The procedure for calculating the path of a baseball in ball parks at different altitudes is described. A link to a computer program for making these calculations is included. The calculations show that, indeed, a baseball travels further in a high altitude park.

I am not much of a sports fan and when the newspaper comes each morning, I usually toss the sports section in the recycling bin. But, when I am working a wind tunnel test out of town, there is usually quite a bit of dead time where I end up reading everything from recipes to lost dog notices and even sports columnists. So, it happened one day that I was perusing The Dallas Morning News and I came across the assertion that a baseball will travel nine percent further in the new Denver stadium than at sea level. If the writer had just said that it would travel further, I probably would have mumbled something about the air density being lower and passed on. But, when it said that it will travel nine percent further - not eight percent, not ten percent, but nine percent further - I had to know just where that number came from. Of course, my interest was also professional, because the essential thing you will need if you are going to make calculations of the air resistance is a procedure for computing the density of air at any altitude.

Almost all elementary textbooks on physics or mechanics show the procedure for calculating the trajectory of an object that is given a certain initial velocity. The solution is always done with the assumption that air resistance is negligible and gravity is the only force acting on the object. This assumption makes it possible to solve the equations and produce a solution that lets you compute the position and velocity of the object at each point along the path. You can compute the horizontal distance travelled, the maximum height reached halfway along the path, and even show that you get the best range if you throw the object at an angle of 45 degrees.

How do we compute the trajectory if we do not neglect air resistance? As we learn in elementary mechanics, we identify all of the forces acting on the object, write F=ma for both the horizontal and vertical directions and then proceed to solve the resulting set of differential equations. Writing the equations is not too hard, but the solution is not easy because the aerodynamic drag force varies with velocity. The direction of the drag force changes as the object travels along its path, always being opposite in direction to the velocity. If you want to go through the details of the dynamical analysis and the numerical solution of the differential equations, go to the Numerical Procedure page.

To perform the calculations, we must specify the initial conditions (altitude and velocity) of the ball and then follow the trajectory until the altitude returns to the initial value. If we do this for Denver with an altitude of 1609 m, and several possible initial velocities, we get the following results:

Initial Velocity (m/s) |
Sea Level Range (meters) |
Denver Range (meters) |
% gain |

35 | 71.0 | 75.2 | 5.9 |

40 | 83.1 | 88.7 | 6.7 |

45 | 94.4 | 101.6 | 7.6 |

50 | 105.0 | 113.6 | 8.2 |

55 | 114.9 | 124.9 | 8.7 |

It turns out that the ball does indeed travel further in the less dense air of Denver. It also turns out that the percent improvement varies with initial velocity. The high velocity balls that will travel a long distance will gain more range than the slower ones. So, I guess I will go along with the Dallas News and say that the guys who hit really long balls can get a nine percent improvement in distance travelled. For us wimps who have a hard time getting 30 m/s, the benefits are significantly smaller.

If we get longer fly balls in Denver, why not build stadiums in Mexico City or even La Paz, Bolivia? For our ultimate fantasy, lets imagine a stadium near the summit of Mount Everest and another one in which all the air has been removed and the game is played by androids or humans with scuba packs. If we repeat the calculations, we get the following results:

Initial Velocity (m/s) |
Sea Level Range (meters) |
Denver Range (meters) |
Mt. Everest Range (meters) |
Vacuum Range (meters) |

35 | 71.0 | 75.2 | 94.4 | 124.9 |

40 | 83.1 | 88.7 | 115.5 | 163.3 |

45 | 94.4 | 100.6 | 136.8 | 206.6 |

50 | 105.1 | 113.6 | 157.7 | 255.1 |

55 | 114.9 | 124.9 | 178.2 | 308.7 |

Now we are getting some really long balls. Our 55 m/s ball that would travel 115 meters at sea level will now go 309 meters in the vacuum ball park. WOW! Two and a half times as far. That shows that air resistance is quite important.

The solution for zero air resistance is easily seen to get a maximum range when the initial angle of the velocity vector is 45 degrees from the horizontal. When air resistance is taken into account, this may not be the optimun angle. The computing program I have supplied lets you investigate this question. We know from anecdotal stories from artillery people that the best angle is less that 45, somewhere around 40 degrees. If we run these calculations again for altitude of zero and initial velocity of 50 m/s, we get the following results:

angle | range |

36 | 104.90 |

38 | 105.14 |

40 | 105.02 |

42 | 104.53 |

44 | 103.71 |

46 | 102.53 |

48 | 101.01 |

50 | 99.17 |

With smaller values of initial velocity, the optimum angle moves back closer to 45 degrees. You can play around with these calculations if you are interested. For each altitude and initial velocity, there is an optimum initial flight path angle. However, it doesn't vary very much. I used 40 degrees for all of the calculations previously shown, except for the vacuum calculations, where 45 degrees was used.

That is actually true, but the effect is very small. From the solution for zero air resistance, we see that the range is proportional to 1/g. At sea level, g=9.8066 m/s^2, while in Denver it is 9.8017 and atop Everest it is 9.7794. With velocity and angle held fixed, the ball with a range of 100m at sea level has a range of 100*9.8066/9.8017 = 100.05 m in Denver and a range of 100*9.8066/9.7794 = 100.28 m atop Everest. We can't explain much from this effect.