Gah! This is why understanding math is important!

[barnabas_truman]

I just realized that "Body Mass Index" is defined as mass divided by height squared;
BMI = m / h^2.
This leads me to wonder if Adolphe Quetelet was an idiot (hint: certainly not, but I can't think of a better explanation), and, more to the point, why anybody ever thought his formula was a good idea.

Allow me to explain. This is going to involve a bit of math and physics, so buckle down and be prepared to look things up or ask questions as needed.

Suppose, for the sake of the argument, that every body has the same proportions and the same density.

Then the mass of a body is its volume times density:
m = d * V
where d is density and V is volume.
But volume is proportional to height cubed:
V = k * h^3
where k is some proportionality constant.
We're assuming for the moment that both d and k are the same for everyone.

If we substitute these into Quetelet's formula, we find that
BMI = d * k * h^3 / h^2
Two of the h's cancel out, leaving
BMI = d * k * h
In other words, BMI is equal to a constant times one's height!

Body Mass Index is a measure of height, not of obesity! What were they thinking??


Of course, the assumptions that d (density) and k (proportions) are the same for everyone isn't true at all in reality. So BMI is really measuring the product of density, height, and a constant involving body proportions. But why allow height to be a factor? The end result of this is that, all other things being equal, tall people have higher BMI. Isn't the whole point of dividing by height to cancel out the influence of height entirely?

On the other hand, if BMI is defined as mass divided by height CUBED, the extra h-factor also cancels, and BMI becomes the product of d and k, which actually means something. Why isn't it defined this way to begin with?

Alternatively, mass divided by (height squared times shoulder width) might work even better--that at least attempts to scale it individually for different body structures, something that BMI ignores entirely.


P.S. It should also be noted that Quetelet's goal was to apply rigorous methods from statistics to the study of populations, and that he never intended BMI to be used to diagnose individuals. However, I still can't imagine why he wouldn't at least take the square-cube law into account!

Comments

[mutive.livejournal.com]

So weight = density * volume = a constant (assuming constant density) * height * width * other width

BMI = weight/height^2 = density * width * other width/height

Unless you are a freak of nature, your, um, width and other width = < your height. So...taller people should actually have lower BMIs in general, correct?

The problem with your math, I think, is that our volume is not height cubed...that would assume that we're spheres. If anything, we're more like cylinders (i.e. h*r^2), although this isn't entirely accurate, either...

Still, I rather like BMI as a rough approximation. If your BMI is 22, you're pretty normal. If it's 40, you probably either are quite plump or are a crazy body builder. It's hardly perfect, but it's better than "if you are 120 lbs as a female, you are over weight", without acknowledging that a very tall woman is bound to be heavier than that.

[barnabas-truman.livejournal.com]

Unless you are a freak of nature, your, um, width and other width = < your height. So...taller people should actually have lower BMIs in general, correct?

No, because if we assume identical proportions, then width and other width both scale up as height increases, so taller people would have larger width and other width too.

For ANY shape, volume is proportional to height (or any linear measurement, really) cubed. What the shape determines is the constant of proportionality (k, in my notation).

And if we assume (for the sake of the argument) that all people have identical proportions, then both width and other width are proportional to height, so we could say that
width = c₁ * h
and
other width = c₂ * h
so your formula of
density * width * other width / height
becomes
density * c_1 * height * c_2 * height / height
Again, one of the heights cancels out, leaving
density * c_1 * c_2 * height
which ends up being, once again, a constant of proportionality times height. Note that c_1 * c_2 ends up being what I called k anyway.

Any formula that compares mass to area is not a "rough approximation"; it's just wrong.

[mutive.livejournal.com]

Well, width and "other width", should scale, but not in direct proportion to height. So Weight = h * c1 * c2 * density.

BMI then = c1 * c2 * density/height

If height > c1 or c2 (which I'd assume is the case, and density is assumed constant), then taller height = lesser BMI, as height is the inverse.

Now you do get into proportions, as c1 and c2 are dependent on height. But still, I doubt that it's a direct proportion. I'd assume that it's more like h^(1/3) or h^(1/10) or whatever. (It says something that petite and long clothing are not typically scaled. You're still considered a 12 whether you're 5' or 6'. So not that a tall person is not expected to be wider than a short person, but he/she is not expected to be *that* much wider. And let's be frank...even if we assume that both C1 and C2 are proportional to the root of height, we still end up with the BMI formula. It is only if we assume that both are proportional to something > h^(1/2) that we find that BMI is unfair towards the tall).)

I do agree that a comparison of mass to area is incorrect. BUT, that said, it's an approximation. I'd hardly say "hey, you have a BMI of 25.2...you're overweight and MUST lose a few pounds!" But, again, it's convenient to be able to say "you have a BMI of 40...time to watch what you're eating." or "you have a BMI of 16, you're probably anorexic" or whatever.

[barnabas-truman.livejournal.com]

I suppose it must also be very convenient for insurance providers that they can use a faulty approximation to deny insurance to their customers.

[mutive.livejournal.com]

Well, for better or worse, unless you use correlations, insurance is close to meaningless. *shrugs* I'm still pissed off that as a woman I have to pay significantly more than a man does because I *might* become pregnant. But, eh, there it is.

[witless-nerd.livejournal.com]

The density of different tissues, namely fat, flesh and bone are markedly different, to the point that some weight loss clinics used density as a measure of fatness.

Secondly is that proportions do vary to some degree among the population, having it as a factor would allow for normalization of the data across the population and accounting for the percentage of bone in a body.

With height as a good approximate for 'ideal' volume, and k to normalize data, and the actual density of a person we can have a rough index of flesh to fat proportions, using weight and height and assuming everyone has the same k, we get an even more rough idea of the composition of someone's body.

Now of course a urine test for fatty acids is a far better indicator that weight clinics now use.

[serpent-star.livejournal.com]

Great discussion. Absolutely wonderful application of mathematicks to practical problems!