Based on a 2002 paper by Chuang et al., I'm trying to understand the math for calculating 'drop-cutter' toolpaths (Julian Todd also has a lot to say about the drop-cutter approach). The basic idea is that the x,y position of the cutter is known, and it is dropped down along the z-axis until it touches a triangulated surface. A complex surface is built up of many many small triangles, but the algorithm stays the same.

For each triangle there are seven items the cutter might be touching: three vertices, three edges, and one facet (the triangle surface). It looks like a contact point with a vertex is the easiest to calculate, so I'm starting with that. I'm hoping to post part 2/3 and 3/3 of this post soon. They will deal with the facet and the edges.

I'm using this fairly simple cutter definition **C(R,r)** where **R** denotes the shaft diameter and **r** the corner radius. The three basic shapes that can be defined this way are shown in the picture. A more elaborate model would include tapered cutters, but I think I won't bother with that now... A cutter thus consists of a cylindrical part or a toroidal part, or both. That means I need six different functions in total for this algorithm:

- Cylinder against vertex
- Toroid against vertex
- Cylinder against facet
- Toroid against facet
- Cylinder against edge
- Toroid against edge

Here's how I think no 1 and 2 work.

Assume the vertex we are testing against is at (**x**, **y**, **z**) and say the cutter is at location **xe,ye. **We are looking for the cutter z-coordinate **ze** so at the end when the cutter is in contact with the vertex it will be located at (**xe**,** ye**, **ze**) .

Calculate the distance in the xy-plane from (**xe**, **ye**) to (**x**, **y**):

**q** = sqrt( (**xe**-**x**)^2 + (**ye**-**y**)^2)

Now if **q > R** the vertex lies outside the cutter and we should report an error or just skip to the next triangle/vertex.

If **q **<= **R**-**r** the vertex lies under the cylindrical part of the cutter and we set **ze** = **z**

If (**R**-**r**) < **q** <= **R** the vertex lies under the spherical/toroidal part of the cutter. This picture should explain the geometry:

The cutter can be positioned a distance z1 lower than z. To calculate z1 we need z2. It can be found by noting that (**x** ,**y**, **z**) should lie on the toroidal surface:

(**q**-(**R**-**r**))^2 + **h2**^2 = **r**^2

or **h2** = sqrt( **r**^2 - (**q**-(**R**-**r**))^2 )

so now **h1**=**r**-**h2** and we found **ze** = **z**-**h1**

A quick test in matlab seems to confirm that this works:

here a ball-nose cutter is dropped down along the dashed line into contact with all three vertices (red, blue, and green stars), and finally it's positioned at the highest **ze** found (red dot).

Well, that was easy. Now onto the facet and edge tests!

## One thought on “Drop Cutter part 1/3: Cutter vs. Vertex”