View Full Version : Math Formula For DOC For Equal Size Flats?

Willy

01-09-2009, 11:08 AM

What is the formula for depth of cut, when cutting round stock into a hex for example?

Or for that matter into equal size flats cut at differing degree angles.

I remember I used to have it here somewhere but I just can't wrap my head around it this morning. I think my brain is getting frozen from all of the snow shoveling. Oh well at least no bugs!

lynnl

01-09-2009, 12:15 PM

Well I'm sure that's written somewhere, but I don't know what the exact method is.

Nevertheless, since no one else has answered I'll take a crack at it. Here's the way I'd go about working it out:

The formula for a chord (c) of a circle of radius r is: c = 2 x sq root of [h(2r-h)], where h is the height of the arc, or in your case that DOC you want.

...or h = r - 1/2 sq root of [4(r sq'd) - c sq'd]

So, now we need to find c.

By dividing the circle into 6 equal pie shapes, we can see each pie subtends an angle of 60 deg in the center. And since a triangle has 180 deg the sum of the two outer angles is 120 and since they're also equal each one is also 60 deg. Sooo..., the chord (c) of those six pie segments is one side of an equilateral triangle whose sides all equal r, the radius of the circle.

So for a hex, that second formula becomes h= r - 1/2 sq root of [3(r sq'd)]

NOTE: that first formula I found in my notes from years ago. So presumably I'd worked it out correctly at that time. I invite someone else to check my work and point out any errors, glaring or otherwise. :D

Where:

N = number of vertices

R = radius of rod

DOC = depth of cut

DOC = R - (R * cos(360/2N))

In proofing my own typing I found I had sin in there instead of cos.

Willy

01-09-2009, 03:15 PM

Thanks guys for helping the mathematical challenged.

Another sticky note to put on the mill.

Mama always said..."if you're not going to be smart at least hang out with smart people".

Dennis, don't feel bad about the edit, it seems every post I make I preview it first to be sure it's correct, and then when I look at it an hour later... all kinds of mistakes have mysteriously appeared!

oldtiffie

01-09-2009, 03:21 PM

If you already have the round and want to put flats on it, the depth of cut is the:

((diameter of the round) - (width across the flats)/2)

For example:

if you have a 1" round that you want to put 2 flats 3/4" apart on, the depth of cut will be ((1 - 0.750)/2) = 0.250/2 = 0.125

In conversational terms: for 3/4 across the flats on a 1" bar set a depth of cut of 1/8" and machine two opposite flats that are 3/4" apart.

It becomes a math problem if there are odd numbers of flats or if there are inscribed or circumscribed circles or arcs involved

The puzzle here is interesting - the solution I provided gives the largest hex or what ever is desired that will fit in the round. If you know the final size of the desired hex, say 1/2" between opposite flats as might fit in a 1/2" wrench, and wish it to be centered on the rod axis then the DOC is found with basic math.

DOC = (D - S) / 2 where D = diameter, and S = space between opposite flats. Assumes an even number of flats, of course.

Edit: I should have refreshed - Tiffie beat me to it :)

oldtiffie

01-09-2009, 08:46 PM

What is the formula for depth of cut, when cutting round stock into a hex for example?

Or for that matter into equal size flats cut at differing degree angles.

I remember I used to have it here somewhere but I just can't wrap my head around it this morning. I think my brain is getting frozen from all of the snow shoveling. Oh well at least no bugs!

Willy,

I decided to have a look at what I've got and this seems to meet your requirements. There figures are for a circle with a diameter of 1.

Just multiply any figure (constant) by the circle or diameter that you wan tor have to use.

For example: for anything on a 2" circle/diameter just multiply the required amount by 2

This will eliminate any need for any math as well as any chance of a slip-up.

http://i200.photobucket.com/albums/aa294/oldtiffie/measuring/Multi-side1.jpg

Willy

01-11-2009, 02:43 AM

Thanks for the chart Tiffie, I'll have to print that one out so that I can misplace it somewhere in the shop.:D