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## Fun with Wind-Resistance (Part 1)

It’s finally happened. After a decade of dealing with frictionless slopes, massless strings, perfect vacuums and other spherical cows, I’m ready to complicate my model. What follows is a simple model for wind resistance, as outlined in University Physics by Young and Freedman. We’ll then have a look at the effect of air resistance on throwing a cricket ball (or baseball, if you must.)

In the absence of wind resistance, the equation of motion for a projectile is quite simple: $a_x = 0$ $a_y = -g$

In the x-direction (horizontally), the ball moves with whatever horizontal velocity the thrower  gave it to start with. In the y-direction (vertically), the ball is pulled downwards, its vertical velocity changing at the constant rate of 9.8 m/s/s.

Wind resistance adds an extra force, one that pushes in the opposite direction to the way the ball is going. The magnitude of the force (for sufficiently large Reynolds number) is $F_D = \frac{1}{2}\rho v^2 C_d A$

where $latex v$ is the speed of the ball, $\rho$ is the density of air $(1.2 kg/m^3)$, $A$ is the cross sectional area of the ball and $C_d$ is a dimensionless factor called the drag coefficient.

Because the drag force increases with velocity, a falling ball will accelerate until it reaches terminal velocity, where the drag force balances gravity. Thereafter, the ball falls with a constant velocity. The terminal velocity is given by: $v_{t} = \sqrt{\frac{2mg}{\rho A C_d}}$

In practice, we use this formula in a different way. The terminal velocity is measurable, so we can use it to constrain the drag coefficient $C_d$. E.g. for a cricket ball, the terminal velocity is 123 km/h.

We now have all the pieces we need. The equation of motion is not solvable analytically, but is easily handled by any good numerical ODE solver. I’ll be using those of Matlab.

Let’s start with a few trajectories. I’m assuming that the thrower releases the ball from 1.8m. The black, red and green lines are for balls thrown at 120, 140 and 160 km/h respectively. The initial angle of the trajectory is chosen to maximize the distance thrown. As expected, when wind resistance is taken into account, the balls trajectory drops more sharply after the ball reaches it’s peak. The difference in the range is quite significant – more than a factor of two for the ball thrown at 160 km/h.

The longest ever throw of a cricket ball seems to be about 140 yards, or about 128 metres. Given that the fastest baseball pitch is 169km/h, the model predicts a range of 103 metres. I’m not sure what the reason for the discrepancy is. An extra 10 km/h gets another 7 metres, though I think it’s unlikely that anyone can throw faster than baseball pitchers, even with the aid of a run-up. I’ve ignored the effect of spin – a ball thrown at 160km/h would have considerable backspin. This would increase the range, as slice does for a tennis ball. I’m not sure how to calculate the magnitude of the effect. Wind assistance and the favourable effects of high altitude are also possible contributors.

The dependence of the range (previous caveats notwithstanding) on the initial speed of the throw is shown below: The range increases as the velocity squared in the absence of wind resistance. With wind resistance, we see this behaviour at low velocity but a slower, linear increase at velocities above about 50 km/h. Thus, a 220 metre throw only goes about 100 metres.

Incidentally, if you threw a ball that fast on the moon, with no wind resistance and lower gravity,  the ball would travel about 1.4 km.

Continued in …

Part Two: Optimal throwing angle

Part Three: Optimal Mass

Part Four: Hitting at altitude

### 7 Responses

1. […] Last time, I showed a few trajectories of cricket balls (or baseballs) thrown in the presence of wind-resistance. I noted that I had chosen the angle of the throw in order to maximise the range of the throw. This optimal changes as the throw speed changes, as shown below. […]

2. on September 27, 2012 at 2:11 am | Reply David W. Hogg

I think the reason for the discrepancy between the range and the pitch speed is that the pitch speed measurement is not a “muzzle velocity” but rather a mean speed from pitcher to plate. That’s a *lot* slower than the muzzle velocity.

3. on September 27, 2012 at 7:31 am | Reply lukebarnes

In cricket, bowling speeds are measured with a radar gun, presumably pointed at the release point, rather than calculated as an average over the flight. This creates a line-of-sight problem – the direction that the radar gun is pointed won’t be perfectly aligned with the initial velocity of the ball, so the measured speed will be less than the true speed.

I’m not sure how baseball pitches are measured, though a quick Google suggests that a radar gun is involved:

http://www.popularmechanics.com/outdoors/sports/physics/how-the-105-mph-fastball-tests-the-limits-of-the-human-body

As my model stands, a 128m throw would require an initial speed of 210km/h = 130 mph. So I still think the model is missing something at high velocities.

4. […] Our model of the flight of a ball (with wind resistance) has a number of parameters: mass and radius of the ball, density of air, gravitational acceleration, drag coefficient and the initial conditions of the throw – height, speed and launch angle. The density of air varies approximately linearly with altitude in the regime in which we are interested. We will vary this parameter only. […]

5. […] Fun with Wind-Resistance (Part 1) Fun with Wind-Resistance (Part 3) – Optimal mass […]

6. […] Part One: Fun with Wind-Resistance […]

7. […] Our model of the flight of a ball (with wind resistance) has a number of parameters: mass and radius of the ball, density of air, gravitational acceleration, drag coefficient and the initial conditions of the throw – height, speed and launch angle. The density of air varies approximately linearly with altitude in the regime in which we are interested. We will vary this parameter only. […]