Three speed automatic turntable replacement

[isw]

In article <k2sk18$icl$1@dont-email.me>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

Get out your microscope and look at the grooves and compare
to the playback waveforms. Complex waveforms are ok but
something like a square wave groove would be best.

Do you know of a disk that has a square wave cut on it?
I have a lot of test disks, but I don't remember one. I'll have to look.

Think about what the stylus would have to do. That's why you won't find
a test disk with a square-wave groove on it.

I assume you mean the cutting stylus.

No; I mean the playback stylus. In the groove, the rising and falling
edges of the waveform would be perfectly perpendicular to the motion of
the disk, and so there would be no possible way for the stylus to follow
it. And if they were not perpendicular, then it wouldn't be a square
wave.

It is possible to send a signal to a
cutting head that would, in principle, produce a square wave on the blank.
For example, you could cut a 200Hz square wave with 36dB of pre-emphasis at
12.8kHz, which is not unreasonable.

That still would not be truly a "square" wave, because the rise and fall
times would not be infinitely fast. You could do it by stopping the
rotation of the disk while the rising and falling edges were being cut,
for example, but no matter; the groove still could not be tracked by a
mechanical stylus.

Isaac

 

[William Sommerwerck]

It is possible to send a signal to a cutting head that would,
in principle, produce a square wave on the blank. For example,
you could cut a 200Hz square wave with 36dB of pre-emphasis
at 12.8kHz, which is not unreasonable.

That still would not be truly a "square" wave, because the rise and fall
times would not be infinitely fast. You could do it by stopping the
rotation of the disk while the rising and falling edges were being cut,
for example, but no matter; the groove still could not be tracked by a
mechanical stylus.

I don't know what the logical fallacy is involved here (other than
intellectual petulance), but it's impossible to have zero rise time on any
square wave. There is no infinite bandwidth in either the electrical or
mechanical realms.

You're saying that because we cannot generate perfect ("infinite-bandwidth")
square waves that there is no such thing as a square wave. I suggest you
look at the output of a square-wave generator on a 'scope, and tell us
whether a reasonable person would consider the waveform "square".

In practical terms, if the highest harmonic in a square wave is
significantly higher than the bandwidth of the DUT, then the DUT should
behave as if a "real" square wave were being applied.

 

[Dave Platt]

In article <k2v77i$5or$1@dont-email.me>,
William Sommerwerck <grizzledgeezer@comcast.net> wrote:

You're saying that because we cannot generate perfect ("infinite-bandwidth")
square waves that there is no such thing as a square wave. I suggest you
look at the output of a square-wave generator on a 'scope, and tell us
whether a reasonable person would consider the waveform "square".

In practical terms, if the highest harmonic in a square wave is
significantly higher than the bandwidth of the DUT, then the DUT should
behave as if a "real" square wave were being applied.

That's a very good working approximation.

And, again, in practical terms, once you make the "angle" in the
(near) square wave you have cut/pressed sharp enough, a mechanical
stylus just won't track it. The angle of the groove's motion
side-to-side, the V-shape of the groove, and the shape of the stylus
(typically an elliptical cone) will be such that most of the
mechanical force generated by the pressure of the stylus on the groove
will be upwards rather than sideways. The stylus tip will accelerate
upwards more rapidly than it will accelerate sideways, and it'll pop
out of the groove entirely.

The Sheffield Direct to Disc pressing of the 1812 Overture was
notorious for this... the "fire the cannon" passage would pop almost
any cartridge out of the groove. It wasn't quite at the point where
it was literally impossible to track... but it was a *really* severe
test of a cartridge's ability to follow a groove.

--
Dave Platt <dplatt@radagast.org> AE6EO
Friends of Jade Warrior home page: http://www.radagast.org/jade-warrior
I do _not_ wish to receive unsolicited commercial email, and I will
boycott any company which has the gall to send me such ads!

 

[William Sommerwerck]

And, again, in practical terms, once you make the "angle" in the
(near) square wave you have cut/pressed sharp enough, a mechanical
stylus just won't track it. The angle of the groove's motion
side-to-side, the V-shape of the groove, and the shape of the stylus
(typically an elliptical cone) will be such that most of the
mechanical force generated by the pressure of the stylus on the groove
will be upwards rather than sideways. The stylus tip will accelerate
upwards more rapidly than it will accelerate sideways, and it'll pop
out of the groove entirely.

The Sheffield Direct to Disc pressing of the 1812 Overture was
notorious for this... the "fire the cannon" passage would pop almost
any cartridge out of the groove. It wasn't quite at the point where
it was literally impossible to track... but it was a *really* severe
test of a cartridge's ability to follow a groove.

I think you're thinking of the Telarc disk. (I don't remember a Sheffield
"1812".) I don't know what the waveform "looked like", but I doubt it
approximated a square wave.

I do remember it knocking the pickup out of the groove. Not only did you
need a good-tracking pickup, but the arm-mass / stylus-compliance resonant
had to be very low -- so low that it would not normally be considered a
reasonable resonant point.

 

[isw]

In article <k2v77i$5or$1@dont-email.me>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

It is possible to send a signal to a cutting head that would,
in principle, produce a square wave on the blank. For example,
you could cut a 200Hz square wave with 36dB of pre-emphasis
at 12.8kHz, which is not unreasonable.

That still would not be truly a "square" wave, because the rise and fall
times would not be infinitely fast. You could do it by stopping the
rotation of the disk while the rising and falling edges were being cut,
for example, but no matter; the groove still could not be tracked by a
mechanical stylus.

I don't know what the logical fallacy is involved here (other than
intellectual petulance)

There's no fallacy. That is, in fact, one way to cut a groove in a vinyl
record that is a true, accurate square wave. Basically, you run the
lathe like a circular plotter. It's sort of the limit case for the
practice of half-speed mastering (which, done right, is a Very Good
Thing to do).

but it's impossible to have zero rise time on any
square wave. There is no infinite bandwidth in either the electrical or
mechanical realms.

Yup. Mathematics, now, can handle them just fine. But if you want to
claim that a band-limited square wave still should be called a "square
wave", I disagree. If the wave's fundamental is, say, 15 kHz, then what
gets carved into the vinyl will be pretty close to a sine wave. And I
think not a lot of folks would describe that as a "square wave".

You're saying that because we cannot generate perfect ("infinite-bandwidth")
square waves that there is no such thing as a square wave. I suggest you
look at the output of a square-wave generator on a 'scope, and tell us
whether a reasonable person would consider the waveform "square".

Having compensated my share of "high bandwidth" 'scope probes (and built
a few special-purpose probes along the way) I'd have to say, nope, those
things with the horns on the edges (Gibbs' phenomenon) don't look very
square to me.

In practical terms, if the highest harmonic in a square wave is
significantly higher than the bandwidth of the DUT, then the DUT should
behave as if a "real" square wave were being applied.

Generally, that is true. There are situations, however, where the
"functional bandwidth" of the DUT seems to predict one thing, while the
"trouble-causing bandwidth" causes something rather different to happen.
Ever see an audio amp that works (and measures) fine with the expected
(and designed in) rolloff above the audible range, but then the gain
pops back up again at a few megahertz (or more), which cause the thing
to go all unstable under certain conditions?

One reason for testing with the fastest-edged waves you can get is to
discover things like that.

Isaac

 

[isw]

In article <k2vt3q$k74$1@dont-email.me>,
"William Sommerwerck" <grizzledgeezer@comcast.net> wrote:

And, again, in practical terms, once you make the "angle" in the
(near) square wave you have cut/pressed sharp enough, a mechanical
stylus just won't track it. The angle of the groove's motion
side-to-side, the V-shape of the groove, and the shape of the stylus
(typically an elliptical cone) will be such that most of the
mechanical force generated by the pressure of the stylus on the groove
will be upwards rather than sideways. The stylus tip will accelerate
upwards more rapidly than it will accelerate sideways, and it'll pop
out of the groove entirely.

The Sheffield Direct to Disc pressing of the 1812 Overture was
notorious for this... the "fire the cannon" passage would pop almost
any cartridge out of the groove. It wasn't quite at the point where
it was literally impossible to track... but it was a *really* severe
test of a cartridge's ability to follow a groove.

I think you're thinking of the Telarc disk. (I don't remember a Sheffield
"1812".) I don't know what the waveform "looked like", but I doubt it
approximated a square wave.

I do remember it knocking the pickup out of the groove. Not only did you
need a good-tracking pickup, but the arm-mass / stylus-compliance resonant
had to be very low -- so low that it would not normally be considered a
reasonable resonant point.

I also think it was a Telarc disk. And the problem with it wasn't fast
rise time, it was that the groove made a huge excursion -- so great
that, on most pickups, the stylus just couldn't move that far. Usually,
the ones that could track it did so because (as you point out) the arm
mass was so low that the whole arm moved sideways, which cut the signal
down, but kept the needle in the groove.

Isaac

 

[William Sommerwerck]

but it's impossible to have zero rise time on any square wave.
There is no infinite bandwidth in either the electrical or
mechanical realms.

Yup. Mathematics, now, can handle them just fine. But if you want to
claim that a band-limited square wave still should be called a "square
wave", I disagree. If the wave's fundamental is, say, 15 kHz, then what
gets carved into the vinyl will be pretty close to a sine wave. And I
think not a lot of folks would describe that as a "square wave".

In the real world, ALL SQUARE WAVES ARE BAND-LIMITED. All complex waveforms
are band-limited. Period.

Do you argue merely for the sake of arguing? We are talking about a
particular situation -- the use of a "real" square wave on an LP to get an
idea of the pickup's mechanical characteristics.

If it's of any interest, I'm working on a fantasy screen play in which
Josiah Gibbs appears.

 

[William Sommerwerck]

I think you're thinking of the Telarc disk. (I don't remember a Sheffield
"1812".) I don't know what the waveform "looked like", but I doubt it
approximated a square wave.

I do remember it knocking the pickup out of the groove. Not only did you
need a good-tracking pickup, but the arm-mass / stylus-compliance
resonant
had to be very low -- so low that it would not normally be considered a
reasonable resonant point.

I also think it was a Telarc disk. And the problem with it wasn't fast
rise time, it was that the groove made a huge excursion -- so great
that, on most pickups, the stylus just couldn't move that far. Usually,
the ones that could track it did so because (as you point out) the arm
mass was so low that the whole arm moved sideways, which cut the
signal down, but kept the needle in the groove.

Stylus, stylus, stylus!

I worked in an audio store at the time, and the best arm for this disk was
the Dynavector, which had very high lateral mass, with relatively low
vertical mass.