I’ll admit it. I don’t watch that much baseball. But check out this throw Rick Ankiel (of the Cardinals) makes all the way from center field.

No one had a radar gun on this to find the speed of the ball. Why would you? But let me see if I can get an estimate of it anyway. Was it as fast as a pitch? I have no idea.

Starting Info

What can I get from the video? First, it looks like he throw the ball pretty close to the back wall. Using Google Earth, this is about 101 meters to third base.

Actually, his second throw was also about 101 +/- 3 meters (it seemed closer to the wall, but further towards the 3rd base side of the field). And what about the time of flight? I could get a rough estimate of that from the video, but only to the nearest second. Using Tracker Video analysis, I get a much better estimate of about 2.84 +/- 0.03 seconds.

Just to be careful, I am including an uncertainty for both of these measurements. For the distance, this is mostly a guess. I don’t know exactly where on the field he threw the ball. For the time, 0.033 seconds is the time between frames in the video. I wasn’t quite sure when he exactly let the ball go and when the ball returned to the same height.

What is uncertainty? You have never seen that before? In short, this is a way of looking at data and knowing that nothing is exact. The distance above says that the actual distance is almost certainly (but not absolutely) somewhere from 98 meters to 104 meters. When using uncertainty with measurements, you always give a range of values. How do you calculate stuff with this? You will see.

First approximation: no air resistance

If I neglect air resistance, then this is a fairly straight forward projectile motion problem. Let me start with a diagram.

The key to projectile motion is that the horizontal (x) motion and the vertical (y) motion are two independent 1-D motion problems. The only thing these two problems have in common is the time. The time to do the horizontal motion is the same time it take for the vertical motion. The other important idea is that the vertical motion has a constant acceleration and the horizontal motion moves at a constant speed. This is because there is essentially only one force acting on the ball while in the air, the gravitational force.

For this case, I know the time and the horizontal distance. This makes it pretty straight forward to calculate the x-component of velocity:

But how do you include the uncertainty in the starting values? There are actually several methods. For this case, I will use the simplest and easiest to see. I can calculate the maximum the x-velocity could likely be and the minimum value it could have.

Notice that to get the maximum velocity, you would use the max-distance and the min-time (since you divide by the time). Once I have both the max and min velocities, I can find the uncertainty in the velocity (which I will call ?v):

This is essentially averages the deviation from the max and min values. Yes, I know. This is not the most accurate method for finding uncertainty. But this is the method I use in algebra-based intro physics labs since students can easily see how this works.

Using the values from above, I get a horizontal velocity of:

Now for the vertical velocity. What do I know about the vertical motion? I know that it starts and ends at the same vertical height. I also know that the vertical acceleration is -9.8 m/s² and the time is the same as above. When an object has a constant acceleration, the following kinematic equation applies:

Since the final and initial y are the same, this becomes:

Since I know both t and g, I can find the vertical velocity. Using both the values and uncertainties from above, I get (assuming a small uncertainty in g so that I can ignore it):

Now that I have both the vertical and horizontal starting velocities, the total velocity can be found by considering the following vectors:

Where v is the magnitude of the total velocity vector. If I know the components, then the total velocity can be found with the pythagorean theorem:

Putting in my values, I get:

So, on the lower end this is a speed of 37 m/s or 82 mph. Impressive, but most fast ball pitches are over 90 mph. But wait. Would he actually have to throw it faster than this since there is air resistance? Really, I know it would have to be faster, but how much faster?

Baseball throw with air resistance

Adding air resistance really makes things quite a bit more complicated. Why? Well, here is a force diagram of the ball at some point in its motion:

The problem here is the air resistance force. Why? Here is a usual model for the force the air exerts on a moving ball (this is the magnitude of the force).

This is a velocity dependent force. That is what makes things tricky. Ok, but maybe this air resistance force is small enough to not really matter. One estimate of its importance is to compare the velocity of the object to its terminal velocity. Terminal velocity is the speed a falling object has in which the gravitational force is equal to the air resistance force. At this speed, the falling object would have a constant velocity.

So, what is the terminal velocity for a baseball? To answer this, you need some stuff. In particular, the constants in the air force model.

• ? is the density of the stuff it is going through. In this case, air. Air has a density around 1.2 kg/m³.
• A is the cross sectional area of the object. The cross sectional area of a baseball is a circle. A baseball has a radius of about 0.0375 meters. This would give it a cross sectional area of about 0.0044 m².
• C is the drag coefficient. It is a property that depends on the shape of the object. The coefficient for a sphere is 0.5. This is probably different than a baseball, let me use a value of 0.4 (experimental values vary).
• The last thing I will need is the mass. The mass of a baseball is around 145 grams.

So, at terminal velocity, the following would be true for a ball falling straight down:

Solving for v:

And using the values from above, this gives a terminal velocity of 36.7 m/s (82 mph). From this, I would say the air resistance does indeed matter.

Now what? How do you deal with a force that changes? You cheat. Here is the momentum principle that is used for these kinds of problems:

Typically, if the forces are constant, you can do things with the average velocity to figure out what you need. That can’t be done here since the force depends on the velocity. Here is the cheat. Instead of trying to solve this problem all in one step, I can break it into smaller pieces of time. If these “time pieces” are small enough, I can pretend like the force is constant and not changing. Really, it isn’t a terrible approximation for small time steps. Oh, but now look at the trouble I am in. Instead of doing one complicated problem, I have thousands of small easy problems. Bummer. No, not bummer. I can get a computer to do this work for me.

And that, in short, is the essence of a numerical calculation. This can be done quite quickly with something like VPython. However, this case is a little different. I am not trying to find the motion of a thrown baseball. Instead, I am trying to find what speed a ball would need to be thrown so that it goes 101 meters in 2.8 seconds. This means that the calculation will have to be run multiple times at different speeds and angles to find the one that works.

Here is a plot of the output. This show how far the ball went horizontally for different starting speeds and angles.

So this says that if you throw the ball at an angle of 10°, then it would have to be going 71 m/s in order to travel 101 meters. If you throw the ball at 40° then it would only have to be thrown with a speed of 44 m/s. So which is it? What was the angle the ball was thrown? From this, you can’t say. Also, it wasn’t really something you could get from the original video. So, what about this? What about the time of flight? I need both the horizontal distance AND the time of flight to be the same as in the video.

Here is a plot of the time of flight for different launch angles in the case that the ball went 101 meters.

From this, it looks like a launch angle of 17.7° should give a time of flight of 2.84 seconds. But what is the launch velocity in this case? Rerunning the calculation at this angle, I get a launch speed of 55 m/s or 123 mph.

So what is the answer?

Did Ankiel throw this ball super fast? Yes. Was it faster than a MLB pitcher? Could be. Here are some important notes:

• Actually, Rick Ankiel was a pitcher. Shows what I know.
• What is the coefficient of drag for a baseball? I used 0.4, but some say it could be as low as 0.3. This would mean less air and a lower speed. With this value, the throw speed would be around 50 m/s (111 mph). Still quite an impressive throw.
• So, could Ankiel be a pitcher? Possibly (or yes since he was a pitcher in the past). However, throwing a ball from center field is a little bit different, motion-wise than a pitch from the mound. First of all, you can only take one step for the pitch. Ankiel takes a couple for his throw. Also, there is more to pitching than speed.
• What about the uncertainty? Yeah, I didn’t do that for the numerical calculation. It can be done though with the Monte Carlo method (as described at SuperFly Physics). Maybe this will be a project for the future.
• What about the lower gravitational field and air density since the game was in Denver? Denver is 1.6 km above sea level. All things equal, this would make a gravitational field that is 99.9% the value at sea level. This ignores effects due to rotation of the Earth and density of the ground material. I don’t think it is a big deal. Not too sure about the density of air. Again, something to consider for later.

In short, it was a hell of a throw. Oh, and he did it twice.

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