When I pitched in Little League, I had a "low octane" fastball, so much so that when I took the mound scouts put away their radar gun and got out a calendar. Yet I was hard to hit because I was really good at hiding the ball. Sometimes the umpires couldn't find it for 20-30 minutes; my favorite spot was behind the water cooler in the visiting dugout but also if you place the ball right on the foul line it totally blends in and umpires will walk right by it.

But this post is not about my Little League career, it's about Brad Kilby and how much phantom velocity a pitcher gets if his delivery makes it difficult for hitters to pick the ball up out of the pitcher's hand. I've done the Math to the best of my ability and here's what I've come up with...Stay with me on the logic and correct me on any faulty Math...

First off, for what distance does a batter normally see the ball out of the pitcher's hand before it gets to the plate? At the moment a pitcher releases the pitch, he's not exactly 60' 6" from the plate and in fact the distance will vary slightly for different pitchers' deliveries. What's important is that at the point of release, the baseball is a little closer, not farther, to the plate than 60' 6", but not much. In other words, the pitcher starts 60' 6" from the plate and is significantly closer to the plate (55"?) when he lands into fielding position, but at the point of release the baseball is very near the pitching rubber, maybe a little in front of it. So I've chosen the distance 60' because it's as good an estimate as any and is a convenient round number.

Let's say a pitcher throws 90MPH. When you convert MPH to feet and seconds, you get that a pitcher throwing 90MPH is also throwing 132 feet per second. That means that if a 90MPH fastball travels for 60 feet, the batter has 0.4545 seconds (60 feet / 132 feet/second) to see the ball before it arrives at home plate.

Now let's say that Brad Kilby throws 90MPH but his delivery hides the ball such that the batter can't pick the ball up for a split second and by the time the batter starts seeing the pitch it has already traveled 3 feet. The batter has only 57 feet to see a pitch traveling at 132 feet/second and that translates to 0.4318 seconds. And if you go back to 60 feet and ask "How hard does a pitcher need to throw to give the batter 0.4318 seconds to react?" the answer is 95MPH.

By my calculations, if you can hide the ball for 3 feet you turn a 90MPH into a pitch the hitter reacts to as a 95MPH. If you can hide it for 5 feet, the batter has only 0.42 seconds to react and it acts like a 97MPH fastball. What's hard to estimate is for many feet can you steal by effectively hiding the ball in your delivery?

All I know is that a ball traveling 132 feet/second has already traveled 1.32 feet in just 1/100 of a second (turning a 90MPH fastball into reaction to a 92MPH fastball). That's 2.64 feet in a 1/50 of a second, and 5.28 feet in 1/25 of a second. Is it not possible that hitters are taking just 1/25 of a second longer to pick up the ball out of Kilby's hand, thus turning his 90MPH into the reaction to a 97MPH fastball? It has kind of looked that way, huh?

And if you ponder nothing else, consider how 1/100 of a second of reaction time is worth about 2 MPH on an average fastball. Hitting is harrrrrrd.