Ok, so on the internet, I have seen equations for solving this, but they require the normal of the plane, and are a lot higher math than I know.
Basically, if I have an x,y,z position (as well as x,y,z rotations) for my ray, and x,y,z for three points that represent my plane, How would I solve for the point of collision?
I have done 2D collisions before, but I am clueless on how this would work in 3D. Also, I work in java, though I understand C# well enough.
Thanks to the answer below, I was able to find the normal of my face. This then allowed me to, through trial and error and http://geomalgorithms.com/a05-_intersect-1.html, come up with the following code (hand made vector math excluded):
Vertice Vertice1 = faces.get(f).getV1();
Vertice Vertice2 = faces.get(f).getV2();
Vertice Vertice3 = faces.get(f).getV3();
Vector v1 = vt.subtractVertices(Vertice2, Vertice1);
Vector v2 = vt.subtractVertices(Vertice3, Vertice1);
Vector normal = vt.dotProduct(v1, v2);
//formula = -(ax + by + cz + d)/n * u where a,b,c = normal(x,y,z) and where u = the vector of the ray from camX,camY,camZ,
// with a rotation of localRotX,localRotY,localRotZ
double Collision =
-(normal.x*camX + normal.y*camY + normal.z*camZ) / vt.dotProduct(normal, vt.subtractVertices(camX,camY,camZ,
camX + Math.sin(localRotY)*Math.cos(localRotX),camY + Math.cos(localRotY)*Math.cos(localRotX),camZ + Math.sin(localRotX)));
This code, mathimatically should work, but I have yet to properly test the code. Tough I will continue working on this, I consider this topic finished. Thank you.
It would be very helpful to post one of the equations that you think would work for your situation. Without more information, I can only suggest using basic linear algebra to get the normal vector for the plane from the data you have.
In R3 (a.k.a. 3d math), the cross product of two vectors will yield a vector that is perpendicular to the two vectors. A plane normal vector is a vector that is perpendicular to the plane.
You can get two vectors that lie in your plane from the three points you mentioned. Let's call them A, B, and C.
v1 = B - A
v2 = C - A
normal = v1 x v2
Stackoverflow doesn't have Mathjax formatting so that's a little ugly, but you should get the idea: construct two vectors from your three points in the plane, take the cross product of your two vectors, and then you have a normal vector. You should then be closer to adapting the equation to your needs.
Related
I have a problem. When i changing a distance between my camera and target, a distance between two objects is changing.
How i can calculate real distance between two objects?
Considering you know the coordinates of both objects, you could use Pythagorean theorem twice (as stated here):
d = sqrt(x^2 + y^2 + z^2)
This is because the way you have set the World Center Mode in Vuforia. You can use the Unity Distance function and it should work fine. If not, try making one object a child of the other, so you can check the distance using the localPosition value of the child object like this:
d = sqrt(CHILD1.transform.localposition.x^2 + CHILD1.transform.localPosition.y^2 + CHILD1.transform.localPosition.z^2);
I want to get a Vector containing a coordinate. I know my beginning coordinates, angle and distance. So far I've tried doing:
Vector2 pos = new Vector2(beginningX, beginningY).add(distance, distance).rotate(angle);
But it doesn't work as I expect it to. When the rotation isn't 0 the coordinates become big, and the ending point isn't where I expect it to be. I know this must be a simple problem, but I just can't solve it.
EDIT:
Tried doing:
Vector2 pos = new Vector2(beginningX, beginningY).add(distance, 0).rotate(angle);
(Adding distance to x only) Still no success.
I'd say you're doing it wrong: you need to rotate the distance vector and add it to the position vector:
Vector2 pos = new Vector2(beginningX, beginningY).add(new Vector2(distance, 0).rotate(angle) );
You might want to read up on vector math but basically it amounts to this (if I correctly understood what you're trying to do):
If you rotate a vector you're always rotating around point 0/0. Thus you'll want to create a vector that covers the distance from 0/0 to your distance on the x-axis:
0---------------->d
Now you rotate that vector by some angle:
d
/
/
/
/
/
0
Then you offset that vector by your starting point, i.e. you add the two vectors (for simplicity I assume your starting point lies on the y-axis):
d
/
/
/
/
/
s
|
|
|
0
You need to rotate only the distance vector, rather than a sum of beginning and distance. Addition is the same in either order (commutative), so you can try this way:
Vector2 pos = new Vector2(distance, 0).rotate(angle).add(beginningX, beginningY);
Advantage: This chained call does not create a temporary Vector2 for the beginning position that would immediately become garbage for the garbage collector. Conserving space and garbage collection time will be important when your code handles millions of vectors.
This is simple vector addition. I'm assuming 2D coordinates, with angles measured counterclockwise from x-axis:
x(new) = x(old) + distance*cos(angle)
y(new) = y(old) + distance*sin(angle)
Be sure that your angles are in radians when you plug them into trig functions.
I'm using the non linear least squares Levenburg Marquardt algorithm in java to fit a number of exponential curves (A+Bexp(Cx)). Although the data is quite clean and has a good approximation to the model the algorithm is not able to model the majority of them even with a excessive number of iterations(5000-6000). For the curves it can model, it does so in about 150 iterations.
LeastSquaresProblem problem = new LeastSquaresBuilder()
.start(start).model(jac).target(dTarget)
.lazyEvaluation(false).maxEvaluations(5000)
.maxIterations(6000).build();
LevenbergMarquardtOptimizer optimizer = new LevenbergMarquardtOptimizer();
LeastSquaresOptimizer.Optimum optimum = optimizer.optimize(problem);}
My question is how would I define a convergence criteria in apache commons in order to stop it hitting a max number of iterations?
I don't believe Java is your problem. Let's address the mathematics.
This problem is easier to solve if you change your function.
Your assumed equation is:
y = A + B*exp(C*x)
It'd be easier if you could do this:
y-A = B*exp(C*x)
Now A is just a constant that can be zero or whatever value you need to shift the curve up or down. Let's call that variable z:
z = B*exp(C*x)
Taking the natural log of both sides:
ln(z) = ln(B*exp(C*x))
We can simplify that right hand side to get the final result:
ln(z) = ln(B) + C*x
Transform your (x, y) data to (x, z) and you can use least squares fitting of a straight line where C is the slope in (x, z) space and ln(B) is the intercept. Lots of software available to do that.
I am new to Mahout, and have been lately transforming a lot of my previous machine learning code to this framework. In many places, I am using cosine similarity between vectors for clustering, classification, etc. Investigating Mahout's distance method, however, gave me quite a surprise. In the following code snippet, the dimension and the float values are taken from an actual output of one of my programs (not that it matters here):
import org.apache.mahout.math.RandomAccessSparseVector;
import org.apache.mahout.common.distance.CosineDistanceMeasure;
public static void main(String[] args) {
RandomAccessSparseVector u = new RandomAccessSparseVector(373);
RandomAccessSparseVector v = new RandomAccessSparseVector(373);
u.set(24, 0.4526985183337534);
u.set(55, 0.5333219834564495);
u.set(54, 0.5333219834564495);
u.set(53, 0.4756042214095471);
v.set(57, 0.6653016370845252);
v.set(56, 0.6653016370845252);
v.set(11, 0.3387439495921685);
CosineDistanceMeasure cosineDistanceMeasure = new CosineDistanceMeasure();
System.out.println(cosineDistanceMeasure.distance(u, v));
}
The output is 1.0. Shouldn't it be 0.0?
Comaring this with the output of cosineDistanceMeasure.distance(u, u), I realize that what I am looking for is 1 - cosineDistanceMeasure.distance(u, v). But this reversal just doesn't make sense to me. Any idea why it was implemented this way? Or am I missing something very obvious?
When two points are "close", the angle they form when viewed as vectors from the origin is small, near zero. The cosine of angles near zero is near 1, and the cosine decreases as the angle goes towards 90 and then 180 degrees.
So cosine decreases as distance increases. This is why the cosine of the angle between two vectors itself can't make sense as a distance metric. The 'canonical' way to make a distance metric is 1 - cosine; it's a proper metric.
How to find if a point exists in which given set of polygons ?
I have coordinates like
polygonA = 1(0,0),2(0,5),3(3,4),4(3,5),5( 2,2)
polygonB = 1(10,10),2(10,15),3(13,14),4(13,15),5(12,12)
I have a point as (6,4) now want to search if this point is in any of this polygon or in both or nearest to which polygon.
How to store such data (polygon) ? is there a system / database / algorithm to do this search ?
Update : Thanks all for such fast response...I think i need to be more specific...
How to search = Yes...got list of algorithms and library for the same.
How to store = based on my research SQL and NoSQL db have their solutions.
NoSQL = MongoDb seems closest what i needed. But issue is I can query like "db.places.find({ "loc" : { "$within" : { "$polygon" : polygonB } } })" But cant make query like db.places.find({ "loc" : { "$within" : { } } })
SQL checked postgre and openGIS for some help. But colud not figureout if its possible.
If someone can help me with that...Thanks in advance.
The basic method (if you have a small number of polygons) is to store all polygons in a collection and loop over the elements to check if a point is inside a polygon.
On the other hand, if you have a considerable number of polygons, I would recommend using an R-tree data structure, which is not available in the standard library. You should check this project, if you want to go with R-tree option: http://sourceforge.net/projects/jsi/.
R-tree allows you to index rectangles (bounding boxes of the polygons in this case). So you can find a small number of candidate polygons very fast using R-tree. Then you can loop over the candidate list to get the final result.
You can use the GeneralPath class to help you with deciding if a point intersects a polygon. First, create a GeneralPath with your coordinates added:
GeneralPath gp = new GeneralPath();
double[] x = ...
double[] y = ...
gp.moveTo(x[0], y[0]);
for (int i =1; i < x.length; i++) {
gp.lineTo(x[i], y[i]);
}
gp.closePath();
if (gp.contains(pointX, pointY)) {
...
}
For the problem of which polygon a point is nearer to, this depends a little on how accurately you need a solution.
For an accurate solution., this amounts (without optimisation) to:
take the shortest distance between the point and each of the lines (segments) connecting the vertices of each of the polygons (Java2D apparently doesn't provide a method for this, but the shortest distance from a point to a line is a fairly simple calculation)
which polygon has the line with the shortest distance to the point?
In practice, you can approximate this process for some applications. For example, you could much more efficiently do this:
take the centre point of the bounding rectangle of each polygon (GeneralPath.getBounds() will give you this)
take the distance between the query point and each of these centre points, and see which is closest.
If you do need an accurate answer, then you can combine these techniques to optimise your search among all the vertices. For example, you could order the polygons by the distance to their "centrepoint" (defined as above). Search from minimum to maximum distance. If the minimum distance to a segment that you have found so far is d, then you can automatically rule out any polygon P where the distance from your query point to its "centrepoint" is d + r, where r is half the length of the diagonal of P's bounding rectangle (in other words, for simplicity, you imagine a bounding circle around that bounding box and check that the distance to that bounding circle is further than the nearest point found so far on other polygons).
I don't quite understand the bit about the database. Your polygons are just defined as a series of points. How you decide to store these in memory/file doesn't essentially make any difference to the algorithm.