I implemented the code from the question "Ball to Ball Collision - Detection and Handling" in Objective-C. However, whenever the balls collide at an angle their velocity increases dramatically. All of the vector math is done using cocos2d-iphone, with the header CGPointExtension.h. What is the cause of this undesired acceleration?
The following is an example of increase in speed:
Input:
mass == 12.56637
velocity.x == 1.73199439
velocity.y == -10.5695238
ball.mass == 12.56637
ball.velocity.x == 6.04341078
ball.velocity.y == 14.2686739
Output:
mass == 12.56637
velocity.x == 110.004326
velocity.y == -10.5695238
ball.mass == 12.56637
ball.velocity.x == -102.22892
ball.velocity.y == -72.4030228
#import "CGPointExtension.h"
#define RESTITUTION_CONSTANT (0.75) //elasticity of the system
- (void) resolveCollision:(Ball*) ball
{
// get the mtd (minimum translation distance)
CGPoint delta = ccpSub(position, ball.position);
float d = ccpLength(delta);
// minimum translation distance to push balls apart after intersecting
CGPoint mtd = ccpMult(delta, (((radius + ball.radius)-d)/d));
// resolve intersection --
// inverse mass quantities
float im1 = 1 / [self mass];
float im2 = 1 / [ball mass];
// push-pull them apart based off their mass
position = ccpAdd(position, ccpMult(mtd, (im1 / (im1 + im2))));
ball.position = ccpSub(ball.position, ccpMult(mtd, (im2 / (im1 + im2))));
// impact speed
CGPoint v = ccpSub(velocity, ball.velocity);
float vn = ccpDot(v,ccpNormalize(mtd));
// sphere intersecting but moving away from each other already
if (vn > 0.0f) return;
// collision impulse
float i = (-(1.0f + RESTITUTION_CONSTANT) * vn) / ([self mass] + [ball mass]);
CGPoint impulse = ccpMult(mtd, i);
// change in momentum
velocity = ccpAdd(velocity, ccpMult(impulse, im1));
ball.velocity = ccpSub(ball.velocity, ccpMult(impulse, im2));
}
Having reviewed the original code and the comments by the original poster, the code seems the same, so if the original is a correct implementation, I would suspect a bad vector library or some kind of uninitialized variable.
Why are you adding 1.0 to the coefficient of restitution?
From: http://en.wikipedia.org/wiki/Coefficient_of_restitution
The COR is generally a number in the range [0,1]. Qualitatively, 1 represents a perfectly elastic collision, while 0 represents a perfectly inelastic collision. A COR greater than one is theoretically possible, representing a collision that generates kinetic energy, such as land mines being thrown together and exploding.
Another problem is this:
/ (im1 + im2)
You're dividing by the sum of the reciprocals of the masses to get the impulse along the vector of contact - you probably should be dividing by the sum of the masses themselves. This is magnifying your impulse ("that's what she said").
I'm the one who wrote the original ball bounce code you referenced. If you download and try out that code, you can see it works fine.
The following code is correct (the way you originally had it):
// collision impulse
float i = (-(1.0f + RESTITUTION_CONSTANT) * vn) / (im1 + im2);
CGPoint impulse = ccpMult(mtd, i);
This is very common physics code and you can see it nearly exactly implemented like this in the following examples:
Find collision response of two objects - GameDev
3D Pong Collision Response
Another ball to ball collision written in Java
This is correct, and it ~isn't~ creating a CoR over 1.0 like others have suggested. This is calculating the relative impulse vector based off mass and Coefficient of Restitution.
Ignoring friction, a simple 1d example is as follows:
J = -Vr(1+e) / {1/m1 + 1/m2}
Where e is your CoR, Vr is your normalized velocity and J is a scalar value of the impulse velocity.
If you plan on doing anything more advanced than this I suggest you use one of the many physics libraries already out there. When I used the code above it was fine for a few balls but when I ramped it up to several hundred it started to choke. I've used the Box2D physics engine and its solver could handle more balls and it is much more accurate.
Anyway, I looked over your code and at first glance it looks fine (it is a pretty faithful translation). It is probably a small and subtle error of a wrong value being passed in, or a vector math problem.
I don't know anything concerning iPhone development but I would suggest setting a breakpoint at the top of that method and monitoring each steps resulting value and finding where the blow-up is. Ensure that the MTD is calculated correctly, the impact velocities, etc, etc until you see where the large increase is getting introduced.
Report back with the values of each step in that method and we'll see what we have.
In this line:
CGPoint impulse = ccpMult(mtd, i);
mtd needs to have been normalised. The error happened because in the original code mtd was normalized in a previous line but not in your code. You can fix it by doing something like this:
CGPoint impulse = ccpMult(ccpNormalize(mtd), i);
Related
After some time searching, I have revised my question.
I have found numerous examples of ball to ball collisions, but the only ones that seem to work use Vector2d or Vector2D.
This is a problem, because I am only allowed to use the regular java library, so my main question is: How do I convert the examples (which I will post below) to use what I can use?
I have several variables, both balls have the same mass, the velocities are broken into different variables, x and y. Also I have access to their x and y pos.
This is the ONLY problem left in my application.
I am at a total loss on how to convert the below example.
// 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.multiply(i);
// change in momentum
this.velocity = this.velocity.add(impulse.multiply(im1));
ball.velocity = ball.velocity.subtract(impulse.multiply(im2));
Here is the URL for the question:
http://stackoverflow.com/questions/345838/ball-to-ball-collision-detection-and-handling
And I have taken a look at his source code.
Thank you for taking the time to read this issue.
SUCCESS!
I have found how to use Vector2d, and it works PERFECTLY!
Will edit later with answer!
I'm implementing my own 3d engine in c# based on a really basic 3d open-source engine in JavaScript called a3. I don't know If I have 100% understand you but It sounds like you can only find examples with Vector2d but you are not allowed to use that class?
I that is the case, as you can imagine javascript does not have native Vector2d types so someone had to implement. Don't be afraid of giving it a try, is just a few high school maths functions, you should be able to implement your own Vector2d class in just a few minutes
The following link contain implementations if vector2d, vector3d, vector4d, matrix3, and matrix4 in javascript: https://github.com/paullewis/a3/tree/master/src/js/core/math hope it helps :)
I am making really simple app in xcode.
And I want to make, if the ball reach the hole the game should finish
So i tried to make.
if (ball . center == hole.center )
and another ways and I failed
and I also tried this
(ball.frame.origin.x == hole.frame.origin.x && ball.frame.origin.y == hole.frame.origin.y)
And as usual failed
Please help.
i just want if the fram of the ball touches the hole The Game FINISH
Problem is that you shouldn't check for a position to be exactly the same, that's not how it works with floating point coordinates (which I guess you are using) and precision of movement of things in games which cannot require to have object on same indentical position.
You should rather check if distance is less than a threshold:
float bx = ball.frame.origin.x;
float by = ball.frame.origin.y;
float hx = hole.frame.origin.x;
float hy = hole.frame.origin.y;
// you don't actually need abs since you are going to raise to the power of 2
// but for sake of soundness it makes sense
float dx = abs(bx-hx);
float dy = abs(by-hy);
if (sqrt(dx*dx + dy*dy) < THRESHOLD) {
// the ball is enough near to center
}
You could use CGRectIntersectsRect (more on CGGeometry) to see if the ball and hole intersect eachother:
if (CGRectIntersectsRect(ball.frame, hole.frame)) {
// Goal reached!
}
... or CGRectEqualToRect the same way (if you want to check if the frames are exactly the same).
My guess would be that you do not want to test whether the two centers are equal, but whether they are close enough to eachother. For example less then epsilon away in the x and y direction.
I've used Nick Vellios' tutorial to create radial gravity with a Box2D object. I am aware of Make a Vortex here on SO, but I couldn't figure out how to implement it in my project.
I have made a vortex object, which is a Box2D circleShape sensor that rotates with a consistent angular velocity. When other Box2D objects contact this vortex object I want them to rotate around at the same angular velocity as the vortex, gradually getting closer to the vortex's centre. At the moment the object is attracted to the vortex's centre but it will head straight for the centre of the vortex, rather than spinning around it slowly like I want it to. It will also travel in the opposite direction than the vortex as well as with the vortex's rotation.
Given a vortex and a box2D body, how can I set the box2d body to rotate with the vortex as it gets 'sucked in'.
I set the rotation of the vortex when I create it like this:
b2BodyDef bodyDef;
bodyDef.type = b2_dynamicBody;
bodyDef.angle = 2.0f;
bodyDef.angularVelocity = 2.0f;
Here is how I'm applying the radial gravity, as per Nick Vellios' sample code.
-(void)applyVortexForcesOnSprite:(CCSpriteSubclass*)sprite spriteBody:(b2Body*)spriteBody withVortex:(Vortex*)vortex VortexBody:(b2Body*)vortexBody vortexCircleShape:(b2CircleShape*)vortexCircleShape{
//From RadialGravity.xcodeproj
b2Body* ground = vortexBody;
b2CircleShape* circle = vortexCircleShape;
// Get position of our "Planet" - Nick
b2Vec2 center = ground->GetWorldPoint(circle->m_p);
// Get position of our current body in the iteration - Nick
b2Vec2 position = spriteBody->GetPosition();
// Get the distance between the two objects. - Nick
b2Vec2 d = center - position;
// The further away the objects are, the weaker the gravitational force is - Nick
float force = 1 / d.LengthSquared(); // 150 can be changed to adjust the amount of force - Nick
d.Normalize();
b2Vec2 F = force * d;
// Finally apply a force on the body in the direction of the "Planet" - Nick
spriteBody->ApplyForce(F, position);
//end radialGravity.xcodeproj
}
Update I think iForce2d has given me enough info to get on my way, now it's just tweaking. This is what I'm doing at the moment, in addition to the above code. What is happening is the body gains enough velocity to exit the vortex's gravity well - somewhere I'll need to check that the velocity stays below this figure. I'm a little concerned I'm not taking into account the object's mass at the moment.
b2Vec2 vortexVelocity = vortexBody->GetLinearVelocityFromWorldPoint(spriteBody->GetPosition() );
b2Vec2 vortexVelNormal = vortexVelocity;
vortexVelNormal.Normalize();
b2Vec2 bodyVelocity = b2Dot( vortexVelNormal, spriteBody->GetLinearVelocity() ) * vortexVelNormal;
//Using a force
b2Vec2 vel = bodyVelocity;
float forceCircleX = .6 * bodyVelocity.x;
float forceCircleY = .6 * bodyVelocity.y;
spriteBody->ApplyForce( b2Vec2(forceCircleX,forceCircleY), spriteBody->GetWorldCenter() );
It sounds like you just need to apply another force according to the direction of the vortex at the current point of the body. You can use b2Body::GetLinearVelocityFromWorldPoint to find the velocity of the vortex at any point in the world. From Box2D source:
/// Get the world linear velocity of a world point attached to this body.
/// #param a point in world coordinates.
/// #return the world velocity of a point.
b2Vec2 GetLinearVelocityFromWorldPoint(const b2Vec2& worldPoint) const;
So that would be:
b2Vec2 vortexVelocity = vortexBody->GetLinearVelocityFromWorldPoint( suckedInBody->GetPosition() );
Once you know the velocity you're aiming for, you can calculate how much force is needed to go from the current velocity, to the desired velocity. This might be helpful: http://www.iforce2d.net/b2dtut/constant-speed
The topic in that link only discusses a 1-dimensional situation. For your case it is also essentially 1-dimensional, if you project the current velocity of the sucked-in body onto the vortexVelocity vector:
b2Vec2 vortexVelNormal = vortexVelocity;
vortexVelNormal.Normalize();
b2Vec2 bodyVelocity = b2Dot( vortexVelNormal, suckedInBody->GetLinearVelocity() ) * vortexVelNormal;
Now bodyVelocity and vortexVelocity will be in the same direction and you can calculate how much force to apply. However, if you simply apply enough force to match the vortex velocity exactly, the sucked in body will probably go into orbit around the vortex and never actually get sucked in. I think you would want to make the force quite a bit less than that, and I would scale it down according to the gravity strength as well, otherwise the sucked-in body will be flung away sideways as soon as it contacts the outer edge of the vortex. It could take a lot of tweaking to get the effect you want.
EDIT:
The force you apply should be based on the difference between the current velocity (bodyVelocity) and the desired velocity (vortexVelocity), ie. if the body is already moving with the vortex then you don't need to apply any force. Take a look at the last code block in the sub-section titled 'Using forces' in the link I gave above. The last three lines there do pretty much what you need if you replace 'vel' and 'desiredVel' with the sizes of your bodyVelocity and vortexVelocity vectors:
float desiredVel = vortexVelocity.Length();
float currentVel = bodyVelocity.Length();
float velChange = desiredVel - currentVel;
float force = body->GetMass() * velChange / (1/60.0); //for a 1/60 sec timestep
body->ApplyForce( b2Vec2(force,0), body->GetWorldCenter() );
But remember this would probably put the body into orbit, so somewhere along the way you would want to reduce the size of the force you apply, eg. reduce 'desiredVel' by some percentage, reduce 'force' by some percentage etc. It would probably look better if you could also scale the force down so that it was zero at the outer edge of the vortex.
I had a project where I had asteroids swirling around a central point (there are things jumping between them...which is a different point).
They are connected to the "center" body via b2DistanceJoints.
You can control the joint length to make them slowly spiral inward (or outward). This gives you find grain control instead of balancing force control, which may be difficult.
You also apply tangential force to make them circle the center.
By applying different (or randomly changing) tangential forces, you can make the
crash into each other, etc.
I posted a more complete answer to this question here.
I am trying to calculate the forces that will act on circular objects in the event of a collision. Unfortunately, my mechanics is slightly rusty so i'm having a bit of trouble.
I have an agent class with members
vector position // (x,y)
vector velocity // (x,y)
vector forward // (x,y)
float radius // radius of the agent (all circles)
float mass
So if we have A,B:Agent, and in the next time step the velocity is going to change the position. If a collision is going to occur I want to work out the force that will act on the objects.
I know Line1 = (B.position-A.position) is needed to work out the angle of the resultant force but how to calculate it is baffling me when I have to take into account current velocity of the vehicle along with the angle of collision.
arctan(L1.y,L1.x) is am angle for the force (direction can be determined)
sin/cos are height/width of the components
Also I know to calculate the rotated axis I need to use
x = cos(T)*vel.x + sin(T)*vel.y
y = cos(T)*vel.y + sin(T)*vel.x
This is where my brain can't cope anymore.. Any help would be appreciated.
As I say, the aim is to work out the vector force applied to the objects as I have already taken into account basic physics.
Added a little psudocode to show where I was starting to go with it..
A,B:Agent
Agent {
vector position, velocity, front;
float radius,mass;
}
vector dist = B.position - A.position;
float distMag = dist.magnitude();
if (distMag < A.radius + B.radius) { // collision
float theta = arctan(dist.y,dist.x);
flost sine = sin(theta);
float cosine = cos(theta);
vector newAxis = new vector;
newAxis.x = cosine * dist .x + sine * dist .y;
newAxis.y = cosine * dist .y - sine * dist .x;
// Converted velocities
vector[] vTemp = {
new vector(), new vector() };
vTemp[0].x = cosine * agent.velocity.x + sine * agent.velocity.y;
vTemp[0].y = cosine * agent.velocity.y - sine * agent.velocity.x;
vTemp[1].x = cosine * current.velocity.x + sine * current.velocity.y;
vTemp[1].y = cosine * current.velocity.y - sine * current.velocity.x;
Here's to hoping there's a curious maths geek on stack..
Let us assume, without loss of generality, that we are in the second object's reference frame before the collision.
Conservation of momentum:
m1*vx1 = m1*vx1' + m2*vx2'
m1*vy1 = m1*vy1' + m2*vy2'
Solving for vx1', vy1':
vx1' = vx1 - (m2/m1)*vx2'
vy1' = vy1 - (m2/m1)*vy2'
Secretly, I will remember the fact that vx1'*vx1' + vy1'*vy1' = v1'*v1'.
Conservation of energy (one of the things elastic collisions give us is that angle of incidence is angle of reflection):
m1*v1*v1 = m1*v1'*v1' + m2*v2'+v2'
Solving for v1' squared:
v1'*v1' = v1*v1 - (m2/m1)v2'*v2'
Combine to eliminate v1':
(1-m2/m1)*v2'*v2' = 2*(vx2'*vx1+vy2'*vy1)
Now, if you've ever seen a stationary poolball hit, you know that it flies off in the direction of the contact normal (this is the same as your theta).
v2x' = v2'cos(theta)
v2y' = v2'sin(theta)
Therefore:
v2' = 2/(1-m2/m1)*(vx1*sin(theta)+vy1*cos(theta))
Now you can solve for v1' (either use v1'=sqrt(v1*v1-(m2/m1)*v2'*v2') or solve the whole thing in terms of the input variables).
Let's call phi = arctan(vy1/vx1). The angle of incidence relative to the tangent line to the circle at the point of intersection is 90-phi-theta (pi/2-phi-theta if you prefer). Add that again for the reflection, then convert back to an angle relative to the horizontal. Let's call the angle of incidence psi = 180-phi-2*theta (pi-phi-2*theta). Or,
psi = (180 or pi) - (arctan(vy1/vx1))-2*(arctan(dy/dx))
So:
vx1' = v1'sin(psi)
vy1' = v1'cos(psi)
Consider: if these circles are supposed to be solid 3D spheres, then use a mass proportional to radius-cubed for each one (note that the proportionality constant cancels out). If they are supposed to be disklike, use mass proportional to radius-squared. If they are rings, just use radius.
Next point to consider: Since the computer updates at discrete time events, you actually have overlapping objects. You should back out the objects so that they don't overlap before computing the new location of each object. For extra credit, figure out the time that they should have intersected, then move them in the new direction for that amount of time. Note that this time is just the overlap / old velocity. The reason that this is important is that you might imagine a collision that is computed that causes the objects to still overlap (causing them to collide again).
Next point to consider: to translate the original problem into this problem, just subtract object 2's velocity from object 1 (component-wise). After the computation, remember to add it back.
Final point to consider: I probably made an algebra error somewhere along the line. You should seriously consider checking my work.
I want to simulate a free fall and a collision with the ground (for example a bouncing ball). The object will fall in a vacuum - an air resistance can be omitted. A collision with the ground should causes some energy loss so finally the object will stop moving. I use JOGL to render a point which is my falling object. A gravity is constant (-9.8 m/s^2).
I found an euler method to calculate a new position of the point:
deltaTime = currentTime - previousTime;
vel += acc * deltaTime;
pos += vel * deltaTime;
but I'm doing something wrong. The point bounces a few times and then it's moving down (very slow).
Here is a pseudocode (initial pos = (0.0f, 2.0f, 0.0f), initial vel(0.0f, 0.0f, 0.0f), gravity = -9.8f):
display()
{
calculateDeltaTime();
velocity.y += gravity * deltaTime;
pos.y += velocity.y * deltaTime;
if(pos.y < -2.0f) //a collision with the ground
{
velocity.y = velocity.y * energyLoss * -1.0f;
}
}
What is the best way to achieve a realistic effect ? How the euler method refer to the constant acceleration equations ?
Because floating points dont round-up nicely, you'll never get at a velocity that's actually 0. You'd probably get something like -0.00000000000001 or something.
you need to to make it 0.0 when it's close enough. (define some delta.)
To expand upon my comment above, and to answer Tobias, I'll add a complete answer here.
Upon initial inspection, I determined that you were bleeding off velocity to fast. Simply put, the relationship between kinetic energy and velocity is E = m v^2 /2, so after taking the derivative with respect to velocity you get
delta_E = m v delta_v
Then, depending on how energyloss is defined, you can establish the relationship between delta_E and energyloss. For instance, in most cases energyloss = delta_E/E_initial, then the above relationship can be simplified as
delta_v = energyloss*v_initial / 2
This is assuming that the time interval is small allowing you to replace v in the first equation with v_initial, so you should be able to get away with it for what your doing. To be clear, delta_v is subtracted from velocity.y inside your collision block instead of what you have.
As to the question of adding air-resistance or not, the answer is it depends. For small initial drop heights, it won't matter, but it can start to matter with smaller energy losses due to bounce and higher drop points. For a 1 gram, 1 inch (2.54 cm) diameter, smooth sphere, I plotted time difference between with and without air friction vs. drop height:
For low energy loss materials (80 - 90+ % energy retained), I'd consider adding it in for 10 meter, and higher, drop heights. But, if the drops are under 2 - 3 meters, I wouldn't bother.
If anyone wants the calculations, I'll share them.