I have a set of images that are 4500px x 3000px in size.
Out of each of these images I need to extract a set of tiles in a 5 columns by 3 rows combination with each tile being 700px wide by 500px high.
What complicates the above is that the starting point for this needs to be at X:650 and Y:450 position from the top left corner of the image.
I have gotten as far as convert image.jpg -gravity NorthWest -chop 650x450 -crop 700x500 tile-%d.jpg
This gets me to a correct start position for creating the tile set but includes the rest of the blank area in the right and bottom edge of the image.
How do I go about solving this?
Within a single ImageMagick command you can crop the initial rectangle that contains all the tiles first, then crop that into the 15 output images. Try something like this...
convert image.jpg -crop 3500x1500+650+450 -crop 5x3# tile-%02d.jpg
I have a JPG image with a known size of 3072x2048. Now I want to rotate that image by any degrees (e.g. 45), while keeping its original size. Thus - using ImageMagick on the command line - I first want to rotate, then crop the image, like this:
convert -rotate 45 -gravity center -crop 3072x2048 +repage original.jpg rotated-45.jpg
By using -gravity center I specify to crop the center part of the image, which is what I want. This operation produces four output images:
rotated-45-0.jpg
rotated-45-1.jpg
rotated-45-2.jpg
rotated-45-3.jpg
The first image rotated-45-0.jpg is exactly the final image I want to get. The other three I don't need. I could delete them, but I think it would be nicer to not generate these "extra" images in the first place. So I thought I could do it with this command instead:
convert -rotate 45 -gravity center -crop 3072x2048+0+0 +repage original.jpg rotated-45.jpg
This only produces one output image, however, now the top-left corner of the image is being cropped. So apparently the -gravity center is not used any longer.
Any ideas what I am missing here?
Using ImageMagick you can rotate an image any number of degrees while keeping the original canvas dimensions using "-distort SRT"...
convert original.jpg -virtual-pixel black -distort SRT 45 rotated-45.jpg
Use "-virtual-pixel" to specify how you want to handle the parts that were outside the canvas before the rotation. In this example I used "black". You can use black, white, background, tile, mirror, or none.
I have 4 (x,y) coordinates between which I want place image as example given below.
The whole image must be placed within this area without cropping.
Using this 800x600 balloon:
You can use a "Perspective Distort" like this:
convert balloon.jpg -matte -virtual-pixel transparent \
-distort Perspective '0,0,50,0 0,599,100,599 800,0,750,100 800,600,500,500' result.png
There are basically 4 pairs of points in the parameters,i.e.
Pt1X,Pt1Y,Pt1NewX,Pt1NewY Pt2X,Pt2Y,Pt2NewX,Pt2NewY Pt3X,Pt3Y,Pt3NewX,Pt3NewY Pt4X,Pt4Y,Pt4NewX,Pt4NewY
So the command above moves point 0,0 to 50,0 and moves point 0,599 to 100,599 and so on.
I have labelled each of the points in red and drawn the path along which each one has moved in green.
I need render text on circle path but no primitive liketexpath in SVG in imagemagick or graphicmagick document. distort is a little bit strange.
I want to put character on circle edge and rotate text with the angle from the point to center of circle. Is there command like: rotate angle x0,y0?
I have been thinking and although you did not reply to my comment to confirm what you wanted this may be what you want:
convert input.jpg -virtual-pixel background +distort SRT "10,10 30" output.jpg
http://www.imagemagick.org/Usage/distorts/#srt
For a project I am trying to create a perspective distortion of an image to match a DVD case front template. So I want to automate this using ImageMagick (CLI) but I have a hard time understanding the mathematical aspects of this transformation.
convert \
-verbose mw2.png \
-alpha set \
-virtual-pixel transparent \
-distort Perspective-Projection '0,0 0,0 0,0 0,0' \
box.png
This code is en empty set of coordinates, I have read the documentation thoroughly but I can't seem to understand what parameter represents what point. The documentation gives me variables and names where I have no clue what they actually mean (more useful for a mathematical mastermind maybe). So if someone could explain me (visually prefered, or give me a link to useful information) on this subject because I have no clue on what I am doing. Just playing around with the parameters just wont do for this job and I need to calculate these points.
Here you will find an easy image of what I am trying to achieve (with CLI tools):
Update:
convert \
-virtual-pixel transparent \
-size 159x92 \
-verbose \
cd_empty.png \
\(mw2.png -distort Perspective '7,40 4,30 4,124 4,123 85,122 100,123 85,2 100,30'\) \
-geometry +3+20 \
-composite cover-after.png
Gives me as output:
cd_empty.png PNG 92x159 92x159+0+0 8-bit sRGB 16.1KB 0.000u 0:00.000
convert: unable to open image `(mw2.png': No such file or directory # error/blob.c/OpenBlob/2641.
convert: unable to open file `(mw2.png' # error/png.c/ReadPNGImage/3741.
convert: invalid argument for option Perspective : 'require at least 4 CPs' # error/distort.c/GenerateCoefficients/807.
convert: no images defined `cover-after.png' # error/convert.c/ConvertImageCommand/3044.
Correction by Kurt Pfeifle:
The command has a syntax error, because it does not surround the \( and \) delimiters by (at least one) blank on each side as required by ImageMagick!
Since there are no links to the source images provided, I cannot test the outcome of this corrected command:
convert \
-virtual-pixel transparent \
-size 159x92 \
-verbose \
cd_empty.png \
\( \
mw2.png -distort Perspective '7,40 4,30 4,124 4,123 85,122 100,123 85,2 100,30' \
\) \
-geometry +3+20 \
-composite \
cover-after.png
Did you see this very detailed explanation of ImageMagick's distortion algorithms? It comes with quite a few illustrations as well.
From looking at your example image, my guess is that you'll get there using a Four Point Distortion Method.
Of course, the example you gave with the 0,0 0,0 0,0 0,0 parameter does not do what you want.
Many of the distortion methods available in ImageMagick work like this:
The method uses a set of pairs of control points.
The values are numbers (may be floating point, not only integer).
Each pair of control points represents a pixel coordinate.
Each set of four values represent a source image coordinate, followed immediately by the destination image coordinate.
Transfer the coordinates for each source image control point into the respective destination image control point exactly as given by the respective parameters.
Transfer all the other pixel's coordinates according to the distortion method given.
Example:
Sx1,Sy1 Dx1,Dy1
Sx2,Sy2 Dx2,Dy2
Sx3,Sy3 Dx3,Dy3
...
Sxn,Syn Dxn,Dyn
x is used to represent an X coordinate.
y is used to represent an Y coordinate.
1, 2, 3, ... n is used to represent the 1st, 2nd, 3rd, ... nth pixel.
S is used here for the source pixel.
D is used here for the destination pixel.
First: method -distort perspective
The distortion method perspective will make sure that straight lines in the source image will remain straight lines in the destination image. Other methods, like barrel or bilinearforward do not: they will distort straight lines into curves.
The -distort perspective requires a set of at least 4 pre-calculated pairs of pixel coordinates (where the last one may be zero). More than 4 pairs of pixel coordinates provide for more accurate distortions. So if you used for example:
-distort perspective '1,2 3,4 5,6 7,8 9,10 11,12 13,14 15,16'
(for readability reasons using more {optional} blanks between the mapping pairs than required) would mean:
From the source image take pixel at coordinate (1,2) and paint it at coordinate (3,4) in the destination image.
From the source image take pixel at coordinate (5,6) and paint it at coordinate (7,8) in the destination image.
From the source image take pixel at coordinate (9,10) and paint it at coordinate (11,12) in the destination image.
From the source image take pixel at coordinate (13,14) and paint it at coordinate (15,16) in the destination image.
You may have seen photo images where the vertical lines (like the corners of building walls) do not look vertical at all (due to some tilting of the camera when taking the snap). The method -distort perspective can rectify this.
It can even achieve things like this, 'straightening' or 'rectifying' one face of a building that appears in the 'correct' perspective of the original photo:
==>
The control points used for this distortion are indicated by the corners of the red (source controls) and blue rectangles (destination controls) drawn over the original image:
==>
This particular distortion used
-distort perspective '7,40 4,30 4,124 4,123 85,122 100,123 85,2 100,30'
Complete command for your copy'n'paste pleasure:
convert \
-verbose \
http://i.stack.imgur.com/SN7sm.jpg \
-matte \
-virtual-pixel transparent \
-distort perspective '7,40 4,30 4,124 4,123 85,122 100,123 85,2 100,30' \
output.png
Second: method -distort perspective-projection
The method -distort perspective-projection is derived from the easier understandable perspective method. It achieves the exactly same distortion result as -distort perspective does, but doesn't use (at least) 4 pairs of coordinate values (at least 16 integers) as parameter, but 8 floating point coefficients.
It uses...
A set of exactly 8 pre-calculated coefficients;
Each of these coefficients is a floating point value (unlike with -distort perspective, where for values only integers are allowed);
These 8 values represent a matrix of the form
sx ry tx
rx sy ty
px py
which is used to calculate the destination pixels from the source pixels according to this formula:
X-of-destination = (sx*xs + ry+ys +tx) / (px*xs + py*ys +1)
Y-of-destination = (rx*xs + sy+ys +ty) / (px*xs + py*ys +1)
(TO BE DONE --
I've no time right now to find out how to
properly format + put formulas into the SO editor)
To avoid (the more difficult) calculating of the 8 required cooefficients for a re-usable -distort perspective-projection method, you can...
FIRST, (more easily) calculate the coordinates for a -distort perspective ,
SECOND, run this -distort perspective with a -verbose parameter added,
LAST, read the 8 coefficients from the output printed to stderr .
The (above quoted) complete command example would spit out this info:
Perspective Projection:
-distort PerspectiveProjection \
'1.945622, 0.071451, -12.187838, 0.799032,
1.276214, -24.470275, 0.006258, 0.000715'
Thanks to ImageMagick Distorting Images Documentation, I ended up with this clean-understandable code:
$points = array(
0,0, # Source Top Left
0,0, # Destination Top Left
0,490, # Source Bottom Left
2.2,512, # Destination Bottom Left
490,838, # Source Bottom Right
490,768, # Destination Bottom Right
838,0, # Source Top Right
838,50 # Destination Top Right
);
$imagick->distortImage(Imagick::DISTORTION_PERSPECTIVE, $points, false);
Please keep in mind that each set of coordinates are separated into two
parts. The first is the X axis and the second is the Y axis .. so when we say 838,0
at Destination Right Top, we mean the X axis of Destination Right Top
is 838 and the Y axis of it is zero (0).