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  • Math / Drafting Question (or Daft Mathing)

    OK, so imagine a ball end mill, 2 inch radius. (I know that's pretty big, but bear along with me a moment.)

    So if I plunge this spherical cutter into a piece of material, say .750 deep, I get a circular depression. The diameter of the edge is 3.1225". Simple enough.

    Now if I were to take this same cutter and allow it to swing on a 4 inch radius, coming down into the material, swinging down to the same .750 depth, and continuing to swing it around, back up and out of the material, I get an oval looking depression in the stock, with a short axis and a long axis, 3.1225" X 4.6637".

    My question is: Is this a true ellipse? Or is it something else? If it is not an ellipse, how do I draw it?

  • #2
    The surface you cut is a section plane of an ellipsoid. An ellipsoid may be generated by rotating an ellipse so any section of an ellipsoid is an ellipse.

    I forgot to mention that the cutter is moving in a circle. A circle is a boundary condition ellipse and a limit condition section of an ellipsoid.
    Last edited by Evan; 02-28-2010, 08:37 PM.
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    • #3
      not an ellipse

      The spherical cutter swinging about a center describes a torus, even though you are using only a portion of it. The intersection of a torus with a plane parallel to its axis is not an ellipse. You can tell this by looking at a limiting case. Take a Krispy Kreme donut and slice it with a knife. If you halve the donut you get two circles. As the intersecting plane gets further from the center the circles become distorted, then joined into a "figure eight." Only the slightest cut will look like an ellipse, but it isn't one, not really.
      Last edited by aostling; 02-28-2010, 08:44 PM.
      Allan Ostling

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      • #4

        Are you quite sure? What do you have if if you take a section of the perimeter of a torus that does not intersect the circumradius and is normal to the outer circumference?
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        • #5
          This is what I mean.

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          • #6
            Your illustration is informative, but consider this:



            This shows five different slices of the torus. Starting at the left, the cross-section consists of two unjoined areas. As the cuts get closer to the edge of the donut, they become increasingly "ellipse-like." But we can deduce that the resemblance, though close, is never exact. The deformation of the shape of the cross-section must vary continuously -- the only slice of the donut which is truly an ellipse is the slice which is no slice at all, the one right at the edge of the Krispy Kreme which leaves the donut intact.
            Last edited by aostling; 02-28-2010, 10:07 PM.
            Allan Ostling

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            • #7
              Look at the left most section. The cross section is a circle. If we take a tube of that same or lesser diameter and project it as I showed it will intersect only the outer hemicircle of the torus. A circle is a limit case of an ellipse and the outer half of a torus can be collapsed to the limit case which generates a sphere. Any intersection of the surface of a sphere by a tube generates an ellipse (or section of an ellipsoid). In the case of the torus not being collapsed to a sphere the only difference is scaling in two axes.
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              • #8
                Originally posted by john hobdeclipe
                If it is not an ellipse, how do I draw it?
                You draw it with SketchUp, which is ideal for problems involving intersections of one shape with another.
                Allan Ostling

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                • #9
                  It's called CSG which stands for Constructive Solid Geometry which I have been using on a computer since about 1985 or so when I bought an Amiga.

                  Incidentally, you can make an elliptical torus with a circular cross section or a circular torus with an elliptical cross section. I am very sure that anything that has a circle in it also has an ellipse.
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                  • #10
                    Here's the equation of your curve:

                    (x2 + y2)2 = dx2 + ey2 + f.

                    This is complements of Wikepedia:

                    http://en.wikipedia.org/wiki/Spiric_section

                    It is not a ellipse. It is called a "spiric section". And as aostling showed, it starts looking somewhat like an ellipse but goes on to becoming two circles.
                    Paul A.

                    Make it fit.
                    You can't win and there is a penalty for trying!

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                    • #11
                      I don't get an ellipse.

                      In the drawing below from my CAD model, the outer (blue) line is the section taken as you describe, and the inner green line is an ellipse of the major/minor dimensions of the original section. Dimensioned at the point of greatest error (or as close as I could get).

                      John, do you mean: how do you draw it on paper? ... or how in a CAD program?


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                      • #12
                        I should have turned up the precision. That is not a very large difference. However, the Spiric section does have another name. It is also called a Cassini Ellipse.

                        http://www.math.ubc.ca/~cass/courses...es/cassini.htm

                        There are also true ellipses to be found in a torus when r=R.
                        Free software for calculating bolt circles and similar: Click Here

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                        • #13
                          An approximation

                          Originally posted by john hobdeclipe
                          OK, so imagine a ball end mill, 2 inch radius. (I know that's pretty big, but bear along with me a moment.)

                          So if I plunge this spherical cutter into a piece of material, say .750 deep, I get a circular depression. The diameter of the edge is 3.1225". Simple enough.

                          Now if I were to take this same cutter and allow it to swing on a 4 inch radius, coming down into the material, swinging down to the same .750 depth, and continuing to swing it around, back up and out of the material, I get an oval looking depression in the stock, with a short axis and a long axis, 3.1225" X 4.6637".

                          My question is: Is this a true ellipse? Or is it something else? If it is not an ellipse, how do I draw it?
                          It is a true ellipse if and only if the depth of varied as the sine of the angler of rotation in the cutter circular path.

                          In other words, the trace of the centre of the end of the cutter would be a simple harmonic motion.
                          http://en.wikipedia.org/wiki/Simple_harmonic_motion

                          Or put a bit more simply, the path of the cutter end must be a cut through a tube that has a diameter of 3.1225" and a length of cut along a tilted face that is 4.6637" long.

                          A straight cut through a cylinder is a conic section - an ellipse. When both axis are equal the ellipse is a circle.
                          http://en.wikipedia.org/wiki/Ellipse

                          Many CAD programs actually take the "quick" way out and plot an oval to approximate an ellipse.
                          http://en.wikipedia.org/wiki/Oval

                          If you were to cut a cylinder the is 3.1225" diameter at an angle of 46.967 degrees (length of cut = 4.6637") and were to cut and "flatten out" the cut tube, the profile of the cut surface would be a sine wave.

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                          • #14
                            Revision needed

                            I need to re-think the math.

                            It is mostly correct.

                            The path of the centre of the hemispherical end of the milling cutter is that of the centre of an elliptical ring with major and minor axis that I need to work out.

                            The elliptical ring can be constructed by drawing the ellipse and using it as a centre of a circle that is produced along the ellipse (as its centre).

                            The shape of the edges of the cut can be modeled by placing a horizontal section plane through the tilted ring at a height/elevation that I need to work out - later.

                            http://www.google.com.au/#hl=en&q=dr...cf0a553a95b437

                            http://www.google.com.au/#hl=en&q=dr...a=N&fp=1&cad=b

                            http://demonstrations.wolfram.com/Dr...rclesAndLines/

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                            • #15
                              I'm aware that my original question describes a section of the outside of a torus, as in aostling's post #6, the extreme right hand image.

                              And I've learned from this discussion that this is not an ellipse.

                              I have Autocad Light 2007, which is strictly 2D, and I have rather limited skills with it. I'm looking for a way to quickly generate drawings of this shape, using different sets of dimensions for the diameter of the cutter, the "swing" of the cutter, and the depth of cut.

                              By the way, I'm making wooden spoons. I'll try to get a couple pix of my "spoonerator" later today.

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