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Calculate ellipse from 4 points? 
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#1
Mar3111, 08:38 AM

P: 2

I have been scouring the Internet and various geometry books trying to figure out an issue I'm dealing with at work (I'm a professional statistician). It's been over 30 years since I've had a geometry class, so my brain is a bit rusty in this area. Here's what I have so far, can someone tell me if I'm on the right track?
Background We are taking 2 measurements of the diameter of a cylindrical object using a simple caliper. The two measurements are taken 90° from each other. I use the word "cylindrical" loosely, as the objects are not perfectly round. There is some eccentricity. Problem What I would like to know is, using only these two measurements, can I calculate the maximum possible eccentricty of the object? I have posed the question as trying to find the major and minor axes of an ellipse. One possibility of course is that I have made my measurements exactly on the major and minor axes. That's easy. But what if my measurements are on two other random lines through the ellipse? Can I use that information to calculate the lengths of the major and minor axes? I've attached a jpg image to illustrate the scenario, exagerating the lengths. Refer to the picture for what follows. I know the length of AB and CD. I know they are perpendicular to each other. Obviously, I don't know theta. I think, given the orientation of my two line segments, an infinite number of ellipses could be drawn through those 4 points. What I'd like to do is assume my major and minor axes are along the X and Y axes and then calculate the lengths of the major and minor axes for various values of theta. For a given theta, I can calculate the coordinates of A, C, D, and B. Questions 1) Am I making this too complicated for myself? Is there something really simple I'm overlooking? 2) Do I iterate theta only through 45° or all the way to 89°? It seems to me I only need to iterate theta from 1° to 45°. 3) So far, as I attempt to calculate the lengths of the major and minor axes as in 2), I get some places where my results "blow up". I.e., Excel returns an error of some kind. It could be my equations are incorrect. I'm checking them for the umpteenth time. Any help or direction would be greatly appreciated! Thank you! 


#2
Apr611, 03:28 PM

P: 2

Anyone?



#3
Apr611, 09:50 PM

HW Helper
P: 925

There is no attachment, btw.
Seems to me 5 pieces of information are needed. I don't think the fact that the two measurements are at right angles helps anything. Can you measure the circumference? That plus the 4 pts might be enough. Also, if you had access to a coordinate measuring machine (CMM), you could take many measurements and perform a least squares fit. 


#4
Apr911, 05:06 AM

P: 360

Calculate ellipse from 4 points?
If I understand you correctly then you can't solve the problem.
4 points could define ellipses with very extreme aspect ratios. Imagine the 4 points are all at the same radius from the center. And you assume the axes of the ellipse are parallel to the sides of the square whose corners are the points. Then the points could be on a circle, or an ellipse with any aspect ratio. However, I think you're saying you want to assume that case isn't happening and only allow the ellipse's axes to be aligned in other orientations which aren't 45deg to the two lines connecting the two pairs of opposite points. If you have 4 points and the orientation angle then that can give you a set of 5 equations by using the equation for an ellipse. That has 5 unknowns so you should be able to solve it uniquely. But I guess it would surely be numerically unstable. Can you show your equations? Oh. I just realised you don't actually know the locations of the 4 points, only the two distances between them. That surely makes it always unsolvable. Still, can you post the equations? 


#5
Apr1111, 06:17 AM

P: 360

Hello bww. I think we have similar problems. If you've solved it or made some progress please post how it's going.
This is my thread. If I can get an ellipse from 4 points it'd be great too. http://www.physicsforums.com/showthread.php?t=487486 


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