How to convert cartesian coordinates to polar coordinates in JS? - javascript

I need to know angle of rotation in polar coordinates using X and Y from cartesian coordinates.
How to do it in JS without a lot of IF statements? I know that I can do it using ,
but I think that it will be bad for performance, because it is in animation cycle.

Javascript comes with a built in function that does what is represented in the image: Math.atan2()
Math.atan2() takes y, x as arguments and returns the angle in radians.
For example:
x = 3
y = 4
Math.atan2(y, x) //Notice that y is first!
//returns 0.92729521... radians, which is 53.1301... degrees
I wrote this function to convert from Cartesian Coordinates to Polar Coordinates, returning the distance and the angle (in radians):
function cartesian2Polar(x, y){
distance = Math.sqrt(x*x + y*y)
radians = Math.atan2(y,x) //This takes y first
polarCoor = { distance:distance, radians:radians }
return polarCoor
}
You can use it like this to get the angle in radians:
cartesian2Polar(5,5).radians
Lastly, if you need degrees, you can convert radians to degrees like this
degrees = radians * (180/Math.PI)

Related

How to have isometric projection be a counter-clockwise rotation of flat 2d view

I am making this isometric projection of a 2D table in a Javascript project. I'm calculating my isometric coordinates (from cartesian) with the following function:
var isoX = x - y;
var isoY = (x + y) / 2;
The problem I have is that this rotates everything clock-wise:
AB
CD
becomes
A
C B
D
Instead, I would like it to become
B
A D
C
Can I adjust the function above somehow to achieve this? If not, is there another way to do this?
One idea is to rotate the table before the projection step.
The 2D rotation matrix for 90 degrees counter clockwise is
[ 0 1]
[-1 0]
Putting it in code, this transformation rotates everything around (0,0) 90 degrees ccw:
var transformedX = y;
var transformedY = -x;
Then, simply use the transformed coordinates for the projection.
var isoX = transformedX - transformedY;
var isoY = (transformedX + transformedY) / 2;

How can I offset a global directional force to be applied over a local axis?

I want to apply a forward force in relation to the object's local axis, but the engine I'm using only allows to me apply a force over the global axis.
I have access to the object's global rotation as a quaternion. I'm not familiar with using quats however (generally untrained in advanced maths). Is that sufficient information to offset the applied force along the desired axis? How?
For example, to move forward globally I would do:
this.entity.rigidbody.applyForce(0, 0, 5);
but to keep that force applied along the object's local axis, I need to distribute the applied force in a different way along the axes, based on the object's rotational quat, for example:
w:0.5785385966300964
x:0
y:-0.815654993057251
z:0
I've researched quaternions trying to figure this out, but watching a video on what they are and why they're used hasn't helped me figure out how to actually work with them to even begin to figure out how to apply the offset needed here.
What I've tried so far was sort of a guess on how to do it, but it's wrong:
Math.degrees = function(radians) {
return radians * 180 / Math.PI;
};
//converted this from a python func on wikipedia,
//not sure if it's working properly or not
function convertQuatToEuler(w, x, y, z){
ysqr = y * y;
t0 = 2 * (w * x + y * z);
t1 = 1 - 2 * (x * x + ysqr);
X = Math.degrees(Math.atan2(t0, t1));
t2 = 2 * (w * y - z * x);
t2 = (t2 >= 1) ? 1 : t2;
t2 = (t2 < -1) ? -1 : t2;
Y = Math.degrees(Math.asin(t2));
t3 = 2 * (w * z + x * y);
t4 = 1 - 2 * (ysqr + z * z);
Z = Math.degrees(Math.atan2(t3, t4));
console.log('converted', {w, x, y, z}, 'to', {X, Y, Z});
return {X, Y, Z};
}
function applyGlobalShift(x, y, z, quat) {
var euler = convertQuatToEuler(quat.w, quat.x, quat.y, quat.z);
x = x - euler.X; // total guess
y = y - euler.Y; // total guess
z = z - euler.Z; // total guess
console.log('converted', quat, 'to', [x, y, z]);
return [x, y, z];
}
// represents the entity's current local rotation in space
var quat = {
w:0.6310858726501465,
x:0,
y:-0.7757129669189453,
z:0
}
console.log(applyGlobalShift(-5, 0, 0, quat));
Don't laugh at my terrible guess at how to calculate the offset :P I knew it was not even close but I'm really bad at math
Quaternions are used as a replacement for euler angles. Your approach, thus, defeats their purpose. Instead of trying to use euler angles, levy the properties of a quaternion.
A quaternion has 4 components, 3 vector components and a scalar component.
q = x*i + y*j + z*k + w
A quaternion therefore has a vector part x*i + y*j + z*k and a scalar part w. A vector is thus a quaternion with a zero scalar or real component.
It is important to note that a vector multiplied by a quaternion is another vector. This can be easily proved by using the rules of multiplication of quaternion basis elements (left as an exercise for the reader).
The inverse of a quaternion is simply its conjugate divided by its magnitude. The conjugate of a quaternion w + (x*i + y*j + z*k) is simply w - (x*i + y*j + z*k), and its magnitude is sqrt(x*x + y*y + z*z + w*w).
A rotation of a vector is simply the vector obtained by rotating that vector through an angle about an axis. Rotation quaternions represent such an angle-axis rotation as shown here.
A vector v can be rotated about the axis and through the angle represented by a rotation quaternion q by conjugating v by q. In other words,
v' = q * v * inverse(q)
Where v' is the rotated vector and q * v * inverse(q) is the conjugation operation.
Since the quaternion represents a rotation, it can be reasonably assumed that its magnitude is one, making inverse(q) = q* where q* is the conjugate of q.
On separating q into real part s and vector part u and simplifying the quaternion operation (as beautifully shown here),
v' = 2 * dot(u, v) * u + (s*s - dot(u, u)) * v + 2 * s * cross(u, v)
Where dot returns the dot product of two vectors, and cross returns the cross product of two vectors.
Putting the above into (pseudo)code,
function rotate(v: vector3, q: quaternion4) -> vector3 {
u = vector3(q.x, q.y, q.z)
s = q.w
return 2 * dot(u, v) * u + (s*s - dot(u, u)) * v + 2 * s * cross(u, v)
}
Now that we know how to rotate a vector with a quaternion, we can use the world (global) rotation quaternion to find the corresponding world direction (or axis) for a local direction by conjugating the local direction by the rotation quaternion.
The local forward axis is always given by 0*i + 0*j + 1*k. Therefore, to find the world forward axis for an object, you must conjugate the vector (0, 0, 1) with the world rotation quaternion.
Using the function defined above, the forward axis becomes
forward = rotate(vector3(0, 0, 1), rotationQuaternion)
Now that you have the world forward axis, a force applied along it will simply be a scalar multiple of the world forward axis.

Calculating the coordinates of a point on circle, along a line between the center of the circle and another point outside it?

Here's an image to demonstrate the question:
Let's say I have Point A at [0,0], and Point B at [50, 30]. I want to find the coordinates of Point X, along a circle of radius 15, with an origin at Point A, which is also on a line between Point A and Point B.
Pointers on the best method to do this?
Since this has been tagged JavaScript, here's a simple implementation:
// disclaimer: code written in browser
function Point2D(x, y) {
this.x = x;
this.y = y;
}
function findCircleInteresction(center, radius, target) {
var vector = new Point2D(target.x - center.x, target.y - target.y);
var length = Math.sqrt(Math.pow(vector.x, 2) + Math.pow(vector.y, 2));
var normal = new Point2D(vector.x / length, vector.y / length);
var result = new Point2D(center.x + (normal.x * radius), center.y + (normal.y * radius));
return result;
}
findCircleInteresction(new Point2D(0, 0), 15, new Point2D(50, 30));
Point2D is just a class to make objects with x and y properties.
findCircleInteresction takes three parameters:
- center the center of the circle
- radius the radius of the circle
- target a point outside the circle
In findCircleInteresction:
- calculate the vector between the center and the target
- get the length of the resulting vector
- compute the normal (normalized) of the vector
- find the point where the vector intersects with the circle by adding the center of the circle plus the normalized vector components multiplied by the radius of the circle
This code could be heavily optimized and it's untested but I think it illustrated the idea.
You would want to think of this as two overlapping triangles, one with sides Bx-Ax and By-Ay. What you want is to find the coordinates of X, which would specifically be a triangle with sides Xx-Ax and Xy-Ay but with known hypotenuse R, which is your radius of the circle. Notice that the angle for both triangles are equal in respect to the x-coordinates-axis.
So to get the angle of the triangle, take the arctan(By-Ay/Bx-Ax) Now with that angle, call it T, you can solve for the smaller legs with your know radius R.
To get the x coordinate you would take Rcos(T)
To get the y coordinate you would take Rsin(T)
Bringing it all together you have that Xx = Rcos(T) and Xy = Rsin(T)
If you are not willing to use a Math library, which this method would use, you can use ratio's (as Pointy commented)

Determine movement vector's direction from velocity

I'm kinda confused with this one.
I have an object and I know it's velocities on axis x and y. My problem is how to determine the angle at which it's moving.
function Object(){
this.velocity = {x: 5, y: 1};
}
Basically I Know that a vector's direction is x_projectioncos(deg) + y_projectionsin(deg), but I don't know how to get those projections since I only have the velocity, as I said I'm really confused.
#EDIT:
in addition to the accepted answer, here's what I did to get a full 360 degree spectrum
var addDeg = 0;
if(obj.velocity.x<0)
addDeg = obj.velocity.y>=0 ? 180 : 270;
else if(obj.velocity.y<=0) addDeg = 360;
deg = Math.abs(Math.abs(Math.atan(obj.velocity.y/obj.velocity.x)*180/Math.PI)-addDeg)
I don't know how to get those projections since I only have the
velocity
Actually, what you seem to be missing is that you already have the projections. That's what x and y are.
x is speed * cos(angle)
y is speed * sin(angle)
So y/x = sin(angle)/cos(angle) which is tan(angle) so angle=arctan(y/x).
That's the angle rotating anti-clockwise starting from the x axis (with x pointing right and y pointing up).
Find the angle between that vector and (1,0) (Right horizontal positive direction).
The math is:
A = (5,1)
B = (1,0)
A.B = |A||B|cos(angle) -> angle = arccos((|A||B|)/(A.B))
Dot product, check geometric definition
Edit:
Another option is to use the cross product formula:
|AxB| = |A||B|sin(angle) -> angle = arcsin((|A||B|)/(|AxB|))
It will give you the angle you need.
There is an easier way to get full 360 degrees. What you're looking for, is Math.atan2:
deg = Math.atan2(obj.velocity.y,obj.velocity.x)*180/Math.PI;

How to change an objects x and y coordinates by degrees?

I'm working on some coordinates function to my canvas in HTML5, and I want to make a function which can move an object by degrees.
My dream is to make a function which works like this:
box.x=10;
box.y=10;
// now the box has the coordinates (10,10)
moveTheBoxWithThisAmountOfDistance=10;
degreesToMoveTheBox=90;
box.moveByDegrees(moveTheBoxWithThisAmountOfDistance,degreesToMoveTheBox);
// now the box.x would be 20 and box.y wouldn't be changed because
// we only move it to the right (directional 90 degrees)
I hope this makes any sense!
So my question is:
How does the mathematical expression look like when I have to turn a degree into to coordinates?
You use sin and cos to convert an angle and a distance into coordinates:
function moveByDegrees(distance, angle) {
var rad = angle * Math.pi / 180;
this.x += Math.cos(rad) * distance;
this.y += Math.sin(rad) * distance;
}
That's it: http://en.wikipedia.org/wiki/Rotation_matrix , all the math is described :)
Also beware that if you need multiple sequential rotations (i.e. you do continuous animation), it's better to recompute x' and y' from initial ones and not just previous. Not doing so will result in rounding errors accumulation, and after some thousand rotations the result will become too rough.

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