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12-Amethyst
Jul 13, 2025 10:54 AM UTC
Solved

Elliptical oblique cone.

Gentlemen, please evaluate my attempts to perform the development of an inclined cone, maybe there are comments and something can be corrected or replaced?

Werner_E
Werner_EAnswer
25-Diamond I
Jul 13, 2025 2:02 PM UTC

No, I don't have a vision, as I've never faced the problem of determining the development/network of an inclined cone with Mathcad 😉

You are still missing a lot of details. For example, a is apparently supposed to be the total expansion in the y-direction and b that in the x-direction. Which is the major and which is the minor axis depends on the values you assign to a and b. I'm used to the term a for half the major axis, but you can use a for the entire minor axis, but you have to specify this clearly.
The apex of the cone is apparently at (d / 0 / h), but nowhere did you mention that the y-coordinate of the cone should apparently be zero.
Such information belongs in plain text in the Mathcad sheet, not added later and then only as an image!
Incidentally, theta is not a coordinate, but simply ONE angle, for which they had also chosen the term t (as an argument of the functions X and Y) shortly before.
And L is not the generating end, but its length for a certain angle theta.
What they mean by g and how they arrive at this calculation is still unclear - as is the exact meaning of dPhi. “Angle incerements” is not enough of an explanation, at least for me. The fact that the range variable i is redefined within the sheet is also unattractive in the way it works, even if I can understand the reason for this. You could also use programs with for loops instead of calculations with range variables.

So I cannot verify the correctness of your calculations.

However, I did quickly find an article on the flattening of an inclined circular cone.

https://de.cdn-website.com/c83401726fef4e2cbd69412a8fbb3714/files/uploaded/ArtikelGallinSchieferKreiskegel.pdf

The values r=5, X=3 and Z=4 given there in that text correspond to a=b=10, d=3 and h=4 in your sheet.

In this text, the maximum angle of the development is given as 130.8°.

Your sheet first gives 271.2° and then 268.8°.

Even if the angle in the linked text is only half the angle, this is still a significant difference. (I increased n in your sheet to 1000 ind the hope for a more accurate result and the max angle was even increasing up to 276°).

 

BTW, have you ever tried to print your plot result to a piece of thicker paper, cat it out and glue together to form the cone?

With a=b=10, d=3 and h=4 it does not look to me that we would get from your plot a base circle with a diameter of 10.

 

12 replies

Werner_E
25-Diamond I
Jul 13, 2025 11:26 AM UTC

More detailed explanations about the meaning of the variables along with sketches showing the location of these quantities might help to understand what exactly you are trying to calculate and what the goal of your attempts is.
The picture you have included is also unclear.
One can only assume that it is possibly supposed to be the unfolding, the mantle of the cone.
But as I said, there is a complete lack of connecting words that could explain your calculations and, at best, make them comprehensible.

 

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12-Amethyst
Jul 13, 2025 12:31 PM UTC

Werner_E,yes, this is a mantle of an inclined elliptical cone, maybe you have your own vision of how to do this?

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Werner_E
Werner_E25-Diamond IAnswer
25-Diamond I
Jul 13, 2025 2:02 PM UTC

No, I don't have a vision, as I've never faced the problem of determining the development/network of an inclined cone with Mathcad 😉

You are still missing a lot of details. For example, a is apparently supposed to be the total expansion in the y-direction and b that in the x-direction. Which is the major and which is the minor axis depends on the values you assign to a and b. I'm used to the term a for half the major axis, but you can use a for the entire minor axis, but you have to specify this clearly.
The apex of the cone is apparently at (d / 0 / h), but nowhere did you mention that the y-coordinate of the cone should apparently be zero.
Such information belongs in plain text in the Mathcad sheet, not added later and then only as an image!
Incidentally, theta is not a coordinate, but simply ONE angle, for which they had also chosen the term t (as an argument of the functions X and Y) shortly before.
And L is not the generating end, but its length for a certain angle theta.
What they mean by g and how they arrive at this calculation is still unclear - as is the exact meaning of dPhi. “Angle incerements” is not enough of an explanation, at least for me. The fact that the range variable i is redefined within the sheet is also unattractive in the way it works, even if I can understand the reason for this. You could also use programs with for loops instead of calculations with range variables.

So I cannot verify the correctness of your calculations.

However, I did quickly find an article on the flattening of an inclined circular cone.

https://de.cdn-website.com/c83401726fef4e2cbd69412a8fbb3714/files/uploaded/ArtikelGallinSchieferKreiskegel.pdf

The values r=5, X=3 and Z=4 given there in that text correspond to a=b=10, d=3 and h=4 in your sheet.

In this text, the maximum angle of the development is given as 130.8°.

Your sheet first gives 271.2° and then 268.8°.

Even if the angle in the linked text is only half the angle, this is still a significant difference. (I increased n in your sheet to 1000 ind the hope for a more accurate result and the max angle was even increasing up to 276°).

 

BTW, have you ever tried to print your plot result to a piece of thicker paper, cat it out and glue together to form the cone?

With a=b=10, d=3 and h=4 it does not look to me that we would get from your plot a base circle with a diameter of 10.

 

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12-Amethyst
Jul 13, 2025 2:14 PM UTC

Werner_E thanks, I will study it.

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Werner_E
25-Diamond I
Jul 13, 2025 3:24 PM UTC

On the other hand - I just made a test using a straight circular cone with radius 3 and height 4.

The flattened cone should be a sector of a circle with radius 5 and an angle of (6*pi)/5 = 216°.

Your first Phi.max gives us 213.3° with n=40, but 215.97° with n=4000. So at least in this special case your calculations seem to be correct.

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Werner_E
25-Diamond I
Jul 13, 2025 8:57 PM UTC

Maybe your calcs are only valid for straight cones.

I gave it a try and could confirm the result from the pdf.

Using your data you should come up with an angle of approx. 99,4°.

Werner_E_0-1752440249555.png

 

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12-Amethyst
Jul 14, 2025 3:23 AM UTC

Werner E, that's great, could you please send me the Mathcad 15 file? Thanks!

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12-Amethyst
Jul 15, 2025 6:11 AM UTC

Werner_E, is it possible to get your file in pdf format?

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Werner_E
25-Diamond I
Jul 15, 2025 7:42 AM UTC

Here you are.

Sheet is equally well documented than yours 😉

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Werner_E
25-Diamond I
Jul 15, 2025 7:43 AM UTC

@NickKemaev wrote:

Werner_E, is it possible to get your file in pdf format?


Why? Aren't you using MC15 or MC14?

 

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