Computing Per-Minute Storage for Full HD TV

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SUMMARY

The discussion focuses on calculating the per-minute storage requirements for uncompressed Full HD video at 60 frames per second, with a resolution of 1920×1080 pixels and 24-bit color depth. The calculations reveal that one minute of such video requires approximately 20.86 gigabytes of storage. This significant storage requirement highlights the necessity for compression techniques, as evidenced by the storage capacities of Blu-Ray discs, which can hold over 9 hours of HD video due to advanced encoding methods.

PREREQUISITES
  • Understanding of pixel resolution and color depth
  • Basic knowledge of data storage units (bytes, kilobytes, megabytes)
  • Familiarity with video frame rates and their impact on storage
  • Awareness of video compression techniques and formats
NEXT STEPS
  • Research video compression algorithms such as H.264 and HEVC
  • Explore the differences between uncompressed and compressed video formats
  • Learn about the storage capacities of various optical media, including Blu-Ray
  • Investigate the impact of frame rate on video quality and storage requirements
USEFUL FOR

This discussion is beneficial for video engineers, software developers working with multimedia applications, and anyone involved in video production and storage optimization.

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Homework Statement



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.

Homework Equations



1 pixel = 3 integers
1 integer = 8 bits

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.
 
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KillerZ said:

Homework Statement



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.

Homework Equations



1 pixel = 3 integers
1 integer = 8 bits

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.

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

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