View Full Version : figuring conical ferrule dimensions

mikem

04-20-2006, 12:33 AM

When trying to make a brass ferrule (it was supposed to be wrapped around the bottom of the leg) for a missing one on a tapered round piano leg, we used paper for a template and cut and tried until it fit. This ferrule looks like what you see on a wood handle for a chisel, except it wasn't made from tubing but rather from flat brass. This took a long time and I was wondering what was the mathmatical method of figuring this out. The paper template was transferred to the flat brass plate and the resultant shape was about 3.14 times the diameter long and both of the long sides were curved slightly. This part I needed was slightly conical, about 1 inch in diameter on one end and .875 inches diameter on the other and about an inch long. When laid flat, the two sides look like they have the same center of radius and the radius looks like it might be somewhere about like 6 inches, only one more and one less in diameter.

When you make a regular cone from flat stock, it looks like a half circle. This example is similar to that except that this is only part of cone and it has a very light taper compared, for example, to a dunce's hat.

Can anybody see what I am trying to do and is there an answer that is easier than doing it by paper template? Thanks--Mike.

CCWKen

04-20-2006, 12:49 AM

It's called a conical frustum. Look here under "Bucket.zip" for a program to give you the measurements/layout.

http://www.myvirtualnetwork.com/mklotz/#shop

Courtesy of Marv Klotz.

Mike

There is a free download for a frustrum calculator for Excel at http://www.ameng.com.au/downloads.htm

I am not sure If it is suitable for your problem or not.

mklotz

04-20-2006, 12:11 PM

CCWKen is close but not quite right. BUCKET is used for calibrating containers

with a conical frustum shape.

The program you really want is CONE. Running it with the dimensions from your

post, it produces an output that looks like:

---------------------------

FLAT PATTERNS FOR CONICAL PARTS

small diameter of cone [3] ? .875

large diameter of cone [5] ? 1

height of cone [10] ? 1

overlap allowance for joining [0.25] ? .125

small diameter = 0.8750

small circumference = 2.7489

large diameter = 1.0000

large circumference = 3.1416

cone height = 1.0000

overlap allowance = 0.1250

included angle of pattern = 23.3497 deg

smaller radius of pattern = 7.0137

chord of smaller radius = 2.8538

arc length for smaller radius = 2.8583

larger radius of pattern = 8.0156

chord of larger radius = 3.2440

arc length for larger radius = 3.2666

length of edge = 1.0020

cone included angle = 3.5763 deg

---------------------------

Instructions for how to use these data to construct a flat pattern are in the

text file included with the download archive.

david_r

04-20-2006, 01:50 PM

Mike,

To get the height of the full cone(the radius of the circle it will be made from), use H*D/(D-d). Once you know that, you use the circumference of the large end divided by the circumference of a circle using the first number times 360 to calc the included angle (or the radiuses or the diameters -- it's just a ratio).

Marv's program is way easier but when I had to make a cone out of 14 gauge steel in the field, I had a piece of soapstone and a calculator to make it happen.

mikem

04-22-2006, 12:24 AM

You guys know everything! God bless Al Gore for inventing the internet and Marv for posting all these calculation programs. This is exactly what I wanted. It would have saved me lots of time and made a much better finished product. Thanks to all who responded--Mike.