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Calculating the flattened design for a conical shape
Message
De
05/10/2006 17:33:22
Jay Johengen
Altamahaw-Ossipee, Caroline du Nord, États-Unis
 
 
À
05/10/2006 16:45:47
Hilmar Zonneveld
Independent Consultant
Cochabamba, Bolivie
Information générale
Forum:
Politics
Catégorie:
Autre
Divers
Thread ID:
01159862
Message ID:
01159877
Vues:
21
a'm also a bit of a mechanical type of guy. I'm thinking that if I know that the diameter of the top circle needs to be 2" and the height of the cone is 10", then I could lay a piece of paper out, draw a 2" circle at the top, go down the paper 12" and draw the bottom part of that circle using the same center point for the (very large) compass. Then just draw straight lines tangent to the sides of both circles, then I would have my shape. But it sounds too easy.

>I am in little bit of a hurry right now, but I think the basic principle is as follows. You would have to extrapolate the figure to where the two straight lines intersect. Then you have two pies, a large one, from which you have to deduct the small one (the cut-off tip). For each of the pies, the formula would be pi() * r^2 * (degrees / 360). (Note: pi() * r^2 is the formula of the circle, so you are getting a fraction of a circle here.)
>
>HTH,
>
>Hilmar.
>
>>Picture a narrow cone with the top lopped-off. More like a tower, but the bottom circumference is larger than the top. Ok, a lighthouse. Now, cut it down the side and lay it out flat. Basically it would be something like a rhombus with a concave top and a convex bottom. How could I calculate/create that design knowing the dimensions I want my final shape to be? Something like this:
>>
>>
>>
>>                    .      .
>>                   . ` .. ` .
>>                  .          .
>>                 .            .
>>                .              .
>>               .                .
>>              .                  .
>>               .                .
>>                 .           .
>>                     `   `
>>
>>
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