I guess I'm just dense, but I'm not following. I don't understand what the bit diameter has to do with the issue.
Full Disclosure: I inadvertently misreported that my spindle set out in front of the gantry beam at about the same distance as the height of the gantry beam. When I wrote that, I was out of town and reporting from a faulty memory. The gantry beam is about 6" high and and the center of the spindle is about 8.5" from the face of the beam.
I believe I did a bad job of explaining the 1:1 ratio thing. It had nothing to do with the bit. I was referring to the ratio of the height of the gantry beam to the distance from the face of the gantry beam to the center of the spindle. If the two distances are equal, there is a 1:1 ratio. With a 1:1 ratio, if I added a shim to rotate the gantry beam .001" forward, the spindle would rotate downward by .001"
In my mind, it's really all about angles. To explain where I'm coming from, let's take the .001" deviation out of it. Rather, let's look at it from an angles perspective. With a 1:1 ratio, rotate the gantry 5 (or any) degrees, and the spindle angle rotates 5 (or any) degrees. Now that I have the true measurements, I dealing with a roughly 1:1.42 ratio (8.5/6=1.42) The angular relationship does not change, but the amount of movement at one end affects the amount of movement at the other end differently. In this case, a .001" deviation at the spoilboard translates to a .0007 at the gantry to get things right. (1" at the spoilboard divided by 1.42= about .0007" at the gantry). Thought of another way, for every .001" rotation at the gantry will result in .00142" rotation at the spindle.
Here's why I'm having trouble relating the bit diameter to the issue at hand. First let's make the assumption there is no error to correct. Second, let's assume that I rotate the gantry by 45 degrees. Finally, let's assume that my DOC is .005". The gantry is at 45 degrees relative to the spoilboard, and so is the cutting face of the bit (in a perfect world, the face of the cutter is parallel to the top and bottom of the gantry beam). Everything rotates the same. Now, I use my touch plate to zero the 60mm bit. Only the lowest edge of the bit can/will make contact with the touch plate. Start the surfacing routine. The bit lowers to -.005" and cuts. Only the leading edge will cut. Most of the bit won't make contact at .005" DOC. We end up with a series of ridges with a .005" deviation. The real deviation will be much more. Now let's change to a 1/2" bit. Same routine. The cut will be .005" deep and most of the bit won't make contact. We end up with a series of ridges with .005" deviation, same as with the 60mm bit. .005" won't tell us the true deviation, because there isn't full contact. Assuming a 60% stepover for both bits, the ridges are exactly the same, except the ridges are closer together with the 1/2" bit.
If we change to a 5 degree rotation, everything works out the same way, except more of the bit (maybe all) will make contact with the spoilboard. Ridges from both bits will be the same, except for the ridge separation distance and a longer taper angle with the 60mm bit.
Note: I ran both the 60mm and 1/2" bits last week. What I observed comported with what I've described. Note also that I had full contact with both bits, which came as no surprise at a .005" cut depth. FWIW, I took a pencil and scribbled lines all over the spoilboard. Surfacing removed all the lines.
So, I'm left not understanding how bit diameter plays into this. I may be overlooking something, but I don't know what that something might be.
Any help you can offer to help clear this up is much appreciated.
Thank you,
Gary