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cutting an elliptic pie

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  • cutting an elliptic pie

    Yesterday I made a maple-pear upside-down cake for my Norwegian relatives. I've always made this recipe in a 9" round Pyrex cake dish, but my brother only had a glass dish in the shape of an ellipse. It was the perfect capacity, but when it came time to cut pieces I wondered how to divide the cake into equal wedges.

    Should the cuts meet in the center, or at one of the foci? And depending upon the position of each piece, how can one determine the angle of the wedge?

    This is shop math, sort of, and I suspect it is a difficult problem, something to chew on during the holiday season.
    Allan Ostling

    Phoenix, Arizona

  • #2
    Pi is of little value when dividing elliptical pie.

    No doubt the ends of the dish are rounded. I'd draw evenly spaced lines parallel to the inner and outer elliptical arcs from one end of the dish to the other. Then I'd divide those lines into the desired number of equal lengths, and then make the cut lines by connecting the dots.

    Roger
    Last edited by winchman; 12-23-2009, 10:44 AM.
    Any products mentioned in my posts have been endorsed by their manufacturer.

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    • #3
      Originally posted by aostling
      .........and I suspect it is a difficult problem, something to chew on during the holiday season.

      What a punny guy !


      Rex

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      • #4
        It shouldn't be too difficult to simply cut it in (two) halves, keeping half for yourself and letting the relatives fight over the other half.

        At least, when it comes to food, that's MY idea of equal division.

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        • #5
          Its the first time I heard of Pi being applied to a pie. But I like John before me I would keep the maths simple, its easy to divide an elliptical pie by 2 or 4. but if there is any more of you you will need more pies.
          Alan

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          • #6
            Id highly recommend attempting to give the math experts here a present and request that the pie (cake?) must be devided to give 10 peices of equal volume.

            You could more or less verify this also by cutting the cake up and weighing the peices.
            Just throw them out if they arnt equal and bake another, your family will understand when you tell them its in persuit of math and equality.
            Play Brutal Nature, Black Moons free to play highly realistic voxel sandbox game.

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            • #7
              Oops, I was thinking about the pan my first wife (an RN) had that was shaped like an emisis basin. It had two elliptical sides joined by rounded ends.

              Could Kepler's equal-area law be of any help here?

              "The line connecting a planet to the sun sweeps out equal areas in equal amounts of time."

              Roger
              Any products mentioned in my posts have been endorsed by their manufacturer.

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              • #8
                They keep getting worse!

                Probably most of us remember hearing repeatedly "Pi R square". Of course, everybody knows that is wrong. Pi are round;cornbread are square!
                Merry Christmas.
                Jim

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                • #9
                  http://www.youtube.com/watch?v=9wO3mRJpg44

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                  • #10
                    Dividing an ellipse into two or 4 equal parts is easy because an ellipse is symetrical around both the major and minor axes.

                    For other subdivisions, orbital mechanics can provide a solution. A body in an elliptical orbit sweeps out equal areas in equal time (Kepler's second law). So, if you look up the parametric equation for an elliptical orbit and divide the time for one orbit into a number of equal areas, and draw lines (cuts) from the position of the orbiting body at those points in time to the center of the orbit (one of the two foci), the pieces will have equal area. Some algebraic substitution would be needed to derive an orbit that had the dimensions matching your confection. Note that you need the parametric equation for an elliptical orbit, not just a parametric orbit for an ellipse which probably wouldn't draw the ellipse with the right velocity vs position.

                    http://en.wikipedia.org/wiki/Kepler%...anetary_motion

                    If this seems like a lot of math, consider yourself lucky. There are other problems involving intersections of tilted ellipses that defy symbolic solution.

                    Another approach would be to take the formula for area of a section of an ellipse and solve for an initial position and the the total area of the ellipse divided by the number of pieces.

                    There may be some easier solutions. Orbital mechanics is just an area where equal area segments of an ellipse is well studied.

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                    • #11
                      At the risk of taking this too seriously , an ellipse is just a circle stretched in one axis.

                      A way to visualise the required result:-

                      Draw on a sheet of rubber a circle with a diameter equal to the short axis of the ellipse.

                      Mark out the lines dividing the circle into equal portions.

                      Stretch the rubber to create the ellipse. It may not be immediately obvious but the lines do still divide the elliptical area equally.


                      Once you can visualise that, it's just a matter of calculating the change in angle of each dividing line caused by the stretch.

                      The maths isn't too complicated - if the angles are measured from the long axis, then the ratio of the tangents of the 'before' and 'after' angles is equal to the ratio of the width and length of the ellipse.


                      Cheers

                      .

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                      • #12
                        You need a Keplerian pie divider.

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                        • #13
                          easiest is to keep the product of the slice length and edge length the same.

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                          • #14
                            Originally posted by Barrington
                            At the risk of taking this too seriously , an ellipse is just a circle stretched in one axis.

                            Draw on a sheet of rubber a circle with a diameter equal to the short axis of the ellipse.

                            Mark out the lines dividing the circle into equal portions.

                            Stretch the rubber to create the ellipse. It may not be immediately obvious but the lines do still divide the elliptical area equally.
                            I was thinking along the same lines and going to post an update when I saw your post. Didn't wait to pursue that approach on the first post because I keep getting interrupted frequently by back pain from shoveling snow. If you imagine the slices being triangles (ignoring the rounded chord part), you can see that triangles centered about the minor axis are stretched in width and the ones centered about the major axis are stretched in height by the same ratio. And generally, stretching an arbitrary shape by a factor of N on one axis multiplies the area by N. Easy to prove informally, harder to come up with a rigorous mathematical proof.

                            As for the math of that approach:
                            x=width/2*cos(theta)
                            y=height/2*sin(theta)
                            where width and height are the major axis and minor axis dimensions and theta is the angle of the imaginary (pre scaling) circle, not the angle of the slices on the ellipse. Simply divide 360 or 2*pi (depending on whether your calculator is in degrees or radian mode) into the number of pieces desired, and solve for x and y at that interval and cut from there to center point. If you prefer to use a protractor, you can then calculate the angles of those x and y coordinates. You can also use a width and height larger than the pie/cake as long as you enlarge both by the same ratio (leaving out the divide by two, for example), allowing your measurements to be made on a piece of paper that extends out beyond the edge of the cake so your cake doesn't end up tasting like Dykem. Or just use your CNC cake cutter or icing printer (yep, the printers exist).

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                            • #15
                              I propose TLAR approach

                              This approach requires that you get out you knife, step back a bit, lick the side of your thumb (left or right depending on whether you are right or wrong handed) hold it out at arms length and then, while checking at least twice, make cuts that look about right.

                              This approach is formally known as the 'That Looks About Right' method. Works fairly well as long as not being unduly influenced by medicines or other "medicinal" preparations.

                              Merry Christmas, Feliz Navidad, Joyeaux Noel

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