```
That makes sense. So I must make a square cone. Would the best way
to make the square be to make the metal blank for the cone as if
it were the regular round cone, then bend it in four equal, flat
sections for the sides of the stone? If there is a detailed
tutorial somewhere about how to do this, I would love to look at
it.
```

I am not aware of any tutorial, which definitely would be helpful

here.

I shall try to describe the process at the risk of proving the old

adage the one picture worth a thousand words.

The best way to proceed is to compute a template, which when bent

would form square pyramid, or prism of you like, of required

dimensions.

Decide on setting height. Let’s take for argument sake 9mm. The

stone width let’s take 6mm for convenience. So our template should

have be composed of 4 triangles, each having 6mm base and 9mm height.

This represents a class of templates which derived from a disk.

We take a disk and draw a line from disk center to it’s perimeter.

This is nothing more than a radius. From the point of intersection,

start laying our 4 distances 6mm each. Now, join all these points

with straight lines, and also join every point with center of the

disk. That is the whole template. It should be cutout and bent along

the lines running from the disk center. One thing I did not mention

is how to determine the disk radius.

This is the only part which requires computations. Once radius is

known, the rest is trivial.

To determine radius, draw a triangle with base 6mm and height 9mm.

It is a good idea to use multiplier of 10 for clarity, so the

triangle will be 60mm base and 90mm height. Mark triangle vertices as

A, B, and C.

Drop a perpendicular from top vertex to the base and call

intersection as D. So vertices would be marked as leftmost as B; top

as A; rightmost as C; and intersection of base and perpendicular as

D. Consider triangle ADC. It is right triangle, which means that we

compute length of AC, which is exactly the radius that we are looking

for. Because it is right triangle, AC is hypotenuse of this triangle

and can be computed using following formula, - AC = sqrt(AD^2 + DC^2)

substituting our dimensions we get, - AC = sqrt (9^2 + 3^2) = sqrt

(81 + 9) = sqrt(90) = 9.486 or approximately 9.5mm.

This is the radius of disk, which yields required template.

There is one complication. We have to account for thickness of

metal.

Here is what we do: Let’s say we will use 0.8mm. The way the pyramid

will be formed is by engraving the lines to triangular profile to

allow for ease of bending.

The depth of these grooves shall be 0.6mm. Since grooves must be 90

degrees, the wall inclination is 45 degrees, which means that width

of such grooves is twice the depth, or 1.2mm. That means that our

template should have not 6mm base, be 7.2m for 2 inner triangles and

6.6mm for outside. For the purposes of calculations 6.6mm must be

used. This of course is thickness depended. Difference thickness

would required different numbers.

If someone would find this post a bit too much, I am in complete

agreement with you. It should have been done with charts and graphs

to illustrate the concept. But I am leaving for Rome and have no time

to do a blog post. May be when I come back…

Leonid Surpin

Studioarete.com