No announcement yet.

aostling -- A day late & a dollar short

  • Filter
  • Time
  • Show
Clear All
new posts

  • aostling -- A day late & a dollar short

    I had the answer the day that you posted the thread but I didn't know how to post a drawing to a thread, so I just stuck with it 'til I got it done.

    If you were to make the drawing above to full scale (9" circle) and project the corners of the segments down to the ellipse, drawn to the same dimensions as the cake dish, then draw from the center of the ellipse to the ends of the projection lines, the resulting segments of the ellipse will have equal areas. If yhou then extended the radiating segment lines a couple of inches beyoun the ellipse you could set the dish with the cake in it on the ellipse & cut from the center of the cake to the radiating segments and each piece of the cake will contain the same volume. This will work with either an even or an odd number of pieces of cake.

    Maybe you could keep this in mind for the next time you get caught in the same or similar circumstances, then again maybe not.


  • #2

    I tried this in CAD, imported DXF to Alibre solid modeler.

    Ellipse 2" x 1". Circle 2" diameter.

    Broke eliipse into 5 pieces.

    Extruded each ellipse section to 1" depth.

    Checked surface area of three sections. 3.096, 3.079, and 3.213 in sq. Close, but not exact.


    • #3
      All out of shape.

      I think that Gumper may be correct.

      The ellipse is a tilted circle in projection. It is - or should be - a conic section.

      The "full circle" should be divided into the required number of parts (bounded by the circle's arc for each part and the radial lines in the circle).

      Project those points (2 x arc/radius intersections and the centre) from the circle to the ellipse - as is the case for each of the 10 segments of the OP's circle - which are projected from the arc to the ellipse.

      A lot of CAD systems use an approximated ellipse.

      Other than that I suspect that calculus (I've forgotten mine) or a reiterative math routine will be needed to resolve it.


      • #4
        Okay, I double checked using another method and it appears the areas are equal.

        Maybe someone else can verify?


        • #5
          Originally posted by GUMPER
          I had the answer the day that you posted the thread but I didn't know how to post a drawing to a thread, so I just stuck with it 'til I got it done.


          This is the answer which Evan posted while I was traveling in Europe in December, reply #30 here: I had to think about it before I was totally convinced, but a little doodling shows that it works for looking at tilted square figures of equal area. It's only a slight stretch to conclude it works for areas of any shape, including pie-shaped wedges.
          Allan Ostling

          Phoenix, Arizona