Assume a Spherical FutureBaby...

Welcome to today's exciting episode of "How Big a Dork Am I?" Today, we'll be discussing the making of unnecessary models:

i-6cc855b80ae87e3cb5ab039cdf6b8855-spherical_baby.jpg

In this graph, the blue points represent the average mass in grams of a fetus at a given week of gestation, while the red line is the mass predicted by a simple model treating the fetus as a sphere of uniform density with a linearly increasing radius.

The "model" was set up by taking the 40-week length reported at BabyCenter, and dividing by two to get an approximate radius for the spherical baby. Then I assumed that the actual radius increased linearly from zero to the final value, calculated the volume of the sphere, and multiplied by a constant density to get reasonable agreement between the model and the data.

If you take the numbers I put into this, and use them to estimate the mass of a cell in this model baby, you find that a cell with a volume of one cubic micron (10-18 m3) would have a mass of about 50 femtograms, which is kind of low, but remarkably good for such a silly model.

Oh, the things I will do to amuse myself...

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That's some serious procrastination you've got going there. Grading? Writing up a paper?

That's some serious procrastination you've got going there. Grading? Writing up a paper?

I was between a couple of meetings, and trying not to grade lab reports.

NERD!

(Also, there's no error bars.)

By John Novak (not verified) on 20 May 2008 #permalink

Here I am being an ambulance-chaser theorist. Why should the radius increase linearly with time? Here's my theory: because the rate at which nutrients can enter the fetus is proportional to the surface area (maybe). According to this theory, we would predict: (where M = mass, A = surface area, and t = time)

dM/dt = k A

where k is an empirically determined constant. Assuming that M scales as the cube of some characteristic length, R, and A scales as the square of R, then we have

k1 d/dt (R^3) = k k2 R^2

where k1 and k2 are two other constants. This
simplifies to

d/dt R = k k2/(3 k1)

We can absorb k1, k2 and the factor 3 into
the unknown constant k to get

d/dt R = k

Do I win? Isn't that even dorkier than your original post?

Should reach 9.2 metric tons in 10 years. Hope your floors are reinforced.

By Anonymous (not verified) on 20 May 2008 #permalink

Congrats, by the way. It's blissfully enjoyable until they learn to talk. Anyway, I actually think that is pretty darn cool, by the way.

Congrats, by the way. It's blissfully enjoyable until they learn to talk. Anyway, I actually think that is pretty darn cool, by the way (the graph).

John Novak: Also, there's no error bars.

They just don't show up on the graph...
The data are specified to +/- 1 gram. I'm sure that makes sense. Really.

dr. dave: Would this be a perfectly smooth, frictionless, spherical baby?

But of course. Can't you tell from the ultrasounds?

(Actually, Kate would probably beg to differ regarding the "frictionless" part...)

Daryl: Here I am being an ambulance-chaser theorist. Why should the radius increase linearly with time?

It's not, really. You can tell from the graph-- the slope of the model is too high early on, and too low later. A somewhat higher power would probably be a better fit-- taking out the early part (which is fairly close to exponential, almost doubling every week) and the last three points, weeks 16-40 fit pretty well to t^4.

Linear growth is the easiest thing to simulate, though.

Yes, I just cranked that into Excel and had it do a power-law fit. God, I'm a dork.

Since your baby is expanding linearly with time, Ω â 0. I'm sure you and your wife are glad that his/her self-gravity isn't significant.