Giancarlo Stanton’s Long Single by Alan Nathan October 15, 2021 Every now and then, something happens in a major league game that arouses my interest as a baseball physicist. In the sixth inning of the recent American League Wild Card game, the Red Sox were up 3-1. With Aaron Judge on first and one out, Giancarlo Stanton hit a laser of a shot that bounced high off the left-center field wall at Fenway, barely missing a home run. As it happened, Judge was thrown out at home plate, while Stanton took second on the throw. It was truly a game-changing play. But what I really want to talk about is Stanton’s shot, particularly the distance the ball would have gone had it reached field level unobstructed. That is the usual meaning attached to “home run distance.” Although not normally done, I want to apply it to a single. As it turns out, there is a wealth of data available that allows us to figure this out. First, we know the Statcast measurement of the launch conditions: Table 1: Giancarlo Stanton Statcast Launch Parameters Exit Velocity (mph) Launch Angle (deg) Spray Angle (deg) Spin Rate (rpm) Spin Axis (deg) 114.9 17.8 -11.9 1115 184 The most important of these parameters, exit velocity and launch angle, are publicly available. Of lesser importance to the calculation are the spray angle, spin rate, and spin axis. Note that the ball is hit very hard and, typical of Stanton, at a somewhat low launch angle. For both those reasons, the spin rate is not particularly large and the spin axis indicates nearly perfect backspin (i.e., very little sidespin). Note also that the small negative spray angle means the ball was hit slightly to the left-field side of center. If we had no additional information other than the launch parameters, the way I would proceed is to simply input those parameters into my Trajectory Calculator and see what we find using the default parameters for drag and lift. However, we can do much better than that since we also have very good information about where the ball impacted the wall (distance, height, and spray angle) as well as the flight time. These data come from two independent (but anonymous) sources, labeled 1 and 2 in the table, which used very different methods to obtain the information. Nevertheless, they are in excellent agreement with each other: Table 2: Just Where Did It Hit the Wall? Wall Impact Extrapolated Landing Data Source D (ft) H (ft) SA (deg) T (sec) D (ft) H (ft) SA (deg) T (sec) 1 358 34.9 -13.8 3.38 424 0 -14.2 4.31 2 360 32.9 -13.5 3.42 422 0 -13.8 4.29 Parameters of the wall impact and the resulting extrapolated landing, where D is the horizontal distance from home plate, H is the height above home plate, SA is the spray angle, and T is the flight time. So, how do we use these data to extrapolate the trajectory? My technique is as follows. First, I use the Trajectory Calculator, with all the correct weather information (temperature, etc.) as well as the initial conditions. Next, I adjust the drag coefficient, the lift coefficient, and the spin axis so that the resulting trajectory impacts the wall at the correct location and at the correct time. I do this independently for each of the two data sources in the second table. This adjustment process utilizes the “Solver” feature of Excel. For the experts out there, Solver utilizes the so-called Levenberg-Marquardt algorithm for non-linear least-squares fitting, an algorithm I have been using since my grad student days. The justification for treating the drag and lift coefficients as fitting parameters is that they are not fixed numbers but are known to vary from ball to ball. Moreover, the spin axis is probably the least well determined of the Statcast parameters, so small adjustments are also not unreasonable. Finally, adjusting these parameters is an indirect way of accounting for any effect of wind. Once the adjustment process has converged, we have the full trajectory for each data set, from home plate to the wall, which is then easily extrapolated to ground level to find the distance. The essential point is that with a little bit of physics and the combined constraints provided by the launch and wall-impact parameters, there is very little wiggle room in the extrapolation. The resulting full trajectories are shown in the graphs, with the landing information summarized in Table 2. Trajectories for the Stanton long single, showing height vs. horizontal distance, with dots indicating the location at 0.5-second intervals. The black vertical line is the location of the wall. The red and blue curves are for data sets 1 and 2, respectively. The top graph shows the full trajectory while the bottom shows an expanded view of the extrapolated segment beyond the wall. Here, some comments are in order. First, The resulting drag and lift coefficients are approximately 8% and 11%, respectively, greater than the default values. These increases may be due to the ball or to an in-blowing wind; from the point of view of the extrapolation, it really doesn’t matter. Moreover, based on lots of trajectory data I have analyzed in the past few years, those changes are not unreasonable. Had I used the default parameters in the Trajectory Calculator, the wall impact would have occurred at a height of 36 feet, a bit higher than the actual height. As a result, the extrapolated distance would have been a bit longer, 431 feet. The amount of backspin and sidespin are very close to the default parameters in the Trajectory Calculator for the given launch and spray angles. In particular, the very small amount of sidespin is expected for balls hit slightly to the pull side of center field. The fact that the spray angle at the wall impact is slightly more negative than the initial spray angle (i.e., the ball hooks slightly toward the left-field foul line) means that the spin axis is less than 180 degrees. This feature justifies allowing the spin axis to vary in the fitting process, since the Statcast value would have resulted in a slice (i.e., a bend toward center field) rather than a hook, which would be inconsistent with the actual data. Finally, the two data sources result in nearly identical trajectories, with the extrapolated distances and flight times identical to within 2 ft and 0.02 sec, respectively. We can therefore say with a very high degree of confidence that the unimpeded ball would have traveled of order 423 feet, with an uncertainty of no more than about two feet. Mission accomplished!