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Milling radius by tilting mill?

ossi

Aluminum
Joined
Mar 6, 2003
Location
Iceland
Hi.
I remember long ago in school I had to mill a radius in a piece. To get the right radius the teacher showed me a method of tilting the milling head.

I would like to learn how to calculate the angle.

For instance, how many degrees I have to tilt a 1" mill to mill a groove for a pipe of 1,043" dia?

Thanks in advance.

Ossi.
 
Sorry, the method only works for fairly large radii and the cutter dia is always greater than the width of part wanted.

You could make a radius for the pipe with a 1" dia mill, but the width would have to be less than 1"

Since you are cutting portions of an ellipse, the wider the part as a relationship to dia of cutter, the less accurate the radius

Great write up in the fifties issue of A Treatise On Milling and Milling Machines by Cincinnati Milling Machine

J.O.
 
I do this all the time... in glass, usually, but have done it in cast iron, too. It's a very common operation in lens and spherical mirror production.

The formula for determining the proper tilt is:

sin(tilt) = d/2R, where "d" is the diameter of the cutter, and "R" is the radius desired on the piece to be machined.

Note that this formula gives you the sine of the angle, not the angle itself. For that, it's

tilt = arc_sin(d/2R)

Just to clarify the above poster's comment... the cutter's diameter must be at least 1/2 the diameter of the part to be machined.

You might need this as well... the formula for the center depth of the curve produced:

depth = R-square_root(R^2 - D^2/4)

R= Radius of curve cut into the material
D= diameter of the spherical cut made (not the cutter!)
 
Johnoder. The piece is just ,551" in with. So its not really a groove...

Dan. You lost me at sin.

Could you write it down like you would type it on a calculator?
 
I laid it out with Dan's calcs and used a 1/2" wide piece (which I think Dan meant to say instead of the other way around).

Spindle would be tilted 16.509 degrees off horizontal using Dan's info for a .5215" R.

The cad layout showed a .519" R, so you may need to fiddle with it.

The .551 width will squeeze down the R a little more than that.

I.E. the wider the part is in relation to cutter dia., the less the AVERAGE radius will be near to the calculated result.

A little more tilt off horizontal ( trending higher than the above figure) will enlarge the radii.

J.O.
 
I do this all the time... in glass, usually, but have done it in cast iron, too. It's a very common operation in lens and spherical mirror production.

The formula for determining the proper tilt is:

sin(tilt) = d/2R, where "d" is the diameter of the cutter, and "R" is the radius desired on the piece to be machined.

Note that this formula gives you the sine of the angle, not the angle itself. For that, it's

tilt = arc_sin(d/2R)

Just to clarify the above poster's comment... the cutter's diameter must be at least 1/2 the diameter of the part to be machined.

You might need this as well... the formula for the center depth of the curve produced:

depth = R-square_root(R^2 - D^2/4)

R= Radius of curve cut into the material
D= diameter of the spherical cut made (not the cutter!)


Don't these formulas for for machining spherical shapes by rotating the work with a rotary table? I thought the OP needed a groove with a radiused bottom, maybe I inferred incorrectly.
 
milling a radius

HELLO,
you can find this formula on
PAT'S BIG LIST OF METALWORKING LINKS.
also a host of other interesting information.
wlbrown
 
Dan Cassaro - I first learned this trick when I was an apprentice Plastic Injection Mold maker. The only difference is the formula I was taught was expressed as d/D = sin of the angle to tilt the head (as you said the dividend is actually arcsin so one must solve for the inverse). d = the cutter diameter, and D = the Radius x 2, or expressed as a diameter. Exactly the same result.

I've seen so many overly complicated answers to this problem, but you've given the best answer. My loupe is off to you. Had it not been for an "old-timer," I'd be ignorant of it as well.

Toolmaker Tim
 








 
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