This is what my brain does when it goes into Flatland withdrawl and I'm procrastinating getting my schoolwork done.
I know that part of this puzzle includes calculating the total surface area of the DVD, and that you will need to include the middle part, but then subtract that from the final answer. I also know that the outter part of the DVD can hold more information because those circles are bigger than the inner circles.
Hm. I don't have time to think about this anymore, but maybe YOU do!!
Since the area of the used portion of the disc increases as the square of the radius, and ignoring the hole in the middle, a disc that looks like it's burned halfway out will only hold about a quarter of the maximum capacity, so, say 1 GB. (That is to say, a disc that's twice as big across will hold four times as much data, assuming uniform density.)
Flipping that around, if you want to find out where the disc burns out to if it's half full, you're looking at 1/sqrt(2) of the radius, or about 70% of the way out - again, ignoring the middle bit.
If you want to take the center hole into account, it gets a little bit messier, but just remember that as the hole size increases, the closer the ratio of burned radius to data capacity approaches 1. The hole in the middle of a CD is actually pretty good sized (compared to, say, an LP), so you probably want to adjust those numbers a bit downward to take that into account.
That all sounds about right to me, but I haven't had any coffee yet, so.
The Bastard Improv For Evil
"new goal: be quoted in Marc's signature." - Jordan T. Maxwell
sara_anm8r wrote:This is what my brain does when it goes into Flatland withdrawl and I'm procrastinating getting my schoolwork done.
I know that part of this puzzle includes calculating the total surface area of the DVD, and that you will need to include the middle part, but then subtract that from the final answer. I also know that the outter part of the DVD can hold more information because those circles are bigger than the inner circles.
Hm. I don't have time to think about this anymore, but maybe YOU do!!
DVD volume burned when 1/2 R is filled = {(pi*[1/2r^2])/(pi*r^2)} * 4.7GB
Radius filled when 1/2 volume is filled = {(r^2)/2}^(1/2)
Of course that's a simplified version - I'm at work. A less simple version would take into consideration the amount of data consumed on the lead-in and lead-out, as well as the radius of the "center" to which no data is written.
I'll actually have hard numbers for you if you can track down the actual dimensions of a DVD.
