heit@meme.Stanford.EDU (Evan Heit) (08/01/90)
I'm trying to figure out how much information is in a one-hour television show, and how much computer memory it would take to store this information. I think the calculations should run something like this. X=bits of information in one frame of videotape Y=number of frames of videotape per second Amount of memory for one hour= X bits/frame * Y frames/second * 60 seconds/minute * 60 minutes/hour. Does this seem right? Can anyone tell me the values for X and Y? Thanks for your help.
naiming@eng.umd.edu (Naiming Shen) (08/01/90)
You even have trouble to put one second of video into memory. the X is about 300-600 and Y is 525. Assume you digitize it with 8 bits per pixel. You have: 400*525 X 30 = 6.3 M Bytes per second. --- -- ***************************************************************************** Naiming Shen *****************************************************************************
mwm@raven.pa.dec.com (Mike (Real Amigas have keyboard garages) Meyer) (08/01/90)
In article <1990Jul31.204803.25358@eng.umd.edu> naiming@eng.umd.edu (Naiming Shen) writes:
You even have trouble to put one second of video into memory.
the X is about 300-600 and Y is 525. Assume you digitize it with
8 bits per pixel.
You have:
400*525 X 30 = 6.3 M Bytes per second.
The referenced article is no longer around, and the relevant
information wasn't quoted. However, for the purpose of playing video
back, you don't store animations raw like that (not if you know what
you're doing, anyway). You store full frames only on "scene" changes,
and delta lists for the frames between those shots.
You double buffer things, so you can build the next image while the
user is viewing the current one (that's a _long_ 1/30th of a second).
With display hardware that makes changing buffers a matter of updating
a pointer, even something as slow as a 68020 at 16Mhz can run 30
frames a second (methinks a 68000 @ 8MHz can do the same, but I'm not
positive).
If you do things right, you only need memory for two full screens
(<.5Meg for both) plus deltas (variable size, but normally small for
each frame) plus software (depends). The net result is that a
relatively old personal computer with 2 to 4 meg of RAM can handle a
seconds worth of video, and then some.
<mike
--
How many times do you have to fall Mike Meyer
While people stand there gawking? mwm@relay.pa.dec.com
How many times do you have to fall decwrl!mwm
Before you end up walking?merlyn@iwarp.intel.com (Randal Schwartz) (08/01/90)
In article <1990Jul31.204803.25358@eng.umd.edu>, naiming@eng (Naiming Shen) writes: | You even have trouble to put one second of video into memory. | the X is about 300-600 and Y is 525. Assume you digitize it with | 8 bits per pixel. | You have: | 400*525 X 30 = 6.3 M Bytes per second. OK, here goes my ignorance, but isn't video transmitted on a 3.58MHz carrier (or whatever you call it... I'm not an analog guy). If so, I had always presumed that you couldn't stuff more than 2 or 3 Mbits/sec (theoretical) down that size of pipe. That'd put your estimate off by a factor of 20. Am *I* really that far off? Also, I've never seen an NTSC with Red in one "pixel" and yellow (blue+green, right?) in the next "pixel" that didn't get sorta smeared together, so I suspect you really have only two or three new bits of information per pixel. Am I close? OK, lets look at it another way. 8mm Exabyte tape, two hour style, holds 2.2 GBytes. That'd put it at 0.3 megabytes/sec, or 2.5 megabits/sec. Pretty close to my numbers. I'm not saying anything about the efficiency of the Exabyte 8mm drive, but it's gotta be close. If there were a way to stuff 6.3MByte/sec into those tapes, don't you think they would have tried? Just a shot in the dark, -- /=Randal L. Schwartz, Stonehenge Consulting Services (503)777-0095 ==========\ | on contract to Intel's iWarp project, Beaverton, Oregon, USA, Sol III | | merlyn@iwarp.intel.com ...!any-MX-mailer-like-uunet!iwarp.intel.com!merlyn | \=Cute Quote: "Welcome to Portland, Oregon, home of the California Raisins!"=/
jindak@surfside.sgi.com (Chris Schoeneman) (08/03/90)
In article <1990Jul31.204803.25358@eng.umd.edu> naiming@eng.umd.edu (Naiming Shen) writes: >the X is about 300-600 and Y is 525. Assume you digitize it with >8 bits per pixel. This X x Y is the resolution. The original post had: X = bits/frame = ? Y = frames/sec = 30 TV isn't a digital device. You have to choose how many bits yourself. However, FYI one channel's bandwidth is 0-6 MHz. Chris Schoeneman | I was neat, clean, shaved and sober, jindak@surfside.esd.sgi.com | and I didn't care who knew it. Silicon Graphics, Inc. | -Raymond Chandler Mountain View, CA | (The Big Sleep)
ccw@nvuxr.UUCP (christopher wood) (08/18/90)
I decided not to quote the earlier articles, anyway.
The question is, how much computer memory do you need to hold video
information?
Lots of x pixels/line by y lines/frame by z frames per second by q
bits/pixel formulas were suggested.
Picking up a handy reference volume, I discover that "the networks" use
"the telepohone network" to carry their video. They do this on a
digital channel that carries 45 MegaBITS per second. I am not sure
about the encoding, but I'm sure that it's clever, but relatively
simple.
Using this number, each minute of video would take up 337 MegaBYTES of
storage.
--
Chris Wood Bellcore ...!bellcore!nvuxr!ccw
or ccw@nvuxr.cc.bellcore.com