I am looking for some assistance on creating an algorithm that is able to calculate a bounding box for an arc. Typically, this would be a simple problem. However, I am required to have one of the edges of the box to be the line that connects the starting point and the ending point of the arc.
I am having difficulty in developing an algorithm that is able to compute the additional two points of the rectangle.
Here is a picture of what I am attempting to calculate:
The two purple dots, I need to develop an algorithm that will determine these two locations. The green dots are the known points that can be inputs. Additionally, I do know the radius and the center of the arc.
I would need the algorithm to handle the different cases for when line AB is vertical, has a + slope, and has a - slope.
I have been considering a few directions. For example, I know that the line through point E is parallel to line AB. Since it is parallel, that means they will have the same slopes and the line from point A to the purple point is perpendicular to this line. I can then consider the intersection point of the line through E and this perpendicular line.
This method seems messy because then I would need to consider the cases for when the slope of line AB is infinite and 0. I am wondering if there is some algorithm that could account for that automatically (or not even consider the slope at all for line AB)
You claim to know points A, B, C, D, E, and that the amplitude of the sustaining angle of the circular arc does not exceed 180° (semi circle).
Let P1, and P2, the two points that complement the bounding box - (in the drawing, P1 is the purple point above A, and P2 the one above B).
Using vectors:
E-C = vector perpendicular to segment AB', i/e subtract C from E; its magnitude is equal to the distance EC.
P1 = A + (E-C)
P2 = B + (E-C)
Bounding Box = Rectangle A, P1, P2, B
If you need a closer fit, you can replace vector (E-C) with vector (D-C) to place the bounding segment P1P2 tangent to D
Related
i am programming a C++ collision detection in my game and am trying to come up with an algorithm:
I have a capsule defined with two center points (C1, C2), length and a radius. Then i have a ray defined with two points (R1, R2). I already know that they are intersecting. I just need to find inner part of ray that is contained in capsule (H1-H2). Thanks in advance for all the help.
First let's take a look at a diagram for reference:
The procedure for calculating H1 and H2 is as follows:
Compute the intersection, if any, between the ray R and the line
segment P1P2. We're only interested in intersections that lie
within the interior of P1P2. Similarly for P3P4. The points P1 to P4 can be calculated easily from the circle centers, C1 and C2, and some vector math. E.g. P1 = C1 + r*nC, where nC is the normal (CCW) of the unit vector from C1 to C2. This
answer on SO provides the necessary math to determine if an
intersection exists between two line segments and, if so, calculate the parameter h, such
that H=R1+h(R2-R1), where H is an intersection point. This step can produce 0, 1, or 2 valid h values, depending on whether the ray intersects neither, one, or both of P1P2, P3P4.
Compute the intersection points, if any, between the ray and each of the 2
circles. Again, an SO answer provides the necessary math for
ray to circle intersection. Each circle can produce 0, 1, or 2
intersections, again represented parametrically.
If no valid h values were generated from steps 1 & 2 then the ray does not intersect the capsule. Otherwise, calculate hMin and hMax, the min and max parameter values of all valid
intersections identified in steps 1 & 2. Note that it's possible that hMin==hMax, in the case where the ray is tangent to one of the circles and does not intersect P1P2 or P3P4. The required intersection point(s) can now be calculated as H1=R1+hMin(R2-R1) and H2=R1+hMax(R2-R1).
I'm afraid my language of choice is Java rather than C++, but hopefully you'll find the code (IDEOne) I put together helpful as a reference. Please be aware that no effort was put it to handle robustness issues resulting from the rounding of double values during calculations.
I have the vertex data of a 2D polygonal region that contains the position of each vertex. My task is to calculate the angles throughout the polygon and replace the vertices with small angles (let's say smaller than 120 degrees) with bigger ones. For example, if three points (ABC) make an angle of 90 degrees, I want to replace the middle point (B) with another one that makes 120 degrees with the other two points.
I want to know if I can write an algorithm to find a new position of the vertex B, given the x-y coordinates of A and C and the 120 degrees angle between ABC (I calculated the current angle so you don't need to write how to do that). Note that, I am not looking for a unique point (there is no unique point), just looking for an angle calculation to replace the point. If you know other methods to do this kind of replacements, they are welcome too.
You have a triangle ABC with B the "small" angle that you want to replace with a larger angle. Call the new location B'. Here is the construction. Let D be the midpoint of AC, i.e., D = (A+B)/2. You want to place B' on the segment BD.
Any location along BD will result in a larger angle at B', with an angle of 180 when B'=D.
Pick a parameter t in [0,1]. Then place B' = (1-t) B + t D.
Now, there is a flaw here (besides that pointed out by spug). It may be that moving B to B' causes AB'C to hit other
parts of the polygon, and so it is no longer simple. You'll have to decide whether you want to compute a t that is guaranteed not to cause self-intersection.
I have a triangle ABC in 3D space and a ray in plane of the triangle given by starting point E (always inside or on the triangle's edge) and direction vector d.
The vertices A, B and C as well as E and d are given in 3D coordinates {x,y,z}.
I am looking for an algorithm to calculate the intersection of the ray with the triangle's edge P.
I can do 3 ray / line segment intersection tests for the 3 edges of the triangle but since I have to do this for a large number of triangles I am looking for a more efficient and faster algorithm.
What is the best way to do this?
Could Barycentric Coordinates help?
Could Barycentric Coordinates help?
It may. It depends a bit on how highly optimized your non-barycentric code is, but I'd say that using barycentric coordinates it's at least easier to write code which is both maintainable and performant.
As far as I can tell, your whole setup is essentially 2d, with both the point E and the direction d contained within the plane spanned by A,B,C. So you have
E = aE*A + bE*B + cE*C with aE+bE+cE=1
d = ad*A + bd*B + cd*C with ad+bd+cd=0
Now you have two subproblems:
How to obtain the barycentric coordinates efficiently
How to find the point of intersection
Let's start with the latter. You simply add to E as many multiples of d until the c coordinate becomes zero.
P = E - (cE/cd)*d
Depending on your setup, you might also be fine using homogeneous coordinates, in which case you could write this as P = cd*E - cE*d.
How do you turn x,y,z coordinates of d and E into barycentric a,b,c? Well, that's just a system of linear equations. You could use the inverse of the matrix formed by the vectors A,B,C. Again, if you are dealing with homogeneous coordinates, you can use the adjunct instead of the inverse.
Here is the homogeneous approach spelled out:
aE = (By*Cz-Bz*Cy)*Ex + (Bz*Cx-Bx*Cz)*Ey + (Bx*Cy-By*Cx)*Ez
bE = (Cy*Az-Cz*Ay)*Ex + (Cz*Ax-Cx*Az)*Ey + (Cx*Ay-Cy*Ax)*Ez
cE = (Ay*Bz-Az*By)*Ex + (Az*Bx-Ax*Bz)*Ey + (Ax*By-Ay*Bx)*Ez
ad = (By*Cz-Bz*Cy)*dx + (Bz*Cx-Bx*Cz)*dy + (Bx*Cy-By*Cx)*dz
bd = (Cy*Az-Cz*Ay)*dx + (Cz*Ax-Cx*Az)*dy + (Cx*Ay-Cy*Ax)*dz
cd = (Ay*Bz-Az*By)*dx + (Az*Bx-Ax*Bz)*dy + (Ax*By-Ay*Bx)*dz
aP = cd*aE - cE*ad
bP = cd*bE - cE*bd
Px = aP/(aP+bP)*Ax + bP/(aP+bP)*Bx
Py = aP/(aP+bP)*Ay + bP/(aP+bP)*By
Pz = aP/(aP+bP)*Az + bP/(aP+bP)*Bz
The first three lines convert E to barycentric coordinates, the next three lines d. Then we compute the barycentric coordinates of P, and turn these back into Cartesian coordinates. In that step we also dehomogenize them, i.e. divide by the sum of the barycentric coordinates.
On the whole, there is a considerable amount of code here. You can reduce the number of operations by moving common expressions to dedicated variables, particularly if your compiler doesn't do that for you. One benefit is that you won't need any divisions except for the final dehomogenization, i.e. the division by (a+b). If you compute 1/(a+b) once, you can make do with but a single division, which is good for performance.
The real benefit of the above code however is likely that you can do many nice things with the barycentric coordinates that you couldn't easily do with other coordinate systems. For example, if you want to check whether the ray hits the line AB on the line segment or somewhere outside the triangle, just check whether aP*bP > 0.
Time complexity is O(1)
It could potentially help.
In case we are talking about 3D and all your points lie on one plane. In cartesian coordinates you need to solve three linear systems to find intersection with three lines, and then determinate whether intersections lies on edge or not.
Or you need to find barycentric coordinates of two points on line. Point A(a1,a2,a3) = E and B(b1,b2,b3) = E+d. Equation of line is
Now you still need to solve three linear systems (where one of the mu = 0 and sum of the others = 1). But it will be 2x2 systems that easer to solve. And then check signes of found roots to detrerminate whether found point lies on edge or not.
Temporarily ignore the axis for which d has the smallest component, and treat the resulting 2D problem, just as we see on your illustration.
You can precompute the equation of the line of support of the ray, in the form ax + by + c = 0 (assuming z was ignored). Plug the coordinates of the three vertices in this expression, and the signs will tell you on what side of the line the vertices are.
The two intersections with the line occur on the edges having different signs at the endpoints.
Then, to determine which is the right edge, the sign of the angle between the ray and the third egde will do (if I am right);
When you know which edge, you can compute the parameter of the parametric equation of the side, which you can deduce from the previously computed signs. For instance, along the edge AB, t = Sb / (Sb - Sa). Also compute 1 - t;
Finally, use the t value to interpolate between the endpoints, this time in 3D.
The computational cost is
evalutation of the three signs Sa, Sb, Sc, taking 6 + and 6 x;
selection of the two edge candidates, taking two or three comparisons;
selection of the single candidate, taking a 2D cross product, 3 + and 2 x and a comparison;
computation of the interpolation parameter, 2 +, 1 /;
final 3D interpolation, 3 +, 6 x.
Will be hard to do shorter.
See the paper
Fast, minimum storage ray-triangle intersection,
by Tomas Möller and Ben Trombone,
in
Journal of Graphics Tools, 2(1):21–28, 1997.
See also this page.
The authors say:
One advantage of this method is that the plane equation need not be computed on the fly nor be stored, which can amount to significant memory savings for triangle meshes. As we found our method to be comparable in speed to previous methods, we believe it is the fastest ray-triangle intersection routine for triangles that do not have precomputed plane equations.
Let P be a simple, but not necessarily convex, polygon and q an arbitrary
point not necessarily in P.
Design an efficient algorithm to find a line segment originating from q that intersects the minimum number of edges of P.
If q is your point in the space of 2D, we can write q(x,y). We know that a polyhedron have the edges(E), vertices(V) and faces(F) which all of this terms are related with the formula V - E + F = 2 from the Euler's theorem but the problem turns out to be easier since it is for a polygon.
So the plain is to find an edge and calcule the direction vector of the point q(x,y) to the center of the edge, doing this (and if the polygon is convex) we are sure that this line segment will only goes through one edge of P.
The calculation will need a little of linear algebra, but it is not that difficult, for example:
1 - Find the distance from a line to a point (so we can find the closest edge to use in this problem):
2 - Also, a good idea would be to find the point on this line which is closest to (x0,y0) has coordinates:
Where the equation of the line is ax + by + c = 0 and the (x0,y0) is an arbitrary point not necessarily in P.
So, now we can find the point k(x,y) on the line that is closest to the initial point q(x0,y0), doing |q-k| = d. Where d is the distance that we just calculate. After that you already have everything that you need to find a line segment originating from q that intersects the minimum number of edges of P.
Some places where you can study and find more about this topic:
Wiki
mathwords
Think of doing this on a map, so you can think of directions like North, South, East, West, and bearings.
If you have a ray going in a particular direction from q out to infinity, whether it intersects a line between points A and B depends only on the bearing of the ray, and the bearings from q to A and B: if the bearing of the ray is within the range spanned by the bearing from q to A and the bearing from q to B, taking this in the direction of the line, then the ray will intersect the line.
So you can reduce this to a simpler problem. Imagine that all the points are in a circle around q, and the lines are arcs of this circle. Now you need to find a point on this circle which is overlapped by the minimum number of arcs. You will also make life very much easier if you divide each arc that spans 0 degrees into two at 0, cutting e.g. an arc from 320 degrees round to 40 degrees into one from 320 degrees to 360=0 degrees, and one from 0 degrees to 40 degrees.
Good candidate points for this are points just clockwise of each point in the circle. Now order each arc so that it is from a low angle to a high angle, sort the arcs by low angle, and work through them in order, incrementing a counter when you see the start of an arc, and decrementing it when you see the end of an arc (you don't need to worry about arcs that wrap across 0=360 degrees because you have just made sure that there aren't any). The point you want to find is the one associated with the smallest value of the counter.
Taking q as the origin of the coordinates, compute the polar arguments of the vertices. This is done in linear time.
Then every edge spans an interval of angles. To handle the wraparound, you can split the intervals that cross the 360° border.
You turned the problem in an instance of 1D intervals overlap. This is readily solved in O(N Log(N)) time.
It does not need to be very accurate. Does anyone know a good way to do this?
any help is much appreciated.
When you say “it does not need to be very accurate” you don’t say how inaccurate a solution you’re prepared to accept. Also, you don’t say how big the geographic region under consideration is likely to be. These two criteria make a big difference to the kind of approach that needs to be taken.
With a small regions (a few kilometres, say), a flat approximation may be good enough (for example, the Mercator projection) and some of the other responses tell you how to do that. With larger regions you have to take the Earth’s sphericity into account. And if you want inaccuracy less than a percent or so, you need to take the eccentricity of Earth into account.
I’m going to assume for the purposes of this answer that a spherical approximation is good enough, and that your points are at similar enough altitudes that we don’t need to worry about their heights.
You can convert a geographical point (ψ, λ) to Cartesian Earth-centred coordinates using the transformation
(ψ, λ) → (a cos(ψ) cos(λ), a cos(ψ) sin(λ), a sin(ψ))
where a is the mean radius of the Earth (6,371 km). So let’s suppose that the two points that define your line are p₀ and p₁; then the shortest line through p₀ and p₁ is a great circle, which defines a plane that slices the Earth into two halves, with normal n = p₀ × p₁.
Now we need to find the border of the circular region. Suppose the centre of this region is at c and that the surface radius of the region is s. Then the straight-line radius of the region is r = a sin(s/a). We’ll also need the true centre of the circular region, c’ = c cos(s/a). (This point is buried deep underground!)
We’d like to intersect the two circles and solve for the points of intersection. Unfortunately, because of numerical imprecision, the chances are that this procedure will never find any solutions because the imprecise circles will miss each other in 3 dimensions. So I suggest the following procedure: intersect the planes of the two circles, getting the dotted line shown below (unless c’ × n = 0 in which case the two circles are parallel and either c’ = o, in which case they are coincident, or else they do not intersect). Then intersect the line with the circular region.
This two-step procedure reduces the problem to two dimensions, and guarantees that a solution will be found even if numerical imprecision makes the two circles miss in 3 dimensions.
If you need more accuracy than this, then you might need to use geodetic coordinates on a reference ellipsoid such as WGS 1984.
I'd say find the closest point on the line to the center of the circle, then determine whether that point is within the circle (i.e. the distance in question is less than or equal to the circle's radius).
Outline for solving the problem: assume the Earth is a sphere of radius one centered at the origin. Convert all three lat, lon points to 3D coordinates. The two points of the line plus the origin define a plane; intersect that plane with the sphere of radius d centered on the other point. If there is no plane-sphere intersection, then the answer is the line does not intersect the region. If there is a plane-sphere intersection, then the problem is simplified to intersecting the circular region defined by the plane-sphere intersection with the shortest circular arc on the plane going between the end points of the line and centered at the origin. This is a straightforward 2D problem if you convert to the coordinate system of the plane.
This question is too vague to be answered precisely. What do you mean by
a line form by 2 geo points (lat, lon)
This can be either a great circle going through them (also called orthodrome) or it a can be a linear function of spherical coordinates (loxodrome).
BTW, I assume your circle is a circle on the surface of the sphere, right?
Assuming line is formed by points (x1, y1) and (x2, y2), and circle has radius r with origin (0,0):
Calculate:
Incidence = r^2 * [(x2 - x1)^2 + (y2 - y1)^2] - (x1 * y2 - x2 * y1)^2
Then, from the value of Incidence, we can determine the following:
Incidence < 0: No intersection
Incidence = 0: Tangent (intersection at 1 point on circle)
Incidence > 0: Intersection
It's likely your circle is not at the origin (0,0), so to fix this, just add the origin coordinates from your line coordinates in the equation above. So, if the circle is at (x3, y3), x1 in the above equation would become x1 + x3. Likewise, y1 would be y1 + y3, and the same goes for x2 and y2.
For more info check out this link