What's the deal with coloring CD's and/or tray mechanism


I just stumbled across an old thread discussing this process, and the sonic advantages. It's intriguing, and I wonder what other members think. I am primarily into vinyl, and don't listen to cd's that often, but if I can improve the audio quality, then I am interested, to say the least.

I am concerned though, about painting the tray of my McIntosh, for fear of damage, and also de-valuing it's re-sale value


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Ah, the proverbial can of worms has been opened once again.
First worm out of the can.

It used to be conventional wisdom that treating the edge of the CD with a green marker would lessen the scattering of the laser's reflection while reading the disc, or something like that. That also led to whole disc covers and some going so far as to treat the insides of the disc assembly with various applications and so on. 

It was a terrible time for audiophiles and left enough of a latent impression which led to the hastening of computer audio, with all it's dos don'ts, and conventional wisdom.

All the best,
Nonoise

The intriguing thing about the whole Green Pen and painting the outer edge of the CD or the CD tray with green ink, the complementary color of red, thing is that the CD laser is essentially invisible light in the near infrared part of the spectrum. Thus theoretically at least green or any other color should have no affect on the sound since infrared light, being invisible, has no complementary color that could absorb it (before it makes its way into the photodetector).

😄
far better tweak is to draw green rings on your power cables
Hey, mods! Randy’s stalking me again!
If edge supposed to reflect infrared then it should be gold marker, but if it supposed to kill reflections it should be black.  Green doesn't make sense.

"If edge supposed to reflect infrared then it should be gold marker, but if it supposed to kill reflections it should be black. Green doesn’t make sense."

green makes sense if you if you want to absorb red light. Reflecting any light doesn’t make sense since reflections will make their way to the photodetector. The scattered light completely fills the transport compartment.

Don't bother with it.  It's an old tweak that died out years ago.  My apologies in advance to anyone still doing it.
geoffkait, black would still absorb red light better than green.  Green makes sense if you want to absorb red light ONLY.
Green indicates other visible wavelengths of light from longer red to shorter blue are absorbed and only green reflected. That’s why you see green. Green is not the complimentary color of red. That would be blue + red which is magenta. So any conclusions based on this incorrect info cannot be correct. It would help to get the science of light correct first before attempting to apply it. Also how visible light is absorbed or reflected indicates nothing about infrared which is longer wavelength than any visible light and absorbed or reflected independently.

You would think a former NASA scientist would know these things. Sheesh. Wikipedia is still your friend.

Also to absorb red light only the color you would see is cyan which is blue + green reflected and red absorbed. What happens with non visible light at longer or shorter wavelengths is again independent of what happens at visible wavelengths blue through green through red.

Even infrared is absorbed best by the black paint.
Kijanki I don't know.     Only that black if red through blue ie all visible wavelengths are absorbed.   In nature water absorbs  IR best and appears black in an image captured with IR frequency light. 

Black absorbs all wavelengths. Special infrared sinking paints, used by military, are all black matte. Cheap version, often used, is Krylon 1602. You can easily test it with infrared led (remote control etc) and cellphone (since they show infrared as visible light).  Thin layer of water would be completely useless.
This was a big deal in the early years of CDS to try to get them to sound better. I colored the outside edge and inner edge with different colors. They actually sold audiophile coloring pens to do this at a premium cost of course. Never really heard any difference. Don't waste your time
Alan

kijanki
Black absorbs all wavelengths. Special infrared sinking paints, used by military, are all black matte. Cheap version, often used, is Krylon 1602. You can easily test it with infrared led (remote control etc) and cellphone (since they show infrared as visible light). Thin layer of water would be completely useless.

Actually green (turquoise) CD trays work better than black. Most stock CD trays are matte black so one can probably conclude black is not really the answer. Besides shining a red laser on the stock black tray illustrates that doesn’t absorb red very well. Probably not other colors either. That’s why complementary colors are used to absorb the various visible colors? Refer to color wheel. Furthermore, green or turquoise or cyan works better than black ink - in terms of SQ - around the outer edge of the CD so, again, one can conclude black is not the answer.

The CD laser also appears red because the primary wavelength at 760 no is just past the red into the near infrared, thus a portion of the laser beam is red. And the photodetector accepts red and near infrared light.

As for military aircraft that exhibit extremely low radar profile I’m reasonably sure you will find that the matte black is not just any old black paint but some highly specialized carbon or graphene concoction or whatever. That’s why it’s secret. We don’t want every Tom, Dick and Harry to know about it.

If I recall correctly there is a small CD company that manufacturers CD with turquoise colored labels. Which makes sense since the color of the label influences the sound, no?

in any case absorbing red light is only a small part of the solution since most of the laser spectrum is invisible, no?

ahendler
This was a big deal in the early years of CDS to try to get them to sound better. I colored the outside edge and inner edge with different colors. They actually sold audiophile coloring pens to do this at a premium cost of course. Never really heard any difference. Don’t waste your time
Alan

unfortunately only certain colors work. There was a reason they sold a lot of those expensive green pens. Maybe now that your system is better than it was 30 years ago you will hear the difference. Green on outer edge, black on inner lip. May the force be with you.
Geoffkait, pure matte black has 100% absorption of all colors. Other colors are used only to reflect only one color. Targets for infrared, for counting number of pulses in rotational speed measurements, are usually gold/matte black (including one I designed: http://himmelstein.com/images/product-datasheets/0807545dc201B8701.pdf )

As I mentioned before you can see infrared with your phone camera and test it for yourself.

kijanki
"Geoffkait, pure matte black has 100% absorption of all colors."

that cannot be true since adding gree to the CD tray improves the sound. Almost all CD trays are black or black matte. As I said you can see for yourself with a red laser poster how much is reflected off the black tray.nalso, of the black matte tray was actually 100% effective the green pen wouldn’t make any difference. Nor would painting the tray cyan. And we know that’s not true.

"Other colors are used only to reflect only one color."

I don’t think we’re on the same page. You want to absorb color not reflect it. For a given color there is a complementary absorbing color.

Furthermore, dare I even say it? Black around the outer edge of the CD actually hurts the sound. 😀

"Targets for infrared, for counting number of pulses in rotational speed measurements, are usually gold/matte black (including one I designed: http://himmelstein.com/images/product-datasheets/0807545dc201B8701.pdf )

As I mentioned before you can see infrared with your phone camera and test it for yourself."

the reason you see the infrared is because it’s near infrared next to the visible part of the spectrum thus part of the signal is in the red zone. Same for a CD laser. Hel-loo!

My head hurts !!

What have I gotten myself into??

As I said you can see for yourself with a red laser poster how much is reflected off the black tray.nalso, of the black matte tray was actually 100% effective the green pen wouldn’t make any difference.

It doesn't, according to many.  You are trying to make up theory based on some silly tweak, that according to many doesn't work at all.

the reason you see the infrared is because it’s near infrared next to the visible part of the spectrum thus part of the signal is in the red zone.
 No, spectrum of the infrared emitting diode is very narrow and it does not project white light.  You still cannot see it with naked eye.  The portion that you can see if you look directly into transmitter will be dark red.  The reason camera can see it is because camera sensors are sensitive to infrared up to around 1100nm.  You see white light because red green and blue sensor filters are not very strong and all of them pass some infrared.

"Other colors are used only to reflect only one color."

I don’t think we’re on the same page. You want to absorb color not reflect it. For a given color there is a complementary absorbing color. 
You still don't get it.  Perhaps this drawing will help you:
http://archive.cnx.org/resources/b3c8a8818e60b70ff638c267249826b1e5c0747f/Figure_27_03_02.jpg
The bandwidth of the laser is not narrow. It’s actually fairly wide, certainly not monochromatic. As is the photodetector bandwidth. That’s why red scattered laser light gets into the detector. Neither one is monochromatic. And that’s precisely why the CD laser appears RED even though it’s wavelength is 760 nm. Which is invisible. Hel-loo! I must be smarter than those other people you’re listening to. You’re following the wrong sheep.

As I explained earlier - spilled light would appear as mostly red but it shows as white (no red at all), because all camera sensors can see infrared and filters are not able to filter it completely. If you cannot understand it I cannot help it. Also LED transmitters have narrow band, like this one: http://www.vishay.com/docs/81009/tsal6100.pdf

As you can see there is no light emitted below 840nm (Fig. 8) while visible light ends at 750nm. If you compare two cameras one with the film and another with digital sensor set to the same sensitivity you’ll find that infrared won’t be visible on the picture made with film camera but will show as white light on the one with digital sensor. It is not because of "spilled" light but because digital sensor is sensitive to infrared and film isn’t. I’m not sure if your understanding of this is as poor as one of the class D amps, but it is getting close. No sense to discuss it any further.

kijanki
"As I explained earlier - spilled light would appear as mostly red but it shows as white (no red at all), because all camera sensors can see infrared and filters are not able to filter it completely. If you cannot understand it I cannot help it. Also LED transmitters have narrow band, like this one: http://www.vishay.com/docs/81009/tsal6100.pdf

As you can see there is no light emitted below 840nm (Fig. 8) while visible light ends at 750nm. If you compare two cameras one with the film and another with digital sensor set to the same sensitivity you’ll find that infrared won’t be visible on the picture made with film camera but will show as white light on the one with digital sensor. It is not because of "spilled" light but because digital sensor is sensitive to infrared and film isn’t. I’m not sure if your understanding of this is as poor as one of the class D amps, but it is getting close. No sense to discuss it any further."

We are not on the same page. Not even close. I define infrared light as above 700 nm. Therefore if there is red in the CD laser light it MUST be because the 760 nm nominal wavelength is NOT monochromatic. It has SOME RED light in it. And that RED LIGHT has wavelenths BELOW 700 nm since that is where visible light ENDS. This conversation can serve no purpose anymore.




Exactly - don't waste your time - or money.

People who do not understand electronics or physics can be convinced to believe it works, just as snake oil was sold to people decades ago.
Post removed 
No matter what the wavelength of the laser used in cd players, wouldn’t any scattered light picked up by the photodiodes be miniscule compared to the direct reflection that the photodiodes read and wouldn’t any halfway well designed cd player be able to ignore the miniscule amounts of scattered light and only send on to the DAC 1s and 0s read from directly reflected light?

tomcy6
"No matter what the wavelength of the laser used in cd players, wouldn’t any scattered light picked up by the photodiodes be miniscule compared to the direct reflection that the photodiodes read and wouldn’t any halfway well designed cd player be able to ignore the miniscule amounts of scattered light and only send on to the DAC 1s and 0s read from directly reflected light?"

The problem is that any scattered light that gets into the detector - scattered light more than 75% of full reflected value is accepted by the detector - is accepted as REAL signal. If it’s less than 75%, the scattered light isn’t accepted and can be ignored. Recall the photodetector is essentially seeing OFF and ON signals, a series of "no signal" and "signal." These ON and OFF signals are converted to meaningful sequences of 1s and 0s later. Thus the data stream is corrupted, even if slightly, by scattered light during the optical reading phase - every so often an OFF is interpreted as ON. The whole issue of course is, how often does the error occur and is it audible. 

There are some players that address this issue by painting the surfaces around the laser assembly turquoise and, as I mentioned earlier, at least one CD manufacturer that paints the CDs labels turquoise. By the way painting the CD label black or painting the outer edge of the CD black ruins the sound. Since most CD player manufacturers are either unaware of this problem with scattered light and sound quality or else don’t address it if they are aware, virtually all players, even high end players, suffer this problem. Furthermore, as I mentioned earlier, even if turquoise or green or cyan is used to counter the RED portion of the laser light the INVISIBLE scattered laser light can still get through, the invisible light that extends from 700 nm to around 850 nm or higher. Since the wavelength of the CD laser is centered at 760 nm most of the scattered light is INVISIBLE.

This is just made up gibberish by someone who doesn't understand the mechanism involved.  

Provide a citation to a study showing the 75%reflectivity.
"Provide a citation to a study showing the 75%reflectivity."

Google it. Google is your friend. Would I kid you?