Latexluv opened this issue on Jun 04, 2010 · 182 posts
bagginsbill posted Mon, 14 June 2010 at 2:27 PM
The yellow slice is tha angle of incidence, alpha. Keeping our point of view of the A tube as before, the B tube has moved along the dotted black arc.
The width of the obstruction is still D. But the width of the gap or transmittance section, T, has been reduced.
The effective spacing because of the viewing angle is marked by the black bar E. It is smaller than the original, true spacing, S. How much smaller is it?
Well, trig tells us that:
cos(alpha) = E / S
Therefore, E = S cos(alpha). But since I made S = 1, that's just cos(alpha).
Now the opacity is the width of the obstructions divided by the width of the effective spacing.
So: opacity = D / E
Voila - effective opacity = D / cos(alpha)
If you're not convinced that the scale doesn't matter, then let's be more explicit.
Letting r represent the actual radius of a fiber, given the chosen density, the radius must be:
r = DS / 2
The actual obstruction is twice the radius, because it's from A and B, so the obstructed area is SD.
And the true effective spacing is S cos(alpha)
The ratio of obstruction to effective spacing is then:
SD / (S cos(alpha))
The S in numerator and denominator cancel, leaving
D / cos(alpha)
This means S doesn't matter.
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