Find centerpoint of sphere if you only have three random points on the surface and a radius

1

Is there a way to find the centerpoint of a sphere, if you have three points in a space and the radius of the sphere.

The 3 points have to be on the surface of the sphere and cant be inside the sphere.

I have searched far and wide but i cant find a solution. below what i am trying to do:

I have 3 points:

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I have a sphere with radius R that is placed on these three points, the points support the sphere on the surface so it cant fall through any point.

I want to find the centerpoint of that sphere:

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2

No, as the 3 points define infinite spheres. You can instead do the exercise with a circle.

Circle.ByBestFitThoughPoints > Circle.Center

If you have 4 points instead you can use the sphere equivalent.

3

Three points on a surface with a given radius define a maximum of two spheres, no?

4

Go grab a balloon, your coffee cup, and a dry erase marker.

Drink any coffee in the cup.

Mark 3 points on the rim of the coffee cup with the dry erase marker. These are the points you’re providing.

Now blow up the balloon but don’t tie it off - just pinch it closed.

Rest the balloon atop the coffee cup. This is a sphere that touches your 3 points.

Now let some air out of the balloon. It’s another sphere touching your 3 points.

Now let some more air out. It’s another sphere touching your 3 points.

Now let out the air slowly and notice how the sphere always fits until it’s small enough to pass though the 3 points.

Now wipe the marks off your cup, fill the cup back up, and get back to work. :wink:

2 Likes
5

Haha yeah when you put it like that its infinite spheres, but my radius is predetermined.
For three points and a given radius, i would need to find the centerpoint.

So with the formula/method im looking for, for any amount of air in the baloon we would be able to find the centerpoint of the sphere

So assume we have a graph, with three fixed points and a radius as an integer slider

6

Ah! Totally missed that. Sorry! Let’s get you squared away.

The logic you need:

  1. Draw a sphere at each radius
  2. Extract the surface of each.
  3. Intersect the first surface with the second surface and take the one item (a circle) out of the list.
  4. Intersect the circle with the 3rd surface.
  5. Any resulting point is a possible sphere with the given radius.
    1. If 0 your radius is too small
    2. If 1 your sphere is the same as the circle which hits all 3 points.
    3. If 2 you’ll need to decide which to work with - positive side of the plane formed by the points or the negative side.

Here is some design script to do the same:

def sphereFinder (pnts: Point[], radius: double)
{
	surfs =
		List.FirstItem(
			Sphere.ByCenterPointRadius(pnts,radius).Explode()@L2<1>
		);
	intersects =
		surfs[0].Intersect(surfs[1])[0].Intersect(surfs[2]);

	return intersects;
};
1 Like
7

Ahh man, wish I left the office a little later haha. Thanks Jacob, I’ll check tomorrow!

1 Like
8

This is so simple yet clever.. I knew i could find the center of a circle on a plane by drawing two circles at the points along the perimeter, no idea why i didn’t realise you could do this with spheres..

Thank you Jacob!

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1 Like
9

Glad it worked out!

10

Just one more follow up..

Can you point me in the right direction so i can get the surface of the circle that is touching the three points?

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I want to drop the sphere and keep only the relevant surface, so it would look something like this:

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11

Likely Geometry.Trim, where the tool is the plane built from the 3 points and the discard side is any point translated slightly towards the center.

That said, it looks like you are building a fabric structure, and those shapes aren’t actually spherical in nature, so you might want to rethink the geometry creation. DynaShape might be a better solution from an end result and product speed standpoint.