Curve Balls

In cricket, the ball is usually regarded to swing more in overcast conditions. As far as i know there has been no scientific reason given for this.
Has the same been noticed in baseball?
I do remember hearing that the Minnesota Twins changed the air-conditiong at their home stadium, dependant on whether they were batting or bowling.
Is this true?
And, if so, was it to gain an advantage for the home pitchers?

The Twins don’t bowl, they field. Or pitch.

Actually, they suck, mostly. :smiley:

Chronos. Since you are my resident physics guru, could you offer a professional opinion on the story that Le Roy told about the softball and Eddie Feigner?

Is is possible for him to have pitched a ball that did what was related? Please note that I AM NOT DOUBTING Le Roy when he says he witnessed this. I just need to understand the physics.

Could he aim the ball slightly downward so that it almost touched the surface at mid-court, and then ascended 10 feet in the next part of the journey?

The Metrodome in Minneapolis has a teflon cloth roof which is held up by air. There are big fans that pump air into the stadium, and when the place first opened some teams claimed the Twins turned the fans up or down depending on which team was at bat. The idea was that the fans caused an air current to blow from the infield towards the outfield, and that it was easier to hit home runs when the fans were turned up. The claim was never substantiated.

It is illegal to change the home field in any way between games or during games to give an advantage to the home team. Bill Veeck tried this once when he owned the Indians. He had the outfield fences put on rollers, and the team would move them in or out depending on the opponent. It didn’t take long for the league to tell him to cut it out and never do it again (not a unique experience for Veeck).

Okay, the lift force acting on the baseball is give by:

F = 0.5pACV*V

where p = air density
A = cross sectional area
C = lift coefficient
V = Velocity

the acceleration due to the lift force

a = F/m, m = mass of the ball

Assuming no initial upwards velocity then d, the rise distance is

d = 0.5 * (a - g) * t * t

solving and simplifying for d

d = (pACV^2*t^2)/(4m) - 0.5gt^2

since V*t = distance traveled and assume the low point of the trajectory is 70 feet or 21.34 meters from the other basket, p = 1.25 kg/m^3, A = 0.0296 m^2, and m = 0.188 kg we have:

d = 22.41C - 0.5gt^2. Note the 22.41C term is in meters and is the distance moved due to lift if gravity is not present.

My research say that C is probably between 0.2 to 0.4 depending on spin rate and release velocity. This means we have d = 4.482 - 0.5gt^2 (for C = 0.2) or d = 8.964 - 0.5gt^2 (for C = 0.4).

Now at V = 70 mph or 31.29 m/s it takes 0.6819 seconds to travel 21.34 meters. This means the 0.5gt^2 term is equal to 2.281 meters.

Finally, d = 2.2010 to 6.6830 meters or a rise of 7.2211 to 21.93 feet.

Someone like “The King” could easily do it.


I once played on a company softball team. Our night computer operator was a fine athlete who had about 4 years earlier been a pitcher in the Mets farm system (an injury made him decide to give up professional baseball). He had an arm that was unbelievable - I doubt that anyone in our league could have thrown a ball more than about 60% of the distance he could.

When he put a lot of spin on a hard-thrown softball, the thing would break so much your eyes could scarcely accept what they saw. I played catcher and he played left field - we often nailed runners at the plate. But I had to get him to “take something off” his hard throws - at full velocity with spin, they were simply uncatchable.

I can believe that a master like Eddie F. could make a throw rise. I’d sure love to have seen it.

Question, LeRoy: You just worked out the calculations for a rising softball; you also stated above that “no human puts enough spin on a baseball to create enough lift force for a fastball to rise.” Looking at your equation above for rise distance d, I note that only A and m (and C, perhaps?) change when considering a baseball vs. a softball. The softball isn’t that much larger than a baseball; surely (A/m)[sub]BB[/sub]/(A/m)[sub]SB[/sub] is > 0.5. So why is the small size change critical to throwing a rising pitch? Is lift coefficient C dependant on size? If so, why? If not, what is it that marks the difference between softballs and baseballs in this calculation?

samclem: Heh. Not slighting your resident physics guru Chronos, but I should point out that, on this subject, LeRoy is the man; he’s probably one of the half-dozen people in the world most qualified to offer a professional opinion. Cecil cited him, fer Crissake. So… good question (I’m wondering about that myself, as you can see), but when you’ve got a relativity question and Einstein’s in the room, he’s the man you ask.

For a baseball (if the ball was to travel 55 feet) the lift distance equation for d is:

d = 8.26C - 16t^2 (here d is in feet)

Note there are a lot of assumptions in this for example, C is assumed constant during a pitch which it is not. C is a function of r, the radius, w, the spin rate, and V, the velocity. C is also dependent on seam height and that varies from ball to ball, little league balls have raised seams so C is larger for them than for major league balls.

The following table shows the drop in the ball due to gravity (again no initial upwards velocity) experienced by the ball if thrown horizontal at the given speed over 55 feet. I am using 55 feet since the pitcher moves forward before releasing the ball. In reality, since the ball is slowing down due to drag this drop is actually greater (time would be longer).

Speed; t; 16t^2
50 0.75 9.0
55 0.68 7.4
60 0.63 6.3
65 0.58 5.3
70 0.54 4.6
75 0.50 4.0
80 0.47 3.5
85 0.44 3.1
90 0.42 2.8
95 0.39 2.5
100 0.38 2.3

Speed in mph, t in seconds, 16t^2 in feet.

C is inversely related to r*w/V, in other words as V goes up C goes down. It may be true, for fastballs, that w goes up as well as V goes up; it typically goes down for curve balls. C for 100 mph pitch with a major league ball, 4-seams, is probably no greater than 0.2 meaning the 8.26C term is 1.65. This still results in a drop of 1.65 - 2.3 = -.65 feet. Now as speed goes down C goes up but so does the time gravity acts on the ball. (Note: 0.2 was highest C measured by Igor Sikorsky in windtunnel tests with spinning major league baseballs for air speeds near 100 mph.)

I will add a disclaimer here: Using a really high raised seam baseball, with a pitcher than can just slow enough with a “hell of a lot of spin” so he can raise C to the .5 or higher. I will back down and say it could happen. But with the pressed seam MLB baseball of today, on this planet, with this gravity and air density, with no drug enhanced robotic spinning pitching arm. I will say it can’t be done with a “baseball” by a human.


Oops. C is ** proportional ** to r*w/V or inversely related to V. The rest makes sense…


Oops. C is ** proportional ** to r*w/V or inversely related to V. The rest makes sense…


Tippy Martinez’s pitches often look like they broke twice in different directions on the way to the plate. I’ve yet to see anything like it except with this basket throwing/catching toy that came out in the late 70s.

With regard to the legality of underhand pitches, how long will it be before we see a top caliber woman’s softball pitcher make it to the majors? IIRC the best of them have struck out many major leaguers in exhibitions.

That must have been what I’d heard then… I hope it is/was true, the Twins became my favorite (note the lack of ‘U’ in ‘favorite’) team in baseball when I heard that…
samclem Ouch… Almost as soon as I posted I realised my mistake…

I already offered my professional opinion that it can, in principle, be done, and with any sort of ball. I gladly defer to Le Roy on the question of how significant it is, for various sorts of balls. Even aside from his greater experience in dealing with this sort of situation, the calculation requires knowledge of quantites determined from experiment, and I have not done those experiments.