I got two routes as polylines as lists of coordinates such as:
[13.072624,52.333508,13.763972,52.679616]
[52.679616,13.763972,52.333508,13.072624]
I need to compare the direction of two polylines. To do so I thought to get the angle between them and return true (going in the same direction) if it is smaller 45°.
example of route comparison
How can we do so?
What you need to compare against each other is the unit vectors of the line segments. I don't know what language you are working in or how the lists are organized but the following pseudo code will get you the bearings.
// create end points of the line segments from lists
point0(x0, y0);
point1(x1, y1);
point2(x2, y2);
point3(x3, y3);
// create direction vectors from points
directionA = point1 - point0;
directionB = point3 - point2;
// normalize the direction vectors to have a length of 1
// assuming they are not degenerate
directionA.unitize();
directionB.unitize();
// get the dot product of the two direction vectors
dot = directionA.dot(directionB);
// Because the direction vectors are both unit length
// the dot product will return a value between -1.0 and 1.0
// The arc cosine will give you your angle between 0 and 180
angle = arc_cosine_degree(dot)
Then compare the angle to what ever tolerance you choose for determining "same" direction. Wikipedia has good explanations of each of the concepts that I use in the pseudo code (i.e. unit vector, dot product, arc cosine).
Good luck! GCB
Related
I have two approximately parallel polylines representing railway tracks, consisting of hundreds (maybe thousands) of x, y, z coordinates. The two lines stay approximately 1.435m apart, but bend and curve as a railway would.
If I pick a point on one of the polylines, how do find the point which is perpendicular on the other parallel polyline?
I take it CAD programs use the cross product to find the distance / point and it chooses the line to snap to based on where your mouse is hovering.
I would like to achieve the same thing, but without hovering your mouse over the line.
Is there a way to simply compute the closest line segment on the parallel line? Or to see which segment of the polyline passes through a perpendicular plane at the selected point?
It isn't practical to loop through the segments as there are so many of them.
In python the input would be something like point x, y, z on rail1 and I would be looking to output point x, y, z on rail2.
Many thanks.
You want the point of the minimum distance to the other track.
If the other track is defined by line segments, each spanning two points with a parameter t going between 0 and 1
pos(t) => pos_1 + t * ( pos_2 - pos_1 )
You need to find the t value that produces the minimum distance to the point. Place a temporary coordinate system on the point and express the ends of each line segment pos_1 and pos_2 in relative coordinates to the point of interest.
The value of t is for the closest point is
dot(pos_1,pos_1) - dot(pos_1,pos_2)
t = ------------------------------------------------------
dot(pos_1,pos_1)-2*dot(pos_1,pos_2)+dot(pos_2,pos_2)
where dot(a,b)=ax*bx+ay*by+az*bz is the vector dot product.
Now if the resulting t is between 0 and 1, then the closest point is on this segment, and its coordinates are given by
pos(t) => pos_1 + t * ( pos_2 - pos_1 )
How would you find the signed angle theta from vector a to b?
And yes, I know that theta = arccos((a.b)/(|a||b|)).
However, this does not contain a sign (i.e. it doesn't distinguish between a clockwise or counterclockwise rotation).
I need something that can tell me the minimum angle to rotate from a to b. A positive sign indicates a rotation from +x-axis towards +y-axis. Conversely, a negative sign indicates a rotation from +x-axis towards -y-axis.
assert angle((1,0),(0,1)) == pi/2.
assert angle((0,1),(1,0)) == -pi/2.
What you want to use is often called the “perp dot product”, that is, find the vector perpendicular to one of the vectors, and then find the dot product with the other vector.
if(a.x*b.y - a.y*b.x < 0)
angle = -angle;
You can also do this:
angle = atan2( a.x*b.y - a.y*b.x, a.x*b.x + a.y*b.y );
If you have an atan2() function in your math library of choice:
signed_angle = atan2(b.y,b.x) - atan2(a.y,a.x)
Given a plane(in my case a triangle) normal N_T and a reference Normal N_R, both have the length 1.
I calculated the rotation_normal
N = N_T x N_R
and now i need to calculate the angle around this rotation_normal, which i get with the following calculation:
angle = acos(<N_T, N_R>), with <x,y> is the dotproduct of x and y
This angle is in the interval of [0°, 180°] and is the smallest angle between both normals.
So my problem is that if i want to rotate my triangle in a manner that its normal is equal to the reference normal, i need to know in which direction (positive or negative) the calculated angle is.
Does anybody know how to get this direction or how to solve this problem in general?
you need to use atan2 (4-quadrant arc tangens)
create reference plane basis vectors u,v
must be perpendicular to each other and lie inside plane
preferably unit vectors (or else you need to account for its size)
so let N=N_T x N_R; ... reference plane normal where the rotation will take place
U=N_T;
V= N x U; ... x means cross product
make them unit U/=|U|; V/=|V|; if they are not already
compute plane coordinates of N_R
u=(N_R.U); ... . means dot product
v=(N_R.V);
compute angle
ang=atan2(v,u);
if you do not have atan2 then use ang=atanxy(u,v);
this will give you angle in range ang=<0,2*M_PI>
if you want signed angle instead then add
if (ang>M_PI) ang-=2.0*M_PI; ... M_PI is well known constant Pi=3.1415...
now if you want the opposite sign direction then just use -ang
I want to calculate the angle between two vectors a and b. Lets assume these are at the origin. This can be done with
theta = arccos(a . b / |a| * |b|)
However arccos gives you the angle in [0, pi], i.e. it will never give you an angle greater than 180 degrees, which is what I want. So how do you find out when the vectors have gone past the 180 degree mark? In 2D I would simply let the sign of the y-component on one of the vectors determine what quadrant the vector is in. But what is the easiest way to do it in 3D?
EDIT: I wanted to keep the question general but here we go. I'm programming this in c and the code I use to get the angle is theta = acos(dot(a, b)/mag(a)*mag(b)) so how would you programmatically determine the orientation?
This works in 2D because you have a plane defined in which you define the rotation.
If you want to do this in 3D, there is no such implicit 2D plane. You could transform your 3D coordinates to a 2D plane going through all three points, and do your calculation inside this plane.
But, there are of course two possible orientations for the plane, and that will affect which angles will be > 180 or smaller.
I came up with the following solution that takes advantage of the direction change of the cross product of the two vectors:
Make a vector n = a X b and normalize it. This vector is normal to the plane spanned by a and b.
Whenever a new angle is calculated compare it with the old normal. In the comparison, treat the old and the current normals as points and compute the distance between them. If this distance is 2 the normal (i.e. the cross product a X b has flipped).
You might want to have a threshold for the distance as the distance after a flip might be shorter than 2, depending on how the vectors a and b are oriented and how often you update the angle.
One solution that you could use:
What you effectively need to do is create a plane that one of the vectors is coplanar to.
Getting the cross product of both vectors will create a plane, then is you get the normal of this plane, you can get the angle between this and the vector you need to get the signed angle for, and you can use the angle to determine the sign.
If the angle is greater than 90 degrees, then it is below the created plane; less than 90 degrees, and it is above.
Depending on cost of calculations, the dot product can be used at this stage instead of the angle.
Just make sure that you always calculate the normals by the same order of vectors.
This is useable more easily if you're using the XYZ axes, and that's what you're comparing against, since you already have the vectors needed for the plane.
There are possbly more efficient solutions, but this is one I came up with.
Edit: clarification of created vectors
a X b = p. This is perpendicular to both a and b.
Then, do either:
a X p or b X p to create another vector that is the normal to the plane created by the 2 vectors. Choice of vector depends on which you're trying to find the angle for.
Strictly speaking, two 3D vectors always have two angles between them - one below or equal to 180, the other over or equal to 180. Arccos gives you one of them, you can get the other by subtracting from 360. Think of it that way: imagine two lines intersect. You have 4 angles there - 2 of one value, 2 of another. What's the angle between the lines? No single answer. Same here. Without some kind of extra criteria, you can not, in theory, tell which of the two angle values should be taken into account.
EDIT: So what you really need is an arbitrary example of fixing an orientation. Here's one: we look from the positive Z direction. If the plane between the two vectors contains the Z axis, we look from the positive Y direction. If the plane is YZ, we look from the positive X direction. I'll think how to express this in coordinate form, then edit again.
I'm trying to make a triangle (isosceles triangle) to move around the screen and at the same time slightly rotate it when a user presses a directional key (like right or left).
I would like the nose (top point) of the triangle to lead the triangle at all times. (Like that old asteroids game).
My problem is with the maths behind this. At every X time interval, I want the triangle to move in "some direction", I need help finding this direction (x and y increments/decrements).
I can find the center point (Centroid) of the triangle, and I have the top most x an y points, so I have a line vector to work with, but not a clue as to "how" to work with it.
I think it has something to do with the old Sin and Cos methods and the amount (angle) that the triangle has been rotated, but I'm a bit rusty on that stuff.
Any help is greatly appreciated.
The arctangent (inverse tangent) of vy/vx, where vx and vy are the components of your (centroid->tip) vector, gives you the angle the vector is facing.
The classical arctangent gives you an angle normalized to -90° < r < +90° degrees, however, so you have to add or subtract 90 degrees from the result depending on the sign of the result and the sign of vx.
Luckily, your standard library should proive an atan2() function that takes vx and vy seperately as parameters, and returns you an angle between 0° and 360°, or -180° and +180° degrees. It will also deal with the special case where vx=0, which would result in a division by zero if you were not careful.
See http://www.arctangent.net/atan.html or just search for "arctangent".
Edit: I've used degrees in my post for clarity, but Java and many other languages/libraries work in radians where 180° = π.
You can also just add vx and vy to the triangle's points to make it move in the "forward" direction, but make sure that the vector is normalized (vx² + vy² = 1), else the speed will depend on your triangle's size.
#Mark:
I've tried writing a primer on vectors, coordinates, points and angles in this answer box twice, but changed my mind on both occasions because it would take too long and I'm sure there are many tutorials out there explaining stuff better than I ever can.
Your centroid and "tip" coordinates are not vectors; that is to say, there is nothing to be gained from thinking of them as vectors.
The vector you want, vForward = pTip - pCentroid, can be calculated by subtracting the coordinates of the "tip" corner from the centroid point. The atan2() of this vector, i.e. atan2(tipY-centY, tipX-centX), gives you the angle your triangle is "facing".
As for what it's relative to, it doesn't matter. Your library will probably use the convention that the increasing X axis (---> the right/east direction on presumably all the 2D graphs you've seen) is 0° or 0π. The increasing Y (top, north) direction will correspond to 90° or (1/2)π.
It seems to me that you need to store the rotation angle of the triangle and possibly it's current speed.
x' = x + speed * cos(angle)
y' = y + speed * sin(angle)
Note that angle is in radians, not degrees!
Radians = Degrees * RadiansInACircle / DegreesInACircle
RadiansInACircle = 2 * Pi
DegressInACircle = 360
For the locations of the vertices, each is located at a certain distance and angle from the center. Add the current rotation angle before doing this calculation. It's the same math as for figuring the movement.
Here's some more:
Vectors represent displacement. Displacement, translation, movement or whatever you want to call it, is meaningless without a starting point, that's why I referred to the "forward" vector above as "from the centroid," and that's why the "centroid vector," the vector with the x/y components of the centroid point doesn't make sense. Those components give you the displacement of the centroid point from the origin. In other words, pOrigin + vCentroid = pCentroid. If you start from the 0 point, then add a vector representing the centroid point's displacement, you get the centroid point.
Note that:
vector + vector = vector
(addition of two displacements gives you a third, different displacement)
point + vector = point
(moving/displacing a point gives you another point)
point + point = ???
(adding two points doesn't make sense; however:)
point - point = vector
(the difference of two points is the displacement between them)
Now, these displacements can be thought of in (at least) two different ways. The one you're already familiar with is the rectangular (x, y) system, where the two components of a vector represent the displacement in the x and y directions, respectively. However, you can also use polar coordinates, (r, Θ). Here, Θ represents the direction of the displacement (in angles relative to an arbitary zero angle) and r, the distance.
Take the (1, 1) vector, for example. It represents a movement one unit to the right and one unit upwards in the coordinate system we're all used to seeing. The polar equivalent of this vector would be (1.414, 45°); the same movement, but represented as a "displacement of 1.414 units in the 45°-angle direction. (Again, using a convenient polar coordinate system where the East direction is 0° and angles increase counter-clockwise.)
The relationship between polar and rectangular coordinates are:
Θ = atan2(y, x)
r = sqrt(x²+y²) (now do you see where the right triangle comes in?)
and conversely,
x = r * cos(Θ)
y = r * sin(Θ)
Now, since a line segment drawn from your triangle's centroid to the "tip" corner would represent the direction your triangle is "facing," if we were to obtain a vector parallel to that line (e.g. vForward = pTip - pCentroid), that vector's Θ-coordinate would correspond to the angle that your triangle is facing.
Take the (1, 1) vector again. If this was vForward, then that would have meant that your "tip" point's x and y coordinates were both 1 more than those of your centroid. Let's say the centroid is on (10, 10). That puts the "tip" corner over at (11, 11). (Remember, pTip = pCentroid + vForward by adding "+ pCentroid" to both sides of the previous equation.) Now in which direction is this triangle facing? 45°, right? That's the Θ-coordinate of our (1, 1) vector!
keep the centroid at the origin. use the vector from the centroid to the nose as the direction vector. http://en.wikipedia.org/wiki/Coordinate_rotation#Two_dimensions will rotate this vector. construct the other two points from this vector. translate the three points to where they are on the screen and draw.
double v; // velocity
double theta; // direction of travel (angle)
double dt; // time elapsed
// To compute increments
double dx = v*dt*cos(theta);
double dy = v*dt*sin(theta);
// To compute position of the top of the triangle
double size; // distance between centroid and top
double top_x = x + size*cos(theta);
double top_y = y + size*sin(theta);
I can see that I need to apply the common 2d rotation formulas to my triangle to get my result, Im just having a little bit of trouble with the relationships between the different components here.
aib, stated that:
The arctangent (inverse tangent) of
vy/vx, where vx and vy are the
components of your (centroid->tip)
vector, gives you the angle the vector
is facing.
Is vx and vy the x and y coords of the centriod or the tip? I think Im getting confused as to the terminology of a "vector" here. I was under the impression that a Vector was just a point in 2d (in this case) space that represented direction.
So in this case, how is the vector of the centroid->tip calculated? Is it just the centriod?
meyahoocomlorenpechtel stated:
It seems to me that you need to store
the rotation angle of the triangle and
possibly it's current speed.
What is the rotation angle relative to? The origin of the triangle, or the game window itself? Also, for future rotations, is the angle the angle from the last rotation or the original position of the triangle?
Thanks all for the help so far, I really appreciate it!
you will want the topmost vertex to be the centroid in order to achieve the desired effect.
First, I would start with the centroid rather than calculate it. You know the position of the centroid and the angle of rotation of the triangle, I would use this to calculate the locations of the verticies. (I apologize in advance for any syntax errors, I have just started to dabble in Java.)
//starting point
double tip_x = 10;
double tip_y = 10;
should be
double center_x = 10;
double center_y = 10;
//triangle details
int width = 6; //base
int height = 9;
should be an array of 3 angle, distance pairs.
angle = rotation_angle + vertex[1].angle;
dist = vertex[1].distance;
p1_x = center_x + math.cos(angle) * dist;
p1_y = center_y - math.sin(angle) * dist;
// and the same for the other two points
Note that I am subtracting the Y distance. You're being tripped up by the fact that screen space is inverted. In our minds Y increases as you go up--but screen coordinates don't work that way.
The math is a lot simpler if you track things as position and rotation angle rather than deriving the rotation angle.
Also, in your final piece of code you're modifying the location by the rotation angle. The result will be that your ship turns by the rotation angle every update cycle. I think the objective is something like Asteroids, not a cat chasing it's tail!