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From: unruh on 3 Dec 2009 23:13
On 2009-12-04, spike1(a)freenet.co.uk <spike1(a)freenet.co.uk> wrote:
> And verily, didst unruh <unruh(a)wormhole.physics.ubc.ca> hastily babble thusly:
>> 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.
>
> Don't think mplayer can crop, you'd need to use mencoder for that (mplayer's
> sister app), or transcode.

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.

From: unruh on 3 Dec 2009 23:15
["Followup-To:" header set to alt.os.linux.]
On 2009-12-04, Anita <me(a)invalid.com> wrote:
> "unruh" <unruh(a)wormhole.physics.ubc.ca> wrote in message
> news:slrnhhg5ud.em0.unruh(a)wormhole.physics.ubc.ca...
>
>> 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.
>
> 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.


>
>
From: Anita on 3 Dec 2009 23:40
"unruh" <unruh(a)wormhole.physics.ubc.ca> wrote in message
news:slrnhhh36m.mqv.unruh(a)wormhole.physics.ubc.ca...

>> 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.


From: Anita on 3 Dec 2009 23:55
I quoted and wrote in message news:7nri6tF3kdfmiU1(a)mid.individual.net...

>>> 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.
I neglected to mention that the original *.mov, as well as the instrument
that captured the "Quicktime" image, will render unacceptable mensuration
parameters into the mix.




From: unruh on 4 Dec 2009 01:12
["Followup-To:" header set to alt.os.linux.]
On 2009-12-04, Anita <me(a)invalid.com> wrote:
> "unruh" <unruh(a)wormhole.physics.ubc.ca> wrote in message
> news:slrnhhh36m.mqv.unruh(a)wormhole.physics.ubc.ca...
>
>>> 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,

Which is why I am converting to png, a lossless format. Cropping is OK
since it is relative changes which are important. Pixels, when
calibrated are a fine measure.

> and pixels are not any sort of accurate measurement of anything other than
> pixels.

You think the pixels move around, or breathe?
>
>
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