Extremely crude approximation
There's a really simple approach:
- estimate flight time
- multiply that by wind speed in the direction of flight
- that's how much distance wind added to the hit
As an example, if the ball is in the air for 5.3 seconds and wind speed is 5 mph (7.3 ft/s), then the wind adds 7.3 * 5.3, or ~39 feet to the total distance. If the ball is in the air for 5.3 seconds and the wind speed is -5 mph (i.e., hitting into the wind), then the wind subtracts ~39 feet from the total distance.
Slightly better approximation
Let's try basic physics. There are two main forces on the ball in flight:
- gravity at a constant m*32 ft/s^2 toward the ground (m = mass of the ball)
- air resistance, modeled by C*v^2 opposite the direction of v where C is a constant and v is the speed in that direction
Thus, when the ball is moving forward, the horizontal force on it is just -C*vx^2 where vx is the horizontal speed. When the ball is moving up (away from the ground), the vertical force is -C*vy^2 - m*32 ft/s^2 where vy is the vertical speed. When the ball is moving down (toward the ground), the vertical force is C*vy^2 - m*32 ft/s^2. Note that the sign changed on the air resistance term.
What is C here? C is a coefficient that depends on the properties of the object moving through the air and the air itself. A good approximation for a baseball is that C = 0.3*(1/2)*p*A where p = air density and A = cross-sectional area of the ball (pi*r^2).
We have our accelerations, so now it's time to pick some initial conditions and get some numbers. Assuming the batter makes really solid contact with the ball in a nearly ideal hit at sea level on a normal day with a standard baseball...
- p = 0.0765 pounds/ft^3
- A = pi * 1.47 in * 1.47 in = 6.8 in^2 = 0.047 ft^2
- m = 3.2 pounds
- launch angle = 40 degrees
- total speed = 100 mph
- vx = 76.6 mph (111.5 ft/s)
- vy = 64.3 mph (94.3 ft/s)
- initial distance = 0 s
- initial height = 3 ft (height of ball when hit)
- wall height = 10 ft
- initial time = 0 s
Combining those, C = 0.00054 pounds/ft.
How does wind factor in? I'm not 100% certain, so I just thought about it for a bit and settled on 'wind changes the effective speed in the air resistance term'. If the ball is moving forward at 10 mph and the wind is moving forward at 10 mph, the ball effectively feels no air resistance. If the ball is stationary and the wind speed is 0 mph, the ball feels no air resistance. If the ball is stationary and the wind speed is 10 mph, the ball feels the same air resistance it would if it were moving at 10 mph with a wind speed of 0 mph. Thus, the force from air resistance in the horizontal direction becomes C*(vx - wind)^2 where 'wind' is the wind speed in the direction of the ball's motion.
From here, just solve the equations of motion given by those forces. I don't know a direct solution (I'd guess one doesn't exist but I'm maybe just rusty on my math) so I simulated it here. To simulate, I just used the initial conditions, applied the accelerations from the forces and constants for 1 ms, updated vx, vy, x, and y, and repeated until y (the height) was less than or equal to the wall height. The code is here if you want to play with it, and here are the results for a variety of wind speeds:
For the 'extremely crude approximation' of a 5 mph wind vs a -5 mph wind, we got a difference of ~78 feet (39 feet in each direction). With this slightly better approximation, we got a difference of ~38 feet. Why the difference?
There are a lot of reasons. For one thing, the wind's effect on the speed isn't instantaneous. It takes some time for the ball to fully change speeds due to it. Further, the ball is changing speed due to air resistance so the force is constantly changing. There are other reasons but this is enough for now. I tried a bunch of initial conditions, and it actually worked out that dividing the extremely crude approximation by 2 was pretty good, so a quick update to that would be 'take flight time, multiply by wind speed, and divide by 2'.
Enough on wind...
Are there more homeruns in the summer?
It turns out that air density depends on a number of factors. One of those is temperature. We can do the same analysis for temperatures. Keeping everything the same except changing air density by temperature and using a wind speed of 0 mph, here is an approximation of how air temperature affects the distance the ball travels:
Theoretically then, you should have more homeruns in the summer than in the winter and in the south than in the north (assuming constant field sizes), and this effect should not really exist in domes.
How does altitude affect baseball distances?
Air resistance in Denver is ~86% of air resistance at sea level it seems. For the same initial conditions as before with no wind, the ball will fly ~20 feet further in Denver than at sea level.
who is the author of this post please? :)
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