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Cartesian Coordinate Smoothing With C#

we are trying to develop an application with RFID readers and using speedway readers so i'm using octane sdk. when we get coordinates from each tag, we need to calculate coordinates. But sometimes we get wrong coordinates and we have to recalculate.
For this calculation, i need an algorithm which is filtering and correcting the multiple points(array of x-y). it should calculate wrong coordiantes with correct coordinates.
I've searched lots of algorithm like kalman, least squares, levenberg-marquardt but i think the problem is i don't have any model for tags and i can't fit any value to it.
I know for this case i have to apply some mathematics but i can't find.

How can you stitch multiple heightmaps together to remove seams?

I am trying to write an algorithm (in c#) that will stitch two or more unrelated heightmaps together so there is no visible seam between the maps. Basically I want to mimic the functionality found on this page :
http://www.bundysoft.com/wiki/doku.php?id=tutorials:l3dt:stitching_heightmaps
(You can just look at the pictures to get the gist of what I'm talking about)
I also want to be able to take a single heightmap and alter it so it can be tiled, in order to create an endless world (All of this is for use in Unity3d). However, if I can stitch multiple heightmaps together, I should be able to easily modify the algorithm to act on a single heightmap, so I am not worried about this part.
Any kind of guidance would be appreciated, as I have searched and searched for a solution without success. Just a simple nudge in the right direction would be greatly appreciated! I understand that many image manipulation techniques can be applied to heightmaps, but have been unable to find a image processing algorithm that produces the results I'm looking for. For instance, image stitching appears to only work for images that have overlapping fields of view, which is not the case with unrelated heightmaps.
Would utilizing a FFT low pass filter in some way work, or would that only be useful in generating a single tileable heightmap?
Because the algorithm is to be used in Unit3d, any c# code will have to be confined to .Net 3.5, as I believe that's the latest version Unity uses.
Thanks for any help!

Okay, seems I was on the right track with my previous attempts at solving this problem. My initial attemp at stitching the heightmaps together involved the following steps for each point on the heightmap:
1) Find the average between a point on the heightmap and its opposite point. The opposite point is simply the first point reflected across either the x axis (if stitching horizontal edges) or the z axis (for the vertical edges).
2) Find the new height for the point using the following formula:
newHeight = oldHeight + (average - oldHeight)*((maxDistance-distance)/maxDistance);
Where distance is the distance from the point on the heightmap to the nearest horizontal or vertical edge (depending on which edge you want to stitch). Any point with a distance less than maxDistance (which is an adjustable value that effects how much of the terrain is altered) is adjusted based on this formula.
That was the old formula, and while it produced really nice results for most of the terrain, it was creating noticeable lines in the areas between the region of altered heightmap points and the region of unaltered heightmap points. I realized almost immediately that this was occurring because the slope of the altered regions was too steep in comparison to the unaltered regions, thus creating a noticeable contrast between the two. Unfortunately, I went about solving this issue the wrong way, looking for solutions on how to blur or smooth the contrasting regions together to remove the line.
After very little success with smoothing techniques, I decided to try and reduce the slope of the altered region, in the hope that it would better blend with the slope of the unaltered region. I am happy to report that this has improved my stitching algorithm greatly, removing 99% of the lines reported above.
The main culprit from the old formula was this part:
(maxDistance-distance)/maxDistance
which was producing a value between 0 and 1 linearly based on the distance of the point to the nearest edge. As the distance between the heightmap points and the edge increased, the heightmap points would utilize less and less of the average (as defined above), and shift more and more towards their original values. This linear interpolation was the cause of the too step slope, but luckily I found a built in method in the Mathf class of Unity's API that allows for quadratic (I believe cubic) interpolation. This is the SmoothStep Method.
Using this method (I believe a similar method can be found in the Xna framework found here), the change in how much of the average is used in determining a heightmap value becomes very severe in middle distances, but that severity lessens exponentially the closer the distance gets to maxDistance, creating a less severe slope that better blends with the slope of the unaltered region. The new forumla looks something like this:
//Using Mathf - Unity only?
float weight = Mathf.SmoothStep(1f, 0f, distance/maxDistance);
//Using XNA
float weight = MathHelper.SmoothStep(1f, 0f, distance/maxDistance);
//If you can't use either of the two methods above
float input = distance/maxDistance;
float weight = 1f + (-1f)*(3f*(float)Math.Pow(input, 2f) - 2f*(float)Math.Pow(input, 3f));
//Then calculate the new height using this weight
newHeight = oldHeight + (average - oldHeight)*weight;
There may be even better interpolation methods that produce better stitching. I will certainly update this question if I find such a method, so anyone else looking to do heightmap stitching can find the information they need. Kudos to rincewound for being on the right track with linear interpolation!

What is done in the images you posted looks a lot like simple linear interpolation to me.
So basically: You take two images (Left, Right) and define a stitching region. For linear interpolation you could take the leftmost pixel of the left image (in the stitching region) and the rightmost pixel of the right image (also in the stitching region). Then you fill the space in between with interpolated values.
Take this example - I'm using a single line here to show the idea:
Left = [11,11,11,10,10,10,10]
Right= [01,01,01,01,02,02,02]
Lets say our overlap is 4 pixels wide:
Left = [11,11,11,10,10,10,10]
Right= [01,01,01,01,02,02,02]
^ ^ ^ ^ overlap/stitiching region.
The leftmost value of the left image would be 10
The rightmost value of the right image would be 1.
Now we interpolate linearly between 10 and 1 in 2 steps, our new stitching region looks as follows
stitch = [10, 07, 04, 01]
We end up with the following stitched line:
line = [11,11,11,10,07,04,01,02,02,02]
If you apply this to two complete images you should get a result similar to what you posted before.

Adding metresc distance to Long/Lat e6

Sorry if this has been asked before. I've looked at posts about Haversine and ellipsoids.
I have two points in 1e6 google maps geopoint format that define a directed vector.
I need to create an OABB (object-aligned bounding box) for the directed vector. Easily done by calculating the normals (-y,x), (y,-x). The only issue is that the length of the normalized vectors defines the width of the OABB.
Say for instance, I want these normals to be 20 km long...
So I need to scale the normals by 20km, but I have no idea how to do this in the 1e6 format that the Geopoint class uses.
If someone could post some code on how to add metre values to geopoints, I would love them.
Cheers.
Craig.

Found the answer here -> https://gis.stackexchange.com/questions/2951/algorithm-for-offsetting-a-latitude-longitude-by-some-amount-of-meters
Not before I worked out the damned thing myself though by using the radius of the earth and some simple trig. While I am a competent enough mathematician to do this, I shouldn't have to.
(rant) What are Google playing at? It's obvious that GMap developers are going to need to do vector arithmetic on Geopoint and Location, they should have built-in vector operators. (/rant)

How can we find out if a point with a certain lat and long is located on road graph?

I am writing a Windows application in C# which should find the nearest path (road) between two points in the city. I needed to access the road data, like the coordinates of the beginning and end of the roads, maybe a complete database.
I did not know how I can access such an information or file. So for simplicity I assumed some certain vertices and the edges (the connection between the vertices). Now I have a point (coordinates of a specific location). I want to see on which edge this point is located. I see by just having information about start and end of vertices and the connection between them (edges=roads) I cannot locate a position of a point on the roads. I need to know if there is any way to find out if a point with a specific coordinates is located on a specific road. I also want to know what kind of information do usually map data contain--is it just the coordinates of two ends of roads, or more info about roads?

You need to calculate the distances between the position and road, then you look at the distance and determine if it's close enough. For the distance from point a to line b you need to calculate point c on b where c to a is perpendicular to b.
To actually calculate these distances you need to do quite some math (take in the rounding of the earth) so I hope you got some skills there. Explanation on how to calculate distances between coordinates and between a line and a point.
I wrote a GPS application where the distance to a point had to be 25 meters for the current postion to be on the point. This extra space it quite important because GPS is 10 meters inaccurate by design, so if you get your data from GPS and you must be 1 meter close to your target, chances are you will never reach it, even if your standing next to it.

How to find a random point in a quadrangle?

I have to be able to set a random location for a waypoint for a flight sim. The maths challenge is straightforward:
"To find a single random location within a quadrangle, where there's an equal chance of the point being at any location."
Visually like this:
An example ABCD quadrangle is:
A:[21417.78 37105.97]
B:[38197.32 24009.74]
C:[1364.19 2455.54]
D:[1227.77 37378.81]
Thanks in advance for any help you can provide. :-)
EDIT
Thanks all for your replies. I'll be taking a look at this at the weekend and will award the accepted answer then. BTW I should have mentioned that the quadrangle can be CONVEX OR CONCAVE. Sry 'bout dat.

Split your quadrangle into two triangles and then use this excellent SO answer to quickly find a random point in one of them.
Update:
Borrowing this great link from Akusete on picking a random point in a triangle.
(from MathWorld - A Wolfram Web Resource: wolfram.com)
Given a triangle with one vertex at
the origin and the others at positions v1
and v2, pick
(from MathWorld - A Wolfram Web Resource: wolfram.com)
where A1
and A2 are uniform
variates in the interval [0,1] , which gives
points uniformly distributed in a
quadrilateral (left figure). The
points not in the triangle interior
can then either be discarded, or
transformed into the corresponding
point inside the triangle (right
figure).

I believe there are two suitable ways to solve this problem.
The first mentioned by other posters is to find the smallest bounding box that encloses the rectangle, then generate points in that box until you find a point which lies inside the rectangle.
Find Bounding box (x,y,width, height)
Pick Random Point x1,y1 with ranges [x to x+width] and [y to y+height]
while (x1 or y1 is no inside the quadrangle){
Select new x1,y1
}
Assuming your quadrangle area is Q and the bounding box is A, the probability that you would need to generate N pairs of points is 1-(Q/A)^N, which approaches 0 inverse exponentially.
I would reccommend the above approach, espesially in two dimensions. It is very fast to generate the points and test.
If you wanted a gaurentee of termination, then you can create an algorithm to only generate points within the quadrangle (easy) but you must ensure the probablity distribution of the points are even thoughout the quadrangle.
http://mathworld.wolfram.com/TrianglePointPicking.html
Gives a very good explination

The "brute force" approach is simply to loop through until you have a valid coordinate. In pseudocode:
left = min(pa.x, pb.x, pc.x, pd.x)
right = max(pa.x, pb.x, pc.x, pd.x)
bottom = min(pa.y, pb.y, pc.y, pd.y)
top = max(pa.y, pb.y, pc.y, pd.y)
do {
x = left + fmod(rand, right-left)
y = bottom + fmod(rand, top-bottom)
} while (!isin(x, y, pa, pb, pc, pd));
You can use a stock function pulled from the net for "isin". I realize that this isn't the fastest-executing thing in the world, but I think it'll work.

So, this time tackling how to figure out if a point is within the quad:
The four edges can be expressed as lines in y = mx + b form. Check if the point is above or below each of the four lines, and taken together you can figure out if it's inside or outside.

Are you allowed to just repeatedly try anywhere within the rectangle which bounds the quadrangle, until you get something within the quad? Might this even be faster than some fancy algorithm to ensure that you pick something within the quad?
Incidentally, in that problem statement, I think the use of the word "find" is confusing. You can't really find a random value that satisfies a condition; the randomizer just gives it to you. What you're trying to do is set parameters on the randomizer to give you values matching certain criteria.

I would divide your quadrangle into multiple figures, where each figure is a regular polygon with one side (or both sides) parallel to one of the axes. For eg, for the figure above, I would first find the maximum rectangle that fits inside the quadrangle, the rectangle has to be parallel to the X/Y axes. Then in the remaining area, I would fit triangles, such triangles will be adjacent to each side of the rectangle.
then it is simple to write a function:
1) get a figure at random.
2) find a random point in the figure.
If the figure chosen in #1 is a rectangle, it should be pretty easy to find a random point in it. The tricky part is to write a routine which can find a random point inside the triangle

You may randomly create points in a bound-in-box only stopping after you find one that it's inside your polygon.
So:
Find the box that contains all the points of your polygon.
Create a random point inside the bounds of the previously box found. Use random functions to generate x and y values.
Check if that point is inside the polygon (See how here or here)
If that point is inside the polygon stop, you're done, if not go to step 2

So, it depends on how you want your distribution.
If you want the points randomly sampled in your 2d view space, then Jacob's answer is great. If you want the points to be sort of like a perspective view (in your example image, more density in top right than bottom left), then you can use bilinear interpolation.
Bilinear interpolation is pretty easy. Generate two random numbers s and t in the range [0..1]. Then if your input points are p0,p1,p2,p3 the bilinear interpolation is:
bilerp(s,t) = t*(s*p3+(1-s)*p2) + (1-t)*(s*p1+(1-s)*p0)
The main difference is whether you want your distribution to be uniform in your 2d space (Jacob's method) or uniform in parameter space.

This is an interesting problem and there's probably as really interesting answer, but in case you just want it to work, let me offer you something simple.
Here's the algorithm:
Pick a random point that is within the rectangle that bounds the quadrangle.
If it is not within the quadrangle (or whatever shape), repeat.
Profit!
edit
I updated the first step to mention the bounding box, per Bart K.'s suggestion.