How to create a rubber thread in Box2D? - c++

Question

Using Box2d, how to create a rubber thread (rubber band / elastic rope) like Parachute Ninja (ZeptoLab)?
-(void) CreateElasticRope {
//=======Params
// Position and size
b2Vec2 lastPos = b2Vec2(4,4); //set position first body
float widthBody = 0.35;
float heightBody = 0.1;
// Body params
float density = 0.05;
float restitution = 0.5;
float friction = 0.5;
// Distance joint
float dampingRatio = 0.85;
float frequencyHz = 10;
// Rope joint
float kMaxWidth = 1.1;
// Bodies
int countBodyInChain = 10;
b2Body* prevBody;
//========Create bodies and joints
for (int k = 0; k < countBodyInChain; k++) {
b2BodyDef bodyDef;
if(k==0 || k==countBodyInChain-1) bodyDef.type = b2_staticBody; //first and last bodies are static
else bodyDef.type = b2_dynamicBody;
bodyDef.position = lastPos;
lastPos += b2Vec2(2*widthBody, 0); //modify b2Vect for next body
bodyDef.fixedRotation = YES;
b2Body* body = world->CreateBody(&bodyDef);
b2PolygonShape distBodyBox;
distBodyBox.SetAsBox(widthBody, heightBody);
b2FixtureDef fixDef;
fixDef.density = density;
fixDef.restitution = restitution;
fixDef.friction = friction;
fixDef.shape = &distBodyBox;
body->CreateFixture(&fixDef);
if(k>0) {
//Create distance joint
b2DistanceJointDef distJDef;
b2Vec2 anchor1 = prevBody->GetWorldCenter();
b2Vec2 anchor2 = body->GetWorldCenter();
distJDef.Initialize(prevBody, body, anchor1, anchor2);
distJDef.collideConnected = false;
distJDef.dampingRatio = dampingRatio;
distJDef.frequencyHz = frequencyHz;
world->CreateJoint(&distJDef);
//Create rope joint
b2RopeJointDef rDef;
rDef.maxLength = (body->GetPosition() - prevBody->GetPosition()).Length() * kMaxWidth;
rDef.localAnchorA = rDef.localAnchorB = b2Vec2_zero;
rDef.bodyA = prevBody;
rDef.bodyB = body;
world->CreateJoint(&rDef);
} //if k>0
prevBody = body;
} //for -loop
}
I use distance and rope Joints, set different values ​​of parameters dampingRatio and frequencyHz, but the effect is far from being an example (my thread for a long time coming to original state, and not so elastic.).

Answer 1

You can simulate springs by applying forces. At each timestep update the forces on the connected bodies (wake the bodies up if necessary too). If one of the bodies is the ground (or a static body) then you don't need to apply any force to the ground just the dynamic body.
A regular spring would apply both tension and compression forces (pull and push) depending on the deflection. In your case you have a bungee so there would be no compression force just tension (pull).
This is the formula you need:
F = K * x
Where F is the force, K is the spring stiffness (force/deflection), and x is the deflection. Deflection is computed as the difference between the initial length and the current length (the distance between connection points). The sign of the F determines if it is pulling or pushing. Once you compute F then you need to apply it along the line connecting two spring connection points. To satisfy force balance you need to apply this force in opposing directions (one of the bodies gets positive the other one gets negative force). This is because Sir Newton says so.
Here is an example (works with pyBox2D but you can easily convert this to C++)
You need spring objects with some properties. Your spring objects need to know their initial lengths, stiffness, body1, body2, connection coordinates (b1x, b1y, b2x, b2y (in local coordinates))
In your case you need to check if length < spr.initialLength, if this is True then you don't apply any force.
body1 = spr.box2DBody1
body2 = spr.box2DBody2
pA = body1.GetWorldPoint(b2Vec2(spr.box2Db1x, spr.box2Db1y))
pB = body2.GetWorldPoint(b2Vec2(spr.box2Db2x, spr.box2Db2y))
lenVector = pB - pA
length = lenVector.Length()
deltaL = length - spr.initialLength
force = spr.K * deltaL
#normalize the lenVector
if length == 0:
lenVector = b2Vec2(0.70710678118654757, 0.70710678118654757)
else:
lenVector = b2Vec2(lenVector.x / length, lenVector.y / length)
sprForce = b2Vec2(lenVector.x * force, lenVector.y * force)
body1.ApplyForce(sprForce, pA)
body2.ApplyForce(-sprForce, pB)

Answer 2

I very much doubt they are using any joints there. They are probably just taking the distance between the current position of the ninja guy, and the middle of the two posts, to calculate a direction and starting impulse... and just drawing two lines between the posts and the ninja guy.

Answer 3

The best physics implementation added to games I have seen was done by a guy with an engineering degree. He used the calculations you would do in physics / engineering translated into C++. Everything from simple gravity, recoil, thrust, to rotational velocities caused by incidental explosions. All the math was separated into a module that was distinct from the animation.
I would suggest looking up formulas for properties of elastics, and also consider that you have three situations for the elastic band:
1) A shaped force is being applied to stretch it back
2) The shape is now driven by the elastic properties of the band
3) The shape is no longer touching the band, and the band is loosely oscillating by its own weight and inertia
The closer you get to using the true physics calculations, the more realistic it will appear. I'm sure you can fudge it to make it easier on yourself, but humans are inherently good at seeing fakeness.