Finding length of tool on spindle rotary

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02 Dec 2019 22:32 #151779 by Stephanisk
Hi everyone,

I am struggling with a math concept.

I heard that what I need to do is not so complicated and involves like a 30 degree turn but I am not getting it.

So what do I need:
I need to determine the distance from tooltip to center of rotation on the spindle.

Some explaining. My spindle has a rotary on it. Means a B axis on the spindle. Rotating from -90 to +90 . The only problem is I need to set the length from tooltip to center of rotation in the postpro for correct milling of the part. Measuring the toollength and such get me close to the correct value but I would like to be able to calculate the exact length by using the z touch-off plate.

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02 Dec 2019 22:58 #151781 by Leon82
Can you touch off the rotary head base?

If it's say 12 inches you would touch off at z6.

The other option is to program it like a horizontal
Top WCS front tool plane

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02 Dec 2019 23:01 - 02 Dec 2019 23:02 #151782 by Leon82
Does your cam software have a right angle head tool path? That is basically what you have if I recall
Last edit: 02 Dec 2019 23:02 by Leon82.

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03 Dec 2019 02:21 #151806 by andypugh

Stephanisk wrote: I need to determine the distance from tooltip to center of rotation on the spindle


Just for one tool at a time? You certainly could do that, but the distance from the pivot centre to the table (and hence top of the touch plate) should never move relative to machine absolute coordinates if you have accurate homing.

If you probe one tool, then turn the head 30 degrees and probe again then the difference in height can be used to calculate the tool length with some simple trigonometry, in theory.
But that does rather assume that you know the length already in order to be able to keep the rotated tool aligned with the probe.
Also, it only works without corrections with an infinitely thin tool. And even if you correct for tool diameter (entirely possible) then the flutes might cause inaccuracy.

I did a quick sketch on an online tool, though I hated it, and can't really recommend it. www.math10.com/en/geometry/geogebra/geogebra.html
And it looks like they want more details than I care to enter in order to share it.

But, the tool centre swings up by (L - Lcos(a)) = L (1 - cos(a))
At the same time the tool corner swings down by (D/2)sin(a)

So delta Z + (D/2)sin(a) = L (1 - cos(a))

L = (deltaZ + (D/2)sin(a))/(1 - cos(a))

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