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Math Help - determining radius of osculating circle

  1. #1
    Newbie zoster's Avatar
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    determining radius of osculating circle

    Hi all!

    ( first of all, i apologize if this thread should be under another sub-forum, like the Analysis, Topology and Differential Geometry, but it is a very conceptual problem that might be too insignificant for that forum )

    Considering this animation, i can't see how the length of the radius is determined in the terminal stage, where the 3 points overlap. Since there are 3 overlapping lines (the normals to the curve), we don't have an intersection of lines, so i don't get how the end point of the radius (meaning the center of the circle) is determined...
    Secondary Question: how is that animation valid when the curve used as example isn't symmetrical anymore to the normal of the given point (and therefore the 3 normals aren't concurrent anymore)

    thanks for the interest!
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by zoster View Post
    Hi all!

    ( first of all, i apologize if this thread should be under another sub-forum, like the Analysis, Topology and Differential Geometry, but it is a very conceptual problem that might be too insignificant for that forum )

    Considering this animation, i can't see how the length of the radius is determined in the terminal stage, where the 3 points overlap. Since there are 3 overlapping lines (the normals to the curve), we don't have an intersection of lines, so i don't get how the end point of the radius (meaning the center of the circle) is determined...
    Secondary Question: how is that animation valid when the curve used as example isn't symmetrical anymore to the normal of the given point (and therefore the 3 normals aren't concurrent anymore)
    In that animation there are no normals at all. The construction used is the fact that there is a unique circle through three (non-collinear) points.

    Maybe what you are thinking about is that in a calculus-based approach to osculating circles it is usual to use normals to the curve. In that case, you can specify the centre of the circle as the point where two normals intersect. The centre of the osculating circle is then the limiting position of that centre as the two points on the curve move closer and closer together.
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  3. #3
    Newbie zoster's Avatar
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    Quote Originally Posted by Opalg View Post
    In that animation there are no normals at all. The construction used is the fact that there is a unique circle through three (non-collinear) points.

    Maybe what you are thinking about is that in a calculus-based approach to osculating circles it is usual to use normals to the curve. In that case, you can specify the centre of the circle as the point where two normals intersect. The centre of the osculating circle is then the limiting position of that centre as the two points on the curve move closer and closer together.

    hey, thanks for the answer!

    i think i was a bit vague in my first post, so i'll try to be a bit clearer.

    I am an architecture student and i think geometry in a graphical manner. I am potent at descriptive geometry but i have hand to mouth knowledge in other branches of mathematics (like trigonometry and such). From what i infer, each descriptive geometry solution has a counterpart in "number mathematics". That being said, if i can't imagine the graphical solution to this problem (in an error-free environment), i can't see how this is a resolvable problem. i don't question that it is, but i just would like to know how it is a determined situation.

    About your comment concerning that the circle in the animation is determined by 3 points, not by 2 normals,it answers my secondary question. thank you! it doesn't though answer my first question, since in the terminal situation where the 3 points overlap, the circle is again impossible to determine (or seems so).

    hope to hear from you and others again.
    thanks for the interest!
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