How to draw a line that passes through the center of a sphere?

Question

I have a Ball, about 10'' in diameter. I want to mark two points on the ball that when connected, will result in a line passing through the center of the ball; or in other words, the diameter.

How can I do this?

I have tried: 1-Suspending the ball from one point, and then locating the extreme point on the other end. 2-Sandwich the ball b/w two parallel blocks and then locating the extreme points.

I also have access to an optical marker based system. Is there any way I can used that to be more precise?

Any other ideas?

Answer 1

Wrap a string around the ball forming as large a loop as possible. Note its length. Then lay off half that length from any point and you'll have the antipodal point.

If you do that in several directions from the same starting point you will reach the same end point all the time, within measurement error. If you're far off, you don't have a perfect (enough) sphere.

Answer 2

Do you have a large pair of dividers - at school my teacher had a large pair for use on the blackboard, I don't know where he got them....

Anyway, I'm recalling a trick we used in shop class to find the centre of a circle, perhaps this can be extended:

Set the dividers so that they are approximately 1/2 the diameter of the ball. Pick a point to be one of the antipodal points, using this as the centre, draw a circle on the sphere.

Now draw circles centred on the new circle - you'll need at least two, but it might be useful to draw 3 or 4 roughly equally spaced around the drawn circular line all the same diameter.

Now go to the opposite end of the sphere. By trial and error, we will find a point which is the centre of a circle which touches the tops of the circles drawn in the previous step, you'll have to guess a starting point and gradually adjust the point so you dividers exactly touch each of the tops of the circles. If you were to draw this circle on the ball, it will be parallel to the first circle you drew.

So, you have two circles parallel on the sphere, so their centres must be the on the same diameter.

Answer 3

Attach markers to any three points on the sphere. The line of intersection of the perpendicular bisecting planes of the line segments joining pairs of these points passes through the center of the sphere. Two such planes are sufficient, although you might want to do a best-fit intersection of all three.

More specifically, let the three points be $\mathbf p_1$, $\mathbf p_2$ and $\mathbf p_3$. The equations of the three bisecting planes are, in vector form, $$\begin{align} (\mathbf p_2-\mathbf p_1)\cdot\mathbf x &= \frac12(\mathbf p_2\cdot\mathbf p_2-\mathbf p_1\cdot\mathbf p_1) \\ (\mathbf p_3-\mathbf p_2)\cdot\mathbf x &= \frac12(\mathbf p_3\cdot\mathbf p_3-\mathbf p_2\cdot\mathbf p_2) \\ (\mathbf p_1-\mathbf p_3)\cdot\mathbf x &= \frac12(\mathbf p_1\cdot\mathbf p_1-\mathbf p_3\cdot\mathbf p_3). \end{align}$$ Solve this system for $\mathbf x$ to get a line through the center of the sphere. One way to do this is to compute a basis for the null space of the matrix that represents this system, which will give you the homogeneous coordinates of a pair of points the join of which is the line of intersection.