%Increased Area, how does it work with conical aoe's?

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Abelarde wrote:

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The formula for the area of a quartercircle is easy, one quarter of a circle. A = .25 * pi * r^2 . Then, the formula for the radius of a semicircle is r = {4 * A / pi)^.5 , which simplifies to r = 2 * (A / pi)^.5 .

Your math is not correct. The area for a quartercircle is indeed A = .25 * pi * r^2 .
But the resulting formula for the radius is r = ((4 * A)/pi)^.5 .


A = .25 * pi * r^2 | * 4
A * 4 = pi * r^2 | / pi
(A * 4) / pi = r^2 | ^.5
((A * 4) / pi)^.5 = r

So, increased Area is not more powerful for conical AoE.

This is correct. For a simple demonstration of this concept, take the example of a circle with a total area of exactly 4. A quarter-circle with the same radius must have area 1, because it covers exactly one quarter of the area of the whole circle. Both have the same radius, which we'll call r1 - we don't need to calculate the value.

If we double the area of the circle, it now has area 8. It also has a larger radius, which we'll call r2. A quarter-circle with the same radius (r2) as this larger circle must by definition have area 2, since again, it's a quarter of the larger circle's area.

So we know that increasing the area of the whole circle with radius r1 by 100% (from 4 to 8 area), results in a circle with radius r2.

If we start with a quarter-circle of radius r1, we know it has area 1. If we increase that area by 100%, we know it will have area 2 - but we've already shown that a quarter-circle with area 2 has radius r2.

So increasing the area of a circle and a quarter circle with the same base radius (r1) results in the same final radius (r2).