Is there a method (or better yet, a website) for calculating the center point of a system of four or more points? By ‘center point’ I mean a point equidistant from all of the initial points. (I realize that this may not be possible in two dimensions, which is what I’m concerned with. In that case I’m curious if my intuition could be confirmed.)
Example:
Jones, Smith, Jackson, and Johnson live in Augusta Maine, Wichita Kansas, Frankfort Kentucky, and Vancouver BC. They would all like to get together. Is there a way to figure out which point (if any) is equidistant from all of them?
Given a pair of points A and B, the set of points equidistant from A and B is a straight line that is perpendicular to line segment AB and contains the midpoint of line segment AB.
Repeat the process for pairs AC and AD. If those lines all intersect at one point then that is the point you want.
If not, then the point does not exist.
by the way, the point you want is the center of a circle that contains the original 4 points. You could also draw the circle that contains 3 of the 4 points and test to see if the 4th happens to lie on the same circle. If not then a solution does not exist.
In two-dimensional space, given two points A at (Ax, Ay) and B at (Bx, By), the set of points that are equidistant between A and B is the perpendicular bisector of A and B. That is, the line that passes through the midpoint of A and B (which is at ((Ax+Bx)/2, (Ay+By)/2) and whose slope is at 90 degrees (i.e., perpendicular) to the line through A and B.
Thus, for two points in two-dimensional space, there is an infinite number of equidistant points.
Given three points in 2D space, you can find a point equidistant to all three by finding the point where the perpendicular bisector of, say, A and B as well as the perpendicular bisector of A and C, say, intersect. This will be exactly one point. However, if A, B, and C are co-linear, their perpendicular bisectors will not intersect at all.
In general, equidistant points will exist for arbitrary points in n-dimensional space only if the number of arbitrary points is less than or equal to n+1.
You do realize that the point that’s equal distance from each of 4 (or 3) other points may not be in any sense a good point for all to meet. It could be very bad Consider a circle and call its radius 100. Now pick 3 points on it that are very close together. Say 1 unit from A to B and another 1 unit from B to C. The center of the circle which is 100 units away is the only point that’s equidistant, but each would have to travel 100 units to get there. On the other hand meeting at B would entail 1 unit of travel for A and C and none for B.
The center point can be made arbitrarily bad by picking a circle with a larger radius, but keeping A and B and B and C only one unit apart.
An equidistant point might not exist, but you could generalize the problem to ask for the point x[sub][/sub] which minimizes the sum of the squared distances from x[sub][/sub] to each of the original points. It’s then easy to show by a first derivative argument that x[sub]*[/sub] is equal to the coordinate-wise average of the original points.
Thanks for the replies.
Geometry was a long time ago, but I think I get it now. I can see how any non-identical points will have infinitely many points from which they are equidistant. And that any triangle (2D) will have exactly one point equidistant from all its apexes. Finally, I think I can see now that some polygons of four or more sides (again, 2D) will have a point equidistant from all their apexes, but that others won’t.
I think what you’re actually looking for is called the “average location” or better known as “center of mass” in physics circles. The average location would have the smallest sum of distances to the 4 points, compared to any other arbitrary location.
On a Cartesian coordinate plane, all you have to do is find the x,y coordinate of each point (or latitude and longitude on the globe), then take the average of the x’s and the y’s to get your average.
Is it really so hard to read six posts?
No it’s not. And I did read them all. Although you did mention taking an average you didn’t say anything about what the method is called. Just fleshing it out a bit.
On a plane, sure… But does it actually work with Lat and Long averaging? I’m having trouble visualizing this, but couldn’t such a straightforward approach lead to absurdities? Like, if all but one of the points was in the western hemisphere…but at Lat 89 North, and the last point was in the eastern hemisphere…and on the equator?
Or…does straightforward averaging work just as well using polar coordinates?
There may be a way to do it with polar coordinates, but I think the easiest way would be to convert into Cartesian coordinates (x,y,z), take the average (which will always be a point inside the sphere), then project that point to the nearest point on the surface.
Good points. :smack:
As long as the points aren’t too far apart, simple averaging should get you reasonably close. Giles method is better, but more complicated. If you really want the exact answer, you need to work with great circle distances, which is going to get pretty messy for even a few points.