TT speed

When I use a protractor to align the stylus I do the alignment at the inside, and then rotate the platter maybe 20 degree when I move the arm to the outside of the LP, or protractor.

On a linear tracking “arm” it would not need to rotate at all.

At 33-1/3, then 15 minutes would be about 500 rotations. And that 20 degrees would be a delay of 18th of a rotation.

So a 1 kHz tone would be about 0.11 Hz below 1000.
It is not much, but seems kind of interesting... maybe?


The OP is correct and his numbers are spot on target. The math to prove it is not difficult.

If the position of the cartridge changes by 20° from beginning to end of the LP, it will lengthen the time of the play back by 100mS: 20°/360°=1/18th of a rev; at exactly 33 1/3 RPM, 1 rev=1.8seconds x 1/18=100mS. The 100mS change in play back takes place over 500 revs (15 min) or 200µS/rev. So each and every revolution will experience a longer playback time of 200µS. 200µS/1.8S (expected time per rev @ 33 1/3)=0.00011111 or 0.0111% longer. Another way to look at it, the playback of each revolution will take 1.800200 seconds instead of 1.8S so the effective speed would be 33.3296 RPM (60Sec/1.80020). This will change a 1kHz tone to 999.888 Hz, assuming the record was cut with a linear tracking cutting head and the speed was held constant at 33 1/3 RPM.

The shift in frequency is insignificant and not audible even to someone with perfect pitch. It is not immeasurable or beyond calculation. It has nothing to do with tangency or FFT algorithms. It is purely a timing issue.

Another way to conceptualize this: Imagine a stylus is playing a 1kHz tone in a locked groove for 15 minutes (500 revs). Over the course of that time, the tone arm is moved in such a way that the tracking error stays within the expected range, but the stylus location finishes at a point that is 20° CW from its starting point in the locked groove. The effect would be the same as what is described by the OP.

My tonearm is 10" effective length. I placed a piece of masking tape on the platter to form a radial "spoke" on the platter.  If I align the stylus tip with the tape at the outer groove radius (146mm) then rotate the arm to the inner groove radius (60mm), it is slightly more than 3/4" CW of the stationary spoke of tape.  At 60 mm radius, 20° represents 0.824" of circumference.  So a 20° shift in location is a reasonable amount. 

I thought this through over the last few days, and I agree, but I would explain it differently. I’ve done a 180, reversal of my prior opinion.There’s more than one path to the same conclusion. Still, I’d like to see a demonstration.

Perform the demonstration in the last paragraph of my previous post.  Keep the platter stationary and move the stylus from the outer groove to the inner groove area and measure the distance from the stylus to the tape when located at the inner groove.  This is the linear distance the stylus will advance over the span of 15 minutes. At 60mm radius inner groove diameter, the circumference is 14.84".  Divide the distance from the tape by 14.84 for the portion of the total circumference and multiply by 360 to give you degrees.

The fact the stylus is CW of the tape proves the length of the groove used for playback increases as the stylus travels towards the dead wax area.  It's fairly easy to crunch the numbers and determine how much it advances on each rev. and therefore the change in pitch that will occur.

Thanks for the clearer explanation @phoenixengr
It is indeed a SFA* amount of shift.


* SFA is a UK/Au expanding to “Sweet F__ All”, which generally means is to the point of not mattering.