- the width and height (or resolution) in pixels (picture elements), and
- the number of colors (depth) per pixel.
A typical resolution might be
800 pixels wide by 600 pixels in height,
Since 65,536 is 2 16, we will need 16 bits for each pixel.
We then compute the amount of video memory required:
with a depth of 65 thousand colors per pixel.
Vertical resolution is sometimes given in lines, so that the above would be denoted as 600
lines, each of which is 800 pixels wide. In this usage, a line is not a real unit; the resolution
is computed by simplying multiplying the horizontal pixel resolution by the vertical line resolution,
and the result is given in pixels. Thus a two dimensional video image is represented as a simple list
of pixels; in this example, the computer would start a new line after every 800 pixels.
- ? bytes = number of pixels * bytes per pixel
- number of pixels = 800 * 600 = 480,000 pixels
- bytes per pixel = 16 bits / pixel / ( 8 bits / byte ) = 2 bytes / pixel
- bytes of video RAM = 480,000 pixels * 2 bytes / pixel = 960,000 bytes
We can again use proportionalities to do additional problems:
- For 1024 by 768 16 bit resolution, we need
960,000 bytes * ( 1024 * 768 pixels / ( 800 * 600 pixels))
= 1,572,864 bytes / ( 2 20 bytes / MB )
= 1.5 MB - a Truecolor (24 bits per pixel) digital camera image which is 2048 by 1536 pixels requires
960,000 bytes * 24 bits / pixel / ( 16 bits / pixel) * 2048 * 1536 pixels / ( 800 * 600 pixels)
= 9,437,184 bytes / ( 2 20 bytes / MB )
= 9 MB
Here we have used powers of 2 for conversions to MB, since we are dealing with storage requirements.
The space computations for audio files can be done in much the same way. A microphone converts sound into changing voltages:
In order to digitize the sound, we need to convert the changing voltages into a series of numbers. To do this, we sample the voltages periodically:
For CD quality sound, the sample rate is 44,100 samples per second. Of course, for stereo recordings we need twice that many samples per second, one for each channel. We also need to decide how many bits are necessary to hold the voltage for each sample:
For CD quality sound 16 bits are used. This corresponds to 2 16 voltage divisions in the graph above. The amount of space required for one minute of CD quality sound is then
? bytes = 1 minute * ( 60 seconds / minute) * ( 44,100 samples / second / channel) *Here, the 44,100 samples per second per channel was represented with units of samples / second / channel. For a 3:21 (3 minutes, 21 seconds or 201 seconds) mono (one channel) audio file with a sample rate of 22050 Hz and a sample depth of 8 bits, we require
2 channels * ( 16 bits / sample) / ( 8 bits / byte)
= 10,584,000 bytes / ( 2 20 bytes / MB)
= 10.09 MB
10.09 MB * ( 201 seconds / 60 seconds) * (1 channel / 2 channels) * ( 8 bits / sample / ( 16 bits / sample))
= 8.45 MB
We can perform a similar computation for full motion video. Using the NTSC standard and VHS quality, full motion video involves a resolution of 352 pixels in width by 240 lines in height for each field; two fields are shown for each frame, and there are 29.97 frames per second. If we use 24 bits for each pixel (TrueColor), we can compute the space requirements for 1 minute of VHS video:
? bytes = ( 352 pixels * 240 lines / field) * ( 2 fields / frame) * ( 29.97 frames / second) * ( 60 seconds / minute ) *Note that the vertical "lines" units were omitted in the computation, as per the discussion above.
( 24 bits / pixel) / ( 8 bits / byte * 2 20 bytes / MB)
= 869.25 MB