Professor Flint O. Thomas Department of Aerospace & Mechanical Engineering

Mechanical Systems Professor Flint O. Thomas Department of Aerospace & Mechanical Engineering Hessert Center for Aerospace Research Selection of Material on the Subject of Projectile Motion given by Professor Thomas for EG 111 - Fall 2000

Mechanical Systems • Today’s Lecture: “Flight Dynamics” • Goal: Develop a model of the flight of the ball. Our analysis commences at the instant the ball leaves the pouch. Trajectory will depend on the initial speed V0 and launch angle q imparted by the launcher.

Consider “snapshots” of the ball taken at equal time intervals Dt = ti+1-ti. At any instant of time the ball will be acted on by two forces: weight and aerodynamic drag. y ti ti+1 t3 t2 t1 t0 yo x

The motion of the ball will be governed by Newton’s second law: • Remember: force and acceleration are vectors which possess both magnitude and direction. Vector addition is by the parallelogram law. Fy F Fx • Newton’s 2nd Law may be written for each component:

Let’s apply Newton’s Second Law to the flight of the ball: • Note that the drag force D always opposes the flight • direction. Also we see that:

Newton’s Law for the x-component forces: • Newton’s Law for the y-component forces: • These equations relate the instantaneous x- and y-component accelerations of the ball to the instantaneous forces acting on the ball during the flight.

Before analyzing the trajectory of the ball, let’s look at the simpler problem of its motion in a vacuum. In this case there is no air resistance and the equations governing the motion can be obtained by setting D = 0: • In this case there is no x-component acceleration so Vx will be constant. The vertical acceleration is constant. The motion of the ball is a superposition of uniformly accelerated motion in y and constant speed in x.

We will divide the flight of the ball into many small sub-flights, each of duration Dt. We start with known initial conditions from launch at time t = 0: • The acceleration is the time rate of change of velocity. Over any of the short time intervals Dt we have: “Recipe” for finding the speed of the ball at t+Dt from the known speed at t.

The average x and y-component speeds of the ball over the time interval Dt are given by: • The new x, y position of the ball is approximated as, • To illustrate consider the following example: Ball Launched with Initial Conditions (time t = 0): V0 = 40 m/s, q = 35 degrees, x0 = 0, y0 = 5 m

= constant Suppose we divide the flight intoDt = 0.1 second intervals. The component speeds of the ball at the next instant of time t = 0.1 second are, The average speed over this interval is,

The position of the ball at t=0.1 second is obtained from, The procedure is now repeated sequentially for t=0.2, 0.3, 0.4 seconds etc. Microsoft Excel is an ideal tool for performing these calculations. A sample spreadsheet is shown below. Initial condition time

The trajectory (and range) of the ball is shown below: • How do we know if this is correct?

!! Whenever possible numerical solution methods should be checked against exact analytical solutions in order to insure the accuracy of the technique. We will apply that approach here. • Recall from your calculus course that: • Acceleration is the time derivative of velocity

Newton’s Second Law becomes: and • These are called differential equations because they involve • derivatives of the velocity of the ball. Note that in a • vacuum the x and y motions are independent, as noted • before. • In order to obtain the velocity Vx(t) and Vy(t) • of the ball from the above equations we need to take • anti-derivatives (i.e. integrate) of both sides of the • equations.

We will integrate with respect to time over the interval from t = 0 to some arbitrary time t during the flight. and So the velocity as a function of time is, Now recall: and

The differential equations for the trajectory of the ball are, and As before, we integrate with respect to time to obtain the trajectory of the ball x(t) and y(t). Remembering that at t = 0, x=0 and y = y0 we get, This is the analytical solution for the trajectory of the ball. How does it compare to our numerical solution ?

Excellent agreement!

With the model validated we can put it to work. Example: For a fixed initial velocity what launch angle gives maximum range? For projectile motion in a vacuum, maximum range occurs at a launch angle of 45 degrees.

Example: What combinations of V0 and q0yield a desired range of 100 meters? Solution... construct a ballistics chart: Numerous possibilities indicated…impossible for V0 = 30 m / s.

We do not live in a vacuum. We are at the bottom of an ocean of air! What is the effect of the air on the flight of the ball? Photograph by F. N. M. Brown, Notre Dame

Visualization of flow over a cylinder: Photo by Thomas C. Corke Where does the energy to put the air in motion come from?

Conclude that the passage of the ball through air will give rise to a loss in the kinetic energy of the ball that must be accounted for in order to realistically model the trajectory. In other words, we must account for aerodynamic drag! Back to the original form of the equations of motion: These equations don’t help us unless we know how to determine D !!

Goal : Determine the aerodynamic drag on the ball and incorporate it into our numerical simulation of the flight. • This requires some basic fluid mechanics (the branch of • mechanics which deals with the dynamic behavior of gases • and liquids). The objective of fluid mechanics is most • often the determination of flow-induced forces. Example: Wing lift and drag:

Select a reference frame: Two ways oflooking at the same problem. The drag would be identical in both cases since it is the relative motion of the air and ball that determine drag. This equivalence is one basis of wind tunnel testing.

How is drag created? • The momentum of the “cylinder of air” is the product of its mass times its velocity. Mass of air = density X volume = Momentum of air = Mass X velocity =

As the cylinder of air and ball collide the momentum • of the air is imparted to the ball. • Another way of writing Newton’s second law is in terms of the time rate of change of momentum.

If we assume all the momentum of the cylinder of air is imparted to the ball during the time interval Dt then the force on the ball D is given by, Momentum of air Transferred during time Dt

Is this result correct? Not necessarily.... Our calculation assumes that all of the momentum of the cylinder of air is transferred to the ball. This may not be the case! What we can say with some certainty is, [ kg/m3] Remember: = air density = area of the body “seen” by the oncoming flow. [m2] = fluid velocity = flight speed [m/s]

To make this an equality the convention is to • incorporate a factor CD/2 (typically determined • from experiment) such that, CD is called the “drag coefficient” • The drag coefficient CD depends on the geometry of • the body and also varies in a complicated way with flow • speed. • Fortunately, for flow over a sphere at the velocities that we’re likely to encounter during the launch project, to good approximation experiments show that:

We will model the drag on the ball by the relation, • Now that we have a model for the drag force we can perform a numerical simulation of the flight of the ball including the effect of drag. • As before, we will divide the flight of the ball into many small sub-flights, each of duration Dt. We start with the known initial conditions from launch at time t = 0:

The acceleration is the time rate of change of velocity. Over any of the short time intervals Dt we have: These provide a “recipe” for advancing the speed of the ball in time…If I know Vx(t) and Vy(t) I can use these equations to find Vx and Vy at time t+Dt later.

The recipe... with drag given by our model, V Vy Vx q What about q ? !!

The angle q depends on the the relative size of the x- and y-components of the velocity of the ball (as shown below) and is therefore a function of time. ‘looks like simple trig to me…”

Now I see the procedure! 1. Using Vx(ti) and Vy(ti) compute the drag force and angle theta. 2. Use “the recipe” provided courtesy of Sir Isaac Newton to find the new velocity Vx(ti+Dt) and Vy(ti+Dt) a time Dt later. 3. Calculate the average x- and y-component velocity for the time interval Dt. 4. Use the average velocity and knowledge of the position of the ball x (ti), y (ti) to find the of the ball x(ti+Dt), y (ti+Dt). Repeat 1-4 NO YES Has the ball hit the ground?

To illustrate, reconsider the example we looked at before: Ball Launched with Initial Conditions (time t = 0): V0 = 40 m/s, q0 = 35 degrees, x0 = 0, y0 = 5 m With the additional information that: m = 300 grams = 0.3 kg A = 2.043 X 10-3 m2 Diameter = 0.051 meters Initial x- and y-component velocities are: • As before we will divide the flight intoDt = 0.1 second • intervals.

Compute the drag force: Compute the new x- and y-component velocities:

Compute the average speed for the interval, • The position of the ball at t=0.1 second is obtained from, • The procedure is now repeated using Microsoft Excel.

You will develop your own spreadsheet in the learning center. q D ti Vy y(ti) x(ti) Vx Initial conditions time How significant is the influence of drag?

A very significant effect on range! The trajectory is no longer parabolic!

Professor Flint O. Thomas Department of Aerospace & Mechanical Engineering