I have 2 points in an x, y plane. I want to rotate one point onto the other point by rotating it about the z-axis.
How can I find the angle to rotate one point onto the other?
Maybe the best thing is to get the angles from horizontal for the two points and then take the difference.
angle_1 = atan2( y_1, x_1 );
angle_2 = atan2( y_2, x_2 );
rotation_angle = angle_1-angle_2;
Well, the sin of this angle is [a, b] / (abs(a) * abs(b)) and its cos is (a, b) / (abs(a) * abs(b)), where [a, b] is a cross product of a and b, (a, b) is a scalar product and abs(x) is the length of the vector x. It is pretty easy to find an angle given its sin and cos.
Related
I am having trouble with finding the X and Y-coordinates of the center of a rotated rectangle through mathematical calculations.
What I have is the width 'A' and height 'B' of a rectangle, as well as its rotation 'R' in degrees.
What would be the formula to calculate those points?
Here is a sketch to further explain what i am trying to do:
Any help is greatly appreciated! :)
if you are trying to do it in pure way, by your self
it think this answer would be great for you
https://math.stackexchange.com/questions/270194/how-to-find-the-vertices-angle-after-rotation
if you want to use a package for this, then I do recommend
flatten-js , geometric ,Ecluid.ts
I doubt that this is what you are after, but here is a solution to a possibly related question.
We assume that the leftmost corner lies on the Y axis, and the bottommost on the X axis. Thus, the coordinates of these corners are (0, B sin R) and (B cos R, 0) respectively. The midpoint of the bottom side is (B/2 cos R, B/2 sin R). To this we add a vector of length A/2 in the orthogonal direction and obtain the coordinates of the center:
(B/2 cos R - A/2 sin R, B/2 sin R + A/2 cos R)
How can I use atan2 to determine the coordinates where a segment ends, knowing the coordinates where it starts (x, y), the angle of the segment and the length of the segment?
Can you tell me a solution it doesn't have to be only with atan2...
You have a right angled triangle and you can assume that the point A is at (x,y), this is much easier to explain with the use of some diagrams:
Now we know what the value for c is already because this is the length of the segment and we want to know the values for a and b because the final point is at (x+b, y+a)
From trigonometry we know what the relations between the angles and the ratios of the sides are
sin(theta) = opposite/hypotenuse and cos(theta) = adjacent/hypotenuse
Substituting in the values we know we get:
sin(theta) = opposite/c
c = opposite * sin(theta)
and
cos(theta) = adjacent/b
b = adjacent * cos(theta)
We know that the end point is at (x+b, y+a) which given the values for the edges we found is (x+adjacent * cos(theta), y+opposite * sin(theta))
I'm having serious problems solving a problem illustrated on the pic below.
Let's say we have 3 points in 3D space (blue dots), and the some center of the triangle based on them (red dot - point P). We also have a normal to this triangle, so that we know which semi-space we talking about.
I need to determine, what is the position on a point (red ??? point) that depends on two angles, both in range of 0-180 degrees. Doesnt matter how the alfa=0 and betha=0 angle is "anchored", it is only important to be able to scan the whole semi-sphere (of radius r).
http://i.stack.imgur.com/a1h1B.png
If anybody could help me, I'd be really thankful.
Kind regards,
Rav
From the drawing it looks as if the position of the point on the sphere is given by a form of spherical coordinates. Let r be the radius of the sphere; let alpha be given relative to the x-axis; and let beta be the angle relative to the x-y-plane. The Cartesian coordinates of the point on the sphere are:
x = r * cos(beta) * cos(alpha)
y = r * cos(beta) * sin(alpha)
z = r * sin(beta)
Edit
But for a general coordinate frame with axes (L, M, N) centered at (X, Y, Z) the coordinates are (as in dmuir's answer):
(x, y, z) =
(X, Y, Z)
+ r * cos(beta) * cos(alpha) * L
+ r * cos(beta) * sin(alpha) * M
+ r * sin(beta) * N
The axes L and N must be orthogonal and M = cross(N, L). alpha is given relative to L, and beta is given relative to the L-M plane. If you don't know how L is related to points of the triangle, then the question can't be answered.
You need to find two unit length orthogonal vectors L, M say, in the plane of the triangle as well as the the unit normal N. The points on the sphere are
r*cos(beta)*cos(alpha) * L + r*cos(beta)*sin(alpha)*M + r*sin(beta)*N
Im trying to find out the angle (in degrees) between two 2D vectors. I know I need to use trig but I'm not too good with it. This is what I'm trying to work out (the Y axis increases downward):
I'm trying to use this code at the moment, but it's not working at all (calculates random angles for some reason):
private float calcAngle(float x, float y, float x1, float y1)
{
float _angle = (float)Math.toDegrees(Math.atan2(Math.abs(x1-x), Math.abs(y1-y)));
Log.d("Angle","Angle: "+_angle+" x: "+x+" y: "+y+" x1: "+x1+" y1: "+y1);
return _angle;
}
These are my results (There constant when providing a constant position, but when I change the position, the angle changes and I can't find any link between the two angles):
Position 1:
x:100 y:100
x1:50 y1:50
Angle: 45
Position 2:
x:92 y:85
x1:24 y1:16
Angle: 44.58
Position 3:
x:44 y: 16
x1:106 y1:132
Angle: 28.12
Edit: Thanks everyone who answered and helped me figure out that was wrong! Sorry the title and the question was confusing.
You first have to understand how to compute angle between two vectors and there are several of them. I will give you what I think is the simplest.
Given v1 and v2, their dot product is: v1x * v2x + v1y * v2y
The norm of a vector v is given by: sqtr(vx^2+vy^2)
With this information, please take this definition:
dot(v1, v2) = norm(v1) * norm(v2) * cos(angle(v1, v2))
Now, you solve for angle(v1, v2):
angle(v1, v2) = acos( dot(v1, v2) / (norm(v1) * norm(v2)) )
Finally, taking the definitions given at the beginning, then you end up with:
angle(v1, v2) = acos( (v1x * v2x + v1y * v2y) / (sqrt(v1x^2+v1y^2) * sqrt(v2x^2+v2y^2)) )
Again, there are many ways to do this, but I like this one because it is helpful for dot product given angle and norm, or angle, given vectors.
The answer will be in radians, but you know that pi radians (that is 3.14 radians) are 180 degrees, so you simply multiply by the conversion factor 180/pi.
Aha! Turns out I just needed to flip my angle and use atan2. This is my final code:
private float calcAngle(float x, float y, float x1, float y1)
{
float _angle = (float)Math.toDegrees(Math.atan2(x1-x, y-y1));
return _angle;
}
Thanks everyone for helping me figure this out and also for helping me to understand what I'm actually doing! :)
Do not take the absolute value of the arguments to atan2. The whole point of atan2 is that it uses the signs of its arguments to work out which qaudrant the angle is in. By taking the absolute values you are forcing atan2 to only return values between 0 and pi/2 instead of -pi to pi.
It looks like Niall figured it out, but I'll finish my explanation, anyways. In addition to explaining why the solution works, my solution has two advantages:
Potential division by zero within atan2() is avoided
Return value is always positive in the range 0 to 360 degrees
atan2() returns the counter-clockwise angle relative to the positive X axis. Niall was looking for the clockwise angle relative to the positive Y axis (between the vector formed by the two points and the positve Y axis).
The following function is adapted from my asteroids game where I wanted to calculate the direction a ship/velocity vector was "pointing:"
// Calculate angle between vector from (x1,y1) to (x2,y2) & +Y axis in degrees.
// Essentially gives a compass reading, where N is 0 degrees and E is 90 degrees.
double bearing(double x1, double y1, double x2, double y2)
{
// x and y args to atan2() swapped to rotate resulting angle 90 degrees
// (Thus angle in respect to +Y axis instead of +X axis)
double angle = Math.toDegrees(atan2(x1 - x2, y2 - y1));
// Ensure result is in interval [0, 360)
// Subtract because positive degree angles go clockwise
return (360 - angle) % 360;
}
It should be :
atan( abs(x1 - x)/abs(y1 - y) )
abs stands for absolute (to avoid negative values)
I believe the equation for the angle between two vectors should look more like:
toDegrees(acos((x*x1+y*y1)/(sqrt(x*x+y*y)*sqrt(x1*x1+y1*y1))))
Your above equation will calculate the angle made between the vector p1-p2 and the line made by extending an orthogonal from the point p2 to the vector p1.
The dot product of two vectors V1 and V2 is equal to |V1|*|V2|cos(theta). Therefore, theta is equal to acos((V1 dot V2)/(|V1||V2|)). V1 dot V2 is V1.xV2.x+V1.yV2.y.
The magnitude of V (i.e., |V|) is the pathogorean theorem... sqrt(V.x^2 + V.y^2)
My first guess would be to calculate the angle of each vector with the axes using atan(y/x) and then subtract those angels and take the absolute value, that is:
abs(atan(y/x) - atan(y1/x1))
Are you using integers? Cast the arguments as doubles, and I would use fabs on the result, not the arguments. The result will be in radians; to get degrees, use:
res *= (360.0/(2.0*Math.PI));
The angle of the second vector relative to the first = atan2(y2,x2) - atan2(y1,x1).
http://www.euclideanspace.com/maths/algebra/vectors/angleBetween/index.htm
I'm using Electro in Lua for some 3D simulations, and I'm running in to something of a mathematical/algorithmic/physics snag.
I'm trying to figure out how I would find the "spin" of a sphere of a sphere that is spinning on some axis. By "spin" I mean a vector along the axis that the sphere is spinning on with a magnitude relative to the speed at which it is spinning. The reason I need this information is to be able to slow down the spin of the sphere by applying reverse torque to the sphere until it stops spinning.
The only information I have access to is the X, Y, and Z unit vectors relative to the sphere. That is, each frame, I can call three different functions, each of which returns a unit vector pointing in the direction of the sphere model's local X, Y and Z axes, respectively. I can keep track of how each of these change by essentially keeping the "previous" value of each vector and comparing it to the "new" value each frame. The question, then, is how would I use this information to determine the sphere's spin? I'm stumped.
Any help would be great. Thanks!
My first answer was wrong. This is my edited answer.
Your unit vectors X,Y,Z can be put together to form a 3x3 matrix:
A = [[x1 y1 z1],
[x2 y2 z2],
[x3 y3 z3]]
Since X,Y,Z change with time, A also changes with time.
A is a rotation matrix!
After all, if you let i=(1,0,0) be the unit vector along the x-axis, then
A i = X so A rotates i into X. Similarly, it rotates the y-axis into Y and the
z-axis into Z.
A is called the direction cosine matrix (DCM).
So using the DCM to Euler axis formula
Compute
theta = arccos((A_11 + A_22 + A_33 - 1)/2)
theta is the Euler angle of rotation.
The magnitude of the angular velocity, |w|, equals
w = d(theta)/dt ~= (theta(t+dt)-theta(t)) / dt
The axis of rotation is given by e = (e1,e2,e3) where
e1 = (A_32 - A_23)/(2 sin(theta))
e2 = (A_13 - A_31)/(2 sin(theta))
e3 = (A_21 - A_12)/(2 sin(theta))
I applaud ~unutbu's, answer, but I think there's a simpler approach that will suffice for this problem.
Take the X unit vector at three successive frames, and compare them to get two deltas:
deltaX1 = X2 - X1
deltaX2 = X3 - X2
(These are vector equations. X1 is a vector, the X vector at time 1, not a number.)
Now take the cross-product of the deltas and you'll get a vector in the direction of the rotation vector.
Now for the magnitude. The angle between the two deltas is the angle swept out in one time interval, so use the dot product:
dx1 = deltaX1/|deltaX1|
dx2 = deltax2/|deltaX2|
costheta = dx1.dx2
theta = acos(costheta)
w = theta/dt
For the sake of precision you should choose the unit vector (X, Y or Z) that changes the most.