NEW GROOVE

New Groove tries to connect a sound medium from the past (the vinyl record) to a contemporary technology (lasercutting). A lasercutter of 2225 ppi (pulses per inch), is not capable of producing soundgrooves for a vinyl record, even though the technical specifications might give the impression that it can.
A vinyl groove is approximately 0.06 mm wide and the generated motion to vibrate the needle is slight.

Microscopic photo of groove

The radius of a laser is wider, depending on the instructions of how deep the cutter has to burn. If we assume that the laser can cut at a width of 0.06 mm (which it can't) then we get the following restriction:

Width of groove in inches:
0.06 mm = 0.00236 inch

Options for lasercutter for positioning a point:
2225 ppi = 0.00236 = 5.251

A vinyl record doesn't know bitrates, so in the case of lasercutting there is a somewhat crude comparison to be made. The solution of the equation, adding up to 5.25 would have to be rounded off to 5 to choose a bitrate. To make a comparison, most mp3's have a bitrate of 192 kbps, or better said 191.995 bits more to choose from.

And seeing as the laser can't produce such a miniscule radius, we can assume that this eventually comes down to 1 bit.

Next to bitrate, there is a sample rate. The number of samples per second for the most mp3's is 44.100. On a constant rotation speed, the vinyl record turns for instance, 33 1/3 rpm, in seconds that is:
(For convenience, let's say 33 instead of 33 1/3)

rps:
33 rpm/60 sec. = 0.55
degrees a second:
0.55 * 360o = 198o

In other words, 198 degrees for 1 second of music, this also means that a second closer to the center of the record is more compressed than towards the outer ring of the record (sometimes the order of the songs was determined by this, because high tones would sounds less good round the end of a record). To translate this back to a sample rate, this would mean that the rate declines as the spiral declines.

Let's say we take a 12' record (outer ring diameter: 300 mm) .

Outline:
3.14 * 300 = 942 mm
lengte 1 seconde:
942 / 360 * 198 = 518 mm

If we want to write all the 44.100 samples from an mp3 on a distance of 518 mm, the distance between each sample would be:
518 / 44.100 = 0.01175 mm

This is of course far too narrow for a lasercutter. It becomes even more impossible when we get closer to the center of the record. Take, for instance, a inner ging with a 120 mm diameter.

Outline:
3.14 * 120 = 377 mm
length 1 second:
377 / 360 * 198 = 207 mm
distance between samples:
207 / 44.100 = 0.00469 mm

Thus, to keep a sample rate of an mp3 in the inner-ring, the distance between samples must be approximately 0.00469 mm.

My goal is to match as closely as possible, the sound of 8bit music. To accomplish this, I would need a groove that has more friction space (bitrate). To increase the sample-rate, I can move from 33rpm to 45 rpm, creating an increase in length of a second to about 70o. The down side is that the length of a record would also decrease drastically. On average this would be 30 minutes with a spiral length of 400 meter, but with this test it was a little over a minute. And due to technical issues, only half of it got recorded, which in theory leaves about 30 seconds.

Because of this technicality, this temporary version has a sinus wave based on a picture in stead of 8bit music. For a better contrast, the sample-rate is quite high in relation to the bitrate. Because of this, the curves are quite steep, causing the needle to jump off and skip sections.

In other words, there's still a long way to go before reaching the goal.

COMMISSIONED BY
Mike Rijnierse

YEAR
2012

NEW GROOVE

New Groove tries to connect a sound medium from the past (the vinyl record) to a contemporary technology (lasercutting). A lasercutter of 2225 ppi (pulses per inch), is not capable of producing soundgrooves for a vinyl record, even though the technical specifications might give the impression that it can.
A vinyl groove is approximately 0.06 mm wide and the generated motion to vibrate the needle is slight.

Microscopic photo of groove

The radius of a laser is wider, depending on the instructions of how deep the cutter has to burn. If we assume that the laser can cut at a width of 0.06 mm (which it can't) then we get the following restriction:

Width of groove in inches:
0.06 mm = 0.00236 inch

Options for lasercutter for positioning a point:
2225 ppi = 0.00236 = 5.251

A vinyl record doesn't know bitrates, so in the case of lasercutting there is a somewhat crude comparison to be made. The solution of the equation, adding up to 5.25 would have to be rounded off to 5 to choose a bitrate. To make a comparison, most mp3's have a bitrate of 192 kbps, or better said 191.995 bits more to choose from.

And seeing as the laser can't produce such a miniscule radius, we can assume that this eventually comes down to 1 bit.

Next to bitrate, there is a sample rate. The number of samples per second for the most mp3's is 44.100. On a constant rotation speed, the vinyl record turns for instance, 33 1/3 rpm, in seconds that is:
(For convenience, let's say 33 instead of 33 1/3)

rps:
33 rpm/60 sec. = 0.55
degrees a second:
0.55 * 360o = 198o

In other words, 198 degrees for 1 second of music, this also means that a second closer to the center of the record is more compressed than towards the outer ring of the record (sometimes the order of the songs was determined by this, because high tones would sounds less good round the end of a record). To translate this back to a sample rate, this would mean that the rate declines as the spiral declines.

Let's say we take a 12' record (outer ring diameter: 300 mm) .

Outline:
3.14 * 300 = 942 mm
lengte 1 seconde:
942 / 360 * 198 = 518 mm

If we want to write all the 44.100 samples from an mp3 on a distance of 518 mm, the distance between each sample would be:
518 / 44.100 = 0.01175 mm

This is of course far too narrow for a lasercutter. It becomes even more impossible when we get closer to the center of the record. Take, for instance, a inner ging with a 120 mm diameter.

Outline:
3.14 * 120 = 377 mm
length 1 second:
377 / 360 * 198 = 207 mm
distance between samples:
207 / 44.100 = 0.00469 mm

Thus, to keep a sample rate of an mp3 in the inner-ring, the distance between samples must be approximately 0.00469 mm.

My goal is to match as closely as possible, the sound of 8bit music. To accomplish this, I would need a groove that has more friction space (bitrate). To increase the sample-rate, I can move from 33rpm to 45 rpm, creating an increase in length of a second to about 70o. The down side is that the length of a record would also decrease drastically. On average this would be 30 minutes with a spiral length of 400 meter, but with this test it was a little over a minute. And due to technical issues, only half of it got recorded, which in theory leaves about 30 seconds.

Because of this technicality, this temporary version has a sinus wave based on a picture in stead of 8bit music. For a better contrast, the sample-rate is quite high in relation to the bitrate. Because of this, the curves are quite steep, causing the needle to jump off and skip sections.

In other words, there's still a long way to go before reaching the goal.

COMMISSIONED BY
Mike Rijnierse

YEAR
2012