The rack appears to be base on 1mm pitch metric spur gears.  This means that there are 3.14159mm between teeth.  The pinion is 16mm in diameter, and has 14 teeth.  The worm has 29 teeth, and is about 31mm in outside diameter.

The circumference of all spur gears is some multiple of Pi times the pitch diameter, which is measured in inches or millimeters.

For each revolution of the control knob, the pinion turns 1/29.  This means that the head moves 14/29 of 3.14159mm.  That is about 1.51663mm, or 0.059709879 inches.  This explains the graduations on the knob.  If the worm gear had 44 teeth, then it would move 14/44 of 3.14159mm, or 0.999597mm.  These are close to whole numbers in either the Inch or Metric systems, but not exact.

If you had to re-create the worm, you would somehow have to find a way to create the ratio of Pi from your stack of change gears.  I don't think that Pi can be created by dividing any two whole numbers.  And when you are dealing with change gears, you have to deal with ratios that are fractions of whole numbers.

How do they create a micrometer position system using spur gears and racks that avoids this problem?

It seems to me that the pitch of the rack should be a whole number fraction of the measurement system you want to work in.  For example, if the rack had 8 teeth per inch, (which IS close to what it actually is, but NOT exact), then the pinion could have, say 14 teeth.  The pinion would then have a pitch diameter of 1/8"x14, which is 1.75/Pi, or 0.5570423".  If the worm gear had 35 teeth, then the head would move 1/35 of 14x1/8", which is 0.050".  The pitch of the worm itself would depend on the diameter of the worm gear.  If the pitch was chosen to be 1/8", then the worm would have a pitch diameter of 1/8"x35/Pi, which would be about 1.3926", or 35.4mm.  Still not exact, but at least you could make it on a lathe without having to find a ratio of change gears that resulted in Pi!  Actually, the outside diameter would be 1/8x37/pi, or about 37.4mm.

To convert the downfeed to metric, you would need a worm gear with either 44 teeth for 1mm per rev of the handwheel, or 22 teeth for 2mm per rev of the handwheel.  Using the same worm, the outside diameter would be 46x1mm which is 46mm.  Likewise for 22 teeth, it would be 24mm.

Contact:  w.j.ward
Updated:  March 4, 2007