Given a non-self-intersecting polygon as a list of points (p1...pn), and a point (A) outside that polygon:
I want to calculate the total angular field of view covered by the polygon from the point, as well as the direction from the point towards the middle of that field of view (as either a vector or angle from horizontal).
Visually, I want the angle Θ and direction of the green vector:
Diagram
I tried finding the minimum and maximum angles from horizontal to each of the polygon points, but I don't know how to tell which is the start of the range and which is the end. Assuming the smaller angle is the start gives incorrect results from the left of a simple box.
I'm guessing the solution will have something to do with whether the polygon points are in clockwise or counterclockwise order.
Whether the polygon goes clockwise or counterclockwise doesn't matter. What matters is that the extent of any edge, as seen from the point, must be less than π radians. That will tell us whether the edge -- as seen from the point -- goes counterclockwise from A to B, or from B to A.
For example, suppose the bearings (in radians) from the point to the vertices are {0, 2π/5, 4π/5, 6π/5, 8π/5}. If the edges are represented by the (unordered) pairs (A,C), (A,E), (B,D), (B,E), (C,D). Then the edges run:
0->4π/5
2π/5->6π/5
4π/5->6π/5
8π/5->0
8π/5->2π/5
So the range of the polygon is [8π/5, 6π/5].
I'd need a formula to calculate 3D position and direction or orientation of a camera in a following situation:
Camera starting position is looking directly into center of the Earth. Green line goes straight up to the sky
Position that camera needs to move to is looking like this
Starting position probably shouldn't matter, but the question is:
How to calculate camera position and direction given 3D coordinates of any point on the globe. In the camera final position, the distance from Earth is always fixed. From desired camera point of view, the chosen point should appear at the rightmost point of a globe.
I think what you want for camera position is a point on the intersection of a plane parallel to the tangent plane at the location, but somewhat further from the Center, and a sphere representing the fixed distance the camera should be from the center. The intersection will be a circle, so there are infinitely many camera positions that work.
Camera direction will be 1/2 determined by the location and 1/2 determined by how much earth you want in the picture.
Suppose (0,0,0) is the center of the earth, Re is the radius of the earth, and (a,b,c) is the location on the earth you want to look at. If it's in terms of latitude and longitude you should convert to Cartesian coordinates which is straightforward. Your camera should be on a plane perpendicular to the vector (a,b,c) and at a height kRe above the earth where k>1 is some number you can adjust. The equation for the plane is then ax+by+cz=d where d = kRe^2. Note that the plane passes through the point (ka,kb,kc) in space, which is what we wanted.
Since you want the camera to be at a certain height above the earth, say h*Re where 1 < k < h, you need to find points on ax+by+cz=d for which x^2+y^2+z^2 = h^2*Re^2. So we need the intersection of the plane and a sphere. It will be easier to manage if we have a coordinate system on the plane, which we get from an orthonormal system which includes (a,b,c). A good candidate for the second vector in the orthonormal system is the projection of the z-axis (polar axis, I assume). Projecting (0,0,1) onto (a,b,c),
proj_(a,b,c)(0,0,1) = (a,b,c).(0,0,1)/|(a,b,c)|^2 (a,b,c)
= c/Re^2 (a,b,c)
Then the "horizontal component" of (0,0,1) is
u = proj_Plane(0,0,1) = (0,0,1) - c/Re^2 (a,b,c)
= (-ac/Re^2,-bc/Re^2,1-c^2/Re^2)
You can normalize the vector to length 1 if you wish but there's no need. However, you do need to calculate and store the square of the length of the vector, namely
|u|^2 = ((ac)^2 + (bc)^2 + (Re^2-c^2))/Re^4
We could complicate this further by taking the cross product of (0,0,1) and the previous vector to get the third vector in the orthonormal system, then obtain a parametric equation for the intersection of the plane and sphere on which the camera lies, but if you just want the simplest answer we won't do that.
Now we need to solve for t such that
|(ka,kb,kc)+t(-ac/Re^2,-bc/Re^2,1-c^2/Re^2)|^2 = h^2 Re^2
|(ka,kb,kc)|^2 + 2t (a,b,c).u + t^2 |u|^2 = h^2 Re^2
Since (a,b,c) and u are perpendicular, the middle term drops out, and you have
t^2 = (h^2 Re^2 - k^2 Re^2)/|u|^2.
Substituting that value of t into
(ka,kb,kc)+t(-ac/Re^2,-bc/Re^2,1-c^2/Re^2)
gives the position of the camera in space.
As for direction, you'll have to experiment with that a bit. Some vector that looks like
(a,b,c) + s(-ac/Re^2,-bc/Re^2,1-c^2/Re^2)
should work. It's hard to say a priori because it depends on the camera magnification, width of the view screen, etc. I'm not sure offhand whether you'll need positive or negative values for s. You may also need to rotate the camera viewport, possibly by 90 degrees, I'm not sure.
If this doesn't work out, it's possible I made an error. Let me know how it works out and I'll check.
I have 2 points and both are 3d points i.e A(x,y,z) and B(x,y,z).
I know the coordinates of point a and the distance between them.
How to find the coordinates of point B?
Thanks a lot
If you only know the absolute distance between the two points, you cannot find the coordinates of point B. If you think about it, knowing one point and a distance defines a sphere.
However, if you know the difference between the coordinates in the x, y, and z axes, its as simple as adding the difference to point A to obtain point B.
I have three vectors in 3D space, one is a light source, one is a ray and one is the point on a circle a ray hits. With this information, how can I work out the vector which points back at the light source from the point the ray hits the circle?
What you really have is two points (light source, circle intersection), and a vector between them, right? The vector is already implied by the two points -- it's the intersection coordinates minus light source coordinations. To reverse it, just negate all the coordinates of the vector!
so i got three variables, my location, my target location and the compass heading.
how can i calculate where the target location should be represented on a virtual radar?
i guess i first must calculate the distance between the two gps points and the angle of them relative to north or so. and then there should be a formula with sin or cos to place that point on a coordinate system...?
ps: in javascript...
Start with simpler problems.
In 2D, try converting back and forth between cartesian and polar coordinates. References are available.
Do the same, but for the polar coordinates use an observer who measures angles from some ray that is not in the X direction.
The same, but using an origin for the polar coordinates that is not at {x=0,y=0}.
In 3D, go back and forth between cartesian and spherical coordinates.
Again, with spherical coordinates in an arbitrary orientation, using an arbitrary origin.
Now convert from GPS coordinates (which are spherical) to cartesian, then to radar-centered spherical.