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Given a sphere and a plane are to intersect so that the plane is mostly inscribed in the sphere (I have attached an image I have drawn as I find difficulty in expressing what I mean by this. Diagram), how can I calculate the ratio between the surface area of the part of the plane inscribed in the sphere and the surface of the whole sphere is?

I know the dimensions d and L in meters (from the diagram). Where d is the distance from the centre of the sphere to the centre of the plane, L is the length of the square plane, and r is the radius of the sphere.

I have created a formula using Pythagoras, though it seems to break down when I use small values of d, I assume this is because L is so large and the sphere then becomes massive, which in turn means that the missing surface area is huge.

If anyone could point me in the right direction that would be great, I have started looking at planes and sphere interactions though I am stuck, to be honest.

Thank you.

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  • $\begingroup$ I don't see any sphere in the diagram, do you mean a cone? And does the plane cuts the cone in a $90$ degrees angle or do you allow skewness? $\endgroup$
    – Levent
    Nov 17, 2021 at 11:51
  • $\begingroup$ What's the $m$ in your diagram? $\endgroup$
    – Magma
    Nov 17, 2021 at 11:56
  • $\begingroup$ Said in a simpler way: you look for the ratio (intersection disk area)/(sphere area) knowing $d$ and $L$. $\endgroup$
    – Jean Marie
    Nov 17, 2021 at 11:58
  • $\begingroup$ Hi @Levent the plane is perpendicular so no skew and the cone is just a section of the sphere, I apologise for being unclear. $\endgroup$
    – a w
    Nov 17, 2021 at 12:04
  • $\begingroup$ Hi @Magma, the m in my diagram is to indicate the dimension is metres. $\endgroup$
    – a w
    Nov 17, 2021 at 12:05

1 Answer 1

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Suppose $r$ is the radius of the sphere, and $d$ is the distance from the sphere center to the plane. For the time being, assume that $d \le r$, so the plane actually intersects the sphere.

The intersection of the sphere and the plane is a disk. By Pythagoras' theorem, the radius of this disk is $\sqrt{r^2-d^2}$, so its area is $\pi(r^2-d^2)$.

The surface area of the the sphere is $4\pi r^2$.

So the ratio you asked about is $$ \frac{r^2-d^2}{4 r^2} $$

If $d=0$, then the disk has the same radius $r$ as the sphere, so the disk area is $\pi r^2$ and the ratio is $1/4$, which is what the formula above gives us.

If $d$ is just slightly less than $r$, then a small change in $d$ will cause a much larger change in the area of the disk, so numerical computation problems are to be expected. You might get subtractive cancellation when you compute $r^2-d^2$. There's nothing wrong with the mathematics, this is just a computer problem.

If $d>r$, then the plane doesn't intersect the sphere, so the disk area is zero, which makes the ratio zero, also.

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  • $\begingroup$ Thank you so much, this is such a detailed and thought-through response. $\endgroup$
    – a w
    Nov 17, 2021 at 12:34
  • $\begingroup$ I don't seem to understand why the radius of the sphere has to follow the pythagorean theorem, I don't see a right angles triangle anywhere? Could anyone please elaborate. $\endgroup$ Nov 14, 2023 at 4:08

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