Giant coffee cup question

R

Ridge

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Joined
Jul 9, 2022
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Hi

I have a real world problem and I am stumped but I'm sure the answer is out there in a maths brain somewhere.

I need to make a truncated cone, measurements I know are;
base diameter
top diameter
the sloping side of the truncated cone (from which even I can calculate the height of the truncated cone if needed)

What I am trying to calculate is the sloping side if it was a full cone, rather than truncated, because then I can easily draw this out on a flat sheet with a tape measure and a pen.

If someone can come up with a simple formula that I can put the figures in to then I would be very grateful, and very impressed.

Please be gentle with me!

Ridge.
 
Ridge said:
Hi

I have a real world problem and I am stumped but I'm sure the answer is out there in a maths brain somewhere.

I need to make a truncated cone, measurements I know are;
base diameter
top diameter
the sloping side of the truncated cone (from which even I can calculate the height of the truncated cone if needed)

What I am trying to calculate is the sloping side if it was a full cone, rather than truncated, because then I can easily draw this out on a flat sheet with a tape measure and a pen.

If someone can come up with a simple formula that I can put the figures in to then I would be very grateful, and very impressed.

Please be gentle with me!

Ridge.
Click to expand...
Please draw a sketch of the coffee cup for us - indicating given dimensions and assumptions.
 
Hi
Hopefully this will be clear. I don't know why I drew it upside down it is just the way my head works.
The dimensions may change but this is what they currently look like being.
I am trying to calculate h2
 
Ridge said:
Hi
Hopefully this will be clear. I don't know why I drew it upside down it is just the way my head works.
The dimensions may change but this is what they currently look like being.
I am trying to calculate h2
Click to expand...
No problem ... when I solved the problem I drew the coffee cup in the same orientation too - spilled coffee.

I assume that cup is made by a good potter and it is symmetric and all the horizontal-sections are circular. All those centers of the sections will line up in a vertical line.

Drop a perpendicular from the apex of the cup to the bottom of the cup. Do you see that the perpendicular will strike the bottom circle at the center (if the potter is any good).

Now let's work with a vertical section through the apex of the cone (cup). It will be an isosceles triangle - with two equal sloping sides and a base and a line for the top of the cup. Also the perpendicular line from the apex of the cone to the center of the cup will be there - we are working with a vertical cross-section (through the apex) of the cone.

Now do you see two similar (vertical triangles) with bases d1/2 and d2/2?

Do you know that the sides of similar triangles are proportional?

Can you use that knowledge and proceed?
 
No I'm not following this at all.
I have a sheet of foam which I am going to cut up and make the giant cup out of, no potter involved! But I'm still none the wiser about what circumferences to use.

A section through the cup (truncated cone) will be 2 right angle triangles and a rectangle won't it?
The hypotenuses of the triangles will be h1 and the base (d1 - d2)/2
The other side (height) of the triangle I can work out with a bit of Pythagoras if needed.
The rectangle will be d2 by the triangle height.

A section through the entire cone would be 1 isosceles triangle with base d1 but I don't know the height or the length of the equal sides. Or 2 right angle triangles with bases d1/2 but I know neither the height or the hypotenuses.
 
Ridge said:
No I'm not following this at all.
I have a sheet of foam which I am going to cut up and make the giant cup out of, no potter involved! But I'm still none the wiser about what circumferences to use.

A section through the cup (truncated cone) will be 2 right angle triangles and a rectangle won't it?
The hypotenuses of the triangles will be h1 and the base (d1 - d2)/2
The other side (height) of the triangle I can work out with a bit of Pythagoras if needed.
The rectangle will be d2 by the triangle height.

A section through the entire cone would be 1 isosceles triangle with base d1 but I don't know the height or the length of the equal sides. Or 2 right angle triangles with bases d1/2 but I know neither the height or the hypotenuses.
Click to expand...
I did not realize that you are making a cup out of a foam sheet.

But going back to the cone - pretend it is not truncated and it is solid (not a hollow one made out of sheet ) . Now if you take vertical section - you can imagine two similar vertical right angled triangles.
One is the big one with base r1 = d1/2 and slope side (h2 +h1). We will calculate h2.​
One is the small one with base r2 = d2/2 and slope side (h2).​
These two triangles are similar. So (by the theorem of proportionality of the sides of similar triangles, we get).

\(\displaystyle \frac{h_2}{r_2} = \frac{h_1+h_2}{r_1}\)

Can you solve for h2 from here?
 
If you ever want to know anything about theatrical costumery, my specialism, I am going to make you work for it!

OK so, in my example
h2/(37.5) = (100 + h2)/55

therefore
h2 x 55 = (100 + h2) x 37.5
h2 x 55 = (37.5 x h2) + 3750

(h2 x 55) - (h2 x 37.5) = (h2 x 37.5) - (h2 x 37.5) + 3750

17.5 x h2 = 3750

h2 = about 214cm certainly near enough for cutting a bit of foam.
 
Ridge said:
If you ever want to know anything about theatrical costumery, my specialism, I am going to make you work for it!

OK so, in my example
h2/(37.5) = (100 + h2)/55

therefore
h2 x 55 = (100 + h2) x 37.5
h2 x 55 = (37.5 x h2) + 3750

(h2 x 55) - (h2 x 37.5) = (h2 x 37.5) - (h2 x 37.5) + 3750

17.5 x h2 = 3750

h2 = about 214cm certainly near enough for cutting a bit of foam.



Click to expand...
and the formula, using radii instead of diameters

h2 = (r2 x h1) / (r1 - r2)
 
Ridge said:
If you ever want to know anything about theatrical costumery, my specialism, I am going to make you work for it!
Click to expand...
That is about the best compliment I have received - to an engineer from an artisan.
 
I would find the equation of the two slanted lines and see where they intersect. Then use the distance formula.
 
Thanks Steven.
I'm hoping never to have to make one again! Wearable cups of coffee are quite tricky.
I just wanted to show the finished piece to the forum as I had mercilessly used Subhotosh's knowledge.
 
Why did my post from July 9th finally post this past Wednesday?
I guess that is because SK is on the job!
 
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