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