Originally posted by sasquatch672Can't see the forest for the trees
Okay Angie, here we go. Try to appreciate this intuitively first and then we'll go to the math. As the number of circles approaches infinity, the radius of an individual circle will go to zero, and therefore, the areas of the cirlces can reduce to points - where there is no area - space - between them. So the two areas will converge.
Okay, here w ...[text shortened]... pi*(L^2/n^2)*n*n^2/2
The n^2 terms cancel
= L^2 - pi*((L^2*(n/2))
Somebody check me?
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Originally posted by sasquatch672r = L/sqrt n
Okay Angie, here we go. Try to appreciate this intuitively first and then we'll go to the math. As the number of circles approaches infinity, the radius of an individual circle will go to zero, and therefore, the areas of the cirlces can reduce to points - where there is no area - space - between them. So the two areas will converge.
Okay, here w ...[text shortened]... pi*(L^2/n^2)*n*n^2/2
The n^2 terms cancel
= L^2 - pi*((L^2*(n/2))
Somebody check me?
If r = radius, as it must for this equation to work
Sv = L^2 - pi*r^2*n
then r = L/2*sqrt n, no? sqrt n gives the number of circles along one side of the square; dividing L by that number gives the diameter of the circles, not the radius!
If you square n, then r, by definition, must decrease by n^2/2.
If you square n, then r becomes divided by sqrt n.
n = 4, r = L/2*2 = L/4 => m = n^2 = 16, r = L/4*2 = L/8.
n = 9, r = L/3*2 = L/6 => m = n^2 = 81, r = L/9*2 = L/18.
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Originally posted by sasquatch672The spaces become infinitely small but the number of such spaces becomes infinite. The infinities cancel.
Yeah but that area goes to zero. As n--> infinity, r-->0, and as r approaches zero, you get an infinite number of infinitely small circles. Therefore, the space between the circles becomes infinitely small. This is the first day of calc II.
If you think this question is tough - try close packing arrangments that occur naturally in crystals, they'll blow your mind.
The answer is...
- a constant Pi over 4.
Now try the close packing arrangement with the circles arranged in a hexagonal arrangement and you have a real problem. Then take the problem to the third dimension and making the cube into trapeziods, then you might progress onto being good at maths. If this all seems daunting then maths to the experts and get a social life instead of a degree.
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Originally posted by XanthosNZis there one big square and lots of circles fitting as tightly as possible in it?
What?
As n--> infinity r--> 0 is correct however n never reaches infinity and therefore r never reaches zero.
As a circle inside a square that it touches on all sides takes up a proportion of the squares area independant of scale the answer is constant.
Area covered by circles is pi*0.5^2 or 0.785 (3 sf).
... or is there a square chopped into tiny little squares ... and then each square has a circle in it?
... the results are different.
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Originally posted by KneverKnightCorrect.
That would work for spheres and cubes too eh? Volume?
For example, for one sphere in a cube:
Volume of a sphere (radius r): (4 pi r^3) / 3
Volume of a cube (one side 2r): (2r)^3 = 8r^3
Ratio of cube wrt sphere = 8r^3 / [(4 pi r^3) / 3] = pi / 6 = 0.524 (to 3 sf)
r would cancel out regardless of the number of spheres so would also remain constant.
EDIT: specified 4 sf instead of 3 by mistake.
Originally posted by flexmoreWhy are the results different?
is there one big square and lots of circles fitting as tightly as possible in it?
... or is there a square chopped into tiny little squares ... and then each square has a circle in it?
... the results are different.
If you have one circle inside one square, surely the ratio of the area inside the circle to outside the circle is the same regardless of the actual size of the shapes.
Actually, that might be another way of looking at this ...