I am trying to cut a mesh with a plane and render the contour on top of an image (resliced along using the same plane), much like the image below:
The examples at https://gitlab.kitware.com/vtk/vtk/-/blob/v9.1.0/Filters/Core/Testing/Python/TestPolyDataPlaneCutter.py almost do the slicing part that I need
But when viewing the contour from the direction of the normal the contour / polyline is invisible (see screenshot below left side) as opposed to viewing it from the side (right side of the screenshot below)
So I assume I should try to "dilate" the resulting contour somehow? Any suggestions?
Related
I'm extracting door shapes with Cubicasa5k dataset but I don't know how svg draws quarter rounds.
Extracted some nodes of svg files with xml encoding, I found that it had a dictionary including 'd' key and value under of "Panel" id tree node like :
{'d': 'M825.47,986.05 q0.00,-72.92 72.92,-72.92 l0.00,72.92Z'}
It draws like this quarter round on svg image.
I want to draw that shape on raster image with cv2 in Python with that dictionary value.
Read article about 'd' commands but I'm still confused.
How can I draw it?
Okay, I hope I understood that d command of svg clearly now
That arc shape is Bézier curve, and it is drawn concavely based on the outer point, not on the centripetal point.
According to the example above ('M825.47,986.05 q0.00,-72.92 72.92,-72.92 l0.00,72.92Z'), it means drawing process following this:
The starting point of arc shape(called Bézier curve) is (825.47, 986.05).
The drawn reference point exists at a position shifted only to the y-axis. So the reference point is (825.47+0.00, 986.05-72.92) = (825.47, 913.13)
And the end point of arc shape is set following calculation about starting point like 2 : (825.47+72.92, 986.05-72.92) = (898.39, 913.13)
Drawing door shape finishes with drawing straight line that starts from the end point of the curve and ends with l code, it points (898.39+0.00, 913.13+72.92) = (898.39, 986.05)
Simply speaking, the door shape is drawn with arc and straight line, arc starts from left bottom and ends right top, then straight line starts from that right top and ends right bottom.
So if using cv2, should call cv2.ellipse, and set center point from process 4 above, axes(same value will show circle-like shape, so it's radius) from q command above, startangle and endangle(it differs from command above)
I am using three.js and created a basic scene with a few meshses rendered with some basic lighting . My torus geometry was previously pointy with few polygons but I was able to smooth it out by adding more segments but now I would like to straighten the lines out in my cube. They seem jaggy. In the pic I am including my render on the left seems a lot jaggier then the example I am following on the right. I have even set width and height segments to very large values (200). Any Ideas why the aliasing/jagginess?My render on left, example on right
Background
I've moved some code to use matplotlib only, instead of pyplot (The reason is it's generating png files in multi-process with futures, and pyplot isn't thread/process safe that way).
So I'm creating my figure and axis via matplotlib.Figure(), not via pyplot.
I've got some code that draw's a text 'table' on a figure, with one side right justified and the other left. In order to keep the spacing between the two sides constant, I was previously using get_window_extent() on my left-side text artist to figure out the location of the right hand side:
# draw left text at figure normalised coordinates, right justified
txt1 = figure.text(x, y, left_str,
ha='right', color=left_color, fontsize=fontsize)
# get the bounding box of that text - this is in display coords
bbox = txt1.get_window_extent(renderer=figure.canvas.get_renderer())
# get x location for right hand side offset from left's bbox
trans = figure.transFigure.inverted()
xr, _ = trans.transform((bbox.x1 + spacing, bbox.y0))
# draw right side text, using lhs's y value
figure.text(xr, y, right_str,
ha='left', color=right_color, fontsize=fontsize)
Problem
My problem is now that I'm not using pyplot, the above code fails to work, because figure.canvas.get_renderer() fails to return a renderer, as I haven't set one, and am simply using Figure.savefig(path) to save my file when I'm done.
So is there a way to find out the bounding box size of an Artist in a Figure without having a renderer set?
From looking at legend, which allows you to use a bounding box line with variable text size, I'm presuming there is, but I can't find how.
I've had a look at https://matplotlib.org/3.1.3/tutorials/intermediate/artists.html, and also tried matplotlib.artist.getp(txt1) on the above, but didn't find any seemingly helpful properties.
Following Quote from this source:
http://www.cambridgeincolour.com/tutorials/image-projections.htm
Equirectangular image projections map the latitude and longitude
coordinates of a spherical globe directly onto horizontal and vertical
coordinates of a grid, where this grid is roughly twice as wide as it
is tall.
I have a 13312 px width and 6656 pixel height Panorama picture. It's a equirectangular projection of a room and have a 2:1 ratio.
I use following formular to calculate the xPosition:
var xPosition = ( panorama.width / 360 ) * azimuth
Azimuth = Phi = Heading = Angle to the left or right
How do I project this now on a 1366x768px browser screen?
I think my results are wrong, because it's not on the point where it should be.. it could be because the sphere has a distortion on the left and right:
Is there any formular to calculate the position with attention to the distortion and scale it to fit on the browser screen? I looked many (MANY) sources to find a solution for this, but they always just say that equirectangular are just lat and long.. they don't consider the distortion.
Last question: To find a special solution, I tryed to put a plane on the circle and expanded the line which shows the alpha angle. I though with Phytagoras I could find the position.. but this didn't worked either.. maybe I did something wrong? Is this the way even possible or am I doing it wrong?
edit
THIS is what I'm actually looking for: http://othree.github.io/360-panorama/three-2d/
The black grid in the background. What is the name of this? For what do I have to google or look for? When you start the 2D Panorama, if you want to get the coordinations of the top right corner of the window, what do you have to do?
The whole calculation problem was about to create a Google Streetview similiar view from a 2:1 equirectangular image. We already found a solution for this with a great help from Martin Matysiak (https://github.com/marmat | Google).
It's been a while so I can't give a direct answer to what the main solution is, but I can provide a URL to an AddOn Martin wrote for adding the custom Markers that we actually were trying to make.
You can follow https://github.com/marmat/google-maps-api-addons and look for yourself. In the end it helped a lot to solve the main problem and let us continue with our main Framework for Google Business Tours.
If you follow the link in the threejs demo you included, it would take you to the source code.
particularly look at:
https://github.com/mrdoob/three.js/blob/dev/examples/webgl_panorama_equirectangular.html
and
https://github.com/mrdoob/three.js/blob/dev/src/geometries/SphereBufferGeometry.js
not sure if there is distortion though. The distortion comes from the fact that the texture is mapped to the sphere, and the sphere is rendered in 3D (openGL).
I'm using Raphael to draw rectangles. Whoo-hoo!
Is there a way to right align contents of an SVG file?
Not just text, but shapes as well?
I can do the math and get the computed x value, but I'm looking for the lazy-simple solution.
Thank you.
There isn't. Unlike normal web pages where the window is resized and the content flows into it, when a Raphael paper is resized, there is no sort of flow, so aligning is irrelevant. Instead of setting align=right, you just set the right edge to be the same position you set the width of the paper to be. If you enlarge the paper, you can scale the contents with a single operation. Once you've set the position of the right edge, you've essentially set the align position. You don't need to re-set all edge values when the paper changes size, you just scale everything with one command. Hope that helps