The Triangle Calculation

My question is about the triangle, top left, top right, and center of cutting bit.

The calibration process sets the distance between 12 o’clock of the top sprocket teeth, which I presume is meant to be same as shaft center to center distance.

From this the software calculates how much chain on each side to pay out to achieve any x, y location of the bit center point.

Yet the actual top left and right triangle points could be considered the separation point between the chain and the sprocket. If thought about in this manner a different triangle is created.

On setups where the chain first goes over the top of the sprocket from the sled, the separation point moves left and right of shaft center, and up and down from shaft center, depending on the sled position.

Same type of separation point movement on the other sprocket, but only the same as the other sprocket when sled is at left to right center at any height.

This means the “triangle” of the two sprocket separation points and bit center is in total motion at all 3 corner points.

(Same concept for chain under first type of construction, but different specifics.)

Given all of this, my question is to ask if the software in any way calculates / considers this when figuring out how much chain to pay out.

( I understand chain sag and chain elongation are additional adjustments that I’ve seen discussed)

Thanks

Ed

Assuming you’re focused on the triangular Kinematics version, here is a link to the PR that @rjon17469 wrote titled ’ Include Sprocket Geometry in Triangular Kinematics’. There is a link there to a discussion that might make it clear.

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Thank you. I see that this has been considered and put into the software.

I did not see if it is different for the two different chain mounting methods.

Ed

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There is compensation in the code for the two different chain mounting options… How it works, I haven’t figured out but I know something is there.

But to the question regarding triangles, I believe that it calculates the lengths based upon the center of the sprocket and not where the chain comes off the sprocket. Ideally, you would figure out where the point the chain comes off the sprocket and calculate the distance from that and then add the chain wrap to it. But I think its much harder to do and if my calculations are correct, the error from doing it the way it is done is really small.

The error with calculating where the vertex is should be most significant when the sled is at the closest point to the motor (to maximize the difference in the angle). If I did the math correctly, when the sled is at the very top left (or right) corner, the amount of error is ~ 0.1 mm. When its in the very bottom left (or right) corner, the error is ~0.02 mm.