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
Convert?fit=crop&h=128&policy=eyjlehbpcnkioje0otu2oti5mdgsimnhbgwiolsicmvhzcisimnvbnzlcnqixx0%3d&rotate=exif&signature=028b947dee4f6f0eee00f6ec376754a538cd72b99f141959bb67dd2244e37d20&w=128marakanetz
Speaking as an SME owner, their power supply is supposed to lock on to speed, 33 1/3, 45 or 78 and hold that steady regardless as to cartridge drag. Whether that happens or not, I've never bothered testing. But any power supply that is set up to lock onto a given speed should be able to account for the effect of the stylus in the groove. Or so I would hope.
I think what you are saying is the velocity decreases as the stylus moves towards the inner groves of the record. If I'm not mistaken, output is specified by cartridge builder at a specific velocity; however, I don't know where that location may be on a 33-1/3 RPM LP record.
At some point of time on this post I saw response from Al (Almarg) but it dissapeared.
???
Interesting question. Since the variation of tangential velocity that occurs across the record is presumably the same when the record is played back as when it is cut, and since one period of a given signal frequency will extend over a physically longer distance if it is located towards the outer edge of the record than if it is located towards the inner edge, I believe that it all cancels out. Signal frequency will be reproduced identically at the two locations, because the time required to traverse the distance corresponding to one period will be the same in the two cases. Signal amplitude at any instant of time, which is proportional to groove velocity at that instant, will also be maintained for the two locations, because as tangential velocity decreases towards the inner part of the record, a given groove excursion will occur over a correspondingly shorter distance, and therefore in the same amount of time, and therefore with the same velocity.

Regards,
-- Al
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...
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.
Thanks for joining this discussion!
Would you consider a decreased performance on the inner groves normal in some of records especially fully "packed"?
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
Precisely! That's all we are saying
Well, I understood this all along, but the original poster was talking about correlating the cartridge's output voltage with the linear velocity of the record.
This is the correlation I said I didn't understand and I don't believe exists.
Everything else is clear to me. Maybe I misunderstood the post itself, wouldn't be the first time... ;-)
There's variation in sound quality between outer tracks and inner ones, due to the in formation being more densely packed together on the inner grooves, but not "output level" differences. Nobody has to turn the volume up as the inner tracks are played.
Mihaitaa, I understood the OP as asking what the effects of the differences in tangential ("linear") velocity at different points on the record would be on the output of the cartridge, given that the cartridge output is spec'd based on velocity. The answer, as you realize, is that there is no effect, because what is relevant is stylus deflection velocity, not tangential groove velocity, and for a given signal stylus deflection velocity does not change as a function of position on the record.
12-30-11: Marakanetz
Would you consider a decreased performance on the inner groves normal in some of records especially fully "packed"?
Inner groove distortion is certainly recognized as a potential issue, but my analog experience is not extensive enough to make a general comment on how pervasive a problem it might be. You might want to start a separate thread on that subject, that will probably catch the eye of some of the analog experts we have here.

Regards,
-- Al
The answer, as you realize, is that there is no effect, because what is relevant is stylus deflection velocity, not tangential groove velocity
Splendid, that was exactly my point as well.
Happy New Year, everybody.