Additive Angles in Javascript - javascript

I'm trying to calculate the angle between a still and moving vector in javascript. However, I want the angle to be additive based on direction (so if you're moving clockwise, the angle would always increase, whereas moving counterclockwise would cause the angle to only decrease).
I'm storing the coordinates in arrays as start[x, y] and current[x,y] and need to calculate the angle while the current array changes. I'm also currently using the atan2 function, but this is limited to -180 to +180 degrees.
start = [event.clientX - discCent[0], event.clientY - discCent[1]];
current = [event.clientX - discCent[0], event.clientY - discCent[1]];
// Get this to be additive
angleDeg = Math.atan2(current[1] - start[1], current[0] - start[0]) * 180 / Math.PI;
Thanks!

Keep the previous angle and add/subtract a multiple of 360° to get closest to it:
var angleDegPrev = 0.; // initialization at start
...
// compute angle in ]-180,180]
start = [event.clientX - discCent[0], event.clientY - discCent[1]];
current = [event.clientX - discCent[0], event.clientY - discCent[1]];
angleDeg = Math.atan2(current[1] - start[1], current[0] - start[0]) * 180 / Math.PI;
// add multiple of 360 to get closest to previous angle
angleDeg += Math.round((angleDegPrev - angleDeg)/360.)*360.;
angleDegPrev = angleDeg;

Related

What is the easiest way to calculate position of balls on collision?

I'm trying to make some simple pool game in java script. I have made it but I do not love way of checking if two balls will collide in next frame. I would like to have more easier way to calculate coordinates of balls when collision occurs. I found lot of answers base on collision kinematics, how to handle velocities and directions after collision, but no calculating a position when collision occurs.
As you can see in sample diagram, gold ball is moving slower than a blue ball, and with distance that each ball will have to move on next frame will not be considered as collision. But, as you can see, they should collide (dashed lines).
In that cause I have divided each movement into sectors and calculating if distance between the points is equal or smaller than ball diameter, which is slowing down process when many balls (like in snooker) have to be calculated in each frame, plus that way is not always 100% accurate and balls can go in inaccurate angles after hit (not a big difference, but important in snooker).
Is there any easier way to calculate those (XAC,YAC) and (XBC,YBC) values with knowing start positions and velocities of each ball without dividing ball paths into sectors and calculating many times to find a proper distance?
It is worth to precalculate collision event only once (this approach works well with reliable number of balls, because we have to treat all ~n^2 pairs of balls).
The first ball position is A0, velocity vector is VA.
The second ball position is B0, velocity vector is VB.
To simplify calculations, we can use Halileo principle - use moving coordinate system connected with the first ball. In that system position and velocity of the first ball are always zero. The second ball position against time is :
B'(t) = (B0 - A0) + (VB - VA) * t = B0' + V'*t
and we just need to find solution of quadratic equation for collision distance=2R:
(B0'.X + V'.X*t)^2 + (B0'.X + V'.Y*t)^2 = 4*R^2
Solving this equation for unknown time t, we might get cases: no solutions (no collision), single solution (only touch event), two solutions - in this case smaller t value corresponds to the physical moment of collision.
Example (sorry, in Python, ** is power operator):
def collision(ax, ay, bx, by, vax, vay, vbx, vby, r):
dx = bx - ax
dy = by - ay
vx = vbx - vax
vy = vby - vay
#(dx + vx*t)**2 + (dy + vy*t)**2 == 4*r*r solve this equation
#coefficients
a = vx**2 + vy**2
b = 2*(vx*dx + vy*dy)
c = dx**2+dy**2 - 4*r**2
dis = b*b - 4*a*c
if dis<0:
return None
else:
t = 0.5*(-b - dis**0.5)/a ##includes case of touch when dis=0
return [(ax + t * vax, ay + t * vay), (bx + t * vbx, by + t * vby)]
print(collision(0,0,100,0,50,50,-50,50,10)) #collision
print(collision(0,0,100,0,50,50,-50,80,10)) #miss
print(collision(0,0,100,0,100,0,99,0,10)) #long lasting chase along OX axis
[(40.0, 40.0), (60.0, 40.0)]
None
[(8000.0, 0.0), (8020.0, 0.0)]
Regarding to MBo's solution, here is a function in java script that will calculate coordinates of balls on collision and time in which collision will happen:
calcCollisionBallCoordinates(ball1_x, ball1_y, ball2_x, ball2_y, ball1_vx, ball1_vy, ball2_vx, ball2_vy, r) {
let dx = ball2_x - ball1_x,
dy = ball2_y - ball1_y,
vx = ball2_vx - ball1_vx,
vy = ball2_vy - ball1_vy,
a = Math.pow(vx, 2) + Math.pow(vy, 2),
b = 2 * (vx * dx + vy * dy),
c = Math.pow(dx, 2) + Math.pow(dy, 2) - 4 * Math.pow(r, 2),
dis = Math.pow(b, 2) - 4 * a * c;
if (dis < 0) {
//no collision
return false;
} else {
let t1 = 0.5 * (-b - Math.sqrt(dis)) / a,
t2 = 0.5 * (-b + Math.sqrt(dis)) / a,
t = Math.min(t1, t2);
if (t < 0) {
//time cannot be smaller than zero
return false;
}
return {
ball1: {x: ball1_x + t * ball1_vx, y: ball1_y + t * ball1_vy},
ball2: {x: ball2_x + t * ball2_vx, y: ball2_y + t * ball2_vy},
time: t
};
}
}

Get the angle degrees from 2 points on a grid

I have a grid for a flat, 2D map which is a bit different from usual. The top left point is 0,0 and bottom right is 1000,1000. I have 2 points on this map, an origin/anchor point and a destination.
I am looking to figure out the degrees (in javascript) from the origin to the destination. I have looked through many answers and they all don't produce the correct result.
function getAngle(origin_x, origin_y, destination_x, destination_y) {
var newx = destination_x - origin_x;
var newy = destination_y - origin_y;
var theta = Math.atan2(-newy, newx);
if (theta < 0) {
theta += 2 * Math.PI;
}
theta *= 180 / Math.PI;
return theta;
}
This is what I have so far but it doesn't produce the right angle.
Thankyou very much in advance!
image from mdn Math.atan2 doc
It will give the angle relative to the x-axis, not the y-axis. To convert all you would do is
var newAngle = 90 - theta;

Find the Points of Intersection of a Circle with a Line in Javascript

I'm trying to animate a given element to go around a pre-defined radius and I'm having trouble getting the position of the element at a Y point given.
I'm trying to find each point with the circle equation, but I can only get one point out of the two possible ones.
In Javascript, I use Math.sqrt( Math.pow(radius, 2) - Math.pow(y, 2) , 2) to get the point. assuming the center of the of the circle is 0,0.
but then I need to translate it to pixels on the screen since there are no negative pixels in positions on the browser.
All the sizing is relative to the window. so the radius, for example, is 80% of the height of the window in my tests.
Also, I'm trying to calculate what the distance of the element between each frame should be for the duration, but I'm not using it yet because I try to fix the issue above first.
This is what I have(a cleaned up version):
let height = window.innerHeight * 0.8,
radius = height / 2,
circumferance = (radius * 2) * Math.PI,
container = document.getElementById('container'),
rotating = document.querySelector('.rotating'),
centerX = radius - (rotating.offsetWidth / 2),
centerY = radius - (rotating.offsetHeight / 2),
duration = 10,
stepDistance = circumferance / 16;
// Setting the dimensions of the container element.
container.style.height = height + 'px';
container.style.width = height + 'px';
// return positive X of any given Y.
function getXOffset(y) {
return Math.sqrt( Math.pow(radius, 2) - Math.pow(y, 2) , 2);
}
// Setting the position of the rotating element to the start.
rotating.style.top = 0 + 'px';
rotating.style.left = centerX + 'px';
setInterval(() => {
let top = parseInt(rotating.style.top),
y = radius - top;
rotating.style.top = (top + 1) + 'px';
rotating.style.left = (centerX + getXOffset(y)) + 'px';
}, 16);
Here is a fiddle with a bit more code for trying to get the right amount of distance between points for a smoother animation(currently needs fixing, but it doesn't bother me yet.)
https://jsfiddle.net/shock/1qcfvr4y/
Last note: I know that there might be other ways to do this with CSS, but I chose to use javascript for learning purposes.
Math.sqrt would only return the positive root. You'll have to account for the negative value based on the application. In this case, you need the positive x value during the 1st half of the cycle and negative during the 2nd half.
To do that, you should implement a method to track the progress and reverse the sign accordingly.
Here is a sample. I modified upon yours.
edit:
Instead of Math.sqrt( Math.pow(radius, 2) - Math.pow(y, 2) , 2) You can use the full formula to get x if you do not want to assume origin as center, which in this case is Math.sqrt( Math.pow(radius, 2) - Math.pow((actualY - centerY), 2) , 2)
explanation:
The original equation (x-a)² + (y'-b)² = r²
becomes x = √(r² - (y'-b)²) + a
Assuming .rotating box have 0 width and height.
The variable equivalents in your code are centerX = a, centerY = b.
By assuming origin as center you're basically doing a pre-calculation so that your y value becomes the equivalent of (y'-b). Hence x = √(r² - y²) + a is valid.
At initial state top = 0
i.e (y'-b) => height - centerY.
In your code y = radius => height/2.
Now (height - centerY) being equal to (height/2) is a side effect of your circle being bound by a square container whose height determines the y value.
In other words, when you use origin as center, you are taking the center offsets outside of circle equation and handling it separately. You could do the same thing by using the whole formula, that is, x = √(r² - (y'-b)²) + a

Rotating an image to face mouse point using ATAN instead of ATAN2

In order to better understand how trigonometry works in game development, I've been creating little javascript snippets on CodePen.
I managed to create an example that uses Math.atan2() to point a pixel-art shotgun at the mouse cursor.
Now, I am trying to accomplish the same exact thing using the Math.atan() function but it isn't functioning properly.
Here is the logic code I am using:
canvas.onmousemove = function(event) {
Mouse = {
x: event.pageX,
y: event.pageY
}
// These length variables use the distance formula
var opposite_length = Math.sqrt((Mouse.x - Mouse.x) * (Mouse.x - Mouse.x) + (Mouse.y - y) * (Mouse.y - y));
var adj_length = Math.sqrt((Mouse.x - x) * (Mouse.x - x) + (y - y) * (y - y));
var angle_in_radians = Math.atan(opposite_length / adj_length);
//var angle_in_radians = Math.atan2(Mouse.y - y, Mouse.x - x);
angle = angle_in_radians * (180 / Math.PI);
}
The in my draw() function, I rotate the gun to the angle var using:
cxt.rotate(angle*(Math.PI/180));
If you uncomment the line that starts as // var angle_in_radians, everything will suddenly work.
So, atan2 is working, but atan is producing the result I want.
I know that opposite_length and adj_length are accurate, because when i console.log() them, they are the correct values.
You can check out the code being used on CodePen for a live example.
There's a lot of initialization stuff but you only really need to focus on the canvas.onmousemove = function(event) section, starting on line 50. You can also check out my draw function on line 68.
Note that your atan computation is equivalent to
atan2( abs(mouse.y-y), abs(mouse.x-x) )
The screen coordinates have the opposite orientation to the cartesian coordinates. To get a cartesian angle from screen coordinates, use
atan2( y-mouse.y, mouse.x-x )

Get new x,y coordinates of a point in a rotated image

I have Google Maps icons which I need to rotate by certain angles before drawing on the map using MarkerImage. I do the rotation on-the-fly in Python using PIL, and the resulting image is of the same size as the original - 32x32. For example, with the following default Google Maps marker:
, a 30 degrees conter-clockwise rotation is achieved using the following python code:
# full_src is a variable holding the full path to image
# rotated is a variable holding the full path to where the rotated image is saved
image = Image.open(full_src)
png_info = image.info
image = image.copy()
image = image.rotate(30, resample=Image.BICUBIC)
image.save(rotated, **png_info)
The resulting image is
The tricky bit is getting the new anchor point to use when creating the MarkerImage using the new rotated image. This needs to be the pointy end of the icon. By default, the anchor point is the bottom middle [defined as (16,32) in x,y coordinates where (0,0) is the top left corner]. Can someone please explain to me how I can easily go about this in JavaScript?
Thanks.
Update 22 Jun 2011:
Had posted the wrong rotated image (original one was for 330 degrees counter-clockwise). I've corrected that. Also added resampling (Image.BICUBIC) which makes the rotated icon clearer.
To calculate the position of a rotated point you can use a rotation matrix.
Converted into JavaScript, this calculates the rotated point:
function rotate(x, y, xm, ym, a) {
var cos = Math.cos,
sin = Math.sin,
a = a * Math.PI / 180, // Convert to radians because that is what
// JavaScript likes
// Subtract midpoints, so that midpoint is translated to origin
// and add it in the end again
xr = (x - xm) * cos(a) - (y - ym) * sin(a) + xm,
yr = (x - xm) * sin(a) + (y - ym) * cos(a) + ym;
return [xr, yr];
}
rotate(16, 32, 16, 16, 30); // [8, 29.856...]
The formula for rotations about 0,0 is:
x1 = cos(theta) x0 - sin(theta) y0
y1 = sin(theta) x0 + cos(theta) y0
But that's for regular axes, and rotation about 0,0. The PIL rotation is clockwise with "graphics" axes. Plus, it's around the center of the image. The final confusing thing is that the size of the image can change, which needs to be accounted for in the final result.
Procedure: take original point, subtract off center of image, apply "graphics axes" corrected rotation, find new size of image, add back center position of new image.
Rotation using graphics axes is:
x1 = cos(theta) x0 + sin(theta) y0
y1 = -sin(theta) x0 + cos(theta) y0
16,32 - 16,16 is 0, 16. Rotate 30 degrees clockwise rotation (based on your images) gives a point cos(-30)*0+sin(-30)*16, -sin(-30)*0+cos(-30)*16 = -8, 13.86. The final step is adding back the center position of the rotated position.
In an image, downwards is positive Y and rightwards is positive X. However, to apply the rotation formula, we need upwards as positive Y. Therefore, step 1 would be to apply f(x,y) = f(x,h-y), where 'h' is the height of the image.
Let's say the image is rotated with respect to x0,y0. You'd then need to transform your origin to this point. Therefore, step 2 would be f(x,y) = f(x-x0,y-y0). At this stage (i.e. after the two steps), your new co-ordinates would be x-x0, h-y-y0. You're now ready to apply the rotation formula
x1 = x*cos(theta) - y*sin(theta)
y1 = xsin(theta) + ycos(theta)
Use the values of x and y obtained after step two.
You'd get
x1 = (x-x0)*cos(theta) - (h-y-y0)*sin(theta)
y1 = (x-x0)*sin(theta) + (h-y-y0)*cos(theta)
Now, undo transformations done in step 2 and step 1 (in that order).
After undoing step2: xNew = x1 + x0 and yNew = y1 + y0
After undoing step1: xNew = x1 + x0 and yNew = h - (y1 + y0)
This gives you:
xNew = (x-x0)*cos(theta) - (h-y-y0)*sin(theta) + x0
yNew = -(x-x0)*sin(theta) - (h-y-y0)*cos(theta) + (h-y0)

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