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ANSWERED on Thu 14 Oct 2010 - 11:41 am UTC by Roger Browne

Question: I want to make a donut from fabric.

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 13 Oct 2010 23:42 UTCWed 13 Oct 2010 - 11:42 pm UTC 

I want to sew a donut. To do that I presumably need to cut a circle in fabric and then another circle inside that. This gives me one ring. Do it again and then sew the rings together, flip it inside out and I should have a donut.

The question is, if I want the final donut to have a height H, outer diameter D1, and inner diameter D2, what should the diameters of the inner and outer circle be for the rings that I cut from the fabric?




 14 Oct 2010 07:55 UTCThu 14 Oct 2010 - 7:55 am UTC 

And are you then intending to eat it?


Roger Browne 


 14 Oct 2010 11:41 UTCThu 14 Oct 2010 - 11:41 am UTC 

Hi happyengineer,

A donut generally has a ring with a circular cross-section. I assume that's what you want, however if you need an elliptical cross-section please request clarification.

For a circular cross-section, the outer diameter D1 of the finished donut will be the inner diameter plus twice the height H. As you obviously realize, that doesn't mean that the outer cut of the fabric will be of diameter D1.

Consider the circular cross-section of the finished donut. This cross-section will have a diameter equal to H, and a circumference of pi times H. Each of the two pieces of fabric contributes half of this circumference. However, we are specifying the diameter, not the radius, so we need to account for this "half-circumference" twice.

Therefore we have:

 Height \; of \; donut: \; H
 Inner \; finished \; diameter: \; D2
 Outer \; finished \; diameter: \; D1 = D2 + 2H
 Diameter \; of \; inner \; cut: \; D2
 Diameter \; of \; outer \; cut: \; D2 + \pi H

After you fill your donut with stuffing to puff it out to donut-shape, your outer cut will be "pulled in" to the outer finished diameter D1. This will cause the fabric at the outer rim of the donut to bunch up because the outer cut had a larger circumference than it now needs.

If you want a tidy finish, you will need to cut out some slim wedges at regular intervals around your fabric. Overall you need to "waste" an amount of the outer circumference equal to the difference between its circumference as cut and circumference as finished:

Outer \; circumference \; as \; cut: \; C = \pi(D2 + \pi H)
Outer \; circumference \; as \; finished: \; F = \pi(D2 + 2 H)
Length \; of \; circumference \; that \; must \; be \; wasted: \; W = C - F
\; \; \; = \pi(D2 + \pi H - D2 - 2H)
\; \; \; = \pi((\pi - 2) H)

Just divide this waste circumference amongst the number of wedges you are going to cut out.

If you don't want to cut wedges out of your fabric (because of all the extra stitching that would be required) you could fold, pin, or gather the outer edge instead of cutting away the waste. But you need to take it in somehow, or else the donut will tend to flatten as the stuffing redistributes.

Also, when you cut the inner and outer circles, you probably want to leave an extra centimeter or so, in order that the seam can be stitched on the exact diameter.





 16 Oct 2010 15:24 UTCSat 16 Oct 2010 - 3:24 pm UTC 

I would consider making the donut with material circling the stuffing  - the seams radiating from the center of the whole thing.  The pieces would all be the same shape.
The dimensions would need some math.  The individual pieces wrap around and meet at the smallest (inside) circumference of the donut, where they are sewn together.  The ends of the strips would be a fraction of that circumference defined by the number of pieces.  The length of the strips would be the circle around the stuffing; the width of the strips at half their length, a fraction of the the outside circumference of the donut.
I can't visualize the curve of the long edges of the strips, but the varying width of the strips would be defined by the fractions of the intermediate circumferences as the strips go around the stuffing.
I hope that is understandable. 

As Roger points out, all pieces cut with extra material for the stitching.

The last seam between two pieces can be left open and the seam connecting the ends of the strips sewn - creating a curved sausage.  Then the material is turned right side out and most of the seam connecting the ends of the sausage sewn before stuffing, and then the rest of that seam closed.

Obviously, the more pieces, the "smoother" the circle of the donut would be.




 18 Oct 2010 23:32 UTCMon 18 Oct 2010 - 11:32 pm UTC 

It occurs to me now that if it is not easy to calculate the varying circumferences for my plan, one could figure out the varying width of the strips by setting up a model of a section of the donut:
2 pieces of cardboard half the circumference around the stuffing taped to a table to arch up in a semicircle.  Say you want to use 12 pieces of material to make the donut, then the cardboards would be set 30° from each other.  Lines radiating at 30° from the center of the donut would intersect with the cardboard arches, allowing one to measure the the distance between them.

Maybe that is trickier than I think, but I can't do the math.




 26 Oct 2010 10:45 UTCTue 26 Oct 2010 - 10:45 am UTC 

Okay, I have gotten carried away with this problem and made a model with two round disks at 30° from each other, the inner radius being also the same as that of the disks, i.e., the empty center of the donut twice the diameter of the disks (representing what would be stuffed). 
A piece of paper wrapped around the model allowed drawing lines at the edges of the two disks.  Unwrapped, it showed two flat bell curves, as people with better visual perception probably expected.  Indeed, it seemed that the convex and concave curves were identical.  That would be nice, since it would allow cutting the material (plus material for the seams) by offsetting the pieces on the whole cloth, avoiding waste.
I recognized that the greatest width and the ends of the pieces would be cords of the outer and inner circumferences of the donut where my disks touched the circumferences, not fractions of the circumferences, as I said before.  (On my model with an outer radius of ca. 26.4 cm, the circumference would have curved about 1 cm outside the cord.) 

The same method would, of course, be possible with other dimensions, say, a "fatter" donut with a smaller inner circumference, perhaps using a model that was a fraction of the size of the intended donut..

If a researcher thinks this solution of the problem could be of interest to Happyengineer, perhaps an RfC could be posted to call his attention to it.

Thanks and regards,  Myo


Roger Browne 


 26 Oct 2010 20:27 UTCTue 26 Oct 2010 - 8:27 pm UTC 

Hi happyengineer,

I'm posting this Clarification to trigger an email notification to alert you to Myoarin's alternative solution. It looks like it's going to be a bumper year for donuts!


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