Speed of groove.


We all know that record spins at (let's take at this point) 33.5 rpm and while the rotational speed is the same the speed of the actual groove is not.
Our cartridges are specked at output voltage for a specific speed of the grove(cm/c). What happens when
a)the speed is higher than specified at outer grooves
b)the speed is lower than specified at inner grooves
c)the record spins at 45rpm or 78rpm
marakanetz
The linear velocity of the groove at the stylus starts out at about 20.1 inches/second and finishes at about 8.7 inches/second. That is based upon the following: 33.3 rpm, outer groove 5.75" from center and inner groove 2.5" from center. It is a simple formula: V=2Pi*r*33.3333/60 (rev/sec) r is in inches so you get inches/second. Cartridges have a rated output voltage at a given lateral velocity. That is defining the amplitude of the stylus motion. The output isn't really affected by groove speed, but by the amplitude and velocity of the stylus motion. The stylus is coupled to a tiny generator and the voltage output is an AC signal that varies in frequency and amplitude based on the stylus motion which is driven by the record grooves. So a 1 kHz waveform, for example, (the squiggles in the vinyl) at the outer groove of the record will be longer than a 1kHz waveform at the inner groove. The musical information is essentially packed tighter at the inner grooves of the record because the linear velocity is lower. 2.3 times tighter to be exact.
12-29-11: Marakanetz
At some point of time on this post I saw response from Al (Almarg) but it dissapeared.
???
I deleted it, and posted the revised version you can see above. My first submittal wasn't quite right :-)

Regards,
-- Al
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...