Is there a way how to copy/paste both coordinates in Inkscape? At this moment, if I want to say, place two circles over each other, I copy/paste x coordinate, than copy/paste y coordinate.
I was wondering wherther I can do it more efficiently, I am newbie to Inkscape
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Let me preface this by saying i know NOTHING about this kind of maths, so dont feel like you are oversimplifying. I have a directional vector firing out of an object and i need to rotate it by X amount left or right. The co-ordinates range from -1.0 to +1.0 on each axis. up until now i have been treating the maths aspect of this like a black box and have just been using premade functions to deal with it, but now i am totally lost.
Screenshot showing the format of the vector along with some lines showing what i want to do:
i am writing this in CPP and only have access to the inbuilt maths functions, no other libraries.
I have an image with various odd shapes (such as circles and squares) which are coloured pure red (rgb(255, 0, 0) exactly). I want to draw boxes around these shapes, but to do that I need the coordinates of each corner from each box. This is the part I am having difficulty with.
I basically want to go from this:
To this:
I have tried many different ways to achieve this, including parsing the y-axis until I find a shape and boxing it that way, starting from the very corners of the image and moving towards the middle (both methods of which don't work well for multiple shapes) and using external packages such as an OpenCV.
I could use OpenCV to achieve this, but given the constraints I was hoping there was a way to do it which doesn't require an external package.
Can anyone with a bit more machine vision experience point me in the right direction please?
First, use the Hoshen-Kopelman algorithm to determine the connected clusters of pixels with the given criteria (being red), then all you have to do is find their min/max regions (on x and y axes) to wrap them with a rectangle.
I'm working on a piece of code to automatically align x-axis labels for a variable number of subplots. When I started having trouble setting label positions manually, I checked to be sure I could just transform from one set of coordinates to the other without changing anything, with a code snippet like this:
# xaxes is a list of Axes objects
textCoords = [ax.xaxis.get_label().get_position() for ax in xaxes]
newCoords = [ax.transAxes.inverted().transform(ax.xaxis.get_label().\
get_transform().transform(c)) for ax,c in zip(xaxes,textCoords)]
for ax,c in zip(xaxes,newCoords): ax.xaxis.set_label_coords(*c)
In theory, this code doesn't change any coordinates; it just gets the coordinates of each label, maps it to Axes coordinates using the Text object's internally-stored transform, and then sets the position. Yet running this code removes my labels entirely, and a little experimentation shows that they go off the bottom edge of the plot.
Have I just misunderstood the transforms involved here?
You're understanding the transforms correctly, but there's a caveat to using display coordinates before the plot has been displayed.
The short answer is that putting in a call to plt.draw() before your code snippet above will fix your immediate problem.
You're trying to link the different axes display system through display coordinates. However, before the plot has been drawn the first time, the renderer isn't fully initialized yet, and the display coordinates don't have much meaning.
Can you elaborate a bit more on what you're trying to do? There may be an easier way.
Alternatively, if you want to avoid the extra draw, you can link things through figure coordinates before the plot has been drawn. (They're well defined regardless.)
I am using OpenTK(OpenGL) and a general hint will be helpful.
I have a 3d terrain. I have one point on this terrain O(x,y,z) and two perpendicular lines passing through this point that will serve as my X and Y axes.
Now I have a set of 2d points with are in polar coordinates (range,theta). I need to find which points on the terrain correspond to these points. I am not sure what is the best way to do it. I can think of two ideas:
Lets say I am drawing A(x1,y1).
Find the intersection of plane passing through O and A which is perpendicular to the XY plane. This will give me a polyline (semantics may be off). Now on this line, I find a point that is visible from O and is at a distance of the range.
Create a circle which is perpendicular to the XY plane with radius "range", find intersection points on the terrain, find which ones are visible from O and drop rest.
I understand I can find several points which satisfy the conditions, so I will do further check based on topography, but for now I need to get a smaller set which satisfy this condition.
I am new to opengl, but I get geometry pretty well. I am wondering if something like this exists in opengl since it is a standard problem with ground measuring systems.
As you say, both of the options you present will give you more than the one point you need. As I understand your problem, you need only to perform a change of bases from polar coordinates (r, angle) to cartesian coordinates (x,y).
This is fairly straight forward to do. Assuming that the two coordinate spaces share the origin O and that the angle is measured from the x-axis, then point (r_i, angle_i) maps to x_i = r_i*cos(angle_i) and y_i = r_i*sin(angle_i). If those assumptions aren't correct (i.e. if the origins aren't coincident or the angle is not measured from a radii parallel to the x-axis), then the transformation is a bit more complicated but can still be done.
If your terrain is represented as a height map, or 2D array of heights (e.g. Terrain[x][y] = z), once you have the point in cartesian coordinates (x_i,y_i) you can find the height at that point. Of course (x_i, y_i) might not be exactly one of the [x] or [y] indices of the height map.
In that case, I think you have a few options:
Choose the closest (x,y) point and take that height; or
Interpolate the height at (x_i,y_i) based on the surrounding points in the height map.
Unfortunately I am also learning OpenGL and can not provide any specific insights there, but I hope this helps solve your problem.
Reading your description I see a bit of confusion... maybe.
You have defined point O(x,y,z). Fine, this is your pole for the 3D coordinate system. Then you want to find a point defined by polar coordinates. That's fine also - it gives you 2D location. Basically all you need to do is to pinpoint the location in 3D A'(x,y,0), because we are assuming you know the elevation of the A at (r,t), which you of course do from the terrain there.
Angle (t) can be measured only from one axis. Choose which axis will be your polar north and stick to. Then you measure r you have and - voila! - you have your location. What's the point of having 2D coordinate set if you don't use it? Instead, you're adding visibility to the mix - I assume it is important, but highest terrain point on azimuth (t) NOT NECESSARILY will be in the range (r).
You have specific coordinates. Just like RonL suggest, convert to (x,y), find (z) from actual terrain and be done with it.
Unless that's not what you need. But in that case a different question is in order: what do you look for?
I have a bunch of points lying on a vertical plane. In reality this plane
should be exactly vertical. But, when I visualize the point cloud, there is a
slight inclination (nearly 2 degrees) from the verticality. At the moment, I can calculate
this inclination only. Concerning other errors, I assume there are no
shifts or something like that.
So, I want to update coordinates of my point data so that they lie on the vertical plane. I think, I should do some kind of transformation. It may be only via rotation along X-axis. Not sure what it would be.
I guess, you understood my question. Honestly, I am poor at
mathematics. So, please let me know how to update my point coordinates
to lie on the exact vertical plane.
Note: AS I am implementing this in c++ and there are many programmers who have sound knowledge on these things, I am posting this question under c++.
UPDATES
If I say exactly what I have done so far;
I have point cloud data representing a vertical object + its surroundings things. (The data is collected by a moving scanner and may have axes deviations from the correct world axes). The problem is, I cannot say exactly that there is an error on my data or not. Therefore, I checked this with a vertical planar object (which is the dominated object in my data as well). In reality that plane is truly vertical. But, when I fit a plane by removing outliers, then that plane is not truly vertical and has nearly 2 degree inclination. Therefore, I am suspecting that my data has some error. So I want to update all my point clouds (including points on the plane and points which represent other objects) in a way to lay that particular planar points exactly on the vertical plane. Then, I guess, all the points will be updated into their correct positions as in the reality. That is all (x,y,z) coordinates should be updated.
As an example please refer the below figure.
left-represents original point cloud (as you can see, points themselves are not vertical) and back line tells the vertical plane which I fitted and red is the zenith line. as you can see, there is an inclination of the vertical plane.
So, I want to update whole my data in the right figure. then, after updating if i fit a plane again (removing outliers), then it is exactly parallel to the zenith line. please help me.
I may be able to help you out, considering I worked with planes recently. First of all, how come the points aren't coplanar from the get go? I'd make the points coplanar in the first place instead of them being at an inclination (from what origin?), and then having to fix them. Also, having the points be coplanar on your first go would increase efficiency.
Sorry if this is the answer you're not looking for, but I need more information before I can help you out. Also, 3D math is hard. If you work with it enough, it starts to get pounded into your head, where you will NEVER forget it, especially if you went through the headaches I had to go through.
I did a bit of thinking on it, and since you want to rotate along the x-axis, your rotation will be done on the xz-plane, which means we can make this a 2D problem. After doing a bit of research on Wikipedia, this may be your solution.
new z = ((x - intended x) * sin(angle)) + (z * cos(angle)) + intended x
What I'm doing here is subtracting our intended x value from our current x value, so that we make (intended x, 0) our point of origin to rotate around. After the point is rotated, I add (intended x, 0) back to our coordinate so that we get the correct result.
Depending on where you got your points from (some kind of measurement, I guess) and what you want to do with them, there are several different things you could do with your data.
The search keyword "regression plane" might help - there are several ways of finding planes approximating point clouds, and several ways to "snap" points to planes.
Edit: You want to apply a rotation around the axis defined by the cross product of the normal vector on your regression plane and the normal of your desired plane, and a point your choice. From your illustration I take it that you probably want the bottom of your vertical planar object to be the point of reference for the rotation.
So you've got your point of reference, you now the axis around which you want to rotate, and the angle. All you need to do is:
Translation (to get to your point of reference)
Rotation
I read your question again, and hopefully this answer will help you out. If there's anything else I need to know, please tell me.
Now, In order to rotate anything, there must be a center point to rotate around. Now you've already been able to detect the angle of inclination, so now we need a formula for rotating a point a certain angle around an origin. In addition, since this problem only occurs on a 2D plane, we can use this basic formula to readjust the points. For any two axis x and y:
Theta is the angle that you will rotate around in a counter-clockwise direction. x' and y' are your new points. x.origin and y.origin are the coordinates for the point you will be going around. Now I don't know if my math is 100% correct on this but if it's not, hopefully you can change a thing or two and it will work.