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I have a flying car that I want to lose some HP when colliding with objects.
I connected car's RigidBody2D to this function to do that
func _on_Car_body_entered(body):
var force = linear_velocity.length()
var dmg = pow(force / 100, 2) - 0.25
if dmg <= 0: return
Health = Health - dmg
Now, since I don't have to be precise I'm just using current velocity as the force, though this is up for change.
After getting my 'force of impact', I put it into damage calculating formula and if damage is above 0, decrease HP by damage.
This works fine in most cases
BUT
I noticed that if car's going fast horizontally and just barely touch the ground (that's perfectly horizontal), car gets hit with a lot of damage, because I'm using the length of the velocity vector.
Ofcourse, this case can be managed by using just the Y component of the velocity vector, but then it removes any horizontal collisions, and vice versa, and it also leads me on to the path of programming vertical and horizontal collisions, and ofcourse those are not the only 2 directions of colisions I need.
Is there a way to remove the sliding factor from this equation?
You can get the sine of the angle between your velocity and the collision normal, and then take the absolute of that.
# 0 When sliding along the wall. 1 when hitting the wall head on
var slide_factor = abs(cos(vel_last_frame.angle_to(collision_normal)))
This will give you a value from 0 to 1. When you are just sliding along the wall, this value will be 0, and when you hit the wall straight on, it will be 1.
I am using the velocity from the last frame here so that it gets the velocity just before the collision. I get it by setting vel_last_frame to linear_velocity inside the _physics_process function.
You can only get the collision normal inside the _integrate_forces function using PhysicsDirectBodyState.get_local_contact_normal(), so you need to make a variable that can be accessed in this function and the _on_Car_body_entered function. Note that you need to set contact_monitor to true and contacts_reported to at least 1 for this function to work.
var collision_normal
func _integrate_forces(state):
# Check if there is a collision
if state.get_contact_count():
# contact_monitor must be true and contacts_reported must be at least 1 for this to work
collision_normal = state.get_contact_local_normal(0)
Now in the _on_Car_body_entered_function, you can multiply dmg by sliding_factor to scale it less depending on how much you are sliding against the wall.
func _on_Car_body_entered(body):
var force = linear_velocity.length()
# 0 When sliding along the wall. 1 when hitting the wall head on
var slide_factor = abs(cos(vel_last_frame.angle_to(collision_normal)))
var dmg = pow(force / 100, 2) - 0.25
# Reduce dmg depending on how much you are sliding against the wall
dmg *= slide_factor
if dmg <= 0: return
Health = Health - dmg
Found a solution for my problem here
This gives me a predictable force range to work with.
I copied all code for 2D collision, just added damage calculation
Range of forces my objects produce is <3000 for small collisions like scratches and bumps, ~10k for beginner friendly damage, and 20k+ for when I really slam the gas pedal, so I just convert that force to damage that I want.
Best part is that I don't have to use the body_entered from RigidBody2D, because now all my cars have this calculation in them, so when 2 of them collide they both get damaged.
extends RigidBody2D
var collision_force : Vector2 = Vector2.ZERO
var previous_linear_velocity : Vector2 = Vector2.ZERO
func _integrate_forces(state : Physics2DDirectBodyState)->void:
collision_force = Vector2.ZERO
if state.get_contact_count() > 0:
var dv : Vector2 = state.linear_velocity - previous_linear_velocity
collision_force = dv / (state.inverse_mass * state.step)
var dmg = collision_force.length() / 2000 - 2
if dmg > 0:
set_hp(Health - dmg)
emit_signal("Damaged")
previous_linear_velocity = state.linear_velocity
**OLD ANSWER**
RUBBER DUCK HERE
In script for car I added a new variable var last_linear_velocity = Vector2()
Then stored the last velocity in _process
func _process(delta):
last_linear_velocity = linear_velocity
Not in _integrate_forces because if I put it there then the new and last velocities are the same.
And just changed how force is calculated in the function mentioned above, so it looks like this
func _on_Car_body_entered(body):
var force = last_linear_velocity.length() - linear_velocity.length()
var dmg = pow(force / 100, 2) - 0.25
if dmg <= 0: return
Health = Health - dmg
Now I get a nice predicable range of values and can transform that to damage.
NOTE
I noticed that sometimes when collision occures the difference between the last and current velocity lengths is negative, as in - car is accelerating.
Anyway, this works for me for now.
If you find a better solutions do post it, as I couldn't find a solution to this problem online elswhere.
I am trying to draw a clock that works. I am using a 600x600 form. I cant' figure out how to place the oval in the center of the form or how to add the minutes or the seconds tick marks inside the oval. I tried dash but couldn't get it to look right. Any suggestions. Thanks in advance.
This is what I have done so far:
from tkinter import *
canvas_width = 600
canvas_height = 600
master = Tk()
w = Canvas(master, width = canvas_width, height = canvas_height)
w.pack()
oval = w.create_oval(75,75,500,500)
minline = w.create_line(150,150,300,300)
mainloop()
The center of a drawn shape is the middle of the two points specified when it is drawn. Currently, the middle point of your shape (draw from 75, 75 to 500, 500) is 237.5, so, if you want the middle of it to be the middle of your page, and keep the 75, 75 coordinate, you would have to make the other one 525, 525 to completely mirror the first.
As for drawing the shape, you'll need some math in python, so I would first suggest doing an image as the background for the clock, so that less objects are drawn. But, if you must do it without other images, you must first import the math library.
import math
Now, for a mathematic principle: Any point on the circle of radius r can be expressed as the point (r*cosθ), (r*sinθ), where θ is the angle from the center to the point. The reason this is important is that you want each line on the side of the clock face to be pointing towards the center of the circle. To do that, we need the two points to draw the line on that together point towards the center, and fortunately for us this means that both points on the line are on different circles (our circle and one within it) but are at the same angle from the center.
Since we want 12 hour points around the circle, and 4 minute points between each of those (so 60 points in total), and 360 degrees in a circle (so 1 point for every 6 degrees), we will need a for loop that goes through that.
for angle in range(0, 360, 6):
Then we'll want 3 constants: One for the radius of the exterior circle (for the points to begin from), one for an interior circle (for the minute points to add at), and one for an even more interior circle (for the hour points to end at). We'll also want it to choose the more interior radius only every 30 degrees (because it appears every 5 points, and there are 6 degrees between them).
radius_out = 225
radius_in = 0 #temporary value
if (angle % 30) == 0: #the % symbol checks for remainder
radius_in = 210
else:
radius_in = 220
Now, for the conversion into radians (As math in python needs radians for sin and cos):
radians = (angle / 180) * math.pi
Next off, assigning the coordinates to variables so it's easier to read.
x_out = (radius_out * math.cos(radians)) + 300
y_out = (radius_out * math.sin(radians)) + 300
x_in = (radius_in * math.cos(radians)) + 300
y_in = (radius_in * math.sin(radians)) + 300
#the (+ 300) moves each point from a relative center of 0,0 to 300,300
And finally we assign it to a list so we can access it later if we need to. Make sure to define this list earlier outside of the for loop.
coords.append( w.create_line(x_out, y_out, x_in, y_in) )
This should give you your clock lines.
NOTE: Due to the way tkinter assigns x and y coordinates, this will draw lines from the 3 hour line clockwise back to it.
Hope this was helpful! If there is anything you don't understand, comment it below.
EDIT - Thanks for all the answers everyone. I think I accidentally led you slightly wrong as the square in the picture below should be a rectangle (I see most of you are referencing squares which seems like it would make my life a lot easier). Also, the x/y lines could go in any direction, so the red dot won't always be at the top y boundary. I was originally going for a y = mx + b solution, but then I got stuck trying to figure out how I know whether to plug in the x or the y (one of them has to be known, obviously).
I have a very simple question (I think) that I'm currently struggling with for some reason. I'm trying to have a type of minimap in my game which shows symbols around the perimeter of the view, pointing towards objectives off-screen.
Anyway, I'm trying to find the value of the red point (while the black borders and everything in green is known):
It seems like simple trigonometry, but for some reason I can't wrap my head around it. I just need to find the "new" x value from the green point to the red point, then I can utilize basic math to get the red point, but how I go about finding that new x is puzzling me.
Thanks in advance!
scale = max(abs(x), abs(y))
x = x / scale
y = y / scale
This is the simple case, for a square from (-1, -1) to (1, 1). If you want a different sized square, multiply the coordinates by sidelen / 2.
If you want a rectangle instead of a square, use the following formula. (This is another solution to the arbitrarily-sized square version)
scale = max(abs(x) / (width / 2), abs(y) / (height / 2))
x = x / scale
y = y / scale
Let's call the length of one side of the square l. The slope of the line is -y/x. That means, if you move along the line and rise a distance y toward the top of the square, then you'll move a distance x to the left. But since the green point is at the center of the square, you can rise only l/2. You can express this as a ratio:
-y -l/2
——— = ———
x d
Where d is the distance you'll move to the left. Solving for d, we have
d = xl/2y
So if the green dot is at (0, 0), the red dot is at (-l/2, xl/2y).
All you need is the angle and the width of the square w.
If the green dot is at (0,0), then the angle is a = atan(y/x), the y-coordinate of the dot is w/2, and therefore the x-coordinate of the dot is tan(1/a) * (w/2). Note that tan(1/a) == pi/2 - tan(a), or in other words the angle you really want to plug into tan is the one outside the box.
Edit: yes, this can be done without trig, too. All you need is to interpolate the x-coordinate of the dot on the line. So you know the y-coordinate is w/2, then the x-coordinate is (w/2) * x/y. But, be careful which quadrant of the square you're working with. That formula is only valid for -y<x<y, otherwise you want to reverse x and y.
I'm making a SHMUP game that has a space ship. That space ship currently fires a main cannon from its center point. The sprite that represents the ship has a center based registration point. 0,0 is center of the ship.
When I fire the main cannon i make a bullet and assign make its x & y coordinates match the avatar and add it to the display list. This works fine.
I then made two new functions called fireLeftCannon, fireRightCannon. These create a bullet and add it to the display list but the x, y values are this.y + 15 and this.y +(-) 10. This creates a sort of triangle of bullet entry points.
Similar to this:
▲
▲ ▲
the game tick function will adjust the avatar's rotation to always point at the cursor. This is my aiming method. When I shoot straight up all 3 bullets fire up in the expected pattern. However when i rotate and face the right the entry points do not rotate. This is not an issue for the center point main cannon.
My question is how do i use the current center position ( this.x, this.y ) and adjust them based on my current rotation to place a new bullet so that it is angled correctly.
Thanks a lot in advance.
Tyler
EDIT
OK i tried your solution and it didn't work. Here is my bullet move code:
var pi:Number = Math.PI
var _xSpeed:Number = Math.cos((_rotation - 90) * (pi/180) );
var _ySpeed:Number = Math.sin((_rotation - 90) * (pi / 180) );
this.x += (_xSpeed * _bulletSpeed );
this.y += (_ySpeed * _bulletSpeed );
And i tried adding your code to the left shoulder cannon:
_bullet.x = this.x + Math.cos( StaticMath.ToRad(this.rotation) ) * ( this.x - 10 ) - Math.sin( StaticMath.ToRad(this.rotation)) * ( this.x - 10 );
_bullet.y = this.y + Math.sin( StaticMath.ToRad(this.rotation)) * ( this.y + 15 ) + Math.cos( StaticMath.ToRad(this.rotation)) * ( this.y + 15 );
This is placing the shots a good deal away from the ship and sometimes off screen.
How am i messing up the translation code?
What you need to start with is, to be precise, the coordinates of your cannons in the ship's coordinate system (or “frame of reference”). This is like what you have now but starting from 0, not the ship's position, so they would be something like:
(0, 0) -- center
(10, 15) -- left shoulder
(-10, 15) -- right shoulder
Then what you need to do is transform those coordinates into the coordinate system of the world/scene; this is the same kind of thing your graphics library is doing to draw the sprite.
In your particular case, the intervening transformations are
world ←translation→ ship position ←rotation→ ship positioned and rotated
So given that you have coordinates in the third frame (how the ship's sprite is drawn), you need to apply the rotation, and then apply the translation, at which point you're in the first frame. There are two approaches to this: one is matrix arithmetic, and the other is performing the transformations individually.
For this case, it is simpler to skip the matrices unless you already have a matrix library handy already, in which case you should use it — calculate "ship's coordinate transformation matrix" once per frame and then use it for all bullets etc.
I'll now explain doing it directly.
The general method of applying a rotation to coordinates (in two dimensions) is this (where (x1,y1) is the original point and (x2,y2) is the new point):
x2 = cos(angle)*x1 - sin(angle)*y1
y2 = sin(angle)*x1 + cos(angle)*y1
Whether this is a clockwise or counterclockwise rotation will depend on the “handedness” of your coordinate system; just try it both ways (+angle and -angle) until you have the right result. Don't forget to use the appropriate units (radians or degrees, but most likely radians) for your angles given the trig functions you have.
Now, you need to apply the translation. I'll continue using the same names, so (x3,y3) is the rotated-and-translated point. (dx,dy) is what we're translating by.
x3 = dx + x2
y3 = dy + x2
As you can see, that's very simple; you could easily combine it with the rotation formulas.
I have described transformations in general. In the particular case of the ship bullets, it works out to this in particular:
bulletX = shipPosX + cos(shipAngle)*gunX - sin(shipAngle)*gunY
bulletY = shipPosY + sin(shipAngle)*gunX + cos(shipAngle)*gunY
If your bullets are turning the wrong direction, negate the angle.
If you want to establish a direction-dependent initial velocity for your bullets (e.g. always-firing-forward guns) then you just apply the rotation but not the translation to the velocity (gunVelX, gunVelY).
bulletVelX = cos(shipAngle)*gunVelX - sin(shipAngle)*gunVelY
bulletVelY = sin(shipAngle)*gunVelX + cos(shipAngle)*gunVelY
If you were to use vector and matrix math, you would be doing all the same calculations as here, but they would be bundled up in single objects rather than pairs of x's and y's and four trig functions. It can greatly simplify your code:
shipTransform = translate(shipX, shipY)*rotate(shipAngle)
bulletPos = shipTransform*gunPos
I've given the explicit formulas because knowing how the bare arithmetic works is useful to the conceptual understanding.
Response to edit:
In the code you edited into your question, you are adding what I assume is the ship position into the coordinates you multiply by sin/cos. Don't do that — just multiply the offset of the gun position from the ship center by sin/cos and only then add that to the ship position. Also, you are using x x; y y on the two lines, where you should be using x y; x y. Here is your code edited to fix those two things:
_bullet.x = this.x + Math.cos( StaticMath.ToRad(this.rotation)) * (-10) - Math.sin( StaticMath.ToRad(this.rotation)) * (+15);
_bullet.y = this.y + Math.sin( StaticMath.ToRad(this.rotation)) * (-10) + Math.cos( StaticMath.ToRad(this.rotation)) * (+15);
This is the code for a gun at offset (-10, 15).
With the help of the Stack Overflow community I've written a pretty basic-but fun physics simulator.
You click and drag the mouse to launch a ball. It will bounce around and eventually stop on the "floor".
My next big feature I want to add in is ball to ball collision. The ball's movement is broken up into a x and y speed vector. I have gravity (small reduction of the y vector each step), I have friction (small reduction of both vectors each collision with a wall). The balls honestly move around in a surprisingly realistic way.
I guess my question has two parts:
What is the best method to detect ball to ball collision?
Do I just have an O(n^2) loop that iterates over each ball and checks every other ball to see if it's radius overlaps?
What equations do I use to handle the ball to ball collisions? Physics 101
How does it effect the two balls speed x/y vectors? What is the resulting direction the two balls head off in? How do I apply this to each ball?
Handling the collision detection of the "walls" and the resulting vector changes were easy but I see more complications with ball-ball collisions. With walls I simply had to take the negative of the appropriate x or y vector and off it would go in the correct direction. With balls I don't think it is that way.
Some quick clarifications: for simplicity I'm ok with a perfectly elastic collision for now, also all my balls have the same mass right now, but I might change that in the future.
Edit: Resources I have found useful
2d Ball physics with vectors: 2-Dimensional Collisions Without Trigonometry.pdf
2d Ball collision detection example: Adding Collision Detection
Success!
I have the ball collision detection and response working great!
Relevant code:
Collision Detection:
for (int i = 0; i < ballCount; i++)
{
for (int j = i + 1; j < ballCount; j++)
{
if (balls[i].colliding(balls[j]))
{
balls[i].resolveCollision(balls[j]);
}
}
}
This will check for collisions between every ball but skip redundant checks (if you have to check if ball 1 collides with ball 2 then you don't need to check if ball 2 collides with ball 1. Also, it skips checking for collisions with itself).
Then, in my ball class I have my colliding() and resolveCollision() methods:
public boolean colliding(Ball ball)
{
float xd = position.getX() - ball.position.getX();
float yd = position.getY() - ball.position.getY();
float sumRadius = getRadius() + ball.getRadius();
float sqrRadius = sumRadius * sumRadius;
float distSqr = (xd * xd) + (yd * yd);
if (distSqr <= sqrRadius)
{
return true;
}
return false;
}
public void resolveCollision(Ball ball)
{
// get the mtd
Vector2d delta = (position.subtract(ball.position));
float d = delta.getLength();
// minimum translation distance to push balls apart after intersecting
Vector2d mtd = delta.multiply(((getRadius() + ball.getRadius())-d)/d);
// resolve intersection --
// inverse mass quantities
float im1 = 1 / getMass();
float im2 = 1 / ball.getMass();
// push-pull them apart based off their mass
position = position.add(mtd.multiply(im1 / (im1 + im2)));
ball.position = ball.position.subtract(mtd.multiply(im2 / (im1 + im2)));
// impact speed
Vector2d v = (this.velocity.subtract(ball.velocity));
float vn = v.dot(mtd.normalize());
// sphere intersecting but moving away from each other already
if (vn > 0.0f) return;
// collision impulse
float i = (-(1.0f + Constants.restitution) * vn) / (im1 + im2);
Vector2d impulse = mtd.normalize().multiply(i);
// change in momentum
this.velocity = this.velocity.add(impulse.multiply(im1));
ball.velocity = ball.velocity.subtract(impulse.multiply(im2));
}
Source Code: Complete source for ball to ball collider.
If anyone has some suggestions for how to improve this basic physics simulator let me know! One thing I have yet to add is angular momentum so the balls will roll more realistically. Any other suggestions? Leave a comment!
To detect whether two balls collide, just check whether the distance between their centers is less than two times the radius. To do a perfectly elastic collision between the balls, you only need to worry about the component of the velocity that is in the direction of the collision. The other component (tangent to the collision) will stay the same for both balls. You can get the collision components by creating a unit vector pointing in the direction from one ball to the other, then taking the dot product with the velocity vectors of the balls. You can then plug these components into a 1D perfectly elastic collision equation.
Wikipedia has a pretty good summary of the whole process. For balls of any mass, the new velocities can be calculated using the equations (where v1 and v2 are the velocities after the collision, and u1, u2 are from before):
If the balls have the same mass then the velocities are simply switched. Here's some code I wrote which does something similar:
void Simulation::collide(Storage::Iterator a, Storage::Iterator b)
{
// Check whether there actually was a collision
if (a == b)
return;
Vector collision = a.position() - b.position();
double distance = collision.length();
if (distance == 0.0) { // hack to avoid div by zero
collision = Vector(1.0, 0.0);
distance = 1.0;
}
if (distance > 1.0)
return;
// Get the components of the velocity vectors which are parallel to the collision.
// The perpendicular component remains the same for both fish
collision = collision / distance;
double aci = a.velocity().dot(collision);
double bci = b.velocity().dot(collision);
// Solve for the new velocities using the 1-dimensional elastic collision equations.
// Turns out it's really simple when the masses are the same.
double acf = bci;
double bcf = aci;
// Replace the collision velocity components with the new ones
a.velocity() += (acf - aci) * collision;
b.velocity() += (bcf - bci) * collision;
}
As for efficiency, Ryan Fox is right, you should consider dividing up the region into sections, then doing collision detection within each section. Keep in mind that balls can collide with other balls on the boundaries of a section, so this may make your code much more complicated. Efficiency probably won't matter until you have several hundred balls though. For bonus points, you can run each section on a different core, or split up the processing of collisions within each section.
Well, years ago I made the program like you presented here.
There is one hidden problem (or many, depends on point of view):
If the speed of the ball is too
high, you can miss the collision.
And also, almost in 100% cases your new speeds will be wrong. Well, not speeds, but positions. You have to calculate new speeds precisely in the correct place. Otherwise you just shift balls on some small "error" amount, which is available from the previous discrete step.
The solution is obvious: you have to split the timestep so, that first you shift to correct place, then collide, then shift for the rest of the time you have.
You should use space partitioning to solve this problem.
Read up on
Binary Space Partitioning
and
Quadtrees
As a clarification to the suggestion by Ryan Fox to split the screen into regions, and only checking for collisions within regions...
e.g. split the play area up into a grid of squares (which will will arbitrarily say are of 1 unit length per side), and check for collisions within each grid square.
That's absolutely the correct solution. The only problem with it (as another poster pointed out) is that collisions across boundaries are a problem.
The solution to this is to overlay a second grid at a 0.5 unit vertical and horizontal offset to the first one.
Then, any collisions that would be across boundaries in the first grid (and hence not detected) will be within grid squares in the second grid. As long as you keep track of the collisions you've already handled (as there is likely to be some overlap) you don't have to worry about handling edge cases. All collisions will be within a grid square on one of the grids.
A good way of reducing the number of collision checks is to split the screen into different sections. You then only compare each ball to the balls in the same section.
One thing I see here to optimize.
While I do agree that the balls hit when the distance is the sum of their radii one should never actually calculate this distance! Rather, calculate it's square and work with it that way. There's no reason for that expensive square root operation.
Also, once you have found a collision you have to continue to evaluate collisions until no more remain. The problem is that the first one might cause others that have to be resolved before you get an accurate picture. Consider what happens if the ball hits a ball at the edge? The second ball hits the edge and immediately rebounds into the first ball. If you bang into a pile of balls in the corner you could have quite a few collisions that have to be resolved before you can iterate the next cycle.
As for the O(n^2), all you can do is minimize the cost of rejecting ones that miss:
1) A ball that is not moving can't hit anything. If there are a reasonable number of balls lying around on the floor this could save a lot of tests. (Note that you must still check if something hit the stationary ball.)
2) Something that might be worth doing: Divide the screen into a number of zones but the lines should be fuzzy--balls at the edge of a zone are listed as being in all the relevant (could be 4) zones. I would use a 4x4 grid, store the zones as bits. If an AND of the zones of two balls zones returns zero, end of test.
3) As I mentioned, don't do the square root.
I found an excellent page with information on collision detection and response in 2D.
http://www.metanetsoftware.com/technique.html (web.archive.org)
They try to explain how it's done from an academic point of view. They start with the simple object-to-object collision detection, and move on to collision response and how to scale it up.
Edit: Updated link
You have two easy ways to do this. Jay has covered the accurate way of checking from the center of the ball.
The easier way is to use a rectangle bounding box, set the size of your box to be 80% the size of the ball, and you'll simulate collision pretty well.
Add a method to your ball class:
public Rectangle getBoundingRect()
{
int ballHeight = (int)Ball.Height * 0.80f;
int ballWidth = (int)Ball.Width * 0.80f;
int x = Ball.X - ballWidth / 2;
int y = Ball.Y - ballHeight / 2;
return new Rectangle(x,y,ballHeight,ballWidth);
}
Then, in your loop:
// Checks every ball against every other ball.
// For best results, split it into quadrants like Ryan suggested.
// I didn't do that for simplicity here.
for (int i = 0; i < balls.count; i++)
{
Rectangle r1 = balls[i].getBoundingRect();
for (int k = 0; k < balls.count; k++)
{
if (balls[i] != balls[k])
{
Rectangle r2 = balls[k].getBoundingRect();
if (r1.Intersects(r2))
{
// balls[i] collided with balls[k]
}
}
}
}
I see it hinted here and there, but you could also do a faster calculation first, like, compare the bounding boxes for overlap, and THEN do a radius-based overlap if that first test passes.
The addition/difference math is much faster for a bounding box than all the trig for the radius, and most times, the bounding box test will dismiss the possibility of a collision. But if you then re-test with trig, you're getting the accurate results that you're seeking.
Yes, it's two tests, but it will be faster overall.
This KineticModel is an implementation of the cited approach in Java.
I implemented this code in JavaScript using the HTML Canvas element, and it produced wonderful simulations at 60 frames per second. I started the simulation off with a collection of a dozen balls at random positions and velocities. I found that at higher velocities, a glancing collision between a small ball and a much larger one caused the small ball to appear to STICK to the edge of the larger ball, and moved up to around 90 degrees around the larger ball before separating. (I wonder if anyone else observed this behavior.)
Some logging of the calculations showed that the Minimum Translation Distance in these cases was not large enough to prevent the same balls from colliding in the very next time step. I did some experimenting and found that I could solve this problem by scaling up the MTD based on the relative velocities:
dot_velocity = ball_1.velocity.dot(ball_2.velocity);
mtd_factor = 1. + 0.5 * Math.abs(dot_velocity * Math.sin(collision_angle));
mtd.multplyScalar(mtd_factor);
I verified that before and after this fix, the total kinetic energy was conserved for every collision. The 0.5 value in the mtd_factor was the approximately the minumum value found to always cause the balls to separate after a collision.
Although this fix introduces a small amount of error in the exact physics of the system, the tradeoff is that now very fast balls can be simulated in a browser without decreasing the time step size.
Improving the solution to detect circle with circle collision detection given within the question:
float dx = circle1.x - circle2.x,
dy = circle1.y - circle2.y,
r = circle1.r + circle2.r;
return (dx * dx + dy * dy <= r * r);
It avoids the unnecessary "if with two returns" and the use of more variables than necessary.
After some trial and error, I used this document's method for 2D collisions : https://www.vobarian.com/collisions/2dcollisions2.pdf
(that OP linked to)
I applied this within a JavaScript program using p5js, and it works perfectly. I had previously attempted to use trigonometrical equations and while they do work for specific collisions, I could not find one that worked for every collision no matter the angle at the which it happened.
The method explained in this document uses no trigonometrical functions whatsoever, it's just plain vector operations, I recommend this to anyone trying to implement ball to ball collision, trigonometrical functions in my experience are hard to generalize. I asked a Physicist at my university to show me how to do it and he told me not to bother with trigonometrical functions and showed me a method that is analogous to the one linked in the document.
NB : My masses are all equal, but this can be generalised to different masses using the equations presented in the document.
Here's my code for calculating the resulting speed vectors after collision :
//you just need a ball object with a speed and position vector.
class TBall {
constructor(x, y, vx, vy) {
this.r = [x, y];
this.v = [0, 0];
}
}
//throw two balls into this function and it'll update their speed vectors
//if they collide, you need to call this in your main loop for every pair of
//balls.
function collision(ball1, ball2) {
n = [ (ball1.r)[0] - (ball2.r)[0], (ball1.r)[1] - (ball2.r)[1] ];
un = [n[0] / vecNorm(n), n[1] / vecNorm(n) ] ;
ut = [ -un[1], un[0] ];
v1n = dotProd(un, (ball1.v));
v1t = dotProd(ut, (ball1.v) );
v2n = dotProd(un, (ball2.v) );
v2t = dotProd(ut, (ball2.v) );
v1t_p = v1t; v2t_p = v2t;
v1n_p = v2n; v2n_p = v1n;
v1n_pvec = [v1n_p * un[0], v1n_p * un[1] ];
v1t_pvec = [v1t_p * ut[0], v1t_p * ut[1] ];
v2n_pvec = [v2n_p * un[0], v2n_p * un[1] ];
v2t_pvec = [v2t_p * ut[0], v2t_p * ut[1] ];
ball1.v = vecSum(v1n_pvec, v1t_pvec); ball2.v = vecSum(v2n_pvec, v2t_pvec);
}
I would consider using a quadtree if you have a large number of balls. For deciding the direction of bounce, just use simple conservation of energy formulas based on the collision normal. Elasticity, weight, and velocity would make it a bit more realistic.
Here is a simple example that supports mass.
private void CollideBalls(Transform ball1, Transform ball2, ref Vector3 vel1, ref Vector3 vel2, float radius1, float radius2)
{
var vec = ball1.position - ball2.position;
float dis = vec.magnitude;
if (dis < radius1 + radius2)
{
var n = vec.normalized;
ReflectVelocity(ref vel1, ref vel2, ballMass1, ballMass2, n);
var c = Vector3.Lerp(ball1.position, ball2.position, radius1 / (radius1 + radius2));
ball1.position = c + (n * radius1);
ball2.position = c - (n * radius2);
}
}
public static void ReflectVelocity(ref Vector3 vel1, ref Vector3 vel2, float mass1, float mass2, Vector3 intersectionNormal)
{
float velImpact1 = Vector3.Dot(vel1, intersectionNormal);
float velImpact2 = Vector3.Dot(vel2, intersectionNormal);
float totalMass = mass1 + mass2;
float massTransfure1 = mass1 / totalMass;
float massTransfure2 = mass2 / totalMass;
vel1 += ((velImpact2 * massTransfure2) - (velImpact1 * massTransfure2)) * intersectionNormal;
vel2 += ((velImpact1 * massTransfure1) - (velImpact2 * massTransfure1)) * intersectionNormal;
}