Peter O’Brien’s Raw Power: Estimating Batted-Ball Velocities in the Minor Leagues

On May 20th Peter O’Brien hit a massive home run to straight away center clearing the 32 foot tall batter’s eye at Arm & Hammer Park more the 400 feet from home plate.  O’Brien is currently 1 home run behind Joey Gallo, in what looks to be an exciting competition for the minor league home run title.  O’Brien isn’t as highly touted a prospect as Gallo, but he still has some of the most impressive power in the minor leagues.  Reggie Jackson saw O’Brien’s home run and said it was one of hardest hit balls in the minor leagues that he had ever seen (and Reggie knows a thing or two about tape measure home runs).

How hard was that ball actually hit?  It is impossible to figure out exactly how hard and how far the ball was hit from the available information.  You can however use basic physics to make a reasonable estimation.

Below I explain the assumptions and thought process I used to get to an estimate of how hard the ball was hit.  If that does not interest you, then just skip to the end to find out what it takes to impress Reggie Jackson. But, if you’re curios or skeptical stick around.

OBSERVATIONS

I started off by watching the video to see what information I could gather (O’Brien’s at bat starts at the 37 second mark in the video).

TIME OF FLIGHT From the crack of the bat, to the ball leaving the park – it appears to take 5 seconds. If you watched the video, you can tell this is not a perfect measurement since the camera doesn’t track the ball very closely. If you think you have a better estimation, let me know and I’ll rework the numbers.  

LOCATION LEAVING THE PARK  The ball was hit to straight away center. From the park dimensions we know when it left the park it was 407 feet from home plate and at least 32 feet in the air to clear the batter’s eye.

ASSUMPTIONS

COEFFICIENTS OF DRAG (Cd) – The Cd determines how much a ball will slow down as it moves through the air. I chose 0.35 for the Cd because it is right in the middle of the most frequently inferred Cd values for the home runs that Allan Nathan was looking at in this paper.In looking at the Cds of baseballs, Allan Nathan showed there is reason to believe that there is some significant (meaning greater than what can be explained by random measurement error) variation in Cd from one baseball to another.

ORIGIN OF BALL I assume the ball was 3.5 feet off the ground and 2 feet in front of home plate when it was hit.  These are the standard parameters in Dr. Nathan’s trajectory calculator. But what if the location is off by a foot? The effects of the origin on the trajectory are translational. One foot up, one foot higher. One foot down, one foot lower. The other observations and assumptions are more significant in determining the trajectory of the home run.

Using these assumptions and the trajectory calculator, I was able to determine the minimum speed and backspin a ball would need in order to clear the 32 foot batter’s eye 5 seconds after being hit at different launch angles.  The table below shows the vertical launch angle (in degrees), the back spin (in RMPs) and the speed of the balled ball (in MPH).

Vertical launch angle Back spin Speed off Bat
19 14121 101
21 6817 101.9
23 4155 102.75
25 2779 103.69
27 1940 104.7
29 1375 105.89
30 1156 106.5
32 805 107.88
34 536 109.4
36 322 111.1
38 149 112.99
40 4 115.1

The graph shows a more visual representation of the trajectories in the table above (with the batter’s eye added in for reference).

Looking at the graph you will notice that all of these balls would be scraping the top of the batter’s eye.  This makes sense because the table shows the minimum velocities and back spins needed for the ball to exactly clear the batter’s eye.

What is the slowest O’Brien could have hit the ball?

If you were in a rush, looking at the table you would think the slowest O’Brien could have hit the ball would be 101 MPH at 19o. But, not so fast! The amount of backspin required for the ball to travel at that trajectory is humanly impossible.

What is a reasonable backspin?

I am highly skeptical of backspin values greater than 4,000 rpm based on the Baseball Prospectus article by Alan Nathan “How Far Did That Fly Ball Travel?.” The backspin on home runs Nathan examined ranged from 500 to 3,500 rpm, with most falling in around 2,000. The first 3 entries in the table have backspins of over 4,000 and can be eliminated as possibilities. If the ball with the 19o launch angle only had 3,500 rpm of back spin it would have hit the batter’s eye less than 11 feet off the ground instead of clearing it.  Maybe you’re skeptical that I eliminated the 3rd entry because it’s close to the 4,000 rpm cut off.  Think about it this way, if a player was able to hit a ball with over 4,000 rpm of back spin, they would have to be hitting at a much higher launch angle than 23o (Higher launch angles generate greater spin while lower launch angles generate less spin).

The high launch angle trajectories with very little back spin (like the bottom three in the table) are also not very likely.  A ball hit with a 40o launch angle would almost certainly have more than 4 rpm of back spin.  If the ball hit with the 40o launch angle had 1,000 rmp of back spin (instead of 4) it would have been 70 feet off the ground, easily clearing the 32 foot batter’s eye.

Accounting for reasonable back spin, the slowest O’Brien could have hit the ball is 103.69 MPH at 25o with 2,779rpm of backspin.

So what do all these observations and assumptions get us?

We can say that the ball was likely hit 103.69 MPH or harder, with a launch angle of 25o or greater.  103.69 MPH launch velocity is not that impressive, it is essentially the league average launch velocity for a home run.  Distance wise, how impressive of a home runs was it? Unobstructed the ball would have landed at least 440 feet from home plate (assuming the 25o scenario).  The ball probably went further than 440 because it did not scrape the batter’s eye. So, how rare is a 440+ foot home run? Last year during the regular season there were 160 home runs that went 440 feet or further, there were a total of 4661 home runs that season, meaning only 3.4% of all home runs were hit at least that far.

For those of you who wanted to just skip to the end. My educated guess is that the ball went at least 440 feet and left the bat at at least 103.69 MPH.

If you like this, you can read other articles on my blog GWRamblings, or follow me on twitter  @GWRambling

None of this would have been possible without Alan Nathan’s great work on the physics of baseball.  I used his trajectory calculator to do this, and I referenced his articles frequently to make sure I wasn’t way making stupid assumptions. The information on major league home run distance is based off of hittrackeronline.com



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George is always thinking about baseball, and frequently writes down those thoughts on his blog. Follow him on Twitter @GWRambling.


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Alan Nathan
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Alan Nathan

Nicely done. Just to summarize, you have two pieces of information that can be used to determine two unknowns. The information is the height and distance of the ball at a time of 5 sec. From your table, it looks like you fix the launch angle to some specified value, then vary the speed and backspin so that the trajectory passes through your fixed point at 5 sec. Greg R. from hittracker does it a bit differently. He uses a model for the ball-bat collision to come up with a formula for the dependence of backspin on the batted ball speed and launch angle. Within the context of that model, he varies speed and angle to make the reproduce the landing point and flight time, with the backspin being determined by the speed and angle. Thus, hcomes up with an exact solution, rather than a table of solutions. But you did a good job justifying which row of your table corresponds to the best solution.

jim S.
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jim S.

I met Alan at a couple of SABR conferences. He’s a really good guy.

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