Sunday, July 3, 2011

Coriolis Effect on a Home Run

Citizen's Bank Park
I like baseball.  Well, technically, I like laying[3] lying on the couch for three hours half-awake eating potato chips and mumbling obscenities at the television.  But let's not split hairs here.

Anyway, out of curiosity and in partial atonement for the sins of my past [1] I would now like to do a quick calculation to see how much effect the Coriolis force has on a home-run ball.

The Coriolis force is one of the artificial forces we have to put in if we are going to pretend the Earth is not rotating.  For a nice intuitive explanation of the Coriolis force see this post over at Dot Physics.

Let's now consider the following problem.  Citizen's Bank Park (home to the Philadelphia Phillies) is oriented such that the line from home plate to the foul pole in left field runs essentially South-North.  Imagine now that Ryan Howard hits a hard shot down the third base line (that is, he hits the ball due North).  Assuming it is long enough to be a home run, how with the Coriolis force effect the ball's trajectory?

This is a well-posed problem and we could solve it as exactly as we wanted.  But please don't make me. It's icky and messy and I don't feel like it.  So let's do some dimensional analysis!  Hooray for that!

So what are the relevant physical quantities in this problem?  Well, we'll certainly need the angular velocity of the Earth and the speed of the baseball.  We'll also need the acceleration due to gravity.  Alright, so what do we want to get out of this?  Well, ideally we'd like to find the distance the ball is displaced from its current trajectory.  So is there any way we can combine an angular velocity, linear velocity and acceleration to get a displacement?

Let's see.  We can write out the dimensions of each in terms of some length, L, and some time, T.  So:

\[ \left[ \Omega \right] = \frac{1}{T} \]

\[ \left[ v \right] = \frac{L}{T} \]

\[ \left[ g \right] = \frac{L}{T^2} \]

where we have used the notation that [some quantity] = units of that quantity.  Combining these in a general way gives:

\[ L = \left[ v^{\alpha} \Omega^{\beta} g^{\gamma} \right] = \left( \frac{L}{T}\right)^{\alpha}\left( \frac{1}{T}\right)^{\beta}\left( \frac{L}{T^2}\right)^{\gamma} = L^{\alpha+\gamma} T^{-(\alpha+\beta+2\gamma)}\]
Since we want just want a length scale here, we need:

\[\alpha+\gamma = 1~~~\mbox{and}~~~\alpha+\beta+2\gamma = 0. \]

We can fiddle around with the above two equations to get two new equations that are both functions of alpha.  This gives:

\[\beta = \alpha - 2~~~\mbox{and}~~~\gamma = 1 - \alpha. \]

Unfortunately, we have two equations and three unknowns, so we have an infinite number of solutions.  I've listed a few of these in the Table below.

Ways of getting a length

At this point, we have taken Math as far as we can.  We'll now have to use some physical intuition to narrow down our infinite number of solutions to one.  Hot dog!

One way we can choose from these expressions is to see which ones have the correct dependencies on each variable.  So let's consider what we would expect to happen to the deflection of our baseball by the Coriolis force if we changed each variable.

What happens if we were to "turn up" the gravity and make g larger?  If we make g much larger, then a baseball hit at a given velocity will not be in the air as long.  If the ball isn't in the air as long, then it won't have as much time to be deflected.  So we would expect the deflection to decrease if we were to increase g.  This suggests that g should be in the denominator of our final expression.

What happens if we turn up the velocity of the baseball?  If we hit the ball harder, then it will be in the air longer and thus we would expect it to have more time to be deflected.  Since increasing the velocity would increase the deflection, we would expect v to be in the numerator.

What happens if we turn up the rotation of the Earth?  Well, if the Earth is spinning faster, it's able to rotate more while the ball is in the air.  This would result in a greater deflection in the baseball's path.  Thus, we would expect this term to be in the numerator.

So, using the above criteria, we have eliminated everything on that table with alpha less than 3 based on physical intuition.  Unfortunately, we still have an infinite number of solutions to choose from (i.e. all those with alpha greater than or equal to 3).  But, we DO have a candidate for the "simplest" solution available, namely the case where alpha = 3.  Since we have exhausted are means of winnowing down our solutions, let's just go with the alpha = 3 case.

Our dimensional analysis expression for the deflection of a baseball is then

\[ \Delta x \sim \frac{v^3 \Omega}{g^2} \]

Plugging in typical values of

\[ v = 50~\mbox{m/s}~~~(110~\mbox{mi/hr}) \]
\[ \Omega = 7 \times 10^{-5}~\mbox{rad/s} \]
\[ g = 9.8~\mbox{m/s}^2 \]
we get
\[ \Delta x \approx 0.1~\mbox{m} = 10~\mbox{cm}. \]

That's all fine and good, but which way does the ball get deflected?  Is it fair or foul?  Well, remembering that the Coriolis force is given by:

\[ {\bf F} = -2m{\bf \Omega} \times {\bf v} \]

and utilizing Ye Olde Right Hand Rule, we see that a ball hit due north will be deflected to the East.  In the case of Citizen's Bank Park, that is fair territory.

But how good is our estimate?  Well, I did the full calculation (which you can find here) and found that the deflection due to the Coriolis force is given by

\[ \Delta x =-\frac{4}{3}\frac{\Omega v^3_0}{g^2} \cos \phi \sin^3 \alpha \left[1 -3 \tan \phi \cot \alpha \right] \]

where phi is the latitude and alpha is the launch angle of the ball. We see that this is essentially what we found by dimensional analysis up to that factor of 4/3 and some geometrical terms. Not bad!

Plugging in the same numbers we used before, along with the appropriate latitude and a 45 degree launch angle we find that the ball is deflected by:

\[ \Delta x = 5~\mbox{cm}. \]

For comparison, we note that the diameter of a baseball is 7.5 cm. So in the grand scheme of things, this effect is essentially negligible. [2]

That wraps up the calculation, but I'm certain that many of you are still a little wary of this voodoo calculating style. And you should be! Although dimensional analysis will give you a result with the proper units and will often give you approximately the right scale, it is not perfect.

But, it can be formalized and made rigorous. The rigorous demonstration for dimensional analysis is due to Buckingham and his famous pi-theorem. The original paper can be found behind a pay-wall here and a really nice set of notes can be found here. It's a pretty neat idea and I highly recommend you check it out!


Unnecessary Footnotes:

[1]  Once in college I argued with a meteorologist named Dr. Thunder over the direction of the Coriolis force on a golf ball for the better half of the front nine at Penn State's golf course.  I was wrong.  Moral of the story:  don't play golf with meteorologists.  

[2] For a counterargument, see Fisk et al. (1975)

[3] Text has been corrected to illustrate our enlightenment by a former English major as to the difference between 'lay' and 'lie' through the following story: 'Once in a college psych class, a young student said "It's too hot. Let's lay down." A mature student, a journalist, asked, "Who's Down?" '


5 comments:

  1. Could you explain why 5 cm is not significant? 2 inches (at it would be called here in Philadelphia)into the fair zone seems like a pretty good nudge to me?

    Walt

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  2. Hi Walt,

    I guess this does come down to semantics at some point. What I consider to be "too small to matter," others might see as making a crucial contribution. For example, an extra two inches fair may be the main reason anyone outside Cincinnati still remembers the '75 World Series.

    But to really call something "negligible" I need to say that it is much smaller than other effects. Most of these effects I have ignored for the calculation. For example, I have neglected all air resistance. In real life, the air does all kinds of funky things to the baseball. A rotating baseball going through the air slices and hooks and moves around quite a bit. This motion would largely "drown out" any effects due to Coriolis forces.

    Additionally, the Phillies' stadium has the ideal orientation to give the maximum for this effect. Hitting the ball any other direction than due north or due south would make the deflection smaller. So this 2 inch estimate should really be considered an upper bound on the deflection. Under other conditions, the deflection will likely be less.

    So perhaps a better way to say it would be that it would be very unlikely in a real game to see some kind of motion which we could really claim was due to the Coriolis force.

    But, otherwise, it may just be semantics.

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  3. My intuition for coriolis was poisoned by an early problem of finding the coriolis force of the tracks on a northbound train.

    Since then I have to be very careful to differentiate between the a ficticious forces as seen by the motion of free particles w.r.t. an non-inertial frame, and the real forces pushing on a particle constrained to follow a straight line on this earth.

    I didn't look at the google doc derivation, but it is interesting to think that the initial launch angle might match the latitude and make the motion parallel to the axis of rotation with zero coriolis effect, and the majority of the deflection would occur on the fast steep portion of the downward arc.

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  4. I absolutely love footnote #1. Good stuff!

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  5. Hi Corky,
    Great post.

    I have a question on Coriolis force. Not related to the baseball scenario but here goes.

    Consider we launch a cannonball due south from a point @ 45% latitude (Or in other words the point defined with the co-ordinate system shown on page1 of your googledocs derivation https://docs.google.com/file/d/0Bwd5hrDOxWsrOGZmMWZkYzQtM2M2Ny00NjlmLTgyYmMtNTQwZjI1ODU1NWI4/edit?hl=en_US&pli=1). The cannonball travels for 2 minutes (through vaccum - ignore air resistance) and drifts 'right/west'.

    A few nanoseconds before it hits the ground, x-velocity of the cannonball is slower than x-velocity of the ground below. In order for the cannonball to stick/stay at the point of impact, its x-velocity must match the velocity of the ground.

    What effect does this differennce in velocity have on the cannonball on impact. Does it experience a sudden west-east acceleration?

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