What Did I Do Wrong in CD Bit-density Calculation?

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Homework Help Overview

The discussion revolves around calculating the bit-density of a CD based on AFM images and manual counting of bits within a specified area. The original poster presents their calculations and expresses confusion regarding the resulting storage capacity compared to expected values for CDs.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the total number of bits on the CD using a ratio derived from their manual counting. Some participants question the dimensions used in the calculations, particularly the radius of the CD. Others suggest that the interpretation of bits versus bytes may be affecting the results.

Discussion Status

Participants are actively engaging with the calculations and assumptions presented. There is acknowledgment of a potential error in the radius measurement, and some guidance has been offered regarding the conversion between bits and bytes. Multiple interpretations of the bit representation in the AFM images are being explored, but no consensus has been reached on the final calculations.

Contextual Notes

There are indications of confusion regarding the units used (micrometers versus millimeters) and the definitions of bits and bytes, which are critical to the calculations. The original poster's calculations are based on specific measurements that may not align with standard CD specifications.

SarahBoberra
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We took AFM images of a CD in lab. Manual counting provided that there were 224 bits in the 900µm^2 image I have. Based on this I want to determine the bit-density of the whole CD.

The total active area for the disc is the total area minus the inner inactive are of the disc. The total area was calculated using a radius of 58.5µm and was found to be 10,746 µm^2. The inner inactive area of the disc was found to be 1,661µm^2. Therefore, the total active area of the disk was 9,085µm^2.

Shouldn't the next step be to take (224 bits/900µm^2)*9,085µm^2 to get the total number of bits? For this I get 2,261.156 bits, but it can't be right because it would mean that the whole CD had only 0.00215640640258789MB of storage! I know that CDs have ~650MB of storage. What did I do wrong? Please?
 
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SarahBoberra said:
We took AFM images of a CD in lab. Manual counting provided that there were 224 bits in the 900µm^2 image I have. Based on this I want to determine the bit-density of the whole CD.

The total active area for the disc is the total area minus the inner inactive are of the disc. The total area was calculated using a radius of 58.5µm and was found to be 10,746 µm^2. The inner inactive area of the disc was found to be 1,661µm^2. Therefore, the total active area of the disk was 9,085µm^2.

Shouldn't the next step be to take (224 bits/900µm^2)*9,085µm^2 to get the total number of bits? For this I get 2,261.156 bits, but it can't be right because it would mean that the whole CD had only 0.00215640640258789MB of storage! I know that CDs have ~650MB of storage. What did I do wrong? Please?

There is nothing wrong with the calculations you did given THOSE numbers, but -- 58.5 micrometres? That's kind of small for a compact disc, don't you think? If they were really that small, you wouldn't be able to see them! :-p

EDIT: This has been posted in the wrong section. Should be intro physics.
 
SarahBoberra said:
The total area was calculated using a radius of 58.5µm and was found to be 10,746 µm^2. The inner inactive area of the disc was found to be 1,661µm^2. Therefore, the total active area of the disk was 9,085µm^2.

Are you sure that radius isn't 58.5 mm = 58.5 103 µm?
 
Yes, thank you, that was a mistake.

Still though, when I run the numbers with that correction I still only get 2.15641MB. I thought CDs were supposed to have 650MB? So, I think I'm way-off still.
 
SarahBoberra said:
Yes, thank you, that was a mistake.

Still though, when I run the numbers with that correction I still only get 2.15641MB. I thought CDs were supposed to have 650MB? So, I think I'm way-off still.

That factor of 1000 appears in a square, so it actually results in an extra factor of 106.
 
Another important thing to note is that in the symbol MB, the 'B' stands for 'byte' (NOT bit). So there is an extra step you'll need to do to get your answer in MB.
 
If I take it so each dot and dash is one bit, and with the above corrections and advice, I still only get 269.55MB. That's a lot better, but I'm wondering if I have the right bit idea.

Do you know if each "dot" and "dash" hole in my AFM pic counts for one bit each? I thought so, but read somewhere that the dash is worth two bits. Any thoughts on that?

If so, I could estimate that I'd be up to roughly 400MB, but I would till be significantly below the predicted 650MB. Do you know if some CDs only have 400MB or so?

By the way, thank you for the help!
 
Each hole is one bit. You still haven't properly taken into account the extra factor of 1000 that I explained in post #5. You're missing a factor of 10 now.
 
I think I did factor that in.
 
  • #10
I am at the library and it's closing. Thanks for your help, I'll check back tomorrow.
 
  • #11
fzero is right. You are too low by a factor of 10. You also haven't taken into account what I said in my previous post.
 

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