Oval hole in sheet metal that will be round in a tube

Discussion in 'Math' started by tracecom, Nov 30, 2013.

  1. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,871
    1,393
    I want to cut an oval hole in a piece of sheet metal that will be round when I roll the sheet metal into a 1 3/4" I.D. tube. The short dimension of the oval will be 1 3/4", but what will the long dimension be? (Assume the sheet metal has a zero thickness.) Will it be half the circumference?

    I guess this is a math problem. Thanks.
     
  2. MaxHeadRoom

    Expert

    Jul 18, 2013
    10,555
    2,375
    2.748?
    Max.
     
  3. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,871
    1,393
    That is half the circumference, so I'll try that.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    It sounds like your hole is going to be the same diameter as the tube. If that's the case, then the long diameter of the "oval" (I'm not convinced it really is an oval) will extend across half the circumference of the tube.

    Setting up a spreadsheet to produce a plot of the shape of the hole would be a pretty straightforward exercise.
     
  5. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Draw a few forms on paper and roll into a 1&3/4" tube.

    It won't take many back-forth tries to get it very close. Then trace the paper onto the metal.

    Totally non-math but very effective for metalwork. ;)
     
    #12 likes this.
  6. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    If you've got two tubes the right diameter (or two pieces of bar stock the right size) you don't need to do any back and forths. Wrap a piece of paper around one tube and position the other tube perpendicular to it resting on the paper. Now use a rule and place it, on-edge, on the side of the perpendicular tube and then extend it down the side of the perpindicular tube until it touches the paper. That's one of the points on your oval. Just do that around the entire circumference of the tube and you have your pattern.
     
  7. Tesla23

    Active Member

    May 10, 2009
    318
    67
    If my rough calculation is correct, then if the large tube has radius 'r', and the small tube radius 'a', then when the large tube is unwrapped, in the (x,y) coordinates on the flat surface the curve is

    r^2 sin^2\left(\frac{x}{r}\right) + y^2 = a^2

    and quickly plotting

    y = \sqrt{\left(a^2-r^2 sin^2\left(\frac{x}{r}\right)\right)}

    in Excel shows it looks reasonable. The y-coordinate is the short axis.

    This assumes that the small tube intersects on axis and at right angles.
    Check before use!
     
  8. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,363
    I have not checked your math but the shape looks correct.
    I suspected that the shape is not oval but more eye or leaf-shaped with 90-degree angles at the tips for equally sized tubes.

    Edit: The shape looks like a sine wave from 0 to 180 degrees. Short dimension is 1.75" (amplitude of sine wave is half of 1.75"). Long dimension is 2.75".
     
    Last edited: Dec 1, 2013
  9. Metalmann

    Active Member

    Dec 8, 2012
    700
    223
    I used to get paid to do such layout problems.

    Getting too old to remember the formulae.;)
     
  10. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,871
    1,393
    Thanks to all for the input. With sort of a blend of different methods, I came up with a shape that seems to work. It's not mathematically perfect, but it's close enough. See the attachment.
     
    Last edited: Dec 1, 2013
  11. Metalmann

    Active Member

    Dec 8, 2012
    700
    223
    Looks about right to me, but it's been a while.;)
     
  12. Tesla23

    Active Member

    May 10, 2009
    318
    67
    You're right, I didn't read the question to see that they were both the same radii, simplifying my equations:

    r^2 sin^2\left(\frac{x}{r}\right) + y^2 = r^2

    solving for y:

    y = \pm r\sqrt{\left(1- sin^2\left(\frac{x}{r}\right)\right)} = r cos{\left(\frac{x}{r}\right)}

    so the short dimension is 2r and the long dimension is \pi r, or 1.75" and 2.75" as you say.

    It's easy to see that the long dimension is half the circumference as the second pipe effectively cuts the first pipe completely, and the short dimension being the same as the second pipe diameter is straightforward.
     
  13. #12

    Expert

    Nov 30, 2010
    16,321
    6,818
    I used to be able to do the math, but math won't draw a line on a pipe.
    I think I'd set up a jig to hold the pipes and use spray paint to get a starting shape,
    or, think in terms of cutting a half circle out of the abutting pipe first and an ovoid out of the receiving pipe second. Like many analog designs, do the last part in order to find out what the first part has to be.
     
  14. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    If you're willing to use up some metal, just clamp a tube or round bar stock in a mill (or drill press) and bore a hole half way through it that is the same diameter as the tube. Paint the outside of the tube and wrap a piece of paper around it while the paint is wet. You now have your form.
     
  15. Tesla23

    Active Member

    May 10, 2009
    318
    67
    No, but with a modern printer I can use it to draw a line on a piece of paper more accurately than I can do by hand, and I can then wrap the paper around the pipe.

    There's lots of ways to skin this cat!
     
    #12 likes this.
  16. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,871
    1,393
    I have actually already done something very similar to that, which is where I came up with the rough shape that I refined and dimensioned on the computer. The surprising thing to me is that you would suggest such a non-academic process. I am reassured that my rube goldberg approaches are not always completely ridiculous. Thanks. :)
     
  17. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    I got curious as to how much effort it would actually take to do it with Excel. It actually turned out to be much easier than expected. Here's what it looks like:

    [​IMG]

    Of course, the pixels are not exactly square, but you get he idea.
     
  18. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    As one of Murphy's Laws of Combat states, "If it's stupid but it works, it ain't stupid!" :D

    Unlike many of my peers (but by no means all of them), my background includes a lot of non-academic stuff. I've learned a lot of "non-academic" techniques for lots of things and am very willing to draw upon them where it seems to make sense. I particularly love when I can combine the two.

    Very curious, does my pattern above resemble the one you came up with?
     
  19. tracecom

    Thread Starter AAC Fanatic!

    Apr 16, 2010
    3,871
    1,393
    Yes, similar, but yours is better. Mine is attached to post #10.

    ETA: I just looked at the scale on yours. If the scale is in inches, yours is twice as big as it should be, but the shape is correct.
     
  20. MrChips

    Moderator

    Oct 2, 2009
    12,449
    3,363
    Looks right. It is two halves (mirror image) of a cosine wave.
    The angles at the pointed ends are both 90 degrees.
     
Loading...