When a ball moves through the air a force called drag slows the ball down. The drag force acts in the opposite direction of the ball's movement. Here's the equation for drag force:
F=Cd*A*d*(v^2)/2
I got this equation from page 362 of The Physics and Technology of Tennis by Brody, Cross, and Lindsey (BCL). In the equation, Cd is the drag coefficient of the ball, A is the cross-sectional area of the ball, d is the density of the air, and v is the ball's velocity. BCL give values of 1.21 kg/m^3 for d (sea level), and 0.55 for Cd for a new tennis ball. The ball's cross sectional area is pi*r^2, where r is the radius of the tennis ball (roughly 33 mm).
Having fun yet?
There's an even more fun equation a few pages later (p. 388) in the appendix on ball trajectories. In this equation we've ignored the relatively minor vertical component of a ball's velocity and are only looking at the horizontal component. When you hit a ball toward the other side of the net, how much does it slow down? Since I'm interested in the effects of lower density air on the flight of a tennis ball, these are the equations that I care about. Here's the equation:
t = (e^(k*Cd*x)-1)/(k*Cd*v0)
This tells us how long it takes (t, measured in seconds) a tennis ball (with a coefficient of friction Cd) to travel a particular distance (x, measured in meters) when moving through air for a specific initial velocity (v0, measured in meters per second). In addition to the terms just mentioned we have two other terms to define. The term e is simply the base of the natural logarithm. The term k is a bit more complicated. We'll use an equation from BCL (p. 387) to define k,
k=(d*pi*R^2)/(2*m)
Argh. More undefined terms. We used d above, and it's the density of the air. R is just the radius of the ball, roughly 33 mm (.033 m), and m is the mass of the tennis ball (57 g or 0.057 kg). Pi is pi.
So for a sea level standard tennis ball,
k = (1.21*pi*0.033^2)/(2*0.057)
k = 0.0363 m^-1
Back we go to our more interesting equation, armed with a value for k,
t = (e^(k*Cd*x)-1)/(k*Cd*v0)
Let's plug in some numbers and see how long it takes a ball hit at 70 mph to travel 80 ft. We're imagining a player standing a little bit behind his own baseline, perhaps, hitting a forehand to roughly the far baseline.
We have to convert feet to meters and mph to m/s. Google will do that for us, spitting out 31.3 m/s for 70 mph, and 24.4 m for 80 ft.
We already said that Cd for a new tennis ball is 0.55 according to BCL. Let's plug and play:
t = (e^(0.0363*0.55*24.4)-1)/(.0363*0.55*31.3)
t = 1.004
Let's round that to 1.0 sec. How cool is that? Must have picked those numbers to come out that way!
So how much does the air slow down a ball moving horizontally? Well, without air we know how long a ball leaving the racquet at 31.3 m/s would take to travel 24.4 m by dividing the second number by the first.
t = 24.4/31.3 = 0.78 sec
So this forehand took 28% longer to go 80 ft through air than though a vacuum. By playing at sea level instead of outer space (or the moon), we bought ourselves 0.22 sec to get to this ball.
Now for the big question: How much time do we lose in Denver?
The density of air in Denver is not 1.21 kg/m^3. It's about 0.84 times that, or roughly 1.02 kg/m^3.
Back to our equation. Where does the density of air come in? It comes in via that pesky k. Recall
k=(d*pi*R^2)/(2*m)
There's d in there. So we need to recalculate k for Denver rather than sea level.
k = 1.02*pi*0.033^2/(2*.057)
k = 0.0306
Should we call that kd for Denver? Back to the time equation:
t = (e^(kd*Cd*x)-1)/(kd*Cd*v0)
Plug in some numbers and let Google give us a value for long it takes a 70 mph forehand to travel 80 ft in Denver:
t = (e^(0.0306*0.55*24.4)-1)/(0.0306*0.55*31.3)
t = 0.964 sec
We've lost some time by moving up to Denver. But not much. We've only lost about 0.036 sec. That's a loss of 3.6% (again note the easy math because of the 1.0 sec transit time at sea level!).
Trouble. In earlier posts, I've been assuming that the ball slows down 16 to 17% less slowly in Denver than it does at sea level. All good players say the ball travels faster through the air up here than down there. What gives?
What gives, indeed!
It could be completely due to the extra bounce of the ball. The high-altitude balls do tend to bounce a bit more up here than balls bounce at sea level. That 9% livelier ball, plus this roughly 4% less time due to thinner air, may add up to a difference that we all notice. If you can simply add these effects, then 13% or so livelier game probably is a big deal.
Since I screwed up so badly on my prior estimation of the effects of our air on the ball, I'll keep my mouth (or fingers) shut until I've given this some more thought.
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