Determine original framerate of progressive source

[bobp127001]

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[jagabo]

Bob the video then and count the number of unique pictures per second during movement.

http://www.afterdawn.com/guides/archive/digital_video_fundamentals-ivtc.cfm

[bobp127001]

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[manono]

Just count the number of duplicate frames per second and remove them.

You said all your sources are 59.94fps. The most likely number of dupes in a second (usually containing 60 frames) are:

zero: keep it at 59.94fps
thirty: remove every other frame (SelectEven/Odd) to make it 29.97fps.
thirty-six: SelectEven.TDecimate to make it 23.976fps.

I am now capturing a TV movie and I have no idea if it will be film or video.

If it's a movie, by definition it wasn't shot on video but on film, and it'll be 23.976fps.

[bobp127001]

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[jagabo]

So you have a 59.94 fps video that was originally 23.976 fps film.

[bobp127001]

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[jagabo]

In the past, I have done TDecimate(mode=0,cycle=5,cycleR=3). I'm not sure what the difference between those two commands is (perhaps the second TDecimate somehow integrates SelectEven()), so I'm hoping manono's command works as well.

In your case you have:

 00111223334455566777

Each digit represents a frame, the value of the digit is the frame number. Ie, 8 frames are repeated with a 2:3 repeat pattern to become 20.

TDecimate(mode=0,cycle=5,cycleR=3) frist breaks it down into 5 frame groups:

 00111 22333 44555 66777

then removes 3 duplicates from each group

 01 23 45 67

leaving

01234567

SelectEven().TDecimate() first removes every other frame:

 0 1 1 2 3 4 5 5 6 7

breaks the sequence into groups of 5

 01123 45567

then removes one duplicate in each group

 0123 4567

leaving

 01234567

They may occasionally give slightly different results at cadence breaks, the boundary between a moving sequence and a still sequence, etc.

By the way, with progressive frames you just need to count the duplicates, no need to bob first.

[bobp127001]

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