[MEncoder-users] Trying to cut at exact frames (-hr-edl-seek etc.)

[Oded Shimon]

On Sun, May 22, 2005 at 02:15:33PM +0200, Thorsten Pferdek?mper wrote:
> On Friday 20 May 2005 07:09, Oded Shimon wrote:
> > On Thu, May 19, 2005 at 08:55:02PM +0200, Thorsten Pferdek?mper wrote:
> > > Hi,
> > > [...]
> > > Let d be the duration of one frame (i.e. 1/framerate).
> > > Let [a,b] be the frame interval you want to keep. (Starting to count with
> > > 0) Let l be something bigger than the length of the video in seconds Then
> > > you need two entries in the EDL file:
> > > 	0.00		a*d	0
> > > 	b*d		l	0
> > >
> > > At least, this works if b is really bigger than a. with a=b, it does not
> > > work. I would have expected, that I will get exactly one frame in this
> > > case, but I get zero frames. With b = a+1, I get two frames (a and b).
> > > So, what does this mean? It simply means that it is impossible to cut out
> > > exactly one frame (or every second frame or something like that). Perhaps
> > > this is not very important and rather academic, but I thought it was
> > > worth to be mentioned.
> >
> > Just a hunch, try 'b = a+0.5'. With 'a = b' You get an error before the
> > encode even starts i assume?
> >
> Hi,
> ok, I did some further tests. I believe that you get 2 frames whenever 
> 	a < b <= a+1
> Here are my test examples: 
> 
> 1. Without any edl: 1933 frames
> 
> 2. Using the following edl 
> 	0	4		0
> 	4	9999	0
> produces a message because of overlapping intervals. It seems that mencoder 
> ignores the second line. It skips the first 4 seconds. The output has 1833 
> frames. (Btw: frame rate is 25, a=100. This makes the "4".)
> 
> 3. Using the edl you suggested:
> 	0	4	 0
> 	4.02	9999 0
> There is no error message, but it creates 2 frames.
> 
> 4. The same as 3. with 
> 	0	4		0
> 	4.04	9999	0		
> 
> Regards,
> 	Thorsten

Try this.

- ods15