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Steve_B
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« on: November 21, 2010, 08:24:52 AM »

I have a rack & pinion gantry that has a 10’ travel on the "X” axis that is slaved. If the pitch diameter on 1 of the pinions wears a few thousands more than the other and you have 10’ of travel the gantry ends up out of square the farther you get away from the home switch. Is there a way in Mach to comp for 1 side of a slaved axis traveling slightly different than the other on a slaved axis?

Thanks
 
Steve
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Hood
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« Reply #1 on: November 21, 2010, 08:59:54 AM »

As far as I know a slaved axis use the master axis steps per unit so they have to be identical.
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Greolt
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« Reply #2 on: November 21, 2010, 03:05:19 PM »


If the pitch diameter on 1 of the pinions wears a few thousands more than the other and you have 10’ of
travel the gantry ends up out of square the farther you get away from the home switch.



Any error due to wear that you have described will not accumulate along the length of the rack.  Unless miraculously, it wears in a way that it has one less tooth.  Smiley

In simple terms, the rack determines how far it will travel each revolution of the pinion, (for a given tooth count)  not the pinion itself.

By nature the rack will wear much slower than the pinion, provided all the machine use is not in one small spot.

The reason being that as the axis travels the length of the rack, each rack tooth will be engaged once.  Each pinion tooth will engage many times.

If you are concerned about slave travel discrepancies then put your racks side by side and make a careful comparison.  Worry less about your pinions.

Greg
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