Can someone check my calculations?

[FamousPerson]

I transcoded a movie using CCE set to single-pass VBR, with a min of 3000 and a max of 8000.

I want to calculate the bitrate of the resulting m2v file, but am having some trouble. Can someone check my calculations?


Total frames in movie: 182760 (24 fps)
End result .m2v file size: 2916346976 bytes


At 24 fps, the movie works out to 7615 seconds long. Using the file size above, that's 382974 bytes per second:

3916346976 / 7615 = 382974 Bps

There are 128 bytes in a kilobit, so that makes 2992 kilobits per second:

382974 / 128 = 2992kbps

According to my calculations above, the movie has an avg bitrate of 2992. I know this can't be right, though, since I set the min bitrate to 3000.

Can anyone tell me what I'm doing wrong?

[slk001]

Here's what I calculate:
2916346976 bytes x 8 bits/byte x 7615 secs = 3.064 Mbps.

This must not have been a very complex movie to be encoded at near the minimum bitrate. I don't use (in fact, I have never used) a single pass VBR, so I am unfamiliar with the encoding options - I always use a multiple pass (3) VBR, but of course, this will take twice as long as your method. However, the resulting quality is worth the time penalty.

[FamousPerson]

slk001, thanks for the response. Would you please elaborate on how you came up with 3.064 Mbps?

Here's what I get:

2916346976 bytes x 8 bits/byte = 23330775808 bits

Divide that by the number of seconds:

23330775808 / 7615 = 3063792 bit per second

1 megabit has 1048576 bits in it, so I come up with

3063792 / 1048576 = 2.92 mbps

I get 2.92Mbps, which is the same as 2992Kbps, which is what I calculated originally.

I just can't figure it out. Could CCE be ignoring it's min and max?

Incidentally, the quality of the picture is great.

[auenf]

download 'bitrate viewer' and see what it says.

but i get about 3mbit for avg bitrate ;)

Enf...