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.
Related
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.
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.
The operation mentioned in the title is common in many Computer Aided Design (CAD) softwares such as AutoCAD, where it is called fillet. However, I found it is really difficult to implement this function in my own program.
The method I thought of is to use the condition that the distances of the arc center to the tangent lines of the curves are equal to the specified radius. Considering that actual curves are defined with piece-wise nonlinear functions, and the contact points could be anywhere on the curves, it is not easy to get the solution. Anyone good ideas?
Given that you don't describe in enough details the characteristics of the curves, it's hard to come with a specific/specified algo, but let's try a descriptive approach:
take a circle of the given radius and roll it on one curve until the circle touches the other one.
I assume you can parametrize you curves.
To "roll the circle" along the curve you need the tangent (or better said the normal, which of course is normal to the tangent) in the point "rolling track curve"-to-circle tangent point. You have this normal, you know the radius, you can compute your circle. You have the circle, you can see if/where it intersects the other curve.
The idea of "rolling" is to bracket your solution (parameter of the tangent-point on one curve) between a point when the circle does not intersect the other curve and another point where it intersects (possible in more than 1 point).
Once you have the bracket, go with a bisection method (binary search) between the two positions until your circle becomes "tangent enough" to the other curve (i.e the intersection points with the other curve are so close that they fall below your acceptable epsilon).
You will now have two points (one on each curves) and the circle that realizes the solution: just keep the arc on this circle corresponding what make sense (based on the convergence or divergence of the two tangents).
To find the arc center you need two robust and strategic algorithms:
Curve offset
Curve intersection
What AutoCAD does to find the arc center is to offset the two curves of the arc radius distance and intersect them. Depending on the curve offset direction you can easily switch between all possible solutions to the problem.
At this point, trimming the curves at tangent points will be trivial.
I have a list of coordinates (latitude, longitude) that define a polygon. Its edges are created by connecting two points with the arc that is the shortest path between those points.
My problem is to determine whether another point (let's call it U) lays in or out of the polygon. I've been searching web for hours looking for an algorithm that will be complete and won't have any flaws. Here's what I want my algorithm to support and what to accept (in terms of possible weaknesses):
The Earth may be treated as a perfect sphere (from what I've read it results in 0.3% precision loss that I'm fine with).
It must correctly handle polygons that cross International Date Line.
It must correctly handle polygons that span over the North Pole and South Pole.
I've decided to implement the following approach (as a modification of ray casting algorithm that works for 2D scenario).
I want to pick the point S (latitude, longitude) that is outside of the polygon.
For each pair of vertices that define a single edge, I want to calculate the great circle (let's call it G).
I want to calculate the great circle for pair of points S and U.
For each great circle defined in point 2, I want to calculate whether this great circle intersects with G. If so, I'll check if the intersection point lays on the edge of the polygon.
I will count how many intersections there are, and based on that (even/odd) I'll decide if point U is inside/outside of the polygon.
I know how to implement the calculations from points 2 to 5, but I don't have a clue how to pick a starting point S. It's not that obvious as on 2D plane, since I can't just pick a point that is to the left of the leftmost point.
Any ideas on how can I pick this point (S) and if my approach makes sense and is optimal?
Thanks for any input!
If your polygons are local, you can just take the plane tangent to the earth sphere at the point B, and then calculate the projection of the polygon vertices on that plane, so that the problem becomes reduced to a 2D one.
This method introduces a small error as you are approximating the spherical arcs with straight lines in the projection. If your polygons are small it would probably be insignificant, otherwise, you can add intermediate points along the arcs when doing the projection.
You should also take into account the polygons on the antipodes of B, but those could be discarded taking into account the polygons orientation, or checking the distance between B and some polygon vertex.
Finally, if you have to query too many points for that, you may like to pick some fixed projection planes (for instance, those forming an octahedron wrapping the sphere) and precalculate the projection of the polygons on then. You could even create some 2d indexing structure as a quadtree for every one in order to speed up the lookup.
The biggest issue is to define what we mean by 'inside the polygon'.
On a sphere, every polygon (as long as the lines are not intersecting) defines two regions of the sphere. Both regions are equally qualified to be called the inside of the polygon.
Consider a simple, 1-meter on a side, yellow square around the south pole.
You can think of the yellow area to be the inside of the square OR you can think of the square enclosing everything north of each line (the rest of the earth).
So, technically, any point on the sphere 'validly' inside the polygon.
The only way to disambiguate is to select which side of the polygon you want. For example, define the interior to always be the area to the right of each edge.
I know how to implement n log n closest pair of points algorithm (Shamos and Hoey) for 2D cases (x and y). However for a problem where latitude and longitude are given this approach cannot be used. The distance between two points is calculated using the haversine formula.
I would like to know if there is some way to convert these latitudes and longitudes to their respective x and y coordinates and find the closest pair of points, or if there is another technique that can be used to do it.
I would translate them to three dimensional coordinates and then use the divide and conquer approach using a plane rather than a line. This will definitely work correctly. We can be assured of this because when only examining points on the sphere, the two closest points by arc distance (distance walking over the surface) will also be the two closest by 3-d Cartesian distance. This will have running time O(nlogn).
To translate to 3-d coordinates, the easiest way is to make (0,0,0) the center of the earth and then your coordinates are (cos(lat)*cos(lon),cos(lat)*sin(lan),sin(lat)). For those purposes I'm using a scale for which the radius of the Earth is 1 in order to simplify calculations. If you want distance in some other unit, just multiply all quantities by the radius of the Earth when measured in that unit.
I should note that all this assumes that the earth is a sphere. It's not exactly one and points may actually have altitude as well, so these answers won't really be completely exact, but they will be very close to correct in almost every case.