# Particle Systems

A particle system in Scotty3D is simply a collection of non-interacting, physics-simulated, spherical particles that interact with the rest of the scene. Take a look at the user guide for an overview of how to create and manage them.

In this task, you will implement bool Particle::update(const PT::BVH<PT::Object>& scene, float dt, float radius) found in student/particles.cpp. Note that this task is dependent on (at the very least) having a working triangle intersection algorithm (from A3.0). If you would like us to grade with the reference intersection code instead of your intersection code, make sure to specify this in your writeup.

## Kinematics

Each particle is described by a position and a velocity. If we want to know where the particle will be at some future time, we simply have to move it along its velocity vector. If the particle is travelling at constant velocity, this is a very easy expression to compute.

However, if the particle can have arbitrary forces applied to it, we no longer know exactly how it will move. Since we’re interested in physical simulation, we can apply the basic equation governing motion: $$F = ma$$. Since acceleration is the time derivative of velocity, we now have a differential equation governing how our particle moves. Unfortunately, it’s no longer feasible (in general) to compute this motion analytically, so instead we will use numerical integration to step our particles forward in time. To get realistic behavior from approximate but easy to compute steps, we can simply take many small time steps until we reach the desired destination.

$$\frac{dv}{dt} = \frac{F}{m}$$ $$\frac{dx}{dt} = v$$

There are many different techniques for integrating our equations of motion, including forward, backward, and symplectic Euler, Verlet, Runge-Kutta, and Leapfrog. Each strategy comes with a slightly different way of computing how much to update our velocity and position across a single time-step. In this case, we will use the simplest - forward Euler - as we aren’t too concerned with stability or energy conservation. Forward Euler simply steps our position forward by our velocity, and then velocity by our acceleration:

$$x_{t+\Delta t} = x_t + v_t \cdot \Delta T$$ $$v_{t+\Delta t} = v_t + a \cdot \Delta T$$

In Particle::update, use this integrator to step the current particle forward for dt seconds. Note that the only force we will apply to our particles is gravity, so acceleration (a static member of Particle) is a constant -9.8 meters per second squared pointing downward (feel free to play around with this!). Once you’ve added the update rule, you should already see some interesting behavior - create a particle system and see how the particles travel along nice parabolic trajectories.

## Collisions

The more substantial part of this task is colliding each particle with the rest of our scene geometry. Thankfully, you’ve already done most of the work required here for A3: we can use Scotty3D’s ray-tracing capabilities to find collisions.

During each timestep, we know that in the absence of a collision, our particle will travel in the direction of velocity for distance ||velocity|| * dt. Hence, we can create a ray from the particle’s position and velocity to look for collisions during the time-step. If the ray intersects with the scene, we can compute exactly when the particle would experience a collision.

When collision is found, we could just place the particle at the collision point and be done, but we don’t want our particles to simply stick the surface! Instead, we will assume all particles collide elastically (and massless-ly): the magnitude of a particle’s velocity should be the same before and after a collision, but its direction should be reflected about the normal of the collided surface.

Once we have reflected velocity, we can compute how much of the time step remains after the collision. But what if the particle collided with the scene again before the end of the time-step? If we are using very small time-steps, it might be acceptable to ignore this possibility, but we want to resolve all collisions. So, we can repeat the ray-casting procedure in a loop until we have used up the entire time-step, up to some epsilon. Remember to only use the remaining portion of the time-step each iteration, and to step forward both the velocity and position at each sub-step.

### Spherical Particles

The above procedure is perfectly realistic for point particles, but we would like to draw our particles with some non-zero size and still have the simulation appear natural.

We will hence represent particles as small spheres. This means you must take radius into account when finding intersections: if the collision ray intersects a surface, at what distance does the closest point on the sphere incur the collision? (Hint - it depends on the angle of intersection.)

Simply finding closer intersections based on radius will not, of course, resolve all sphere-scene intersections: the ray can miss all geometry while the edge of the sphere would see a collision. However, this will produce visually acceptable results, greatly reducing overlap and never letting particles ‘leak’ through geometry.

Once you have got collisions working, you should be able to open particles.dae and see a randomized collision-fueled waterfall. Try rendering the scene!

Tips:

• Don’t use abs(). In GCC, this is the integer-only absolute value function. To get the float version, use std::abs() or fabsf().
• When accounting for radius, you don’t know how far along the ray a collision might occur. Look for a collision at any distance, and if it occurs after the end of the current timestep, ignore it.
• When accounting for radius, consider in what situation(s) you could find a collision at a negative time. In this case, you should not move the particle - just reflect the velocity.

Finally, note that Particle::update is supposed to return a boolean representing whether or not the particle should be removed from the simulation. Each particle has an age member that represents the remaining time it has to live. Each time-step, you should subtract dt from age and return whether age is still greater than zero.