• anonymous
Problem set 2, PDF 2, Part 2 (d) - question regarding the max number of dimples of certain size. Does anybody understand how to approach this? I definitely get an answer where more than 100 dimples(r = 0.15) may exist on a sphere R=1.5. I tried many things: e.g. dividing sphere area by the dimple spherical cap area or dividing sphere area by a rectangle formed by a dimple, etc. In all cases I get an answer which says more than 100 dimples will fit on a sphere surface without overlapping. confused :-(
MIT 18.01 Single Variable Calculus (OCW)
  • Stacey Warren - Expert
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  • schrodinger
I got my questions answered at in under 10 minutes. Go to now for free help!
  • anonymous
Ah, yeah it's course 18.01 as taught in 2006
  • phi
use the rough approximation that each dimple takes pi r^2 area then the ratio of total area of the golf ball to the total area of dimples is \[ \frac{4 \pi R^2}{100 \pi r^2 }\] with R/r = 10 we get \[ \frac{4 \pi R^2}{100 \pi r^2 } = \frac{4 \pi \cdot 100}{100 \pi}= 4\] so there is 4 times as much surface area as dimple area. This should allow a "boundary ring" |dw:1447284676829:dw| I don't see how 100 would not fit.
  • anonymous
aha, thanks, phi. that's what I thought. I definitely get nearly the same answer as you (a bit less than 400 dimples. I did use spherical cap area formula expressed in terms of R and r which is 2*pi*R^2*(1-sqrt(1-(r/R)^2) ) instead of approximation you have suggested. That said, there might be a mistake in the problem or the r and R given in the original problem were different from 0.15 and 1.5.

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