SEARCH HOME
Math CentralQuandaries & Queries

search

Question from Mike (by fax):

I have been trying to get this cone flat so I can build this column. Can you please help me so I can figure this out? Thanks for your help.

Mike

Mike faxed some drawings:

fax

Hi Mike.

3d

The radius of the column is 11 15/16" .

The radius of the circle at the top is 29 9/16".

The height of the truncated cone is 13 1/4".

This gives the following geometric representation:


Due to similar triangles, corresponding side length ratios are equal:

ratios

Thus, x + 13 1/4 = 22 1/4, which you had already calculated. Knowing x, we can now calculate the really useful radii r and R by using Pythagorus:

r values

So now we know we are cutting from an annulus with inside radius 14 15/16" and outside radius 37".

annulus

Now we need to consider the portion we must cut.

We know the outside circumference we want remaining is the circumference of the larger radius of the truncated cone. That circumference is the circumference of a circle with radius 29 9/16":

whereas the circumference of the outside of the annulus pattern is


The ratios of these circumferences is the same as the ratio of the angles through which they turn. That is, since the full circle of the annulus is 360 degrees, we can calculate what the angle θ is by using the common ratios:

When we solve for θ we get:

and now substitute in our earlier expressions for c and C to find the value of θ

 

Now we have our pattern:

It is a portion of an annulus whose

  • outside radius is 37,
  • inside radius is 14 15/16,
  • missing sector is 72.4°.

 

Cheers,
   Stephen La Rocque

final pattern

 

 

 

 

About Math Central
 

 


Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.
Quandaries & Queries page Home page University of Regina PIMS