Speed of groove.

One could easily tell that I'm not a phono guy, but I fail to understand why a phono cartridge's output voltage will be in any way correlated to the linear velocity of the record.
To me, if one would need to correlate the output voltage to a speed, it would make more sense that that speed would be the speed of the stylus relative to the cartridge itself (*vertical* movement) as that's the movement that induces the voltage to begin with.

To me, if one would need to correlate the output voltage to a speed, it would make more sense that that speed would be the speed of the stylus relative to the cartridge itself (*vertical* movement) as that's the movement that induces the voltage to begin with.

Yes, the output voltage is a function of the velocity with which stylus deflection occurs, relative to the cartridge body, the deflection being lateral for the mono component of the signal, and vertical for the stereo component of the signal.

However, for a given physical excursion of the groove, laterally or vertically, the speed of that stylus deflection relative to the cartridge body will vary depending on the tangential velocity of the groove, which gets smaller towards the inner part of the record. Therefore for a given music signal the physical distance over which that excursion occurs has to be smaller at the inner part of the record compared to the outer part of the record, in order to result in the same stylus deflection speed. Tony put it well: "The musical information is essentially packed tighter at the inner grooves of the record because the linear velocity is lower."

Regards,
-- Al

For sake of simplicity, let's consider the vertical movement only and 1kHz sine wave shaped groove for a constant amplitude of the signal. And let's take only the positive semi-period, thus a nice rounded "bump".
At the outer edge of the record, the bump will be "elongated" whereas towards the center of the record it will be "shortened". But in both cases the time the stylus "climbs" from the bottom to the peak of the bump stays the same. And given the fact the the bump has the same height at its peak in both cases, so does the "climbing speed".
This is the speed I'm referring to and I still fail to understand how it could be correlated to the linear velocity of the spinning record.
The linear velocity was "abstracted away" by the "length of the bump".
I'm only trying to understand here, by no means my knowledge of electronics could even come close to Al's...

You can us the math to help understand it. Take a 1kHz waveform again for an example. At the outer groove, the velocity is 20.1"/s. So 1000 (cycles/sec)/20.1 (inches/sec)= 49.8 cycles/inch. Now at the inner groove the velocity is 8.7"/s so the same 1kHz waveform is generated by 114.5 cycles/inch. The amplitude of the cycles is the same at the two locations of the record for the same volume level because the excursion of the stylus must be the same. That makes the inner grooves more demanding because the stylus has to track these 2.3 times more dense undulations in the grooves.

The linear velocity was "abstracted away" by the "length of the bump".

Precisely! That's all we are saying. As you correctly put it, "at the outer edge of the record, the bump will be 'elongated' whereas towards the center of the record it will be 'shortened.'" That shortening or lengthening results in the velocity of the stylus deflection being the same for a given signal at different points on the record, despite the fact that tangential ("linear") velocity is not the same.

Regards,
-- Al