Trajectory of a Projectile with Air Resistance

A sphere falling through the air is acted upon by only two forces, namely the aerodynamic force and gravity. We can write the equation of motion as

drag + gravity = mass × acceleration

This is a vector equation and we must be careful with signs. Altitude is measured positive upwards and a body that is falling has a negative velocity (velocity is rate of change of altitude and altitude is decreasing). Upward forces are positive and downward forces are negative. The aerodynamic drag force acts opposite to the direction of velocity. If the body is falling, then drag is positive; if a body is thrown upward, the velocity is positive and the drag force is negative. The force of gravity is always negative. The gravitational force on an object decreases with increasing altitude because the gravitational field of the earth has an inverse square relationship with altitude.

g=g_{0} × (R/(R+z))**2

where z is the altitude and R is the radius of the earth (more than 6 million meters). This is a small, but noticable effect. At 36 km altitude, g is 9.6965 meters per second squared compared to a value of 9.8066 at sea level. Because the drag is a non-linear function of velocity, we will not be able to find an analytic solution and a numerical solution will be necessary. Hence, we have a second order non-linear differential equations. You need to be able to solve equations of this sort with ease, because most of the real problems of the engineering world do not yield to simple linear analysis.

The usual technique is to replace the second order differential equation with two first order equations. The two variables are the altitude and velocity. Call them z and v. The two equations are

dz/dt = v

dv/dt = (1/mass) × (drag + gravity)

Aerodynamic Drag of a Sphere

Everything is quite simple except for the calculation of the drag force. The traditional way to express the drag of an object such as a sphere is through a drag coefficient defined as the drag force divided by the product of the frontal area and the dynamic pressure. The dynamic pressure is defined as one-half the density times the square of the velocity. This gives a model of the varying atmospheric force on the sphere as the velocity and altitude change.

To estimate the drag coefficient of a sphere, I have included a procedure taken from Chow, An Introduction to Computational Fluid Mechanics. I have added an estimate of wave drag for a sphere of CD = 1.0 and have programmed it to begin appearing at M = 0.6, increasing smoothly until it reaches its full value at M = 1. I recommend Chow's book for some examples for further study.

To use this routine, you need to know the Reynolds number of the flow which requires the density of the air as well as the viscosity of air. You can see the routines in the program that compute these quantities and eventually return the drag coefficient to the procedure that computes the acceleration. You have to know velocity, elevation, air density, drag coefficient, gravity and it all comes together in the calculation of acceleration. The atmosphere procedures are taken from the Standard Atmosphere program.

Solution of the Differential Equations

Each of the variables is a function of time. To perform the numerical integration, we take time steps of finite size and advance the solution. The most straightforward way to proceed is to use a fourth-order Runge-Kutta algorithm. The best way to do this is to use a variable step-size differential equation solver that computes the step size as it develops the solution. However, for this simple problem, I felt it would be confusing to introduce variable step size. So the calculations are done with a simple fixed step size algorithm. It is not obvious what time step one should use and the usual approach is to use a very small value and increase it until you see a difference in the solution. A value of 0.1 second should be small enough.

Correction of the final point

At each step along the trajectory, we check to see that the altitude is greater than the initial altitude. Eventually, of course, we will arrive at a point where this is no longer true. This is our clue to terminate the calculation. The fixed step size in time will most likely step over the point where the altitude is zero and give a point with negative altitude as the final point. To make the solution look good, it is best to do an interpolation in the final interval and land right on zero altitude.

Source code for the computer program

Finally, you can look at the source code of the computer program. This program is written without input data. To run different cases, you update the source code and recompile and rerun. You will need a Fortran compiler.