Apologies in advance for my feeble maths.
I'm trying to be able to find the corners of a plane in space based on the equation of that plane. Here's what I know. I know three points on the plane and I know where they fall in the 2d coordinate space of the plane (x,y) and where they are in 3d space. I know the width and height of the plane and I can now calculate the equation of the plane. The plane sits on the inside of a large sphere that surrounds the origin so, in theory, it should more or less face where the camera is (though in my diagram it doesn't face the origin as it's just for illustrative purposes)
But it's not clear to me how I can use that to figure out another point. One thought I had was to find the transform that moves the plane parallel to the xy axis and rotate it round one of the points (so it stays in the same place), find the position of the new point, and then rotate it by the inverse of that transform. But it's not clear to me how I would find that transform matrix or how to use it. Could I do this using the normal and vector maths? I understand what normals are, but I'm fuzzy about how to use them.
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I currently have two images of a plane in real life from straight above. One to use as a reference image, and another when the plane has undergone a rotation fixed at the centre of the plane thus changing its orientation. The camera stays at a constant position.
I was wondering if I found the homography matrix of this rotation in opencv and then decomposed the homography matrix in order to find the rotation matrix whether this would yield accurate results and I would be able to find the three angles needed to describe the planes rotation in euclidean coordinates to a reasonable degree of accuracy.
Thanks
I'd like to add "flying" (3D) arcs to my orthographic projection, as shown here, but with clipping instead of the fade effect. This seems difficult since the arcs are created independently of the projection. (Each arc is defined by three points obtained from the projection--the start, end, and great circle midpoint extended along a line from the center of the canvas--but the arc itself is drawn using "2D" cardinal interpolation on the corresponding points on the svg canvas.)
My first thought was that I might need to do some spherical geometry to get the coordinates where the clipping happens, but now I'm wondering if there's a more straightforward way to accomplish this (I'm new to D3).
This is what my map looks like without clipping:
I'm also very green to d3, but fortunately I'm also fresh from my own search for a decent solution for clipping flight lines in orthographic. The demo you link to is clever in more ways than one:
The arc is drawn from three points interpolated with a Catmull-Rom curve in the projected 2D coordinates that happens to visually approximate a true circular curve in 3D nicely
The line fades with proximity of either of its points to the clipping plane, as you've pointed out
Drawing the spline in the projected, 2D coordinates eliminates any option to split the line before projection and get visual smoothness for cheap, even if d3 had the functionality natively (which I haven't been able to find anyways). That means that interpolation will have to be a lot more manual.
My first thought was that I might need to do some spherical geometry to get the coordinates where the clipping happens, but now I'm wondering if there's a more straightforward way to accomplish this
I eventually settled with what I consider the most obvious option, which unfortunately you're aiming precisely to avoid:
Obtain the coordinates of the clipping point by the cross product of the current globe center with the normal to the great circle plane of the arc. Given your origin and destination Po and Pd respectively and the globe center C, you're looking for C x (Po x Pd) normalized
Interpolate coordinates between your origin and destination using something like d3.geoInterpolate
Project interpolated point at the right scale (read: elevation) above the ground for that fraction of the flight line
Draw one [smoothed] line from the origin to the clipping point along the interpolated points in between, and another from the clipping point to the destination, moving one accordingly to the background. Watch the cases where the whole line is in front or behind the clipping plane.
To figure out where in the flight path you need to splice your clipped point, you will probably need to compare the great angles of your clipping point to one end vs. end-to-end. Note also that performance takes a hit, but you may still be satisfied with the number of flight lines you've drawn in your example.
It seems that there are lots of information both in Google and here, that speak a lot about many different conversions of latitude,longitude.
So what i'm asking is for you to be simple as possible, and try not sending me to other places to seek an answer.
I am trying to put the entire world in to 2D square, where each point represent the distance(in meters) from a point which I choose to define it (0,0),
Can you give me a mathematical algorithm to do so.
You could either use a azimuthal equidistant or two-point equidistant projection.
Of these the azimuthal equidistant is easiest. To do this, just start at your reference point on the world, and put this in the center of your map. Then proceed outward in concentric circles on the map, and for each new circle plot all the points on the world at the appropriate distance and angle.
After doing this, your map should look like a circle, and all of the points will be the correct distance from your center point.
So we have such situation:
In this illustration, the first quadrilateral is shown on the Image Plane and the second quadrilateral is shown on the World Plane. [1]
In my particular case the Image Plane has 3 quadrilaterals - projections of real world squares, which, as we know, have same size, lying on the same plane, with same rotation relative to the plane they are lying on, and are not situated on same line on plane.
I wonder if we can get rotation angles of Image Plane to World Plane knowing stuff described?
In my case as input I have such data structures: original image (RGB pixels), objects (squares) with angles points in pixels (x,y) on Image Plane.
Take a look at Sections 2 and 3 of Algorithms for plane-based pose estimation.
The methods described there assume that you know the (x,y) coordinates of the features in question - in this case the red squares.
The problem you are describing is generally known as pose estimation - determining the 3D orientation and position of an object relative to a camera from a 2D view. For you, the object is a plane. Googling 'pose estimation plane' should give you more sources.
As a followup to my previous question about determining camera parameters I have formulated a new problem.
I have two pictures of the same rectangle:
The first is an image without any transformations and shows the rectangle as it is.
The second image shows the rectangle after some 3d transformation (XYZ-rotation, scaling, XY-translation) is applied. This has caused the rectangle to look a trapezoid.
I hope the following picture describes my problem:
alt text http://wilco.menge.nl/application.data/cms/upload/transformation%20matrix.png
How do determine what transformations (more specifically: what transformation matrix) have caused this tranformation?
I know the pixel locations of the corners in both images, hence i also know the distances between the corners.
I'm confused. Is this a 2d or a 3d problem?
The way I understand it, you have a flat rectangle embedded in 3d space, and you're looking at two 2d "pictures" of it - one of the original version and one based on the transformed version. Is this correct?
If this is correct, then there is not enough information to solve the problem. For example, suppose the two pictures look exactly the same. This could be because the translation is the identity, or it could be because the translation moves the rectangle twice as far away from the camera and doubles its size (thus making it look exactly the same).
This is a math problem, not programming ..
you need to define a set of equations (your transformation matrix, my guess is 3 equations) and then solve it for the 4 transformations of the corner-points.
I've only ever described this using German words ... so the above will sound strange ..
Based on the information you have, this is not that easy. I will give you some ideas to play with, however. If you had the 3D coordinates of the corners, you'd have an easier time. Here's the basic idea.
Move a corner to the origin. Thereafter, rotations will take place about the origin.
Determine vectors of the axes. Do this by subtracting the adjacent corners from the origin point. These will be a local x and y axis for your world.
Determine angles using the vectors. You can use the dot and cross products to determine the angle between the local x axis and the global x axis (1, 0, 0).
Rotate by the angle in step 3. This will give you a new x axis which should match the global x axis and a new local y axis. You can then determine another rotation about the x axis which will bring the y axis into alignment with the global y axis.
Without the z coordinates, you can see that this will be difficult, but this is the general process. I hope this helps.
The solution will not be unique, as Alex319 points out.
If the second image is really a trapezoid as you say, then this won't be too hard. It is a trapezoid (not a parallelogram) because of perspective, so it must be an isosceles trapezoid.
Draw the two diagonals. They intersect at the center of the rectangle, so that takes care of the translation.
Rotate the trapezoid until its parallel sides are parallel to two sides of the original rectangle. (Which two? It doesn't matter.)
Draw a third parallel through the center. Scale this to the sides of the rectangle you chose.
Now for the rotation out of the plane. Measure the distance from the center to one of the parallel sides and use the law of sines.
If it's not a trapezoid, just a quadralateral, then it'll be harder, you'll have to use the angles between the diagonals to find the axis of rotation.