[math-fun] Human body shape, size, mass
Tall people are *not* just scaled up versions of short people. Hence, human body weight does *not* scale proportionally to height^3. The so-called BMI (body mass index) suggested by health and fitness scumballs, would indicate that the mass of "healthy weight" adults should scale proportionally to height^2. For children of both genders, median heights are related to median weights at all ages 2-19 by the following empirical law found by Nick Korevaar http://www.math.utah.edu/~korevaar/ACCESS2003/bmi.pdf for median height and median weight USA boys and girls ages 2-19: MedianWeightInPounds = 0.002384649 * MedianHeightInInches^2.593828 fits extremely well. So for example I'm 68.5 inches = 174 cm so if I were age 19 and median among both genders, I "should" be weight 137.7 pounds = 0.002384649 * 68.5^2.593828 = 62.5 kg. Unfortunately I'm considerably older than 19 and currently weigh 155 lb = 70.3 kg. In metric, Korevaar's formula is MedianWeightInKg = 14.848467 * MedianHeightInMeters^2.593828. Other fitters using other data sets for people below age 20 in different countries have found p's throughout the interval [2.59, 2.93]. Why the exponent p=2.59383? Korevaar's formula also would predict (at least naively) that waist circumference should be proportional to Height^q where q=(p-1)/2=0.79691. And empirical fits to data for people below age 20 indeed have found q's throughout the interval 0.68<q<0.82. FAILURE #1: If a human were shaped like a height=H, radius=R cylinder of constant material, and wished to just-survive "thin column Euler buckling" under a load proportional to her weight and hence to H*R^2, then H*R^2 = const * R^4/H^2 hence R is proportional to H^1.5, hence q=1.5. FAILURE #2: Same, but human sprints at stride duration proportional to height^(1/2) [ideal pendulum] and hence ground speed also proportional to height^(1/2) if constant leg angle. This agrees with the fact the fastest sprinters are tall. Usain Bolt is 6'5". Each stride the force exerted scales like weight*height/height^(1/2) = H^(3/2)*R^2, not weight = H*R^2. Hence redoing the Euler buckling argument we find H^(3/2)*R^2 = const * R^4/H^2 so H^(7/2) propto R^2 hence q=7/4=1.75. Well... that only made it worse. I think your body is in no danger of Euler buckling so those two models were irrelevant. We need truly limiting factors for human bodies. FAILURE #3: You generate heat presumably proportional to your mass, which must be dumped thru your surface. Hence you want your surface area which behaves like H*R to scale like your mass H*R^2, hence R=constant, hence q=0. FAILURE #4: Actually, your body heat does not scale proportionally to your mass, if you believe animals have smaller metabolic rates. In fact animal's heat production scales like mass^(3/4) under a fractal-blood-vessel-network model, see http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2936637, which agrees with empirical data on animals. So redoing that we find H*R scales like (H*R^2)^(3/4) hence H^(1/4) = R^(1/2) hence R=H^0.5 hence q=0.5. That is better but still too low, outside the observed interval 0.68<q<0.82 for humans. This q agrees with the "body mass index" kludge though. FAILURE #5: To prevent your knees and hips breaking, you need their cross sectional area R^2 to be proportional to your weight H*R^2 (stationary standing) or to H^(3/2)*R^2 (sprinting model). Both would imply H=constant. Wrong. FAILURE #6: To prevent your pelvis breaking, modeling pelvis as "lintel" type beam of length R, thickness also proportional to R (i.e. pelvis assumed to have constant shape and constant material properties), you need your moment-load = weight*R to scale like R^4 (stationary stand) implying q=1/2 (wrong, same as in fail#4 and for BMI) -- or instead... SUCCESS ON TRY #7: ...with the sprinter load model, H^(3/2)*R^2 = R^4 for q=3/4=0.75. Hey, that finally agrees superbly with the empirical data for humans under age 20, which says 0.68<q<0.82. And this q=3/4 would predict p=2.5. OK, so we finally were able to predict the correct power law q for humans in the growing-age range under a model where the limiting factor is the risk of your pelvis breaking when you sprint. This also explains why women are shorter than men. For birth, women need to have wider pelvises. Given that is so, they are forced under our model to be shorter than men. Now once you are an adult, you stop growing and your pelvis and height stay invariant. But your weight and waist circumference need not stay fixed. The "optimum" weight is the one at which (among people your height & gender) mortality rate is minimized. Optimum weights were computed from mortality statistics by Metropolitan Life Insurance co in 1943 with revision 1983. See http://www.ncbi.nlm.nih.gov/pubmed/6707396 http://www.ncbi.nlm.nih.gov/pubmed/8332284 http://www.ncbi.nlm.nih.gov/pubmed/6362514 Here are the metlife tables, where you are warned they are expected to be noisy at especially tall or short heights since less data there. MORTALITY-MINIMIZING WEIGHTS (lb) FOR WOMEN AGES 25-59, HEIGHT SMALL, MEDIUM, LARGE FRAMES 4'9" 102-111 109-121 118-131 4'10" 103-113 111-123 120-134 4'11" 104-115 113-126 122-137 5'0" 106-118 115-129 125-140 5'1" 108-121 118-132 128-143 5'2" 111-124 121-135 131-147 5'3" 114-127 124-138 134-151 5'4" 117-130 127-141 137-155 5'5" 120-133 130-144 140-159 5'6" 123-136 133-147 143-163 5'7" 126-139 136-150 146-167 5'8" 129-142 139-153 149-170 5'9" 132-145 142-156 152-173 5'10" 135-148 145-159 155-176 5'11" 138-151 148-162 158-179 All weights overstated to include 3lb of clothing. MORTALITY-MINIMIZING WEIGHTS (lb) FOR MEN AGES 25-59, HEIGHT SMALL, MEDIUM, LARGE FRAMES 5'1" 128-134 131-141 138-150 5'2" 130-136 133-143 140-153 5'3" 132-138 135-145 142-156 5'4" 134-140 137-148 144-160 5'5" 136-142 139-151 146-164 5'6" 138-145 142-154 149-168 5'7" 140-148 145-157 152-172 5'8" 142-151 148-160 155-176 5'9" 144-154 151-163 158-180 5'10" 146-157 154-166 161-184 5'11" 149-160 157-170 164-188 6'0" 152-164 160-174 168-192 6'1" 155-168 164-178 172-197 6'2" 158-172 167-182 176-202 6'3" 162-176 171-187 181-207 All weights overstated to include 3lb of clothing. Oh wonderful, it seems I am in the min-mortality weight range for my height (assuming medium frame). Metlife says you can assess your frame size by elbow measurement: "Bend forearm upward at a 90 degree angle. Keep fingers straight and turn the inside of your wrist toward your body. Place thumb and index finger of other hand on the two prominent bones on either side of the elbow. Measure space between your fingers on a ruler.(A physician would use a caliper.) Compare with tables below listing elbow measurements in inches for medium-framed men and women. Measurements lower than those listed indicate small frame. Higher measurements indicate large frame." ELBOW MEASUREMENTS FOR MEDIUM FRAME MenHeight ElbowBreadth WomenHeight ElbowBreadth 5'1"-5'2" 2.5-2.875 4'9"-4'10" 2.25-2.5 5'3"-5'6" 2.625-2.825 4'11"-5'2" 2.25-2.5 5'7"-5'10" 2.75-3.0 5'3"-5'6" 2.375-2.625 5'11"-6'2" 2.75-3.125 5'7"-5'10" 2.275-2.625 6'3" 2.825-3.25 5'11" 2.5-2.75 I get 2.8 inches so I'm medium frame. Fitting the medium-frame men optimum weights (now with the 3lb overstatement removed) to a function of height, I find by log-linear-fit of the data (61,log(133)), (63,log(137)), (64,log(139.5)), (65,log(142)), (67,log(148)), (68,log(151)), (70,log(157)), (72,log(164)), (73,log(168)), (75,log(175.5)) that WeightLb = 0.4881 * HeightInch^1.3605 by which formula my ideal weight is allegedly 153.5 lb. Wow, I'm only 1.5lb off! A good linear fit, but it seems slightly worse than the above power law, is WeightLb = 2.93*HeightInch - 45.6. Fitting the medium frame women optimum weights (now with the 3lb overstatement removed) to a function of height, I find a linear fit works very well, WeightLb = 2.93*HeightInch - 55.9 which conveniently has the exact same slope as for the men. Plots: https://dl.dropboxusercontent.com/u/3507527/IdealHumanWeights.png You can't really deduce a good power law from the metife adult data because it spans only a small domain of heights, unlike the ages 2-19 data, so I don't see how to deduce the right exponent without having a huge error bar. Now, you can't control your height (well... certain people apparently are feeding their children HGH to make them taller... which as we'll see seems to be a really stupid idea) but nevertheless we can be envious of those whose heights are optimal. The "optimum" height for your gender is the one that produces the longest lifespan. As far as I can tell from these tables http://rangevoting.org/HeightWeightUSA2008.pdf the optimal heights are about MEN: 5'5" to 5'6" (about 4 inches below median among USA men ages 20-30) WOMEN: 5'2" to 5'3" (about 2 inches below median for 20s-ages USA women).
CORRECTION... I no longer believe in my "successful model." MacKay also published a model http://arxiv.org/abs/0910.5834 based on pelvis bone "sag distance" relative to pelvic breadth being equal for all size people, but that makes no sense to me. Relative sag seems irrelevant evolutionarily. Breaking your pelvis does matter. It is about strength, not relative sag distance. So... now I'm back to not believing in any model including any of my own. So then, examining MacKay's model, he actually models the human body as height=z > breadth=x > depth (front to back) = y. The pelvis is assumed to have (up to constant factors) breadth (hip to hip) x, height z, and thickness (front to back) y. The load on the pelvis as a torque-moment (standing stationary) then is BodyWeight*x, which is x*y*z*x. Everything will be up to constant factors. The ability of the pelvis to withstand that load without cracking is proportional to I/z where I=area moment of inertia=y*z^3. Hence we'd want x*y*z*x and y*z*z to be proportional, i.e. z = x^2. If instead of standing stationary we sprint using my sprinting model, the max load on the pelvis as a torque-moment then is BodyWeight*x*sqrt(z), which is y*x^2*z^(3/2). The ability of the pelvis to withstand that load without cracking again is proportional to y*z^2. Hence we'd want these to be proportional, i.e. z^(1/2) = x^2 or z = x^4. Meanwhile MacKay finds z = x^1.5 using his relative sag criterion which I claim makes no evolutionary sense. MacKay then offers arguments about bipedal balance and heat conservation that perhaps y and z should be proportional. It has in fact been argued one of the great evolutionary wins of humans is that they can SHED heat well, which is why we are not covered with fur, hence can outrun prey without dying of heat stroke ("persistence hunting" via long distance running). Humans are actually one of the best animals at long distance running. This is exactly the opposite of MacKay's heat argument and I suppose would tell us that y should be proportional to x not z? Be nicer if we had data on that. With y propto z we then would find: MacKay fixed-relative-sag Model: weight = height^(8/3) * Stationary standing break-strength model: weight = height^(5/2) * Sprint break-strength model: weight = height^(9/4) With y propto x instead, we then would find: MacKay fixed-relative-sag Model: weight = height^(7/3) * Stationary standing break-strength model: weight = height^2 Sprint break-strength model: weight = height^(3/2) assuming I have not made more mistakes. I starred(*) the ones seeming not absurdly in disagreement with real data. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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Warren D Smith