algebra problem?

Question

George Lucas pioneered the use of digital movie cameras with the most recent Star Wars film. Assume that Lucas's camera has a swappable 80 GB hard drive (remember: 1 GB = 1024 MB, 1 MB = 1024 KB, 1 KB = 1024 Bytes, and 1 Byte = 8 bits), that it records each pixel in x-bit color, that there are 26 frames recorded per second, and that each frame is recorded in 1600x1200 resolution.

1.Create a polynomial function that gives the number of minutes of video that can be recorded before swapping in a new hard drive, as a function of x (the number of bits used to encode the color of each pixel)

2.Evaluate this function for x = 32 (32-bit true color)

All I can get up to is....First, each frame would require x bits for each of the 1600*1200 = 1,920,000 pixels.
The number of bits required per second would therefore be
(x bits/pixel)(1,920,000 pixels/frame)(26 frames/second)
= 49,920,000x bits/second

any help would be appreciated, I have been racking my brain all night. THANKS!!!

Answer 1

You're on the right track so far. From there, just convert what you have into GB/min:

(49,920,000x b/s)(60 s/min)(B/8 b)(KB/1024 B)(MB/1024 KB)(GB/1024 MB) = 0.349x GB/min

Your function would then be:

Time per drive = 80/(0.349x) min
= 229.432x min

So for 32-bit color, Lucas would be burning (0.349)(32) GB/min, or 11.158 GB/min, and he'd have to swap every 7.17 minutes.

The math may be a bit easier if we used shortcuts for the KB->GB conversion. Instead of using 1 KB = 1024 B (which is correct), we can use 1 KB = 1000 B (which may work out well too - hard drives are sold in GBs but not all manufacturers adhere to the 1 KB = 1024 B calculations - it's easier to sell them as "80 GB HDs" and just round things off). So if we did it this way, we'd have:

(49,920,000x b/s)(60 s/min)(B/8 b)(KB/1000 B)(MB/1000 KB)(GB/1000 MB) = 0.3744x GB/min

Your function would then be:

Time per drive = 80/(0.3744x) min
= 213.675/x min

If we used this, Lucas would run out of space in 213.675/32 = 6.677 minutes.

That seems pretty low - filling up an 80 GB drive in just under 7 minutes?! But you're shooting at a high bitrate and a high resolution, so you need more data per frame. And this doesn't account for any compression that could be possible. So the time-to-swap number, although outrageously low, isn't entirely out of the question.