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Math Help - Circle / Ellipse Tangent

  1. #1
    mdb
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    Circle / Ellipse Tangent

    I'm trying to figure out an equation for the distance between the midpoints of a circle and an ellipse while they are tangent to each other as the ellipse rotates. The ellipse and the circle are centered on the same vertical axis and can only move relative to each other along that axis. See attached picture.
    Attached Thumbnails Attached Thumbnails Circle / Ellipse Tangent-circle-ellipse.jpg  
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  2. #2
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    As I understand the question, you have an ellipse ‘rolling’ around a circle.
    You want to express the distance D between their centers as a function of \theta the angle of rotation of the ellipse.
    Under this understanding the answer is straightforward.
    Suppose that R is the radius of the circle, a is the half the length of the major axis of the ellipse, and b is the half the length of the minor axis of the ellipse.
    Then D=R+\sqrt{(b\cos(\theta))^2 + (a\sin(\theta))^2}

    With reference to your drawing.
    When \theta=0 then D=R+b.
    When \theta=\pi/2 then D=R+a.
    Last edited by Plato; July 16th 2009 at 09:30 AM. Reason: correction a typo
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  3. #3
    mdb
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    The ellipse does not rotate around the circle. It rotates about it's own center. The circle and ellipse can only move relative to each other's center along the vertical axis. The attached drawing shows the ellipse with a theta of 30 degrees.
    Attached Thumbnails Attached Thumbnails Circle / Ellipse Tangent-circle-ellipse-3-.jpg  
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  4. #4
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    Quote Originally Posted by mdb View Post
    The ellipse does not rotate around the circle. It rotates about it's own center. The circle and ellipse can only move relative to each other's center along the vertical axis. The attached drawing shows the ellipse with a theta of 30 degrees.
    In other words, the point of tangency on the circle is stationary.
    The answer is still the same.
    The two examples I gave still hold.
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  5. #5
    mdb
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    No, the tangent point on the circle is not stationary. See attached drawing with theta = 60 degrees.
    Attached Thumbnails Attached Thumbnails Circle / Ellipse Tangent-circle-ellipse-4-.jpg  
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  6. #6
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    Quote Originally Posted by mdb View Post
    No, the tangent point on the circle is not stationary. See attached drawing with theta = 60 degrees.
    I not sure that you understand you own question.
    Is it not true that the center of the circle and the center of the ellipse are on a fix vertical line?
    If that is not true what is the rule for the motion.
    If it is true then the point of tangency is fixed on the circle.
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  7. #7
    mdb
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    The tangent point on the circle is at the bottom (6 o'clock) when the ellipse is at theta = 0 deg. When the ellipse is at theta=60 deg the tangent point on the circle is at about 5 o'clock.
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    I think that the difficulty here may be the fact that the line joining the centre of the ellipse to the point (a\cos\phi, b\sin\phi) on the ellipse does not make an angle \phi with the x-axis. In fact, the angle \theta between these lines is given by \tan\theta = \tfrac ab\tan\phi. So when the ellipse has rotated through an angle \theta, the distance between the centres of the ellipse and the circle will be given by R+\sqrt{b^2\cos^2\phi + a^2\sin^2\phi}, where \phi = \arctan\bigl(\tfrac ba\tan\theta\bigr). (A messy formula, but I don't think it can be simplified.)
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  9. #9
    mdb
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    Thanks for the response but I can't reconcile it with the measured values. I think your equation is close...

    If R=3, a=2.5, b=1.25...

    your equation:

    theta = 0: 4.25
    theta = 15: 4.28
    theta = 30: 4.39
    theta = 45: 4.58
    theta = 60: 4.89
    theta = 75: 5.28
    theta = 90: 5.50

    measured value (from CAD drawing):

    theta = 0: 4.25
    theta = 15: 4.31
    theta = 30: 4.51
    theta = 45: 4.82
    theta = 60: 5.14
    theta = 75: 5.40
    theta = 90: 5.50
    Last edited by mdb; July 16th 2009 at 05:47 PM.
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  10. #10
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    Quote Originally Posted by mdb View Post
    Thanks for the response but I can't reconcile it with the measured values.
    Ah, I see another difficulty! Your drawing with \theta=60^\circ shows that the point of tangency is not on the line joining the centres of the circle and the ellipse. I think that both Plato and I had overlooked that. So the problem is more subtle than we realised. However, the measured distance should be the straight line distance between the centres, so it ought to be smaller than the distance given by the formula. In fact, it seems to be greater. I don't understand that.

    I don't have time at present to think about how these difficulties might be overcome.
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  11. #11
    mdb
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    Well if you get time and want a challenging problem, have at it...

    Thanks
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    Quote Originally Posted by mdb View Post
    The ellipse does not rotate around the circle. It rotates about it's own center. The circle and ellipse can only move relative to each other's center along the vertical axis. The attached drawing shows the ellipse with a theta of 30 degrees.
    A few ways of describing the ellipse (cylindrical coordinates) use:

    r^2=\frac{a^2}{1-e^2*sin^2(\theta)}=\frac{b^2}{1-e^2*cos^2(\theta)}=\frac{a^2(1-e^2)}{1-e^2*cos^2(\theta)} =\frac{a^2b^2}{a^2*sin^2(\theta)+b^2*cos^2(\theta)  }

    where a is the semi major axis and the eccentricity e can be calculated from:

    \frac{b^2}{a^2}=1-e^2

    Now I think you will find that the theta in your problem and the theta in the eqn above are the same so r is the distance from the centre of the ellipse to the point where the circle and the ellipse touch.

    The distance between 'point where the circle and the ellipse touch' and the centre of the circle is simply R.

    Now you have a triangle formed by the centre of the ellipse, centre of circle and point of contact. You know one of the internal angles (90-theta) and you know the length of two sides.

    I will leave it to you to work out the trigonometry to try to find the length of the third side (your answer).
    Last edited by Kiwi_Dave; July 18th 2009 at 01:21 AM.
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  13. #13
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    I got it wrong the theta in your question and the theta in my equation are not equal.

    Some interesting things happen when you consider the special case b=0. The distance is independent of 'a' up until some limit on theta and then the distance suddenly becomes dependent on 'a'.

    This is an unusual property for a function. So I guess the answer will probably be a function that is not defined for b=0 or a=0.
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  14. #14
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    Okay, here's another picture.

    E is the centre of the ellipse (with semi-axes a and b), C is the centre of the circle (with radius R) and P is the point of contact. (If it makes you happier, you can rotate the whole picture so that the line of centres EC is vertical. But that won't affect the geometry.)

    The key point that I was missing before is that \theta is the angle between the minor axis of the ellipse and the line EC (not the angle between the minor axis of the ellipse and the line EP).

    Now let P = (a\cos\phi,b\sin\phi). Then C is the point at a distance R along the normal to the ellipse at P. The normal is in the direction (b\cos\phi,a\sin\phi), and it follows that C is the point (a\cos\phi + \tfrac{bR\cos\phi}N, b\sin\phi + \tfrac{aR\sin\phi}N), where N=\sqrt{b^2\cos^2\phi+a^2\sin^2\phi}. Thus the distance between the centres is given by EC = \frac{\sqrt{\cos^2\phi(aN+bR)^2 + \sin^2\phi(bN+aR)^2}}N, and the angle \theta is given by \tan\theta = \frac{(aN+bR)\cos\phi}{(bN+aR)\sin\phi}.

    What you would like to do is to solve that last equation to give \phi in terms of \theta. You could then express the formula for the length EC in terms of \theta. But it looks as though there is no way to do this (because of the awkward fact that \phi appears in the definition of N).

    My conclusion is that it will not be possible to provide an analytic solution to this problem, and you will have to make do with the CAD simulation.
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  15. #15
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    Quote Originally Posted by Opalg View Post
    My conclusion is that it will not be possible to provide an analytic solution to this problem, and you will have to make do with the CAD simulation.
    Quite the problem...quite the analysis, Opal.
    Sure agree with your diagram; let the circle do the walking!
    Questions:
    1: are you saying it cannot possibly be done by iteration?
    2: what additional "given" is required to permit analytic solution?
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