I need to create flight routes in Google Earth. Example from point A to point B, How do i get the equivalent middle point for both and along point A to B, there are also many different coordinates joining so that the line would be a curve.
Generally you should use the great-circle distance to compute the distance of two point on the surface of a sphere, like in this case the Earth. Even WolframAlpha uses this to compute direct travel times.
This would also define the midpoint for you uniquely.
Use e.g., Perl's Math::Trig module. It comes complete with the great circle functions you need.
Related
In part of an Artificial Neural Network matlab code, I want to find nearest points of two convex polygons.
I saw
dsearchn(X,T,XI)
command's description here, but that finds closest points between two sets of points, and polygons (like convexs) have infinites points.
So can you suggest any way/idea?
Notice: I'm using MATLAB 2014a. I have the coordinate of each convex's vertex point.
If you are not happy with what is provided by dsearchn, then, If I were you, I would do one of two following:
Find Nearest Neighbours on the vertices (for example which vertex of
polygon A is the NN of a given vertex of polygon B).
Pick a random point inside polygon A (you may want to compute the
convex hull of A, but you may skip that and take into account only
the vertices you know already). That random point is the query. Find
an NN of that point from the vertices of polygon B.
You may want to ask in Software recommendations for more.
Edit:
Another approach is this:
Create a representative dataset of polygon A. Set the size of the dataset yourself and fill it with samples of points that lie inside the polygon. Choose them uniformly randomly inside the polygon.
Then take a point of polygon B (either a vertex or a random point inside polygon B) and that's the query point, for which you will seek Nearest Neighbour(s) inside the representative dataset of polygon A.
Of course that's just an approximation, but I can't think of something else now.
Notice that you can of course do the same for polygon B.
With This File in File Exchange, I've find the solution.
With a little code modification, I have drawn a perpendicular bisector which I wanted. Of course, this way is time consuming.
There are a lot of similar questions but I can't get a clear answer out of them. So, I want to represent latitude and longitude in a 2D space such that I can calculate the distances if necessary.
There is the equirectangular approach which can calculate the distances but this is not exactly what I want.
There is the UTM but it seems there are many zones and letters. So the distance should take into consideration the changing of zone which is not trivial.
I want to have a representation such that i can deal with x,y as numbers in Euclidean space and perform the standard distance formula on them without multiplying with the diameter of Earth every time I need to calculate the distance between two points.
Is there anything in Matlab that can change lat/long to x,y in Euclidean space?
I am not a matlab speciallist but the answer is not limited to matlab. Generally in GIS when you want to perform calculations in Euclidean space you have to apply 'projection' to the data. There are various types of projections, one of the most popular being Transverse Mercator
The common feature of such projections is the fact you can't precisely represent whole world with it. I mean the projection is based on chosen meridian and is precise enough up to some distance from it (e.g. Gauss Krueger projection is quite accurate around +-500km from the meridian.
You will always have to choose some kind of 'zone' or 'meridian', regardless of what projection you choose, because it is impossible to transform a sphere into plane without any deformations (be it distance, angle or area).
So if you are working on a set of data located around some geographical area you can simply transform (project) the data and treat it as normal Enclidean 2d space.
But if you think of processing data located around the whole world you will have to properly cluster and project it using proper zone.
I need to evaluate the proximity of a Point to a LineString using MongoDB.
Because the $near operator can only compare a Point to another Point, I need to generate a polygon out of the LineString, so I can use the $within operator. The distance between the LineString and the edges of the polygon should represent the radius I want to search in, such as represented in red below:
What might be a useful algorithm in order to accomplish this?
I think much easier would be to write your own function
To find (perpendicular) distance between point and line and then creating thickness of poly-line by polygon means.
Where:
P0,P1 are line endpoints
P is point
d is distance between them
Line is defined as: p(t)=P0+(P1-P0)*t where t=<0.0,1.0>
So the function should do this:
create perpendicular line
q(t)=P+DQ*u where u=(-inf,+inf)
DQ is perpendicular vector to (P1-P0)
In 2D you can obtain it easily like this (x,y) -> (y,-x). In higher dimensions use cross product with some non coplanar vectors.
compute line vs. line intersection
there are tons of stuff about this so google or solve the equation yourself here you can extract mine implementation.
now after successful intersection
just compute d as distance between P and intersection point. Do not forget that parameter t must be in range. If not (or if no intersection) then return min(|P-P0|,|P-P1|)
[hints]
t and u parameters can be obtained directly from intersection test so if the perpendicular vector to (P1-P0) is normalized to size = 1 then the abs(u) parameter of intersection point is the distance
[notes]
I am not familiar with mongodb so if you have no means to use own tests inside then this answer is of coarse obsolete.
Unfortunately, MongoDB provides very basic geospatial query, so you should create the buffer by your own. You can read how to do it here: Computing a polygon that surrounds a multi-point line
If you have longitude/latitude coordinates like WGS84 you must adjust this code; Read here how to calculate distance between point on a sphere https://en.wikipedia.org/wiki/Haversine_formula
I am developing an app that animates a motion on a UIBezierPath (made of several curves).
In some use cases I need to place an item so it will start moving from some point on the route, and not from its beginning. E.g put item in the middle or 2/3 point of the path. How can I calculate the location of such point?
Thanks!
A Bezier curve is parametric curve a http://en.wikipedia.org/wiki/B%C3%A9zier_curve meaning you have two functions of a parameter T over a range. One function generates the X coordinate, the other generates the Y coordinate. If you know the two functions, just pick a value of T halfway or 2/3rds of the way between the endpoints of the range and plug that into the two functions to get the X & Y coordinates of the desired point.
I have an array of CGPoints (basic struct with two floats: x and y). I want to use OpenGL ES to draw a textured curve using these points. I can do this fine with just two points, but it gets harder when I need to make a line from several points.
Currently I draw a line horizontally, calculate its angle from the points given, and then rotate it. I don't think doing this for all lines in a curve is a good idea. There's probably a faster way.
I'm thinking that I can "enlarge" or "constrict" all the points at once to make a curve with some sort of width.
I'm not positive what you want to accomplish, but consider this:
Based on a ordered list of points, you can draw a polyline using those points. If you want to have a polyline with a 2D texture on it, you can draw a series of quadrilaterals (using two triangles each, of course). You can generate these quadrilaterals using an idea similar to catmul-rom spline generation.
Consider a series of points p[i-1], p[i], p[i+1]. Now, for each i, you can find two points each an epsilon distance away from p[i] along the line perpendicular to the line connecting p[i-1] and p[i+1]. You can determine the two points generated for the endpoints in various ways, like using the perpendicular to the line from p[0] to p[1].
I'm not sure if this will be faster than your method, but you should be caching the results. If you are planning on doing this every frame, another type of solution to your problem may be needed.