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The pattern for dropping frames in 29.97 drop-frame SMPTE timecode is well documented (2 frames at every minute except when minute is a multiple of 10).

I was wondering how it works with 59.94 fps drop-frame timecode? I guess it needs to be twice the amount of frames dropped, but what is the exact pattern?

1 Answer 1

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Apparently, the frames are dropped at the same cadence, but with double frequency.

From FFmpeg's timecode drop function,

if (fps == 30) {
    drop_frames = 2;
    frames_per_10mins = 17982;
} else if (fps == 60) {
    drop_frames = 4;
    frames_per_10mins = 35964;
} 
....
d = framenum / frames_per_10mins;
m = framenum % frames_per_10mins;

return framenum + 9*drop_frames*d + drop_frames*((m - drop_frames)/(frames_per_10mins/10));

and the string is

ff = framenum % fps;
ss = framenum / fps      % 60;
mm = framenum / (fps*60) % 60;
hh = framenum / (fps*3600) % 24;
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  • Great idea. So from that code it really looks like the pattern of dropping every minute excepts tens of minutes is the same but dropping 4 instead of 2 frames.
    – Sebastian
    Commented Nov 5, 2017 at 10:34

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