This function calulates the velocities after a 2D collision vaf, vbf, waf and wbf from information about the colliding double e coefficient of restitution which depends on the nature of the two colliding double ma total mass of body double mb total mass of body double Ia inertia for body double Ib inertia for body vector ra position of collision point relative to centre of mass of body a in absolute coordinates (if this is
#VECTOR 2D COLLISION CODE#
+ rax 2 rby 2/IaIb - 2 rax ray rbx rby/IaIb Code /** rby 2/maIb - rby 2/mbIb + ray 2 rbx 2/IaIb ray 2/maIa - ray 2/mbIa - rax 2/mbIa - rbx 2/mbIb Jy = - (e+1)/k * (Vaix - Vbix) (rax ray / Ia + rbx rby / Ib) + (e+1)/k * (Vaiy (e+1) * (Vaiy - Vbiy)= Jx (rax ray / Ia + rbx rby / Ib) + Jy ( 1/ma - rax 2/Ia We want to solve this for J, to do this we need to separate out the J terms (e+1) * (Vaiy - Vbiy)= Jx rax ray / Ia + Jy ( 1/ma - rax 2/Ia) + Jx (e+1) * (Vaix - Vbix)= Jx ( 1/ma - ray 2/Ia) + Jy rax ray / Ia + Is in absolute coordinates, not local body coordinates, so this will be a Position of particle relative to centre of mass (vector) - Note: this Inertia scalar such that torque = I * alpha (only true when object is
Vp = linear velocity of a particle on a solid body Solid bodies, these are the points on the bodies that actually collide. Of mass, they are the velocities of points (in absolute coordinates) on the However, in this case, the velocities are not the velocities of the centre Restitution which is the ratio of converging and diverging velocities. + ray 2 rbx 2/(Ia*Ib) + rax 2 rby 2/(Ia*Ib)Īs with the 1D case we can start with the definition of the coefficient of Jy = - (e+1)/k * (Vaix - Vbix) (rax ray / Ia + rbx rby / Ib) + (e+1)/k (e+1)/k * (Vaiy - Vbiy) (rax ray / Ia + rbx rby / Ib) This is just an overview, we will define the terms and give the derivation later: Then the equations to use this to calculate the new velocities. Here are equations for calculating the two components of impulse (J) in the x and y dimensions. If you would like to see a practical example of this see this page (sent to me by Kevin Pegrume) which demonstrates this and highlights some of the counter intuitive issues. Calculate the effect of this impulse on the rotation and linear motion.Reverse the approach velocity, depending on the coefficient of friction. Calculate the minimum impulse to prevent the objects intersecting, or to.Determine the collision point and the normal direction at this point.The main method we will use to do this involves the following stages.
#VECTOR 2D COLLISION HOW TO#
One difficult bit is understanding how to work with objects that have both rotation and linear motion. These issues are all discussed on the page above this and it may be worth reading that page before this. Modeling collisions involves a lot of assumptions and approximations, also the concept of an 'impulse' is not always intuitively obvious. We will therefore start analysing this in terms of individual scalar components, this may make the equations quite long but there are less mathematical concepts to learn. We can't use identical equations to the 3D case because the vector cross product does not apply to 2D vectors. Although we live in a 3 dimensional world we can think of this 2D example as a special case where the objects are constrained so that their velocities can be represented in a plane. Here we are looking at the collision of 2D rigid objects.
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