and work out the mass.

Of course this isn't accurate in the slightest, but at least it's an answer.

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or possibly get someone to push something with their head using the muscles in their neck then get them to move their head using what they feel is the same amount of force and hey presto, f/a=m –  George Appleton Jul 12 '13 at 14:56

Float free on water. Now lift your head clear of water. Measure displaced water. (Take account of water was displaced by head)

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Similar solutions have been proposed and it was already established that you have to know the average density of "head matter" to apply Archimede's principle... –  Neuneck Jul 25 '13 at 9:41

You should be able to do it by jumping off a table onto a spring scale. Simply film the jump with a high speed camera and look for the shockwave that moves through your body as you are landing and note when it reaches your neck. At the same time note the distance the spring has moved (call it point x). Then note how much further the spring moved afterwards to measure your full weight. Now drop a mass off the table that equals moving the spring to point x. Deduct that weight from your weight and you have hour head weight.

ps. you should be able to get rid of the camera and calculate the shockwave/compressionn wave speed using 9.9m per s per s and the height to your neck and height of the table.

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Archimedes:

Get a bucket big enough to contain your whole body at the most shrinking position (like a baby in the womb).

Take a note for these following weight scale:

1. Strip off all your clothes - fully naked.
2. Weigh your naked body - Record it as $W_n$
3. Weigh the bucket - Record it as $W_b$
4. Put the bucket on the weigh scale
5. Fill that bucket with water until flow over a little bit. That's how I define 'full'. Record = Record it $W_m$ (Bucket filled up with water).
6. Get into the bucket and submerge your body but leave your head above the water surface - Record it as $W_t$
7. Get out of the bucket very slowly so that you don't spill any more water and record the weight as $W_r$. Your Head Weight will be represented as these following formula:

$$W_h = W_n - (W_t-W_r)$$ Meaning that your head's weight is equal to your naked weight deducted by the weight of overflowed water that flowed out when you enter the bucket.

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...at the most shrinking position... usually referred to as the fetal position. Anyways, this is pretty much the same as this other answer. –  Kyle Kanos Aug 10 at 1:12

## protected by Kyle KanosAug 10 at 1:12

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