I'm not the best in Maths, but for what I am doing now I need to calculate the angle of the vector which is shown as arrow in the picture below:
I have a point A and a point B in a 2D plane. I need to calculate the following:
The angle in which the arrow must be rotated in order to point to B
atan2(yB-yA, xB-xA), assuming your library has atan2. Otherwise you need to use atan, which will return the correct answer if B is to the right of A, and will be 180 degrees off otherwise. Also note that the return value is in radians, you can convert radians to degrees by multiplying by 180/pi if necessary.
Wikipedia has a detailed explanation of the geometry.
arctan((A.y - B.y) / (A.x - B.x)) and note the special case where A.x = B.x
Related
I'm making a game in pygame and I want a spaceship to move in the direction it's facing. I have the angle and the magnitude, how can I get the direction in the form of a normalized vector that I can then add to the spaceship's x and y to make it move?
What you want to do is change polar coordinates to Cartesian coordinates.
So as I see it you are in 2D. Now you can not avoid using trigonometry.
Let's assume that you are facing with an α angle beginning from the x axis, and growing towards y. Let d be the magnitude.
Vx = cos(α) * d
Vy = sin(α) * d
Now this is not normalized, as I have already multiplied it with d, the magnitude of the speed.
Please check what type of angle your sin and cos functions take, and give matching inputs (it can be radians or degrees). An easy way to check this is to print out sin(30). If it is 0.5, you are in degrees. If it is -0.988... you are in radians.
Who is a math wiz and wants to check my math?
I'm trying to find the placement of a point(x,y) on the circumference of a circle in degrees. If I use x=radius*cos(Q)+xOrigin, I end up with arccos((x/2)-xOrigin) = Q which gives me the angle in radians. Then I go D = Q(180/Pi) to get degrees, but I keep coming up about 3-5 degrees off of target.
What's up with that?
Your formula for Q should be
Q = acos((x-xOrigin)/radius)
Note that acos returns an angle between 0 and pi; if the angle is in the third or fourth quadrant you'll need to subtract the angle from 2 * pi.
Without knowing the actual values that are coming up wrong, it's hard to tell if that's the only problem. Your conversion from radians to degrees is correct.
Most modern programming languages have atan2() somewhere, which takes a rise and a run and spits out the angle in radians in all four quadrants.
I have two 3D vectors called A and B that both only have a 3D position. I know how to find the angle along the unit circle ranging from 0-360 degrees with the atan2 function by doing:
EDIT: (my atan2 function made no sense, now it should find the "y-angle" between 2 vectors):
toDegrees(atan2(A.x-B.x,A.z-B.z))+180
But that gives me the Y angle between the 2 vectors.
I need to find the X angle between them. It has to do with using the x, y and z position values. Not the x and z only, because that gives the Y angle between the two vectors.
I need the X angle, I know it sounds vague but I don't know how to explain. Maybe for example you have a camera in 3D space, if you look up or down than you rotate the x-axis. But now I need to get the "up/down" angle between the 2 vectors. If I rotate that 3D camera along the y-axis, the x-axis doens't change. So with the 2 vectors, no matter what the "y-angle" is between them, the x-angle between the 2 vectors wil stay the same if y-angle changes because it's the "up/down" angle, like in the camara.
Please help? I just need a line of math/pseudocode, or explanation. :)
atan2(crossproduct.length,scalarproduct)
The reason for using atan2 instead of arccos or arcsin is accuracy. arccos behaves very badly close to 0 degrees. Small computation errors in argument will lead to disproportionally big errors in result. arcsin has same problem close to 90 degrees.
Computing the altitude angle
OK, it might be I finally understood your comment below about the result being independent of the y angle, and about how it relates to the two vectors. It seems you are not really interested in two vectors and the angle between these two, but instead you're interested in the difference vector and the angle that one forms against the horizontal plane. In a horizontal coordinate system (often used in astronomy), that angle would be called “altitude” or “elevation”, as opposed to the “azimuth” you compute with the formula in your (edited) question. “altitude” closely relates to the “tilt” of your camera, whereas “azimuth” relates to “panning”.
We still have a 2D problem. One coordinate of the 2D vector is the y coordinate of the difference vector. The other coordinate is the length of the vector after projecting it on the horizontal plane, i.e. sqrt(x*x + z*z). The final solution would be
x = A.x - B.x
y = A.y - B.y
z = A.z - B.z
alt = toDegrees(atan2(y, sqrt(x*x + z*z)))
az = toDegrees(atan2(-x, -z))
The order (A - B as opposed to B - A) was chosen such that “A above B” yields a positive y and therefore a positive altitude, in accordance with your comment below. The minus signs in the azimuth computation above should replace the + 180 in the code from your question, except that the range now is [-180, 180] instead of your [0, 360]. Just to give you an alternative, choose whichever you prefer. In effect you compute the azimuth of B - A either way. The fact that you use a different order for these two angles might be somewhat confusing, so think about whether this really is what you want, or whether you want to reverse the sign of the altitude or change the azimuth by 180°.
Orthogonal projection
For reference, I'll include my original answer below, for those who are actually looking for the angle of rotation around some fixed x axis, the way the original question suggested.
If this x angle you mention in your question is indeed the angle of rotation around the x axis, as the camera example suggests, then you might want to think about it this way: set the x coordinate to zero, and you will end up with 2D vectors in the y-z plane. You can think of this as an orthogonal projection onto said plain. Now you are back to a 2D problem and can tackle it there.
Personally I'd simply call atan2 twice, once for each vector, and subtract the resulting angles:
toDegrees(atan2(A.z, A.y) - atan2(B.z, B.y))
The x=0 is implicit in the above formula simply because I only operate on y and z.
I haven't fully understood the logic behind your single atan2 call yet, but the fact that I have to think about it this long indicates that I wouldn't want to maintain it, at least not without a good explanatory comment.
I hope I understood your question correctly, and this is the thing you're looking for.
Just like 2D Vectors , you calculate their angle by solving cos of their Dot Product
You don't need atan, you always go for the dot product since its a fundamental operation of vectors and then use acos to get the angle.
double angleInDegrees = acos ( cos(theta) ) * 180.0 / PI;
I am trying to understand trigonometry and the short answer is that I do not.
I drew a little triangle to mess around with and I asked myself the question, "If I know the length of the hypotenuse and the angle, how do I find the length of the other edges?".
Then I started reading. Apparently, the sine of angle A is supposed to equal the length of the opposite side divided by the length of the hypotenuse. So I figured that, using a right triangle, multiplying the length of the hypotenuse by the sine of the angle would yield the length of the opposing side.
1.414 / 1 = .707blahblah * 1.414 = 1 on my calculator.
But in every programming language I try sin(45.0) equals .8somethingsomething. I tried c++, c#, java, php, and lua.
Is the input not being interpreted as degrees? What unit is being used and how do I convert it? I've been seeing the word Radians, it would be helpful if someone could explain what a Radian is.
Radians are units of angular measure, like degrees, except that while there are 360 degrees in a circle, there are 2*pi (about 6.28) radians in a circle. You can convert degrees to radians by multiplying by pi (3.14159) and dividing by 180.
The formula works if the triangle is a right triangle, and yes, most programming languages expect radians rather than degrees as arguments to functions like sin() and cos().
Regarding the argument in the comments below: if you fix angle <BAC, side AB, and side BC, you can see that there are two possible positions for point C which preserve the the length D2 for side BC. Therefore <BAC, D1, and D2 do not fully determine a triangle.
The input to sin functions generally is expected in radians, not degrees. For example, in the Java documentation for sin it's stated that:
Parameters:
a - an angle, in radians.
Convert the angle in degrees to radians first, by multiplying it by pi/180
A radian is the distance of the radius of a circle along its circumference. Since a circle's circumference is 2 times pi times its radius, there are 2 times pi radians in one complete circle.
Yes, you are correct. Those functions all take their input in radians, not degrees.
You can convert degrees to radians by multiplying the degrees by π/180.
Convert to radians: Radian = degree/180*Pi
In order to convert from degrees to radians, divide the number in degrees by 180 and multiply by pi.
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!