Another point to consider is that baseballs aren’t flying. Lift is negligible or nonexistent, so the ball falls back to Earth. If there were lift, then denser air would increase it, but drag is much more significant than lift for a baseball, so dense air decreases the range.
bibliophage, another question, given that formula. It has a variable for height above sea level, but it seems to me that one must also consider why one is that height. If we just lopped off the Rocky Mountains at sea level, and had a balloon hovering a mile above the surface, I think that that formula would be accurate, but as it is, we’ve got the “hovering balloon” gravitational force, plus the gravitational force from all that rock between your feet and sea level. The way to check this would be to look up the measured value of g at Denver, but so far, all I’ve been able find are places which apparently used the CRC formula to get 9.796616 ms[sup]-2[/sup] . Does anyone know the exact measured value for Denver?
Okay, I finally had enoug time to do some reading and calculating today.
octothorpe says
There is lift on a baseball, but only if it’s spinning. The phenomenon is the Magnus effect, the same reason a curve ball curves and the same reason a ball batted toward left or right field tends to curve toward the outside. A well-batted ball typically has a backspin of about 2000 rpm, which generates a modest lifting force (never more than a third of the weight of the ball, and usually more like a fifth). The drag force tends slows down the rotation rate so the ball will be rotating at only a few hundred rpm by the time it reaches the ground. This lift can add about 20 feet to the range of a ball at sea level, compared to a non-rotating ball. As a general rule, the Magnus effect is about two-thirds as strong in Denver’s thinner air, so I would expect backspin there will add around 13 feet to the range. This difference of only seven feet is taken into account by Adair (author of The Physics of Baseball) when he says a ball would go 40 feet farther in Denver than at sea level. Evidently the reduced drag in Denver is several times more important than the reduced lift.
Now I’m really confused. I went back to the CRC Handbook and discovered I may have made a mistake. I had assumed H in the formula I posted above was height above sea level. Now I see it says “as a function of latitude and height above the earth’s surface” [my emphasis]. Now I’m not quite sure what to make of that. Denver is on the surface, so should H be zero?
Of course you’re right that the rock under Denver should increase gravity, but I hadn’t really taken it into consideration before. I’m pretty sure the effect isn’t enough to make Denver’s gravity stronger than Philadelphia’s, unless the rock under Denver is unusually dense. As an upper limit, I calculate that if there were a shell of rock (ρ=2800 kgm[sup]-3[/sup]) a mile thick completely surrounding a spherical earth, the higher mass would almost compensate for the higher radius, but not quite. Then gravity at Denver would then be only about one part in ten thousand less than at Philadelphia, not (as I calculated before) about 5 parts per ten thousand. The rock under Denver doesn’t go all the way around the world of course, so I figure it must be a lot less pronounced than that. Unfortunately, I seem to have lost the math skills I would need to give a better answer than that.
I’d like to get to the bottom of this, even though it’s a negligible effect as far as baseball goes.
bibliophage, SlowMindThinking, Chronos, saoirse, and Pleonast,
Thank you one and all for the stimulating conversation. This is certainly more than i ever needed to know about baseball, but the direction this thread took was well worth the detour.
Nope. When pilots talk about ground effect, they’re discussing the property where air flowing past the wings and getting swirled over them builds up below the wings, providing a cushion of air. This cushion of air increases lift and reduces drag, making it easier to get off the ground, but you can’t get out of ground effect until your speed increases enough to climb into the thinner (no cushion) air.
Gosh, I go home for the weekend, and look what happens.
bibliophage, if your CRC book doesn’t explain that equation, perhaps you might contact Peter Wahr at the University of Colorado. At least he was there 15 years ago! He is a geophycisal theorist of first order, and should know how that equation is derived. Compounding this issue, at least for me, is I believe I have read that the Rocky Mountain region is an area of anomalously low density. But, perhaps that is due to the gap between my ears. (I live in the region.)
Irishman, thanks for the explanation of ground effect. I assume you meant air swirled under wings, and not over.
saoirse, thanks for the quote. Does your name stand for something?
Pleonast, my misunderstanding was even simpler than not knowing how to get a bulged earth. I was picturing more mass beneath someone at the equator, and mass increasing as the radius cubed would overwhelm the radius squared decrease due to distance. Obviously whether you are standing on a pole or the equator, you still feel the mass of the whole earth, and so I’m feeling kind of stupid. (Not as obviously, equator guy is further away than polar girl from much of the “extra” mass, even if some is right under his feet.)
For a sphere of given density, surface gravity is proportional to radius. Of course, rock is less dense than the average density of the Earth, so this is still consistent with your result of “almost no change”.
The CRC formula has to mean “above sea level” for H: Since there’s no longitude contribution to the formula, it clearly can’t know that the Rockies are higher than the Great Planes.
And for calculating the extra gravity of a mountain, the first approximation is to assume that the mountain is spherical. Then, you’ve got the known gravitational field due to the rest of the Earth, and add in the gravitational field due to a mass at the mountain’s center of mass (for a triangular-prism mountain range, the center of mass would be at a height of about a third the total height).