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I'm working on building a chicken army and I'm trying to find out how much metal or kevlar (still deciding) I need to make armor for the chickens. this measurement does not need to be exact I'm just trying to get an estimate for how much I will need. You will be spared when my chickens take over the world if you give me a working answer.

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    $\begingroup$ I’m voting to close this question because making clothes for animals is a solved problem that doesn’t really involve any physics. $\endgroup$ – rob Mar 24 at 15:53
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    $\begingroup$ +1 vote for reopen. This is a completely legitimate and interesting physics problem. What if the chicken is a fractal ... $\endgroup$ – gandalf61 Mar 24 at 16:24
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    $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Mar 24 at 18:49
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    $\begingroup$ Please take further back-and-forth about the on-topic-ness of this question to this meta discussion. Please do not vandalize this question. $\endgroup$ – rob Mar 25 at 20:12
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    $\begingroup$ Does this answer your question? How do I experimentally measure the surface area of a rock? $\endgroup$ – Jonas Mar 25 at 21:10
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Assume a spherical chicken, of some radius $r$. The surface area of said bird would then be $4\pi r^2$, which would be a good approximation of the amount of metal required to cover it completely.

If you assume $r\approx 30\text{ cm}$, that would come to an area of around $$A \approx 1\text{m}^2/\text{chicken}.$$

If the metal has a thickness $h$, this would be a total volume of roughly $V = 4\pi r^2 h.$ If $h\approx 5\text{mm}$, that means $$V \approx 5 \times 10^{-3} \text{m}^3/\text{chicken}.$$


How much would such an armour weigh? Unless you have some sort of "super-chicken" (though I immediately regret giving you the idea), you would require the weight to not be too substantial. If you are using some sort of metal (say, Aluminium or Steel) its density would be around $\sim 5 \times 10^3 \text{ kg/m}^3$. As a result, the mass of a metallic armour would be around: $$M = \rho \times V\approx 25 \text{ kg}/\text{chicken}.$$

Pretty heavy for the average chicken, I would think. Using Kevlar would reduce the mass by about a factor of $5$ (its density is around $1000 \text{ kg/m}^3$), so that would make it a slightly more bearable $5 \text{ kg}$ per chicken.

P.S: Come the revolution, my house will be the one flying the chicken flag. Inform the troops.

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    $\begingroup$ I disagree with the use of the phrase "said bird" after you assumed a spherical chicken. $\endgroup$ – Winston Mar 25 at 9:15
  • $\begingroup$ A chicken of this size would be immense: $60$cm across. $\endgroup$ – ZeroTheHero Mar 26 at 21:52
  • $\begingroup$ @ZeroTheHero I was surprised myself to learn that the average chicken stands $70\text{cm}$ tall, which is what led me to choose $30\text{cm}$, though I admit I was thinking of an army or rather large roosters. A more reasonable estimate might be around $10-15\text{cm}$, granted. $\endgroup$ – Philip Mar 27 at 5:34
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The astrophysics answer.

Take a representative chicken, put it in a cold room that is lined with infrared detectors measuring the flux in several wavelength bands.

Assume the chicken is a blackbody and fit a Planck function to estimate both temperature and emitting surface area.

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    $\begingroup$ Now, all we need is an answer to this question using every field in physics. I'm stilling waiting for the answer where we use black hole physics to measure the surface area of chicken. $\endgroup$ – Buraian Mar 25 at 9:28
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$\ldots$ another way is to use the physics of thermal radiation. Here the basic pieces of physics we need are Kirchoff's law, the Stefan Boltzmann law, and some standard thermal physics.

Kirchoff's law states that the absorptivity and emissivity are equal (for a given body in a given internal state): $$ \epsilon = \alpha $$ This is useful because one can measure absorptivity $\alpha$ more easily than emissivity $\epsilon$. Having done so (see below), one then uses that in thermal equilibrium the radiative emission (power per unit surface area) is $$ F = \epsilon \sigma T^4 $$ where $\sigma$ is the Stefan-Boltzmann constant.

Now here is the method. First, we will want to use the chicken in a state of thermal equilibrium, so this is not possible for a live chicken. But if we are careful then it may be possible to eat the chicken at the end. The first step is to kill and cook the chicken. You have to cook it, because later on we do not want chemical processes inside the chicken; they will introduce unknown amounts of internal energy changes or an unknown internal state. So, having cooked the chicken (but see later for some further measurements that have to be made) we place it in a fridge and measure its temperature as a function of time. For best results, remove the air so as to eliminate conduction and convection. Then the rate of loss of energy is $$ \begin{array}{rcl} P = \oint F dS &=& \oint \epsilon \sigma T^4 - \alpha \sigma T_{\rm fridge}^4 dS \\ &=& \oint \alpha( \sigma T^4 - \sigma T_{\rm fridge}^4 ) dS \\ &=& \sigma ( T^4 - T_{\rm fridge}^4 ) \oint \alpha dS \\ \end{array} $$ where in the second line I used Kirchoff's law, and then I assumed the temperature was uniform across the chicken and walls of the fridge. I am allowing that $\alpha$ may vary across the surface of the chicken. This surface integral is going to give us the surface area we want, but we should note what surface we will obtain: the area we will get from this method is in fact the area of the convex envelope of the chicken. This is the minimum area that an elastic membrane wrapped around the chicken would need to entirely cover the chicken. But this is just what we want to know. This method is good for getting a lower bound for the chicken armour calculation.

I will explain how to measure $\alpha$ in a moment. First let's complete the argument. The chicken temperature will fall as a function of time in such a way that the total energy leaving the chicken equates to the loss of its internal energy: $$ P = \frac{\partial U}{\partial t} = C_p \frac{\partial T}{\partial t} $$ where one should take constant pressure as understood in the partial derivatives. So we need to know the constant-pressure heat capacity of our chicken. To find this out, we shall need to heat it up again, supplying known amounts of heat and measuring the temperature. To supply a known amount of heat, one way is to put an electric heater element near the chicken and thermally isolate the whole. Here we must confront a food safety issue. Re-heating cold chicken causes multiplication of any residual salmonella and consequently is a bad idea if the chicken is intended to be eaten. So there are two options.

Option 1. Do not eat the chicken. In this case one can safely reheat the chicken, measuring its heat capacity as a function of temperature, and also check that it is fixed at any given temperature because chemical and biological processes have ceased.

Option 2. We wish to eat chicken. To make this possible one could measure the heat capacity during the cool-down. It is awkward because now there are two measurements going on at once. At each temperature one must briefly isolate the chicken, raise the temperature a little using a known amount of energy supplied to a heating element, and then allow the cooling to continue. Thus one gets a series of heat capacity measurements along with the overall time-constant for the radiative cooling.

Finally, we have to measure the absorptivity $\alpha$. This can be done by illuminating the chicken with infra-red radiation and detecting the scattered (reflected) radiation if there is any. This may be one of the more difficult parts of the method, from an experimental point of view. Absorptivity may also be a function of temperature, but I think it will not be a strong function of temperature for a cooked chicken. One can either measure the absorptivity of each small region of the surface, or else measure the average over the whole surface. It is just this average $\bar{\alpha}$ we need, for we have $$ C_p \frac{\partial T}{\partial t} = \sigma( T^4 - T_{\rm fridge}^4 ) \bar{\alpha} A $$ so the area of the convex envelope of the chicken is $$ A = \frac{C_p}{\sigma( T^4 - T_{\rm fridge}^4 ) \bar{\alpha} } \frac{\partial T}{\partial t}. $$ This is the minimum area of armour that will be needed to make similar such chickens ready for combat.

P.S. While I was composing this answer the same concept was also proposed more briefly by ProfRob.

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    $\begingroup$ I don't understand why this answer was donvoted $\endgroup$ – Buraian Mar 25 at 16:41
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    $\begingroup$ @Andrew Steane: sorry but won't the cooking dehydrate the bird and make it shrink? This will lead to underestimates. $\endgroup$ – Winston Mar 25 at 16:51
  • $\begingroup$ @Winston good point. $\endgroup$ – Andrew Steane Mar 25 at 17:18
  • $\begingroup$ @AndrewSteane I think your arguement works even if chicken isn't in thermal equib [see here ](physics.stackexchange.com/a/584219/236734) $\endgroup$ – Buraian Mar 26 at 10:51
  • $\begingroup$ @Buraian The emitted flux is $\epsilon \sigma T^4$ for some but not all non-equilibrium states. Consider an LED or a laser for example, also bioluminescence and many chemical processes. $\endgroup$ – Andrew Steane Mar 26 at 13:06
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Use the ancient Japanese art of Gyotaku. This method was used to record the catches of Japanese fishermen. They would paint the fish with ink and then roll the fish onto rice paper to get a print. enter image description here

  1. Dye the chicken in harmless food coloring.
  2. Carefully roll one side of the dyed chicken onto a large piece of paper.
  3. Wash the chicken before returning it to the coop.
  4. Estimate the area of the image on the paper using polygons. Multiply the result by 2 (to account for the other side of the chicken).

An advantage to this method is that it will also provide a template for stamping steel sheet to be formed on a rigid chicken armory dummy, or to cut the Kevlar prior to sewing.

I embrace my chicken overlords and submit to their rule!

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  • $\begingroup$ Isn't this method using that fish are convex, while chicken aren't? $\endgroup$ – Norbert Schuch Mar 25 at 9:12
  • $\begingroup$ Could you explain why this works only for convex bodies? $\endgroup$ – Buraian Mar 25 at 9:14
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    $\begingroup$ For a rigid, concave body like a soup can, this method would vastly under-approximate surface area, since it wouldn't account for the surface on the interior of the can. A fish has plenty of concavities, but it is somewhat flexible, so one can roll it on paper and still get most of the surface. A chicken has even more concavities, so this method would also under-approximate surface area, but you could get a good "convex hull" approximation for armor estimation. $\endgroup$ – Connor Garcia Mar 25 at 16:40
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The trouble with existing answers is that they are subject to the classical limit of measurement precision. To get quantum-enhanced measurement precision you could illuminate the chicken with a specially-prepared light beam in a far-from-classical state such as a squeezed state or one of a class of entangled states called NOON states, or certain cleverly constructed superpositions of Gaussian states. I won't bother you with the technical details. In this way you are able to measure the distribution $|\psi|^2$ of the chickenfunction $\psi$ to improved accuracy.

You could alternatively gather many chickens and use classical averaging for the measurement, which can get, in principle, arbitrarily high precision for properties of an average chicken. But the above method, correctly carried out, causes minimal disturbance of the chickenfunction, and therefore it satisfies all the requirements of animal welfare legislation. But given your plans for world domination perhaps you don't care about that.

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    $\begingroup$ Given OP plan to make chicken take over the world, it seems they are more concerned with animal welfare than humans' and as such I believe your advice to be on point. $\endgroup$ – Winston Mar 25 at 8:48
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Archimedes' principle can also be used to measure rate of change of volume as the chicken is lowered into liquid. Sedate the chicken before doing this. Plot a curve of volume vs depth. This tells you the area of a cross section of the chicken at the surface of the liquid at each depth. Each cross-sectional part of the chicken can be approximated pretty well as a circular disk whose top surface has the corresponding area. Calculate the area of the chicken as the sum of all the circumferences times the thickness of the cross sections. Of course the leg cross sections will need to be treated as pairs of disks; and you'll need to treat the wings separately. Just make "gloves" that fit the wings, and measure the area of fabric required to make the gloves. You should end up with a surprisingly accurate estimate.

A possibly more accurate way would be to wrap a string around the chicken at different heights to get the length of the perimeter of whatever shape the chicken is at each height. Plot those perimeters as a curve, with the height axis along the x-axis. Calculate the area under the curve, as height times the spacing between the heights. This would handle any convex cross sections, and give you the area of a soap bubble covering the chicken parts. Again, wings would present an easily solvable problem.

And finally, if you plan to put the chicken in stew, first wring its neck, then wrap its body and wings separately with shrink-wrap plastic you can get from the butcher. Use a heat gun to shrink the plastic tight against the chicken. With a razor blade, make a long cut so you can peel the shrink wrap off the chicken. That has the area of the chicken, and even records the shapes of the parts you need to make for the armor.

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Use Archimedes' method to measure the volume of the chicken and then, assuming a spherical chicken (as suggested in the comments) you can calculate its surface: $$ V=\frac{4}{3}\pi R^3, S=4\pi R^2,\\ S=\alpha V^{\frac{2}{3}}, $$ where $\alpha $ is a chicken-specific coefficient.

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    $\begingroup$ and how do you measure that coefficient? The coefficient is important, not a minor detail. $\endgroup$ – user253751 Mar 26 at 11:54
  • $\begingroup$ @user253751 It is a coefficient of order $1$, so as an estimate it is quite sufficient. Moreover, once the coefficient is known for one chicken, we will have a formula that will apply with great precision to all of them, even though they may ahve different sizes. $\endgroup$ – Roger Vadim Mar 26 at 12:48
  • $\begingroup$ @Vadim Chicken don't vary in size by a significant amount - and if they do, the coefficient might well vary. $\endgroup$ – Norbert Schuch Mar 26 at 17:48
  • $\begingroup$ @NorbertSchuch they do vary in size. Now we can look for the data and analyze intra-species and inter-species variation for chickens... but it will only add the merit to the OP ;) $\endgroup$ – Roger Vadim Mar 26 at 19:30
  • $\begingroup$ @NorbertSchuch no offense, but it seems to me that your claim about the chickens not varying in size or the objections against the use of scaling in this case are of just as low quality as the OP. $\endgroup$ – Roger Vadim Mar 27 at 6:36
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Gather the chickens, say 20 of them (the more, the merrier). Slaughter (oops!) and spread skin on graph paper. Of course, make cuts where the skin is joint as answer will be inaccurate if skin is not lay flat. Count graph paper squares.

If only I could explain how I roll as I say this...

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