July 05, 2020, 12:45:57 AM
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Topic: Colloidal Particles on the Head of a Pin  (Read 8682 times)

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Offline Gosseyn

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Re: Colloidal Particles on the Head of a Pin
« Reply #15 on: March 15, 2009, 12:13:46 PM »
Sorry, it would be 2, my mistake. I was thinking about diameter, not about radius.

And 1.73 is just a square root of 3, isn't it? So for side 2 that's correct.

http://en.wikipedia.org/wiki/Equilateral_triangle

OK, since we're just looking at percentages, we can just assume that the three circles cover the whole of the pin, and then take half of pi and divide it by the total area of the triangle.  So 1.57/1.73 = 90.79%, which is about what I think you said it would be in an earlier post.  And it doesn't matter how many particles there are, we'd get the same percentage of coverage and empty space.  Thanks!!

Offline Borek

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Re: Colloidal Particles on the Head of a Pin
« Reply #16 on: March 15, 2009, 01:10:06 PM »
Exactly :)
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