# Is Verlet method suitable for to solve projectile motion with quadratic drag?

Consider the following non-dimensional equations of projectile motion in presence of air resistance,

I understand it can be numerically solved using the mid-point method, but can I also use Verlet method? Why midpoint method is more favourable in this case? Is this because the truncation error in velocity equation is of order $$O(t^2)$$ and $$O(t^3)$$ for position when midpoint method is used, which are $$O(t^2)$$ and $$O(t^4)$$ respectively when Verlet method is used? Am I thinking in the right direction?

• Did somewhere claim that it was better to use the mid-point method, or was that just the method they chose to mention? – JMac May 23 '19 at 20:58
• The Verlet method is nice for mechanics problems because it conserves energy. – HiddenBabel May 23 '19 at 21:03
• @JMac I am reading this book called "Numerical Methods for Physics" by Alejandro L. Garcia and he mentions that mid-point method is better for projectile motion. While he didn't relate Verlet method with projectiles, he did it for pendulum. Now I am working on this assignment which asks the question why or why not Verlet method is suitable for projectiles and I couldn't find a straight answer anywhere. – Rafiul Nakib May 23 '19 at 21:17
• Related, if not dupe of, physics.stackexchange.com/q/239621/25301 – Kyle Kanos May 24 '19 at 11:22

## 1 Answer

The reason Verlet's algorithm is less suitable than midpoint method is due to the form of the force. Verlet's algorithm requires the acceleration at time step $$t+\Delta t$$ in order to update velocity at time $$t+\Delta t$$ (I refer to the so called "velocity" form of the algorithm, although the same considerations hold for the "position" form too): $$\begin{eqnarray} x(t+\Delta t) &=& x(t) + v(t)\Delta t +\frac12 a(t) \Delta t^2 ~~~~~~~~~~~~~~[1]\\ v(t+\Delta t) &=& v(t) + \frac12\left( a(t) + a(t+\Delta t) \right) \Delta t ~~~~~~~~[2] \end{eqnarray}$$ If the force, and then the acceleration, depends on position only, $$a(t+\Delta t)$$ appearing in equation $$[2]$$ can be explicitly evaluated from the new position evaluated at $$[1]$$. However, if forces and then acceleration depend on velocities, equation [2] becomes an implicit equation for the velocity $$v(t+\Delta t)$$. It is simple to solve such equation in the case of linear forces (linear viscous friction or magnetic Lorentz force om a charged particle) but in the case of non-linear dependence on velocity things become more complex (and in 2D or 3D, much more complex) and usually other methods become preferable.

• So it has nothing to do with the order of error whatsoever? – Rafiul Nakib May 24 '19 at 8:50
• The error is the least of the problem. However, it works the other way around: Verlet's algorithm (I am referring to the velocity form) provides a reasonably good $dt^2$ global behavior for both position and velocity, therefore for energy. Midpoint formula is definitely poorer: linear decrease of the global error on velocity. Even more important, Verlets' algorithm is time-reversible and symplectic. Mid-point algorithm is neither symplectic nor time-reversible. Everything would point to Verlet's as the best choice, but velocity dependent forces don't make it explicit. – GiorgioP May 24 '19 at 11:51