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A technique of vital importance at all levels in physics is estimation. This is obvious from the first chapter in any introductory physics textbook, but is also related to the working physicist. Checking orders of magnitudes during research presentations is common practice - I've seen many good questions with good followup answers that started with "If I estimated that value I would get something much different". In general, the actually result is not the interesting thing - it's what individual things will affect the result. There are even famous examples of this: Fermi's piano tuner problem, and the Drake equation. Apparently, Fermi was so good at this that he estimated the size of the Trinity nuclear bomb test to within a factor of 2 (see the wikipedia article for a discussion of that).

In this spirit, I would like to see someone try and estimate the number of hairs on the human head. The answer must include the basic assumptions so we can see where the major unknowns lie, and the best answer is one which requires no specific knowledge

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closed as off topic by Mark Eichenlaub, Chris White, Manishearth Mar 2 '13 at 11:37

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+1. I like the idea behind this question. I don't think it will stay on physics.stackexchange very long though. (Since the question isn't about physics) –  Mew Mar 2 '13 at 4:49
I agree; it's Fermi's piano tuners in Chicago. –  levitopher Mar 2 '13 at 4:54
What did you try? Please show some effort. –  Bernhard Mar 2 '13 at 10:38
@levitopher In my opinion, applications of estimation skill to physics problems would be on-topic. But estimation is a tool. It is not physics in-and-of-itself, and this question isn't a physics question. –  dmckee Dec 16 '13 at 16:26
I disagree - there is a long and rich history of estimation in the physical sciences that does not exist in other sciences. "How many piano tuners ...?" is question which a biologist is not trained to answer, whereas a physicist actually is (even if it may not be a good use of our time). There is a reason B.S. degrees in physics mean something, whereas in the social sciences they generally do not. Computer skills are part of that, but physicists have analytic skills in general where other disciplines do not. I will give up my crusade for this question, but I think the site suffers for it. –  levitopher Dec 16 '13 at 22:38
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5 Answers 5

I'll take a go at it - as with the piano tuners in Chicago, I take the approach as if I have "no facts to go on". Your head has a surface area of $4\pi r^2$, the fraction of it which is covered with hair is $\gamma$. The density of hairs per unit area is $\sigma$, and the number of hairs is then

$N=4\pi r^2 \gamma \sigma$

Hairs per unit area is obviously the main guesswork involved here. Most heads look like hair, which I will interpret as "when projected to your skin, over 50% of what is seen is hair." If your average hair length is $l$, average diameter $d$, the density of your hair is then


(obviously, this breaks when the hair is so long it leaves your scalp, but our hair length is usually 1/10-2 times the size of our head, so we are still within an order of magnitude. Also hairs from other parts of your head cover your skin as well, so this might be an underestimate). My final answer

$N=2\pi \frac{r^2\gamma}{ld}$

For $r=10$ cm, $\gamma=0.4$, $l=6$ cm (size of my head), and $d=0.1$ mm I get


Seems kinda low, but 419 is certainly too small, and 41900 seems maybe too large, so I am comfortable with this as an estimation.

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Nice work. I think the average person has about 100 000 hairs on the head based on some quick research. –  Mew Mar 2 '13 at 5:16
I understand why \sigma is affected by width, by I'm not sure why it is effected by length? –  Mew Mar 2 '13 at 5:23
depending on the human the diameter of the human hair starts from 17 to 180 microns.en.wikipedia.org/wiki/Hair –  anna v Mar 2 '13 at 5:24
we are not supposed to solve completely homework problems –  anna v Mar 2 '13 at 5:25
Chris: Well, intuitively if my hair is 1cm long, then it will cover a 1cm x 1cm square of my head. But if my hair is 2 cm long, it could cover a 1 cm x 2 cm rectangle of my head when it's combed. –  levitopher Mar 2 '13 at 16:49
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I just went to a mirror to count the linear hair density of my head. I found that in about $1 cm$ there are $15 hairs$, thus the linear hair density is about $\lambda=15 hairs/cm$. So density of hair per unit area is

$\sigma=\lambda^2=225 hairs/{cm}^2$

And assume that this hair density is roughly constant. I found that it takes about 6 times the area of my hand to cover my scalp (2 for top, 2 for back, and 1 each for left and right part of my head). The area of my hand is about $A_{hand}=8cm\times 15cm=120{cm}^2$. Putting them together, the total number of hair is

$N=\sigma\times6\times A_{hand}=162000hairs$

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I'll take a slightly different approach to the others. I just got a close haircut (not for science, but why waste a good opportunity right?) and managed to keep something like 90% of the hair. So I can use the fact that $N$ hairs of diameter $d$, length $\ell$ and density $\rho$ have a mass

$$ M = N \frac{\pi}{4} d^2 \ell \rho. $$

Accounting for the fact that I caught a fraction $\eta \sim .9$ I can estimate the number of my head hairs as

$$ N \sim \frac{4 M}{\pi\eta d^2 \ell \rho}. $$

Now I'll give some very rough error bars on the measurements but not carry through the error analysis. I leave that as an exercise. :) The measured mass of the hair was $M = 22\pm1\ \mathrm{g}$. I'll take $\eta = 0.9 \pm 0.05$. The average length of my hair was about $\ell = 3 \pm 0.5\ \mathrm{cm}$.

I have precision calipers but I can't for the life of me remember where they went to, so I'll have to guess the diameter of my hair. Ask anyone I know - I have luxuriously thick silky hair - like a gopher. So I'll go for a bit over the mean value given by wikipedia $\ell = 90\ \mathrm{\mu m}$ with a rather substantial error of say 20%.

According to the impressive sounding book by Clarence Robbins, Chemical and Physical Behavior of Human Hair, the density of human hair varies a bit depending on humidity. I'll take a middle of the road value (Table 9.8 ibid) of $\rho = 1.3\ \mathrm{g/cm^3}$ with an error on the order of 2%.

Putting it all together gives

$$ N \approx 100000 $$

Note that the undertainty in the diameter $d$ dominates the error or this estimate - 20% error in $d$ translates to about a 40% error in $N$!

So yes I basically chose $d$ to give the value I wanted to get. :) I need to find my calipers...

Edit: Just remembered I have a laser pointer, so I can do a diffraction measurement. Watch this space...

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This is old, but I would LOVE to see your diffraction measurement.... –  levitopher Dec 10 '13 at 4:49
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Firstly, I assume that we have 300 hairs per square cm on our head. This can be tested by waxing an area of 1cm^2 on your scalp and counting the number of hairs that are removed.

Step 2, we must calculate the area of the scalp, and we assume 100 hairs per square cm applies to the whole area of the scalp.

I assume the radius of my head is sphere. I measured the circumference to be 60cm.

$C = 2\pi r$

$r = \frac{C}{2\pi} = \frac{60}{2\pi} = 9.55cm$


$A = \pi r^2 = \pi \times 9.55^2 = 286.4 cm^2$

Now I will assume that only 4/5 (slightly more than half) of this ball is covered in hair.

Therefore area covered in hair = 286.4*0.8 = 214.72 cm^2.

Finally we calculate the number of hairs to be:

textNo. of Hairs = 214.72*300 = 64416 hairs

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up vote 0 down vote accepted

First estimate roughly the no. of hairs in 1mm^2 and consider the distance between two hairs is uniform all over the head and calculate the area of the whole head and subtract the area of the head having no hair.then multiply that with the hair contained in 1mm^2. hair is supposed to distribute uniformly.

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