I need a milling fomrula

I need a formula that calculates an angle for an apparent radius of a cut using a standard endmill, let me explain what I need.

When an endmill is exactly 90 degrees from the work piece the mill creates a flat pass, however as you angle the endmill a radius forms, the closer you are to 90 degrees the bigger the radius. When the end mill is at 0 degrees or parallel to the work piece the radius is that of the endmill.

My problem is that I have a 1" endmill and I need to create a 2" radius length wise in a 3/4" x 3/4" x 6" bar, so I need the formula described above to know the angle at which I should mill.

Any help would help.

WGB

Reply to
Wd Gre
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The arc milled out by a tilted end mill isn't circular, it's elliptical. Think of the shape you get when you intersect a right circular cylinder with a plane. If the plane is perpendicular to the axis of the cylinder then the intersection is a circle, else it is an ellipse.

Presumably you want to mill something as close as possible to a 2" radius by approximating with a tilted 1" end mill. How wide do you want this indentation to be? Is it the full 3/4"? Then your indent would be 0.073" at its deepest.

If you need it, I can figure the angle you'd need to mill an elliptical indent

3/4" wide and 0.073" deep with a 1" end mill. It wouldn't be too hard to cut and try, though. Tilt it over some, take a pass, measure width and depth. Divide 0.750" by the width of the pass and call this result M. Multiply the depth of your pass by this result M and see if it's 0.073". If it's bigger, then tilt it less, if it's smaller, then tilt it more. Repeat.

An easier method would be to stand your workpiece on end and bore it straight down using a boring head. Or make a 2" flycutter. Or rough it out and sand it close enough using a 2" sanding drum.

Grant

Wd Gre wrote:

Reply to
Grant Erwin

The projection on a plane of an oblique cylinder is an ellipse, not a circle. The angle you seek and the radius of the circle that fits such an elliptical section with least RMS error will depend on the width of the region you are "radiussing". Would that be 3/4"? (I'm not sure what you mean by a lengthwise radius)

I'm also not sure I could figure it out if I knew, but the foregoing is definitely so.

Reply to
Don Foreman

For a best- fit ellipse to a 2" radius with a 3/4" width, the angle is

60.3962 degrees. The max deviation of the resulting ellipse in the 3/4" span is less than .001" from a 2" radius.
Reply to
Don Foreman

Trying to picture in my mind what you exactly want to do - and my mind is a little low on film.... :-) Are you trying to mill a groove down the length of the bar? or are you trying to "dish out" a hollow in the center of the bar across the 3/4" width? A little more explanation would help. Ken.

Reply to
Anonymous

Grant is correct. The cut will be a portion of an ellipse not a portion of a circle. However, I believe that Grant's calculation is for a 2" diameter groove rather than a 2" radius.

If you want to cut to the correct depth (0.0354" for a 2" radius), then a 1" cutter should be set at 77.9 degrees measured from the horizontal. This will produce a groove that is a first approximation to a 2" radius. The groove will be slightly more flat-bottomed than a true radius it will also have slightly sharper side walls. Another way to view this error is that the volume of material that is removed will be greater than the volume removed if it were a true radius.

To get a groove that is a better approximation to a radius, the cutter should be set at a shallower angle to the horizontal and the cut should be deeper at the center.

The equation for the depth of cut as a function of cutter angle is:

Depth=0.16928*cos(alpha)

Where alpha is the angle from the horizontal.

Or, if you prefer: Depth=0.16928*sin(beta), where beta is the angle from the vertical

HTH, George.

Reply to
George

No argument with this, and you *did* ask for a formula. What I did was use the "solver" in Excel to minimize the error between a 2" radius and an ellipse -- which means it cuts proud in some areas and shy in others so the max + and - deviations are nearly the same.

After playing with it a bit, it found an angle of 60.44597 degrees gives peak plus and minus errors of about .00012" -- yes, that's 1.2 tenths or 120 microinches. Feed the cutter in depth until the cut is

0.7500 wide, or depth of cut is .03547.

The error curve looks like a W, with max + deviation at the extremes and in the middle, max - deviations at about +/- .265" from centerline, and zeros near +/- .155 and .340 from centerline.

Realism: If you're off on the angle by 0.5 degree the max error is still only about +/- half a thou if the swath is still .7500 wide.

Reply to
Don Foreman

Marv Klotz has a great site with formulas for many things, including just this problem. Go to

formatting link
down load "RADIUS.ZIP", expand it and try the formula out. Steve

Reply to
Steve Steven

Wow, he sure does. Lotsa good stuff there. Thanks, Marv!

Reply to
Don Foreman

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