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View Full Version : How to determine if AR of movie is 16/9 or 2.35 ?


buzzqw
9th January 2006, 14:07
Using dgindex and avisynth how can i determine if AR in movie is 16/9 or 2.35 ?

I can read from d2v if is 4/3 or 16/9 but if movie is 2.35 in d2v i read

Aspect_Ratio=16:9
NOT
Aspect_Ratio=2.35

there are any possibility (using filters...) ?

thansk !

BHH

dbzgundam
9th January 2006, 14:29
Assuming your source is a DVD, you can most likely check the box... If this isn't available, the MPEG2 should have telltale signs.

-If anamorphic, and 16x9, it should take up the whole frame.
-If non-anamorphic, and 16x9, it should only take up a 16x9 part of a 4x3 frame.
-If anamorphic, and 2.35:1... Your eyes sort of have to tell, but it should SORT of look like a 16x9 movie in a 4x3 frame, *when it's not properly displayed*
-If non-anamorphic, and 2.35:1, this follows the same operation as a non-anamorphic 16x9 image... Obviously, it's still just letterboxed into a 4x3 frame.

Wilbert
9th January 2006, 14:49
4:3 and 16:9 refer to the DAR, and 2.35 to the PAR. You can't determine whether a movie is 2.35 without looking at the amount of black borders. In contrary to the DAR which is written in the stream itself.

@dbzgundam,
There is no such thing as "non-anamorphic 16x9 image". 16:9 is the same as anamorphic. Fullscreen and letterboxed widescreen are 4:3.

buzzqw
9th January 2006, 15:23
4:3 and 16:9 refer to the DAR, and 2.35 to the PAR. You can't determine whether a movie is 2.35 without looking at the amount of black borders. In contrary to the DAR which is written in the stream itself.


Amount of crop... :goodpost:

but ... how to know how is enough... ?


i can
1' case) take the crop value (let's say 72 top, 72 bottom)

720/(576-72-72)=1.666666 -> 16/9

2' case) take crop value (let's say 136 top, 136 bottom)

720/(576-136-136)=2.3684 -> 2.35


could be correct ???

BHH

dbzgundam
9th January 2006, 15:51
4:3 and 16:9 refer to the DAR, and 2.35 to the PAR. You can't determine whether a movie is 2.35 without looking at the amount of black borders. In contrary to the DAR which is written in the stream itself.

@dbzgundam,
There is no such thing as "non-anamorphic 16x9 image". 16:9 is the same as anamorphic. Fullscreen and letterboxed widescreen are 4:3.

That's what I'm referring to, sorry for not using correct terminology! :)

Wilbert
9th January 2006, 21:46
2.35 is not PAR of course, it's just the real aspect ration or whatever you call this.

i can
1' case) take the crop value (let's say 72 top, 72 bottom)

720/(576-72-72)=1.666666 -> 16/9

2' case) take crop value (let's say 136 top, 136 bottom)

720/(576-136-136)=2.3684 -> 2.35
No. You need to know the DAR before you can calculate the real aspect ratio. Suppose your source is 720x576 with DAR 16:9, and black bars 2*72. The real aspect ratio is given by (non ITU compliant):

16/9*576 x 576-2*72 = 1024 x 432 and 1024/432 = 2.3704

If your DAR would be 4:3 you would get

4/3*576 x 576-2*72 = 768 x 432 and 768/432 = 1.7778

I hope it's clear now :)

buzzqw
10th January 2006, 08:18
yep ... it's is clear...

(mumble... mumble...) i will look for a clear solution !

thanks !!!

BHH