Sorry, the old Ctrl + Enter hotkey got me again...
I think an important thing here is to look at the command that was being run:
oiiotool in.exr --ch R,G,B --resize 50% --attrib PixelAspectRatio 2.0 -o
nonsquare.jpg
There is no non-uniform scaling being applied, and changing the pixel aspect
ratio of the image should not change its physical resolution at all. As such,
the result of this should be an image that, when viewed with proper pixel
aspect ratio correction, should appear to be twice as wide as the original.
Now, about those metadata tags...
If you look at the EXIF tag definitions for XResolution and YResolution, the
language is a bit strange:
The number of pixels per<ResolutionUnit> in the<ImageWidth> direction. [...]
If you reorient your thinking so the "resolution" of the image is actually
whatever<ResolutionUnit> is (defaults to inches), it starts to make some sense. For a 512x512
square image at 72 dpi, the "resolution" is actually 7.111111 x 7.1111111.
Now, if the<XResolution> and<YResolution> values of that same image are 2400 and 1200, respectively,
and we take the language of the EXIF tags as-is, those 7.111111 inches of resolution are not going to be uniformly
mapped to screen pixels in both dimensions(remember, "pixels per<ResolutionUnit>"). Instead, the
result will be a "pixel image" twice as wide as it is tall.
In practice, the resolution tags are treated as a ratio, and the image's pixel
resolution is read directly from the file.
Finally, even if you disagree with all of this (which I wouldn't really fault
you for), the fact is that Nuke, RV, and Adobe products currently all agree on
how they should be handled, so I think it might be best to try and stay
consistent.
-Nathan
-----Original Message----- From: Larry Gritz
Sent: Wednesday, January 28, 2015 11:04 PM
To: OpenImageIO developers
Subject: Re: [Oiio-dev] aspect ratio from oiiotool
(WARNING: this whole explanation depends on your viewing with a fixed-width
font.)
I think maybe this is a disagreement between the different apps on what aspect
ratio means.
I very well could have screwed this up, so let me explain my thinking, and
people can tell me if it makes sense or if I botched it.
First of all, when we talk about the FRAME aspect ratio of a whole film image, we say the aspect is
1.85, or 16/9, or 2.35, or whatever, all of which are varying degrees of wider than they are tall.
"Wider than tall" means a frame aspect ratio of greater than 1.0, "taller than
wide" means a frame aspect ratio of less than 1.0. Right? So I'm gonna assume that the same is
true of pixel aspect ratio.
OK, here's my cartoon of a 2x2 image with square pixels. Let's make up some densities,
say the image is supposed to print 1 cm wide and 1 cm tall, so XResolution = 2,
YResolution = 2, ResolutionUnit = "cm".
+---+---+ ^
| * | * | |
+---+---+ 1cm
| * | * | |
+---+---+ v
<- 1cm ->
We agree that this is a 1.0 aspect ratio, I assume. (I do hope you're viewing
with a fixed width font)
So let's say we want to cut the density in half horizontally, giving us wide
pixels.
+-------+ ^
| * | |
+-------+ 1cm
| * | |
+-------+ v
<- 1cm ->
There are still 2 vertical samples per cm, but only 1 horizontal sample per
pixel. In other words, XResolution = 1, YResolution = 2.
What is the PixelAspectRatio?
Here's the shape of just one pixel:
+-------+
| |
+-------+
Remembering what we said about the aspect ratio of a whole frame, I would argue that the aspect ratio of a
pixel that is wider than it is tall also should be a number greater than 1. For the above pixel, its
PixelAspectRatio is 2.0, i.e. YResolution/XResolution. Also known as ydensity/xdensity, because note that in
this terminology, "resolution" means "dots per length", like printer's resolution, NOT
the faux "resolution" we use to describe the number of pixels in a whole image.
Nuke wrote an image that it says is 1047x858, with 2400 horizontal pixels per
inch (let's say; the units are undefined), so the image is 0.43625 inches
wide, and at a y density of 1200 pixels per inch, it should be 0.715 inches
tall:
<-0.436"->
^ +--------+
| | |
| | |
0.715"| |
| | |
| | |
| | |
| | |
v +--------+
Is that what you expect? It's a tall skinny image?
Or, do you expect a wide image? If you expect wide, then I'm going to go out
on a limb and claim that Nuke is totally botching the meaning of the density
fields, and thus the aspect ratio. Maybe rv is also getting it backwards,
either coincidentally having made the same mistake, or else purposely backwards
in order to match Nuke's broken output.
Somebody let me know if I'm totally borked in my thinking about this. Maybe I'm
the one who got it all wrong.
-- lg
PS. https://www.youtube.com/watch?v=Bt9zSfinwFA
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