How to calculate viewing BBox of an image in pdf?
I am trying to calculate the Showing image coordinates. But the actual image is showing bigger than showing below(fig1). But we can able to see only part of the image only. I want to calculate how the matrices are transforming(Calculation for shown image coords).
fig1
The content stream looks like below
The coords I am getting when I multiplied first q cm with second q cm is
[-122.196, 356.535, 484.061, 759.372]
But these are full image coord. How the 're' will change the calculation for part of image?
After removing the 're' and 'W*'
Need Another answer on the same scenario.
What I tried
0.24 0 0
0 -0.24 0
0 850 1
the 're' calculate to above CTM and it will gives
[595.92 0 0 7.05]
the nex cm instruction become CTM and looks like
1 0 0
0 1 0
262 404 1
the resulted matrix will be what How can I calculate it?
For the sake of brevity I'm going to round the numbers a bit here.
Case 1
Let's simply analyze your content stream excerpt:
Let's assume that the preceding instructions left the user space coordinate system and the clip path in its default state, so we can assume an identity current transformation matrix (CTM) and a clip path encompassing the whole page.
The first instruction
.24 0 0 -.24 0 850 cm
then changes the CTM to
0.24 0 0
0 -0.24 0
0 850 1
Thus, the rectangle path defined and used as a clip path thereafter
169 349.49 1038.37 1670.15 re
has, in the default user space, the coordinates (lower left, upper right):
[40.56 365.29 289.77 766.12]
Then the next cm instruction
2517.74 0 0 -1670.15 -504.99 2019.64 cm
changes the CTM to
604.26 0 0
0 400.84 0
-121.2 365.29 1
So the following bitmap image
/Im0 Do
is drawn, in the default user space, at the coordinates (lower left, upper right):
[-121.2 365.29 483.06 766.13]
This area partially is outside the clip path, so we get the visible image area in user space coordinates by intersecting those coordinates with the clip path, resulting in
[40.56 365.29 289.77 766.12]
So these are the coordinates you appear to be looking for.
Beware, in general clip paths can have arbitrary forms, and the CTM at image drawing time may not only scale, mirror, or translate (resulting in a rectangle parallel to the axis) but also rotate or skew (resulting in a rhomboid or something not parallel to the axis). Thus, calculating the intersection and making sense out of the result in general is more complicated.
In a comment you ask
But still I need bit more clarity to, after 're' how you got [40.56 365.29 289.77 766.12] this. How the calculation is happening.
I got those by applying the CTM to two diagonally opposed corners of the rectangle.
To get two such corners of
169 349.49 1038.37 1670.15 re
I first took the anchor point at 169 349.49 and as second point the anchor point with width and height added 1207.37 2019.64.
Then I applied the CTM to those two points
0.24 0 0
[169 349.49 1] × 0 -0.24 0 = [40.56 766.12 1]
0 850 1
0.24 0 0
[1207.37 2019.64 1] × 0 -0.24 0 = [289.77 365.29 1]
0 850 1
So I get the transformed corners at 40.56 766.12 and 289.77 365.29.
Due to the mirroring the resulting points were not lower-left to upper-right but instead upper-left to lower-right. Thus, I normalized the rectangle to [40.56 365.29 289.77 766.12].
Beware, this calculation makes use of the fact that the CTM only scaled, mirrored, and translated. If it also rotated or skewed, I would have had to apply the CTM to all corners of the rectangle (or at least three of them) and then worked with the rhomboid spanned by them.
Case 2
In an edit to your question you added another case:
This example shows that one has to inspect the XObject in question first.
If one assumed that Fm0 was an image XObject, the image would be drawn in a .24×.24 default user space units square, a tiny dot.
But Fm0 is not an image XObject, instead it is a form XObject which in turn shows an image XObject from its own resources. Thus, here is another step in the calculations:
The first instruction
.24 0 0 -.24 0 850 cm
then changes the CTM to
0.24 0 0
0 -0.24 0
0 850 1
Thus, the rectangle path defined and used as a clip path thereafter
0 0 2483.33 3512.32 re
has, in the default user space, the coordinates (lower left, upper right):
[0 7.04 596 850]
Then the next cm instruction
1 0 0 1 262 404 cm
changes the CTM to
0.24 0 0
0 -0.24 0
62.88 750.04 1
Due to
/Fm0 Do
we then have to continue with the XObject Fm0. First of all it has a bounding box entry
[ 0 0 1959 1306 ]
Applying the CTM to this we get a bounding box in the default user space of
[62.88 436.6 533.04 750.04]
which has to be intersected with the clip path.
The relevant content of the Fm0 is
0.72 196.505 1957.892 913.266 re
W* n
q
/GS0 gs
1957.8926 0 0 -1304.21912 0.7203979 1305.13342 cm
/Im0 Do
Q
Thus, the rectangle path defined and intersected with the clip path thereafter
0.72 196.51 1957.89 913.27 re
has, in the default user space, the coordinates (lower left, upper right):
[63.05 483.69 532.95 702.88]
Then the next cm instruction
1957.89 0 0 -1304.22 0.72 1305.13 cm
changes the CTM to
469.89 0 0
0 313.01 0
63.05 436.81 1
So the following bitmap image
/Im0 Do
is drawn, in the default user space, at the coordinates (lower left, upper right):
[63.05 436.81 532.94 749.82]
The effective clip path at this time is the intersection of the rectangles
[0 7.04 596 850]
[62.88 436.6 533.04 750.04]
[63.05 483.69 532.95 702.88]
so it is the rectangle
[63.05 483.69 532.95 702.88]
Thus, the visible area of that drawn image is
[63.05 483.69 532.94 702.88]
(Well, I hope it is, but maybe I somewhere along the path erred in some calculation...)