Capacity of a cd

[scribblec]

basically im doing a coursework on cds and i come to this equation
Data capacity in Mb for an audio-CD
74 min
= 333,000 sectors * 2352 bytes / sector
= 783216000 bytes
= 746.9 Mb

and i dont get how from 783216000 bytes they get 746.9 Mb

i kno theres 8 bits in a byte but when u times 783216000 by 8 u dont get 746.9

[lynx]

Assume 1024 bytes = 1KB, 1024KB = 1MB

783216000 / (1024x1024) = 746.932


Shouldn't this be in hardware world?

[JPaul]

Indeed.

They only call it KB as a matter of convenience.

Are CDs considered hardware.

[scribblec]

this was the only place i could think of to get a reply, but thanks for the answer was exactly what i was looking for, also could you give me any ideas about whats physicsy about cds so far ive got info on pits and grooves of a cd and how much storage capacity it can have but im a bit too out of ideas :(

[JPaul]

This might help (http://electronics.howstuffworks.com/cd.htm)

[ilw]

have you got all the cd trivia stuff like:
the pits were close to the smallest man made thing when first done
the length of the track (spiral) is xxx metres long
cds are partially made out of polycarbonate which is same material as bullet proof glass

Other than that you could talk about manufacture, ie master copy and stamping the others off it. You could talk about the structure of them ie tiny metal layer between plastic sheets. Or you could talk about how cd-r / rw work. You could talk about how the reader works with lasers tracking one reads it by destructive interference etc.
Or you could jsut copy a website like:
http://electronics.howstuffworks.com/cd.htm

edit: arsebiscuits jpaul beat me to the link

[JPaul]

:lol:

[scribblec]

yea ive pretty much got all the trivia stuff. i think im going to talk about angular velocity.

speed of cd surface = 2*pi*r/t
t is time and r radius

but i dont know how many revolutions a cd makes per second, it makes 500 in the center per minute and 200 on the outsides per minute, i want to also talk about

at these high speeds, friction with the surrounding air particles is high. This results in the generation of thermal energy which consequently heats up the cd

and how the material of the cd has to have a higher melting point then the amount the cd heats up but i cant find out how to find out how much it heats up



EDIT: the destructive interference can you elaborate on that a bit more

[JPaul]

"it makes 500 in the center per minute and 200 on the outsides per minute, i want to also talk about "

Woah there tiger, every bit of the CD makes the same number of revolutions per minute. Otherwise it would break up. The outside travels faster than the inside, because it must cover a greater distance in the same amount of time.

[scribblec]

Constant Linear Velocity (CLV) is the principle by which data is read from a CD-ROM. This principal states that the read head must interact with the data track at a constant rate, whether it is accessing data from the inner or outermost portions of the disc. This is affected by varying the rotation speed of the disc, from 500 rpm at the center, to 200 rpm at the outside. In a music CD, data is read sequentially, so rotation speed is not an issue. The CD-ROM, on the other hand, must read in random patterns, which necessitates constantly shifting rotation speeds. Pauses in the read function are audible, and some of the faster drives can be quite



That is where i got that information from, if its the same speed Jpaul then what would that speed be?

[JPaul]

Velocity CAV v CLV


CD-ROM drives that are over 16x will generally have max written next to it. When max is written next to the speed or on the drive, this indicates that the CD-ROM uses CAV, or Constant Angular Velocity, to access the data on the CD. Originally, CD-ROM drives utilized CLV, or Constant Linear Velocity, where the information was accessed slower / faster by adjusting the speed of the motor which allows the data to be transferred at a steady flow.

With CAV, the CD rotates at a constant speed without adjusting when at a different location on the CD, which means a 32x speed CD-ROM, for example, will be able to access the data 32x on the outside layer, however, when approaching the middle of the CD-ROM drive, the access speed can decrease close to 20x.

http://www.computerhope.com/issues/ch000219.htm

[JPaul]

Constant Linear Velocity (CLV) is the principle by which data is read from a CD-ROM. This principal states that the read head must interact with the data track at a constant rate, whether it is accessing data from the inner or outermost portions of the disc. This is affected by varying the rotation speed of the disc, from 500 rpm at the center, to 200 rpm at the outside. In a music CD, data is read sequentially, so rotation speed is not an issue. The CD-ROM, on the other hand, must read in random patterns, which necessitates constantly shifting rotation speeds. Pauses in the read function are audible, and some of the faster drives can be quite



That is where i got that information from, if its the same speed Jpaul then what would that speed be?

See my last, I think CAV has superceded CLV in newer drives.

I think it depends on the max speed of the CD Rom and the disk being used. I'll try to find out what speeds they are talking about.

[JPaul]

Feck, there P-CAV (partial) and Z-CLV (zoned) as well.

http://www.cdspeed2000.com/go.php3?link=faq_general.html

[scribblec]

looks very confusing

[JPaul]

It certainly appears to be.

I would just rip stuff from the "howthingswork" site. There's actually quite a lot there.

[thewizeard]

...but there is a special quantum algorithm, which saves you doing all of that...


edit:- http://www.filesharingtalk.com/vb3/t112206.html

[ilw]

re destructive interference

the data is stored in pits and non-pits (land?)
the laser is shone onto the track and is reflected back either from hte pit or land. Laser light is coherent ie all in phase. the distance between the laser and track is set up so that reflections from a non-pit are in phase with the incident light. The depth of a pit is set up to be 1/4 wavelength of the laser light so for the light to go down into the pit and come out again it will travel an extra 1/2 a wavelength and be 180degrees out of phase with the incident light and this causes light reflected from a pit to destructively interfere.
Therefore to the sensor bright = land and dark = pit.
Pretty bad explanation, but it should give you info you can google

[scribblec]

thanks alot guys ended up getting an A on the coursework and probably due to the extra topics brought up for me to include, the teacher said it was one of the most interesting he had read this year :D

[thewizeard]

haha .. nice one scribblec