Update after version 14.0.0.0, the bitrate of the video is incorrect
I usually process videos from DMM.com, and the coding mode of DMM is about 6000 bitrate average. In the previous WMV era, 5744+256=6000kbps was a standard, so we can judge whether the video source is correct or not. Many of my friends and I think that Windows attributes are absolutely not as accurate as the metadata of MediaInfo and Potplayer, so I think it would be better to quote the Text of MediaInfo.
Your point is clear.
We are dealing with the Variable bitrate. So it changes in different parts of video.
On the other hand, we need a single number to display.
I am just trying to understand which value is more correct to use.
Your file indeed contains:
bitrate=5744
vbv_maxrate=5744
I can try to find a way to extract that data from the file.
The question is - is it reliable for other videos?
The video length is 312,18 sec.
In your case we can try to calculate the overall size of the file like that:
(5 740 000 + 256 000) * 312,18 / 8 = 233 978 910 bytes.
It is larger than 231 533 364. The difference is 2 445 546 bytes (~2.35 MB).
In my case:
(5 670 755 + 256 000) * 312,18 / 8 = 231 276 796 bytes.
It is less than 231 533 364. The difference is 256 567 bytes (~0.24 MB).
So my estimation is ten times closer to the real file size.
So I would say in your case 5744 is the max bitrate, not average.
What do you think?
We are dealing with the Variable bitrate. So it changes in different parts of video.
On the other hand, we need a single number to display.
I am just trying to understand which value is more correct to use.
Your file indeed contains:
bitrate=5744
vbv_maxrate=5744
I can try to find a way to extract that data from the file.
The question is - is it reliable for other videos?
The video length is 312,18 sec.
In your case we can try to calculate the overall size of the file like that:
(5 740 000 + 256 000) * 312,18 / 8 = 233 978 910 bytes.
It is larger than 231 533 364. The difference is 2 445 546 bytes (~2.35 MB).
In my case:
(5 670 755 + 256 000) * 312,18 / 8 = 231 276 796 bytes.
It is less than 231 533 364. The difference is 256 567 bytes (~0.24 MB).
So my estimation is ten times closer to the real file size.
So I would say in your case 5744 is the max bitrate, not average.
What do you think?
Maybe I don't know the real coding mode of DMM, but you can try to add options: calculate the average, or follow the bitrate like Mediainfo. I usually download DMM videos, and sometimes when the network fluctuates, the finished .mp4 can't find the 5744 bitrate data anyway, which can prove that .mp4 is absolutely broken.admin wrote: ↑06 Jun 2022, 16:55Your point is clear.
We are dealing with the Variable bitrate. So it changes in different parts of video.
On the other hand, we need a single number to display.
I am just trying to understand which value is more correct to use.
Your file indeed contains:
bitrate=5744
vbv_maxrate=5744
I can try to find a way to extract that data from the file.
The question is - is it reliable for other videos?
The video length is 312,18 sec.
In your case we can try to calculate the overall size of the file like that:
(5 740 000 + 256 000) * 312,18 / 8 = 233 978 910 bytes.
It is larger than 231 533 364. The difference is 2 445 546 bytes (~2.35 MB).
In my case:
(5 670 755 + 256 000) * 312,18 / 8 = 231 276 796 bytes.
It is less than 231 533 364. The difference is 256 567 bytes (~0.24 MB).
So my estimation is ten times closer to the real file size.
So I would say in your case 5744 is the max bitrate, not average.
What do you think?