tech question that google can't answer...

[Linksys120n]

We all know the standards of CD and DVD ... A CD can hold about 700 Megabytes or 800 if you really push it's limits... a DVD 4.7GB per layer... But what's always bothered me is that 12 inch LaserDiscs were "analog".... But what I want to know is the hypothetical what if scenario of a 12 inch digital CD or DVD... Ignoring the fact that read/write speeds at the outer edges would make this pretty much impossible... How much raw data could you cram onto something like that?

 

Could we fit like 20+ Gigs on a single 12 inch CD? or a terabyte on a 12 inch DVD? What about bluray? Could you have 10 terabyes of data on a 12 inch bluray?

 

As noted before, The speed at witch the laser would have to read the outer grooves at would make this impossible... but what if it wasn't?

[alex_dose_stuff]

this is a really good idea and i would like to see it myself

[Droidbot]

im no mathematician 

but shit, this is easy

 

DVDs are 12cm, lets take a double sided DVD for example, which is 9.4GB. 

12" is 30.4cm. 

0.783333 GB of data per CM

0.78333333333 (which was the original answer) x 30.48cm (12in) = 23.875GB

 

blurays are also 12cm, let's take a doublesided 100GB BD-R as an example

100 divided by 12 is 8.33GB/cm

8.33 x 30.48 = 253.89GB

 

 

[ARikozuM]

The outside of a disk would be the fastest as there is far more data there than on the inside, but the slowdowns commonly associated with HDD's has to do [where] realigning the head with various tracks unless they are usable from outside-in or inside-out.

[Linksys120n]

1 minute ago, Droidbot said:

im no mathematician 

but shit, this is easy

 

DVDs are 12cm, lets take a double sided DVD for example, which is 9.4GB. 

12" is 30.4cm. 

0.783333 GB of data per CM

0.78333333333 (which was the original answer) x 30.48cm (12in) = 23.875GB

 

blurays are also 12cm, let's take a doublesided 100GB BD-R as an example

100 divided by 12 is 8.33GB/cm

8.33 x 30.48 = 253.89GB

 

 

Nooo you forget that there is more data on the last outer cm then on the first inner cm...

[Droidbot]

3 minutes ago, Linksys120n said:

Nooo you forget that there is more data on the last outer cm then on the first inner cm...

Aaaaaa I hate physical media

[ARikozuM]

4 minutes ago, Linksys120n said:

Nooo you forget that there is more data on the last outer cm then on the first inner cm...

He's accounting for the entire disk on average. If you want a more exact answer, I'd call MIT. 

[Linksys120n]

What we need to solve this with is the density of data per cm of circumference...

[cj09beira]

4 minutes ago, Linksys120n said:

What we need to solve this with is the density of data per cm of circumference...

using the radius and the inner radius (where there is no data ) you can figure it out using the area

[ARikozuM]

6 minutes ago, Linksys120n said:

What we need to solve this with is the density of data per cm of circumference...

That's essentially what he did... I don't know if he eliminated the spindle's "empty space" but he drew the density to GB/cm².

 

DVD = .0208 GB/cm²

.0208 x 2903 = 60.33GB

 

Edit: It seems @Droidbot used diameter rather than area. By no means is my answer more correct since I'm not going to subtract the empty space form my calculation.

[Linksys120n]

Just now, cj09beira said:

using the radius and the inner radius (where there is no data ) you can figure it out using the area

the inner radius of no data would be the same as a regular DVD

[cj09beira]

Just now, Linksys120n said:

the inner radius of no data would be the same as a regular DVD

yes

[Enderman]

15 minutes ago, Linksys120n said:

Nooo you forget that there is more data on the last outer cm then on the first inner cm...

That's because the circumference on the outside is larger than the circumference on the inside...

Taking the area accounts for that.

This is just grade 8 math.

[cj09beira]

the inner diameter is 43mm the outer is 120mm

area=39.288cm^2

density for blueray=0.068722GB/cm^2

[Linksys120n]

2 minutes ago, ARikozuM said:

That's essentially what he did... I don't know if he eliminated the spindle's "empty space" but he drew the density to GB/cm².

 

DVD = .0208 GB/cm²

.0208 x 2903 = 60.33GB

 

Edit: It seems @Droidbot used diameter rather than area. By no means is my answer more correct since I'm not going to subtract the empty space form my calculation.

I'll use this and knock off about 5GB for unusable parts of the disk so about 55GB for a 12 inch DVD... Cool... It's interesting how just using a better laser (BluRay) can give almost double that on a normal sized disc...

[ARikozuM]

1 minute ago, Linksys120n said:

I'll use this and knock off about 5GB for unusable parts

Why are knocking off 1/12th of the disk when the empty space is only 1/15th-1/16th of the disc?

[SuperCloneRanger]

a CD and DVD have 86.05cm2 of usable area. for a 700mb cd that works out to 8.13mb per cm2. for a 4.7gb dvd thats 54.62mb per cm2.

a single side laser disc has about 678.6cm2 of usable area.

 

a CD the size of a laser disc will hold 5,517mb

 

a DVD the size of a laser disc will hold 37,065mb

 

 

[Linksys120n]

Just now, ARikozuM said:

Why are knocking off 1/12th of the disk when the empty space is only 1/15th-1/16th of the disc?

because I like nice round numbers.. lol I guess you could say 56 because 56 is a number that's used a lot in computers...

[SuperCloneRanger]

obviously double the numbers for a double sided laser disc.

[vanished]

I feel like this very simple question was struggled with more than it should have been

[mariushm]

If you want to be super accurate about it, you'd have to download the specification of the format and use some simple math

 

for example for BD-R you have the specs [here]

 

see page 4 , where you have track pitch of 0.32 um (micrometers) and minimum mark pitch 0.149um

so you can go fro here two ways..

1.

measure the distance from the start of active area of the disc to the last usable track

figure out number of tracks in that area.

for each track figure out the length of the track (it's a circle with circumference known)

divide length of track by 0.149 to estimate the number of bits per track

add bits of all tracks

 

2.

measure active area of the disc (area of big circle minus area of interior circle, center hole plus plastic)

just simplify and say each bit will use a rectangular slice of 0.32um x 0.149um - how many of these chunks can you fit in that area?

 

-

 

the further you are from center the slower the data would  move under the laser. if throughput isn't a concern then you can do big discs

but blurays are meant to have up to something like 144mbps bitrate so they have to be small and spin fast. big discs would wobble under their own weight and the laser wouldn't lock on tracks or the disc itself could crack/explode in pieces. also fast speeds means friction even between air inside unit and plastic of disc so plastic can warp if it gets too hot

 

they could make discs using some stronger substrate to prevent warp/bend whatever but that makes discs more expensive to make (SDxc cards and low endurance flash memory good enough for ~10 writes can be made very cheap) and less environmental friendly not to mention they could become deadly projectiles if some critical failure happens.