The weighty matter of baseball

Cecil's Columns/Staff Reports
1

In Can a baseball be hit farther at altitude (I hope I entered that link right!) I’m afraid the Straight Dope staff has led Cecil astray. (Alas, I have trouble making physics humourous. Only my attempts at understanding it are humurous.)

The Earth’s gravity at Miami is not greater than that at Denver because Miami is closer to the equator. Yeah, the surface of the earth, or more accurately, sea level, bulges near the equator. But that is because the force of gravity “bulges” there also. The reason sea level is an important reference altitude is because the force of gravity is a constant along the sea. If gravity were stronger at Miami’s latitude, the ocean would flow towards that latitude until the ocean surface there was under the same gravitational pull as the surface at Denver’s latitude. (In fact, it already has, hence the bulge.) The ocean surface is a surface of equipotential, with respect to gravity, just like a metal is an equipotential with respect to electricity. All that stuff about “centrifugal force” etc, shuld likewise be ignored.
Of greater import is the composition of the earth under Denver (or Miami). I believe, that mountains tend to lie over less dense material, which is one reason mountains form, so the Earth’s gravitational pull at Denver might well be weaker than at Miami. But you won’t calculate it based on Denver’s altitude.

I’ll barely mention that gravity gets weaker like the distance squared, so a mere 1 mile of altitude would certainly negligible for baseball.

2

This is a common mistake among folks starting physics, confusing field and potential. Sea level is (approximately) a gravitational equipotential, but that does not mean that the magnitude of the gravitational field is constant at sea level, and it’s the field we’re interested in here. The gravitational field is, indeed, weaker at lower latitudes.

Likewise, there’s no reason to ignore centrifugal force (note lack of quote marks). It’s a real effect which needs to be taken into consideration, if one wants to use a rotating frame of reference such as the Earth. Centrifugal force is only “wrong” if you’re in an inertial frame of reference, but who wants to discuss a baseball game in a frame of reference where the field is moving at 700 mph?

3

Welcome to the Straight Dope Message Board, Slow, glad to have you here. An interesting first post indeed!

And thanks for the reply and clarification, Chronos! (FYI, Slow, Chronos is our Straight Dope Staff Astronomer Extraordinary)

4

So now i’m both enlightened and confused; along comes Chronos’ clarification on centrifugal force (with the ‘note lack of quote marks’ aside) which forces me to pull out my trusty Webster’s Collegiate Dictionary and i find that, once again, i’ve been flim-flammed by my own trusting nature. It was explained to me, in no uncertain terms, by my college physics instructor that there is no such thing as centrifugal force; merely lack of centripetal force. Had i, 'lo those many years ago, availed myself of the information supplied in the above referenced dictionary, i would have discovered that, though the term centripetal force originated in 1709, the term centrifugal force has been in use since 1721 and, as such, is no johnny-come-lately it regards proper usage. Instead, i now find myself prostrated at the feet of Chronos, humbly thanking him for dragging this poor ignorant fool into the actinic glare of knowledge. Which makes me wonder why i wasted so much time and money on school when i could have found all my answers right here on the SDMB?

My question in regards the original Staff Report is thus:

Bibliophage mentions that, when the game is played in more humid climes, there is a deadening effect on the ball with the effect being that it will not travel as for when acted upon with the same force as a ‘dry’ ball. While this may be so, isn’t there also more drag in moist air as the moisture displaces air? If i understand this correctly, the seams of the ball, when spinning, create lift which allows the ball to stay aloft longer (as well as exaggerating any hooking/slicing) while moist air creates enough drag on the ball to prevent it travelling as far. On the other hand, the air at higher elevations is less dense and, thus, would provide less lift. It seems logical that, to get the greatest travel, it would be best to play in a cold, dry location.

5

I’m not entirely convinced by bibliophage’s calculation. (Off topic: bib’s had a great series of staff reports. Running for successor of Uncle Cecil? :slight_smile: )

When calculating the gravitational field at various latitudes and altitudes, was a spherical-earth model being used? I understand ignoring gravitational anomalies due to local geology. My issue is whether bib assumed all the earth’s mass is at the center. As we all know, this works fine for a spherical body. But what about the mass of earth that’s bulging out at the equator? It’s not right to assume it’s at the center of the earth.

This is probably a small effect, but we’re already talking about parts per thousand. If pressed, I could do the numbers myself, but I’d like to know what went into the original calculation.

6

I used the formula found in the CRC Handbook, which should take care of both the non-spherical shape and the centrifugal force

where g is acceleration due to gravity (in ms[sup]-2[/sup]), L is latitude and H is height above sea level in kilometers.

7

Ok, I was totally wrong about the centrifugal force. I can only claim haste and a slow, caffine besotted mind as my excuse. I was “thinking” that sea level would reflect the centrifugal force, which is just stupid for a former physicist.

For octothorpe, the centrifugal force is sometimes referred to as a fictious force. To paraphrase Newton, an object traveling along a straight line will continue to travel along a straight line, unless some force acts on the object. That statement is correct, as long as the observer is in what is called “an inertial reference frame”. Rotating frames are not inertial, so if you roll a ball in the back of a truck making a turn, the ball appears to follow a curve. (The ball will move towards the side of the truck on the outside of the curve.) In fact, the ball is traveling in a straight line, but sitting in the truck, you can’t tell. You perceive a force, which is the centrifugal force. It depends on the velocity of the ball, which I conveniently forgot when I said to ignore it. The ocean, which is, more or less, at rest, relative to the ball field, does not feel the same force.

As far as humidity goes, octothorpe, you must remember that moisture is a component of air. Changes in air density affect the movement of the ball, as Bibliophage said. You are picturing a baseball game with and without lots of moisture. Adding moisture to air of a given density in a container will increase the density of the air, because you have put more air in the same container, not because the moisture is somehow “denser” than the air. A ball game played in a given density of air is a ball game played in that density of air, regardless of the moisture content of that air. I do not know if humid days tend to have denser air than nonhumid days.

My point about local geology dominating the latitude still stands. I understand that Bibliophage couldn’t possibly calculate it, but I believe Denver sits on an anomalous weak spot in the earth’s field, and that weakness is greater than what we should apparantly expect from altitude. You just have to ask a geodisist what’s been measured.

Now, about that gravitational weakening with latitude. Pleonast, that CRC formula is probably based on a “multipolar” expansion of the Earth’s field. A sphere is a monopole - so you can picture the gravitational field as emanating from a point. The next step is a quadrapole, in which you picture the gravitational field as arising from four masses placed in particular ways. Far be it from me to question the CRC, but I don’t get it. I know that geodesists look for gravitational anomalies by measuring sea level from satellites. I have to admit, I was thinking that g would be the same anywhere on the ocean. I’m going to have to take my time thinking this through.

8

bibliophage, What! You didn’t calculate the numbers from first principles? :slight_smile: Yep, the CRC formula was exactly what I was looking for. I assume latitude is in radians?

SlowMindThinking, the water in the ocean is trying to minimize the gravitational potential energy. Thus sea level approximates an equi-potential surface. The local acceleration due to gravity is the gradient of the potential. A higher acceleration means the potential changes faster. This is independent of the value of the potential along which the sea level lies.

I can explain it in simpler terms if I’m not clear enough.

9

SlowMindThinking, Hmmmm… Once again i find that i have taken an idea that is clear in my mind and muddied it to the point of incomprehensibility on translation to (electronic) paper.

I’ll give you what i got on the humidity thing and see if this doesn’t make a bit more sense:

In my younger days i worked for an uncle who was an ag pilot (crop-duster). He told me that the worst days to fly were those that were hot and humid and on those days he carried a lighter load. The way he explained it was that the air was thinner on hotter days and thus, provided for less lift. Humidity (moisture) displaces air as well as creating drag which, again, provides for less lift. I didn’t mean to lump humidity in with temperature in my comments about air density, but in regards lift (at least the way my uncle explained it to me) the best flying conditions are cold (i.e. dense air), dry (i.e. less drag) days.

I’m not sure why you mentioned the ‘container’ as i wasn’t looking at a closed system (i always picture baseball as played in outdoor stadiums not indoor monstrosities).

10

octothorpe, I must be misunderstanding what you are misunderstanding. Let me see if I can explain why I brought up a container. Suppose you play baseball in one of those indoor monstrosities, only let’s make this one airtight. (Anyone else old enough to remember when the Astrodome was the “eighth wonder of the world.?”) If you played baseball at 70 degrees in this stadium you would be playing in air of a certain density and pressure. If you were to heat the stadium up to 80 degrees, the density would be the same, because the mass of the air has not changed, but the air pressure will have increased, because the air molecules are moving faster. According to bibliophage (if I remember correctly!) it is the density and not the pressure that affects the flight of a baseball.

Suppose, n the other hand, if it were 70 degrees and bone dry in one game, and then you brought in a large quantity of water, which would raise the humidity. Now the air would be more dense, because there is more mass in the same volume. Baseballs now would not fly as well, because the air is more dense, not because of any special property of water vapor.

Bibliophage is telling us that it is not the moisture in the air, and not the ambient pressure, but the density of air that matters. So other affects of moisture (perhaps due to do water’s peculiar charge distribution) do not matter as much as the mass of air that gets in the way of the baseball. And the amount of matter does not depend directly on the moisture content of the atmosphere. Adding moisture to the air increases the density only if you are in a container. In the great outdoors, where all games - including hockey - are best, adding moisture may or may not increase the density of the air, because the atmosphere can always expand. So, the calculations are based on “at a given density.”

11

octothorpe, I must be misunderstanding what you are misunderstanding. Let me see if I can explain why I brought up a container. Suppose you play baseball in one of those indoor monstrosities, only let’s make this one airtight. (Anyone else old enough to remember when the Astrodome was the “eighth wonder of the world.?”) If you played baseball at 70 degrees in this stadium you would be playing in air of a certain density and pressure. If you were to heat the stadium up to 80 degrees, the density would be the same, because the mass of the air has not changed, but the air pressure will have increased, because the air molecules are moving faster. According to bibliophage (if I remember correctly!) it is the density and not the pressure that affects the flight of a baseball.

Suppose, n the other hand, if it were 70 degrees and bone dry in one game, and then you brought in a large quantity of water, which would raise the humidity. Now the air would be more dense, because there is more mass in the same volume. Baseballs now would not fly as well, because the air is more dense, not because of any special property of water vapor.

Bibliophage is telling us that it is not the moisture in the air, and not the ambient pressure, but the density of air that matters. So other affects of moisture (perhaps due to do water’s peculiar charge distribution) do not matter as much as the mass of air that gets in the way of the baseball. And the amount of matter does not depend directly on the moisture content of the atmosphere. Adding moisture to the air increases the density only if you are in a container. In the great outdoors, where all games - including hockey - are best, adding moisture may or may not increase the density of the air, because the atmosphere can always expand. So, the calculations are based on “at a given density.”

12

Sorry for the double post guys, my connection claimed that it time out, so I resubmitted.

Pleonast, I think understand equipotential surfaces, let me explained where my thinking is muddled. The earth’s gravitational field does not have any components along the ocean surface. Otherwise, you could let a ship fall from Miami to London. The gravitational force must be orthogonal to the ocean surface.

So, I’m sitting at a ballgame in Tampa complaining about the price of beer and wondering why gravity seems so weak. Yes, I’m further from the center of the earth, but there might also be more mass under me. The amount of mass, and hence the strength of the field, is a volume (r cubed) thing, while the gravitational strength dropping with distance is an r squared thing. So, weaker gravity must mean lower density, or I would be wondering why everything seems so heavy. That is what I don’t quite understand. The mantle and the core can flow, so it is not obvious to me why the density of the earth is enough less to weaken gravity.

13

OK SlowMindThinking, i’m gonna get through this one way or the other as it is obvious that, though we’re both speaking English, we’re talking in different languages.

In a non-closed system:

Air density: Leave humidity entirely out of the equation. Cold air, more density, greater lift, ball stays aloft longer, ball travels farther. Hot air, less density, ball will not stay aloft as long, ball does not travel as far.

Humidity: Displaces air, creates drag, ball will not stay aloft as long, ball does not travel as far.

Hot, moist air: Less density, more drag, poor flight.

Now, i do understand that the presence of moisture still provides lift (after all, gas or liquid, it’s still fluid dynamics), but the introduction of moisture into the air creates drag sufficient to prevent the ball from travelling as far.

In actuality, there would be more lift generated in a liquid as it is less compressable than gas, but the drag would require that much more power be expended for the same amount of travel: so actually, i erred when i mentioned that there is less lift in moist air it is more correct to say that any benefit gained by the increased lift is more than cancelled out by the introduction of greater drag.

Now, to get back to my uncle and his plane, on hot, humid mornings, on take off, he would quickly get into the ‘ground effect’, but transitioning to true flight would take the entire runway (we actually had to take the fence down at the end of the runway after his landing gear took out the top strand of barb-wire on one such morning). On cool dry mornings, with the same load, he would use about 2/3 of the runway and basically ‘fly through’ the ground effect.

As to the matter of centrifugal force, my comments concerned my mistaken belief that centrifugal force was the incorrect term for lack of centripetal force. I actually have no problem following your arguments regarding centrifugal force.

By the way, thanks for your attempts to exorcise my stupidity demon.

14

Yes, so that means the gradient is perpendicular to the surface. That gradient is a function of latitude and altitude. What’s true for the ocean also applies to the core and mantle, so I’ll stick with the ocean. Let’s have an example with numbers.

From the CRC formula above, at sea level at the equator, local gravity is 9.780356 m/s^2. At the north pole (also at sea level) it is 9.832079 m/s^2. Since both locations are at sea level, they have the same gravitational potential. Let’s use this as our zero potential (i.e., 0 J/kg). So far, you should understand all this?

Now pretend that just enough water is added to the ocean so that the new sea level is at a potential of 1 J/kg. The ocean is still at a constant potential, so no falling from Miami to London. Let’s calculate the how high the water has risen at the equator and pole.

At the equator, the water will rise (1 J/kg) / (9.780356 m/s^2) == 0.102246 m. At the pole, (1 J/kg) / (9.832079 m/s^2) == 0.101708 m. The water rises more at the equator because the gradient is less. This means you have to go higher to reach the same gravitational potential. (And notice how the equatorial bulge is maintained.)

There is a complicated interplay between the gravitational potential and the gravitational acceleration. They are related, but independent, so you can have a constant potential even while the acceleration is varying.

15

Ok, octothorpe , I think we just about have this. Thanks for the compliment by the way. Frankly, I’m much happier doing this than wrestling with my demons of stupidity, i.e., doing my job.

What you need to remember is that moisture in the air is not liquid. It is a gas. That means that interactions between water molecules are negligible in the air, so they should not affect drag in that manner.

When water evaporates, it does not displace air, it becomes air. In fact, water is one of the less massive constituents of air. (The oxygen and nitrogen molecules are about twice as massive as the water molecules.) So, at constant pressure and temperature, a given volume of air with moisture is less dense than a given volume of dry air. Since drag is density related, I would think moist air would give you less drag and not more. I would guess that lift is proportional to pressure, and not density, so at the same pressure and temperature a plane would fly better in humid air than dry air. In real life, however, humid days tend to be hot days with low pressure, so planes would fly poorly on hot, humid days. (High pressure generally implies good weather, if I remember the weather channel correctly.)

As far as your uncle’s plane is concerned, I suspect the issue was the temperature and not the humidity, at least not directly. On a hot day, the ground will be considerably warmer than the surrounding air. The air near the ground heats up, and thus expands. Because it expands, it rises, which might help lift the plane. This must be the ground effect you are talking about. However, to attain true flight, you need lift from the wings. And the less dense air works against you, so it takes longer to take off.

Less you think I am blowing steam, I went and talked with a former navy pilot who is a coworker. If any one knows about flying in heat and humidity it is those guys. He said they talk about humidity, but never bother to take it into account in their calculations.

16

Pleonast, I was trying to relay that I understand how gravity can be greater at sea level in one place than another. That I was in error originally.

The question I am now asking is, why is gravity less at the equator than at the north pole? Or, why is that CRC equation correct? A spinning, nondeformable earth can not expand at the equator, by the definition of nondeformable. However, we know the earth is deformable and it does expand. If the earth expanded at constant density, which could happen, then gravity would be greater at the equator, because there would be enough more mass under you at the equator to more than make up for the greater distance from the center of the earth.

If the CRC equation is correct, the earth is less dense at the equator. Since the expansive force is orthogonal to the axis of the earth’s rotation, I could buy that. Except, I would think that would create a pressure gradient forcing mass towards the equatorial plane. Equilibrium of a totally free flowing fluid would form a disk, like a galaxy or the solar system, would it not? So, it is not obvious to me why gravity is weaker at the equator.

17

For our purposes here, earth is a constant-density, deformable object. The earth bulges at the equator because the earth is spinning. Basically, it’s centrifugal force that makes it bulge.

This is not true. There is not enough extra mass to make up for the extra distance. (I say this because the CRC formula is correct. Measurements have been.)

No, the fluid surface will follow an equipotential surface. In general, this will be an oblate spheroid. (Surface tension could play a role in an arbitrary fluid, but is not important for the earth.)

Let’s see, another way to look at this… Look at a small object on the surface of the earth. There are two real forces on the object: gravity and the normal force (the support from the earth itself on the object). The force of gravity (and I am not including centrifugal effects) points towards the center of gravity (which happens to be the geometric center as well). The normal force is by definition perpendicular to the surface. If the earth were not rotating, these two forces would be equal and opposite and thus cancel exactly.

However, the earth is rotating. The object is in circular motion around the polar axis. Circular motion implies centripetal acceleration, and thus we require that the net force on the object point towards the axis of rotation. We can do this by slightly tipping the normal force towards the pole. Therefore, the force of gravity and the normal force will not quite cancel out, supplying the necessary centripetal acceleration.

Since the normal force is perpendicular to the surface, by tipping it towards the pole, we are in effect pushing out an equatorial bulge. Try sketching a picture. This is how to explain the bulge from an inertial frame of reference.

18

After reading bibliophage’s mailbag article, I only have one thing to ask:

What the hell does “Chimborazo, Cotopaxi, they had stolen my weight away!” mean?!?

19

I had always heard that humidity was a red herring as far as baseball is concerned (crop dusters would be a different matter). Baseball players swear by it, but there’s no evidence that it affects the flight of the ball. From my son’s Dorling Kindersley Tim McCarver’s How Baseball Works : “The effect of humidity on the flight of the ball is negligible, unless there is 100% humidity, in which case the ball remains in the locker until it stops raining.”

20

It’s a reference to Turner’s poem “Romance”