I need a milling fomrula

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

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:

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.

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

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.

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.

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

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.

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

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