## Friday, May 20, 2022

### Today's Analytic Geometry Problem

When you see my question, you will see why math was only non-A grades in high school and why Harvey Mudd had the good sense to put me on the waiting list.

I need to mill an inside cylinder.  I am using a round endmill to get a smooth circle.  The problem is depth of cut as I move from yStart to yEnd.  Call them 0.0 and 1.0 for this word problem.

The equation for a circle is x^2 + y^2 = r^2.  I am leaving out the displacement from 0,0.

Just to cause pain, y in this equation will be the depth of cut: z on a mill.  x in the equation will be the distance across the workpiece: so y on the mill.  So z = sqrt(r^2-x^2) in equation form, but sqrt(r^2-y^2) for the mill.  This is a displacement from the surface, so -(r-z)?

y   z           -(r-z)

0   5          0

.1  4.999  -.001

.2  4.996   -.004

.3  4.991   -.009

So far, so good,  Now we cross the midline at 0.5.  Of course, half way across the cut, we need to start rising again.  Do we start using sqrt(r^2-((y-midpoint)^2)?  I tried that with y=.6.   z = -.001--clearly wrong.

1. I'm not sure what you're trying to do; I don't understand what you mean by "inside cylinder." Are you trying to cut a hemi-cylinder trench into the surface of your block of material, trying to cut a circular well into the block, or something else? Also, is z - "depth" - the height (or depth) of the cylinder, or is it the height/depth of the circle with the circular cross-section being on the mill's y-z plane?

In any case, if the circle of the circular cross-section is centered at (0,0) then shouldn't y run from -1 to 1 rather than from 0 to 1? (and I'm presuming r=1)

1. Hemicylinder trench. Changing the y from 0 to 1 to -.5 to .5 is a very good solution.

2. Define x-axis horizontal parallel to trench axis, y-axis horizontal perpendicular to trench axis, z-axis vertical perpendicular to trench axis. Trench axis defined by y=0 and z=0.
Using ball-end mill with radius r. Want to cut trench with radius R.
You need to account for radius of endmill itself. Note when tip of end mill at depth z=-r, end mill is cutting hemisphere radius= r at z=0 (on surface of part). To cut a semicircular trench radius = R, you need to have the center of the end mill's ball move along an arc of radius = R-r. So y^2 + z^2 = (R-r)^2. Define R, r and for each value of y (horizontal table location -- position of endmill axis perpendicular to trench) calculate corresponding z (depth of cut).
y=sqrt((R-r)^2-z^2)
BTW, this may be your only option for milling, but it is not recommended. At the bottom of the trench, the mill will be cutting at its tip with zero cutter velocity (velocity = radius*rpm and at tip, radius approaches zero). Also there will be ridges between each transverse pass due to pass basically cutting semicircular trenches r. You can smooth things somewhat by doing finish cuts parallel to trench axis (moving mill in x for each given y,z) but this will leave a grid of roughness. Might be able to then file or sand away?
If possible setup with mill axis parallel to trench axis (requires open end trench though.)

1. I am using a ball endmill. With a small enough movement across, you can get a relatively step free trench.