THE BOOK--Playing The Percentages In Baseball

I replied to Tango’s comment in the last couple paragraphs of my piece today, which is linked in my name.  The basic result is that, assuming a .2 second reaction time, which is probably an incorrect assumption, a hitter makes his decision to swing when the pitch is 27 feet away for a Beckett fastball, 20 feet on a Wakefield knuckleball and 21 feet on a Rich Hill curve. 

If you add .1 seconds to the reaction time to account for the time the swing actually takes once the hitter recognizes the pitch, which I’m also not sure is accurate, the “reaction distance” increase to 29 feet for Wakefield, 41 feet for Beckett, and 31 feet for Hill.

Good job Joe.  I will defer to the physicists and baseball players to establish the “point of no return”.  It’s not that important to decide if it’s .20, .25, .30 or whatnot.  As long as we pick on one standard number, the resulting values (41 feet, 29 feet, etc) are far easier to grasp than milliseconds. 

I know my eyes glaze over when I see those milliseconds, and I love numbers!  “Reaction Distance” is a great name!

According to this site:
http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/Jon_Drobnis/bmb.html

They are saying .20 seconds for decision, and .15 seconds for swing, which is .35 seconds. According to Joe’s original article, that point would be 50 feet for the 100mph pitches.  That seems too hard to believe, that a (MLB) batter must decide almost as soon as the ball is out of the pitcher’s hand to decide whether to swing (knowing it’s a fastball).  He needs to first figure out what kind of pitch it’s going to be!

I’d be happy to make the standard at .25 seconds, maybe .30.

I do not believe you are thinking about this the right way. Consider this:

You are given a very important 30 question test and are allowed one hour to complete it. Each question will take you 1 minute to read and fill in an answer. Would you fly through the test, making sure you have something written down and finish it in 30 minutes or would you use all of your time to better answer each question to make sure they are correct?

Of course, you’d take the alloted time to make sure you’ve answered all of the questions to the best of your abilities.

Well, why would a batter be any different? Would he see the ball for .2 seconds and then decide to swing right there if he has more time to make his decision? He would certainly wait as long as possible before deciding to swing or to let the pitch go.

I do not know how to do the specific math for the rest of this but I believe we can assume with fair accuracy…

If it takes .4 seconds for a 90 MPH pitch to reach home plate, we need to be aware that the time is not evenly distributed. The ball slows down as it approaches the plate so the first 30 feet is not covered in .2 seconds but roughly .18 seconds. The final 30 feet will be covered in .22 seconds and the final 15 feet will be covered in about .125 seconds.

After .2 seconds, the ball will be roughly 25 feet from home plate.

If the batter requires .15 seconds to swing, that would equate to about 18 feet.

The batter can use the first 42 feet (.25 seconds) to analyze the pitch’s speed, spin, location etc. and then begin his swing.

I don’t think a lot of batters make the decision whether to swing or take just on what they observe after the pitch is released. Obviously they need to figure out the velocity and location, but they also arrive in the batter’s box with some prior expectations that may affect how they react to what they see. And they have an at-bat strategy that affects how and how quickly they make a decision whether and how to swing at the next pitch.

Consider the following as just some of the auxiliary considerations that come into play.

1) The count;

2) The man on base situation and who is on base;

3) Is there a “play” on—e.g., hit-and-run, bunt sign;

4) What kind of pitches does a pitcher have in his repertoire (or what kinds does he have command of today), and what kind does he like to throw in certain situations (information that may help the batter know what to look for);

5) What kind of pitch and in what location the pitcher threw to him in his last plate appearance, or last time in a similar situation;

6) Whether the pitch is hittable (even if it’s not a strike; but even if it looks like a strike, if it’s not already 2 strikes against him, the hitter may pass on the pitch if he can’t get good wood on it);

7) Whether to try to hit the ball in fair territory or just foul it off.

All of these things and more are in the batter’s head before the pitcher winds up, and all may affect the decision to whether (and how) to swing as well as how long it takes him to reach that decision (for good or for ill);

Seems to me someone has to work through the batter’s logic, not just his ability to pick up the pitch and get a bat on it.

Dwight,

I think we’re saying the same thing...the numbers I gave are the minimum distance from the plate where the batter must recognize the pitch, react and swing in order to still make contact.  If it takes .2 seconds for the ball to reach that point, than yes, I am saying he will make his decision to swing based on just that quick look at the pitch...he needs to or he won’t make contact at all.

I’m not totally sure what you’re saying in your math section, but I think the reason you’re getting different answers is that you’re using 60 feet as the release point for part of it and 50 feet for other parts. 

A 90 MPH pitch takes .4 seconds to reach home from a point 50 feet away, but .5 seconds to reach home from a point 60 feet away.  50 feet is a proxy for the release point, and where the initial parameters of the gameday data are measured from, so I used that in my analysis. 

When you say that ‘The batter can use the first 42 feet (.25 seconds) to analyze the pitch’s speed, spin, location etc. and then begin his swing.’, I think you used both release points.  From 50 feet, a 90 MPH pitch will be 18 feet from home plate after .25 seconds, but the batter will have only tracked it for 32 feet.  After .25 seconds a 90 MPH pitch from 60 feet will be 28 feet from home, also having been tracked 32 feet.

Fargo,

The hardest thing to quantify when looking at this is the reaction+swing time of the hitter.  I had a hard time finding a baseline reaction time and wouldn’t know where to begin to try to quantify the rxn time based on whatever is going through a hitter’s head during his at-bat, so all I can say right now is that this is the distance when a decision needs to be made.  If you could figure out the impact of each of those factors it would be pretty sweet though.

I’m not going to bother with any math or anything this time especially since I dont even know how to do it. My main point was that the ball slows down as it approaches the plate.

If a ball covers X feet in Y seconds, it does not cover X/2 feet in Y/2 seconds. It will take longer for the 2nd half of the pitch (the final X/2 feet) than for the 1st half of the pitch. This needs to be brought into the equation.

The equations of motion account for drag and lift force, so the “slow down” effect is accounted for.

I have a question regarding the accelerations provided in the xml file - I don’t know who would be best suited to answer.

Is az relatively constant across pitches?  An object accelerates whenever a force acts on it.  In principle, the acceleration along the z-axis should be due only to the force of gravity.  It would be interesting to know if, in practice, 1) the lift/drag/Magnus forces induce an acceleration in the z-direction 2) how big this acceleration is compared to acceleration due to gravity and 3) what is the magnitude of this difference (wrt to gravity) for pitches with the greatest vertical breaks.

Sal

That is the theory. Gravity (az) should be of the 32 ft/sec^2. In reality if you look at the data it ranges from about 10-35 ft/sec^2—so a wide range of different values. I agree with you that you’d have intuitively thought az would be constant, and = to gravity.

I’m not sure what is causing the different value. I suppose you could argue that the magnus force is modifying gravity. However, given how pitch f/x works I do not know how az is calculated.

Pitch f/x, as far as I am aware, works based on three cameras that tracks the position of the pitch. I can only assume that from that trajectory it back calculates an az acceleration component.

Assuming the az component is accurate you can do what you suggest from the data. I might take a look at Hudson later, where I have the analysis and report back.

John,
If I understand correctly, pitch f/x tracks the trajectory and then computes the best-fit a and v paramaters based on the flight of the ball (the time-position data).

I don’t have an intuitive feel for such things, but I am very surprised to learn that az can be as low as 10 ft/sec.  That’s an acceleration of -25 ft/sec^2, or about .85 g!  Can that be correct?

Not sure if we are quite on the same page.

g = -32 ft/sec^2

so a g = -10 = 0.3g .... not sure how you work out .85g.

Anyway, I looked at Hudson and Smoltz

Hudson ax varies mostly between -24 and -29 but with readings as low as -15. It does sort of segment by pitch type with four seam fastball having lower az and the slider, apparently have higher az ... these are on small samples.

For Smoltz the az variance is much higher from -8 to -37.

I think you are right on the best-fit hypothesis. It could be that the az parameter isn’t particularly sensitive.

John
It doesn’t really matter, but what I was getting at was that

a_z = a_g + a_m where a_m = magnus/lift/drag, etc and a_g is the acceleration due to gravity.  (More formally, F_z = F_g + F_m, but F = ma and all the masses all cancel out)

if a_g = 32 ft/s2, then a_m must be - 22 ft/s2 in order to get az = 10 ft/s2.  So the magnitude of the a_m is 22/32 = . 7 (uh....85 was sloppy mental math), or a_m = .7g.  Like I said, I find it remarkable that the lift/drag/magnus forces can have such a great magnitude.

Or, as you said, the fitting may be relatively insensitive to a_z.  That’s probably something worth investigating.

Salb918/John, take a look at Dr. Nathan’s Physics of Baseball site, linked in Tango’s post.  He deals with all these questions about the forces on the ball.  Particularly interesting to you might be his analysis of the drag force and Magnus force that’s linked at the bottom of that site.

Thanks Mike.

According to Dr Nathan the az component =

d^2z/dt^2 = -K(CD)(vvz) + K(CL)(vvy) cos(phi) - g

where K = 5.5x10^-3; g = 32.3 ft/s^2; (CL)varies be 0 and 0.7; phi is the angle between the axis of spin and the x-axis. Phi = 90 deg is equal to spin pointing up (i.e. sidespin), phi = 0 is equal to top spin

We can ignore vz so to get a az < g the second term in the above equation has to be positive

We know that vy is negative as the plate is the orgin of the coordinate system. When there is top spin cos(phi) ~ 1 so we should see az > g, this fits with practice as curveballs must have a downward component of acceleration to drop as much as they do.

If there is backspin on the ball then cos(phi will be negative). If CL = 0.1, cos phi = -1, vy= v= -130 (ft/sec); K = 0.0055 then we get the second component = 10 or so, for a az of -22 ... in line with the Smoltz numbers.

I have no idea if that calc is at all realistic or whether the assumptions v=vy is correct (the Nathan paper is ambiguous on that) but I guess it feels sorta plausible.

John,
After considering for a moment, yes, it is okay to assume that v_y >> v_z since v_y/v_z is something like the ratio of the distance the ball travels on those axes, so 60/3.  What this also implies is that drag is unimportant and that the Magnus force dominates.

Nathan’s fitting algorithm is rigorous, though, and doesn’t make the assumption that v_y >> v_z.

Nathan says that you can make the above approximation if I read the paper correctly.

On further reading my reply above was *slightly* wrong. CL is between 0.1 to 0.2 (the calc is right) as evidenced by fig 4b. Nathan confirms that FM (magnus force) can be anywhere from 0 to 0.7mg ... consistent with the data

Some more cool charts from Dan Fox:

http://www.rmsabr.org/?p=46

I think that one reason the reaction times seem unreasonable (e.g. deciding on a Zumaya fastball after only 10ft. of travel) is that the actual process probably does not work the same way as, say, a drag racer waiting for the green light. 

I think it’s more likely that a batter begins the swinging process early, and decides halfway through to stop it if he doesn’t want to swing.  This is why we see a lot of check swings, why announcers comment on batters who were “taking all the way” (normally not the case), and why timing is seen as so important.

If this is right, then the .15 sec of swinging and the .2 second of reaction actually overlap, they aren’t sequential.  For this reason, I think tango’s right, and it’s not important to know the exact number (it’s probably different for every batter, anyway), but to decide on a standard.

That’s a very interesting point.  There are obviously certain movements a batter makes that he’d be smart to do during the “overlap” phase.  It would be interesting to compare how batters react to down the middle pitches, pitches at the corners, and just really bad pitches.  What “overlap” mechanics are consistent in all cases, and when does he actually bail (i.e., what overlap mechanics are being stopped), without looking like he was actually doing anything to begin with.