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January 28th, 2015, 04:25 PM   #1
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Question reposted: sewing conundrum concerning cones

Accidentally posted this in high school geometry first. Yeah, I think this is way out of their league.

So after some research (i.e., googling "what is topology") I figured it's better suited for ths forum instead.

Quote:
made some pics to explain a bit better...



so, I'm making a costume, and I need to make a dress that's shaped like a cone. I have the materials to make a rigid hoop to sew into the bottom hem to weigh it down and hold a circular shape, but I don't know how to make the cone itself.

the basic shape has to be a circle with a wedge cut out, but there are a few stipulations.

the inner circle needs to have a final circumference of 40 inches, meaning afterthe edges of the wedge are sewn back together, the circle cut out of the middle needs to be 40 inches around.

the distance between the inner circle and the outer circle (the hem) needs to be 24 inches.

basically, these are the measurements I need:
  1. new inner circumference (BEFORE CUTTING AND SEWING)
  2. new outer diameter (BEFORE CUTTING/SEWING)
  3. wedge width (how much I need to cut out to form the cone)

hopefully this makes some sense... I'm totally lost here.
To clarify a bit, I'm not looking for an explanation of how to do it. I just want the answers. This is a personal project, not some college exam. I'm just at a total roadblock in my costume-making project and I'd like to get moving again.
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January 29th, 2015, 06:06 AM   #2
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This isn't really topology, but I'll answer it here rather than shuffle it around further.

Let's say the angle of the sector you cut out is $\alpha.$ Then the circumference of the new inner circle is the circumference of the original inner circle (let's call it C) times $1-\frac{\alpha}{2\pi}$. Setting that equal to 40 we get $C=\frac{40}{1-\frac{\alpha}{2\pi}}$ or $r=\frac{40}{2\pi-\alpha}.$

Now the radius of the large circle is 24 more, or $\frac{40}{2\pi-\alpha}+24.$ This makes the (original) outer circumference $\frac{80\pi}{2\pi-\alpha}+48\pi.$ You'll then cut out the same angle from it, making the new outer circumference
$$
\left(\frac{80\pi}{2\pi-\alpha}+48\pi\right)\left(1-\frac{\alpha}{2\pi}\right).
$$

I'm not sure how you want to measure the wedge width. The 'missing' outer circumference is
$$
\alpha\left(\frac{40}{2\pi-\alpha}+24\right)
$$
and the 'missing' inner circumference is
$$
\frac{40\alpha}{2\pi-\alpha}.
$$

If you prefer to measure in degrees, replace $\pi$ with $180^\circ.$
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