OK, will look into that. If you have a hint as to which options I need
to tickle I would appreciate it.
>
> I don't know if kino can do that or not.
>
>>
>>
>
mplayer -vo png/tga/jpeg
Florian
--
<http://www.florian-diesch.de/software/pdfrecycle/>
Use the following to extract the video.mov to jpg frames:
mplayer video.mov -vo jpeg
Use the following to extract 1000 frames from the 1178'th second.
mplayer video.mov -ss 1178 -frames 1000 -vo jpeg
If jpg's are not usefull, you can convert them to tiff with convert.
See 'man mplayer' and 'man convert'.
Convert is part of ImageMagic. You might need to install that, first.
--
Any time things appear to be going better, you have overlooked
something.
Thanks. I did finally find the png/jpeg/.., options in that huge list of
mplayer options but thanks.
>
>
>
> Florian
Since one of the purposes is to do measurements on the frame for a
scientific experiment I do not want layers of lossy compression so I
would probably use png. . How are the frames stored in a .mov file? I
assume that they are probably stored with some sort of lossy
compression.
.mov movie frames stored
Thanks. Since we are making measurments on the frames, I suspect it
would be better to convert to png (lossless compression) rather than
jpeg, unless the files are stored as jpg, and mplayer simply copies the
frames. Also only need a narrow slice of the picture, but I assume
mplayer cannot do that ( ie crop down to some small slice) while it is
copying out the frames. I can certainly do that with convert, but the
intermediate files are going to far larger than they need to be (slowing
down the conversion time).
(Ie, on the 1080p (1960x1080) frame we really only need a 1960x50 slice of the
frame to do the measurements on.
>
There are the -vf crop options crop=w:h:x:y although I do not know what x
and y represent (middle of the picture of one of the corners. the
"Position of the cropped picture, defaults to center."
but they are a bit obscure.
>> It begs the question: precisely what "measuremenst" are required?
>
> Since you asked.
> There is a bright line in the frame which is a laser sheet reflected from
the tops of
> the waves. We need to measure their height very accurately.
And how will you accomplish that sort of "very accurate" measurement using
the method you (very loosely) suggest? Cropping the image, converting image
format et. al. will most assuredly result in various losses of information,
and pixels are not any sort of accurate measurement of anything other than
pixels.