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Trying to cut a parabolic shape in sheet metal

  1. Sep 19, 2014 #1

    I am trying to cut a parabolic curve in a piece of sheet metal using a numerically-controlled cutting machine (basically a very hard rotating router blade). The router blade is 10mm diameter. I want to achieve a parabolic cut in the sheet metal that adheres to the formula y=x^2

    The trouble is, if I cause the axis of the 10mm router blade to travel along a parabolic path with formula y=x^2 I do not get the resulting top and bottom cuts to be parabolas that conform exactly to the formula y=x^2 This is due to the thickness of the router blade and the fact that it always cuts at a point that is normal to the parabolic path that the axis of the rotating blade is traveling along. I guess that if my router blade was of infinitesimally small diameter, I would then get a faithful cut in the metal that conforms to y=x^2

    Can anyone please tell me the parabolic formula that the axis of a router blade of certain thickness would need to travel along to produce a parabolic cut that conforms to the formula y=x^2


  2. jcsd
  3. Sep 19, 2014 #2

    Simon Bridge

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    You cannot have a point normal to a curve - only a line, a line-segment, or a vector.
    Do you mean that the cut is made some distance from the axis?
    i.e. if you put the axis at (x,y) the cut appears at point (x+a,y+b)?

    You can correct for the thickness of the cut by running the blade so one edge of the cut conforms to the y=x^2 formula. That means one side won't have the exact relationship.
  4. Sep 19, 2014 #3


    Staff: Mentor

    You need to program the router so that the center of the bit is travelling along an arc that is "parallel" to the parabolic shape you're trying to cut. Since you're cutting a parabola whose equation is y = x2, the vertex of the cutting path will be at (0, -5) in mm. The 5 mm represents the radius of the router bit.

    You can calculate another point on the outer parabola by taking a point on the inner parabola, finding the slope of the tangent line at that point, and moving outward 5mm along the perpendicular to that tangent line to get the coordinates of the corresponding point on the outer parabola.

    If the coordinates of this point are (xo, yo) another point on the outer parabola will be at (-xo, yo) by symmetry.

    From these three points you can derive the equation of the outer parabola, and presumably program the router so that the center of the bit travels along that path.

    Caveat: I'm not sure that what I'm calling the "outer parabola" is actually a parabola, but if the 10 mm kerf is relatively small in relation to the dimensions of the parabola, the cut edge should be pretty close to what you want.
  5. Sep 20, 2014 #4

    Simon Bridge

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