# Estimation of data stored on a CD

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1. Nov 18, 2015

### Alettix

Hello! I would be very happy if you could lend me a helping hand with the following problem. :)

1. The problem statement, all variables and given/known data

We know that a CD is read with a laser with a wavelenght of λ = 1 μm.
a) Estimate the data stored on the CD.
b) We know that the CD contains 80 minutes of sound. How fast is it spinning?

2. The attempt at a solution
Firstly, I tried to look up how a CD works and what part of its function is affected by the wavelenght of the light used to read it. The explations said that along the surface of the CD, there is a long spiral path. On the path, there are small bumps with a height of λ/4. When a transition between a bump and "normal land" or vice versa occures, there will be a time when half of the laser beam is on the bump and half on normal land. At this point destructive interference takes place and this corresponds to a 0. The other times correpsonds to 1.

Now, I don't really know how to apply this to my problem. It is generally known that the shorter the wavelenght, the the more data can be stored. However, according to the information above, the only differece this will make is that the bumps will decrease in height. I really don't see how this will allow more bumps (and data) to be stored. Other sources on the other hand said that the wavelenght equaled the distance d between two parts of the spiral (see picture). I don't see the full logic behind this, but if this is the case, I can see that the spiral can be tighter when using light of shorter wavelenght, and consequently the disc can store data.

Which ever the case, I do not know how to estimate the stored data. By using the second explanation, assuming that the path has zero width and estimating the radius of the dics, I believe one could approximate the total lenght of the spiral path. However, unless one knows the lengt of a bit, this does not tell us much about the data. Using the first explanation, neighter the path lenght or the bitlenght can be found, only the height of the bumps, whose connection to the stored data I cannot see. Can anybody tell me which explanation is right and how the stored data should be calculated?

But, let's say that we can calculate the stored data and know the lenght of the path. How is then the speed of the dics calculated? To get a constant sound quality, the same amount of bits should be read each second, shouldn't it? But if the lenght of a bit is constant and we have a constant angular velocity, then less bits will be read when the readinghead is closer to the middle of the dics than when it is on the edge of it. This means that as the readinghead approaches the middle of the dics, the disc should then start to spin faster. But if this is correct, what is b) asking for? The average angular velocity or the tangential speed that should be kept constant?

PS: One of the information sources:

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2. Nov 18, 2015

### CWatters

It's possible you are trying to come up with a more accurate estimate than they intend. In general you can't see/resolve objects that are smaller than the wavelength of light. So that helps you estimate the minimum area of "one bit" of data. Try that.

Last edited: Nov 18, 2015
3. Nov 18, 2015

### CWatters

As for question b).

The data doesn't go all the way to the middle so the variation in bit length might not be an order of magnitude. Somewhere there is an "average" bit length and an average circumference.

4. Nov 18, 2015

### J Hann

Remember that CD is read at the same speed at which recording took place
so that would seem to say that the "bumps" towards the outer edge would be
farther apart than the ones towards the center.
That would seem to imply that the inner "bumps could quite close together.
Think of what would happen if you were listening to music on the CD
while it rotated at a constant speed.

5. Nov 18, 2015

### Alettix

Okay, so that means that the area of a singel bit should be (1*10^-6)^2 = 10^-12 m^2.
If we then approximate the outer radius of the disc to be 6 cm-ers, and the inner to be 1 cm, the total numer of bits is:
$\frac{\pi(0,06^2-0,01^2}{10^{-12}} ≈ 1,1 *10^{10} bit ≈ 1,4 Gb$

6. Nov 18, 2015

### Staff: Mentor

Actual CDs are read at constant linear speed, not constant angular speed. The bumps have the same "spacing" all over the CD.

7. Nov 18, 2015

### Alettix

If I have understood it correctly, these two answeres collide a little, don't they? Of have I missinterpreted something?

Nevertheless, Using the answere from a), we obtain that ca 2,3 *106 bites should be read each second.
However, I am unsure about how to use an "average radius". I could have (6+1)/2 = 3,5 cm-s, but not equal time will be spent at each radius. Does this result in somekind of integral?

8. Nov 18, 2015

### Alettix

So, is it the linear speed I should be looking for in b)?

9. Nov 18, 2015

### Staff: Mentor

The question asks "How fast is it spinning?", which hints at an answer in terms of an angular speed. Maybe the person who wrote the question didn't know that CDs don't always rotate at the same speed.

10. Nov 18, 2015

### Alettix

Is it possible that they are looking for a function ω of R?