# Some Algebra

1. Oct 5, 2009

### KillerZ

1. The problem statement, all variables and given/known data

This question isn't that hard just confusing:

Write a program to compute the per-minute storage requirements for “full
HD” TV at 60 HZ (60 frames per second) where each frame is captured at
a resolution of 1920×1080 pixels using 24-bit color.

2. Relevant equations

1 pixel = 3 integers
1 integer = 8 bits

3. The attempt at a solution

I said:

Number of pixels one frame = (1920)(1080) = 2073600

Number of integers one frame = (2073600)(3) = 6220800

Number of bits one frame = (6220800)(8) = 49766400

Number of bytes one frame = 49766400/8 = 6220800

Number of kilobytes one frame = (6220800)/1024 = 6075

Number of kilobytes one second = (6075)(60) = 364500

Number of kilobytes one minute = (364500)(60) = 21870000

Number of megabytes one minute = (21870000)/(1024) = 21357.42 approx. = 20.86 gigabytes approx. This looks like a very large number for one minute.

2. Oct 5, 2009

### jbunniii

I get the same answer. It IS a very large number for one minute. That's why DVDs and cable TV use such heavy compression. e.g., a dual-layer Blu-Ray DVD has 50GB storage capacity TOTAL, enough to hold 2.5 minutes of uncompressed video, not even including audio. The Blu-Ray FAQ claims that you can store over 9 hours of HD video on a 50GB disc, so that shows how extreme the compression is. (Source: http://www.blu-ray.com/faq/ )

3. Oct 5, 2009

### KillerZ

Ok that makes sense as the question also says "assume no other encoding is done" so it would be uncompressed video.