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  • 1 Post By romsek
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October 26th, 2016, 09:19 AM   #1
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Making a Cone Tent

So, I'm currently designing a cone tent (technically, it's just the tent top), and I'm stuck on some math. It's been years since I've taken these courses and I'm rusty and really need some help figuring out my design.

For the top cone, I have the following measurements:

Slope = 117.56"
Diameter = 183.44"
Height = 73.54"
Bottom Circumference = 588.32"

Now, this cone, will have a design on it (see pic).

((Note: This model is NOT to scale and therefore can't be measured and scaled up.))

This has lead me to this image.

I need to figure out a and b (labeled on the pic) so that I can make a template for cutting out the canvas pieces for sewing. I also need to find the width and height of the diamonds but I didn't mark them.

I'm just completely stumped as to how to find the measurements I need... could someone please walk me through here? I think if I could determine what the angle of the top of the cone is for each segment (8 equal cone segments) then I could solve the rest, but I'm stumped there too...

Last edited by skipjack; December 15th, 2017 at 05:56 AM.
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October 26th, 2016, 03:51 PM   #2
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I can't really make out your picture.

What I can suggest is to draw a circle on the biggest piece of paper you have and remove 1/4 of it.

This gives you an 270 degree segment of a circular disk.

This just happens to be the shape of your cone unrolled. (yours was 269.423)

You can go ahead and lay out any designs you want on this and then make whatever measurements you like, linear and angular, and those measurements will properly scale.

To find the scale divide the length of the actual tent slope by the radius of your circle on paper.

For reference if you have a cone with base radius $r_b$, and height $h$, the unrolled cone will be a segment of a circle with radius

$r = \sqrt{r_b^2 + h^2}$

with angular width

$\theta = \dfrac{2\pi}{\sqrt{1+\left(\frac{h}{r_b}\right)^2} }$
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October 27th, 2016, 11:23 AM   #3
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Firstly, thanks for your reply!

I really wanted to avoid making a scale model if possible, and I knew since I had enough information to make a scale model, I must have enough information to find all the sides and angles I need without making the model.

That being said, your post actually did help me TREMENDOUSLY. I can't thank you enough.

Specifically, you reminded me of the angle of the unrolled cone which was the piece of the puzzle I had been missing (even though it was in front of me cuz it had been previously worked out, the whole time). Once I had that, I simply needed to divide that by 8 (since there are 8 equal cone segments) which gave me the top angle of my cone. From there, it's just been a series of working out angles and line segments. I still haven't finished all the math, but that is mostly because I have a 6 month old baby I'm taking care of and only working on this on the side.

I won't bore everyone with all the math, but needless to say, I think I'm on the right track now.

Last edited by skipjack; October 28th, 2016 at 07:08 AM.
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October 27th, 2017, 07:43 AM   #4
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So rather than a "cone" this really a cone on top of a cylinder?

You say this has "diameter 183.44 inches and circumference 588.33 inches. I take it you mean that only approximately. A circle with diameter 183.44 inches will have a circumference of $\pi d= \pi (183.44)= (3.1416)(183.44)= 576.30$ inches. Are you using 588.33 inches so you will have some overlap to sew together? That's a good idea.

To make the conical top, start by cutting a circle with radius 117.56 inches, your "slope" measurement. That circle will have circumference $2\pi r= 2(\pi)(117.56)= 2(3.1416)(117.56)= 738.65$ inches. You want to fit that to the 576.30 inch circumference of the cylinder bottom, so mark any point on the circumference of that circle, measure off 738.65- 576.30= 162.35 inches, mark that point and, finally, cut along the radius from each mark to the center of the circle, removing the "triangular" section between. (Actually, make that distance between marks a little less than 162.35 inches so there will be a little "overlap" for sewing.)

I would like to check the height. The conical section will have height, from the Pythagorean theorem, $\sqrt{117.56^2- 91.72^2}= \sqrt{5407.8}= 73.528$ inches (a little over 6 feet so most people could stand up in it). That doesn't leave anything for the cylindrical bottom! Did you mean just the cone without the cylindrical bottom in the picture? If so, just ignore my first paragraph!

Last edited by skipjack; October 27th, 2017 at 08:25 AM.
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December 15th, 2017, 01:48 AM   #5
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hi guys
i want to build this how to do it?
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