Two different fences i need them to be about the same in a baseball field

In summary, to have equal difficulty hitting a home run in both right and left field, the right field fence should be between 12.5 and 19 feet high, depending on the distance between the player and the left field fence. This calculation assumes a 45 degree angle for the optimal trajectory of the projectile, and may not be accurate for other angles. Additionally, most parks are not symmetric and do not have fences that are so short.
  • #1
fences
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Theres a right field fence and a left field fence
the left field fence is 15 feet farther than the right field
the fences are 4 feet high in left field i need to know how high to make the fence in right field to have equal difficulty to hit a home run to right field
 
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  • #2
Hi and welcome.
There's not enough data yet, to be able to make a calculation (at least for a non baseball player). Could you, perhaps, suggest the highest and furthest that a baseball can travel and the relevant dimensions of a park (?? right word?).

Two free tickets for the correct solution?
 
  • #3
The closer fence will most likely always be easier to hit a home run on. An out to the warning track in left field will be a home run in right, unless the wall is extremely high. Even then, look at fenway. The green monster is 36'2" high and is only 300-315 feet from home plate, and routine popups to left are home runs or doubles.

Is there any particular reason you want to do this? Most parks generally are not symmetric like this.

edit: I just realized 4 foot fence is really short... You sure you want fences that one can simply reach over (unless it is for little league)?
 
  • #4
Not an expert on the subject, but here is how I would solve it.

Using the simple http://en.wikipedia.org/wiki/Trajectory_of_a_projectile" , you have the following equations:

http://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Height_at_x":

939be4f0fab1698937920cf1208e0044.png


Where y0 = 0 and [itex]\theta[/itex] = 45° ([itex]\theta[/itex] is the angle at which the projectile is launched; 45° is the value that will give the greatest distance). Knowing y (the height of the left field fence, 4ft) and x (the distance between the player and the fence), you can find the initial velocity v.

Using the same equation with the same v but with x that is 15 ft shorter, you can find the new value of y (the height of the fence 15 ft closer for the same projectile path). Re-arranging for your problem, you get:

[itex]y_{R} = \left( x - 15 \right) - \frac{\left( x - 15 \right) ^{2}}{x^{2}} \left( x - 4 \right)[/itex]

Where [itex]y_{R}[/itex] is the height of the RIGHT field fence and [itex]x[/itex] is the distance between the player and the LEFT field fence. Solutions for this equation are:

[itex]x[/itex] = 50 ft --> [itex]y_{R}[/itex] = 12.5 ft;
[itex]x[/itex] = 75 ft --> [itex]y_{R}[/itex] = 14.6 ft;
[itex]x[/itex] = 100 ft --> [itex]y_{R}[/itex] = 15.6 ft;
[itex]x[/itex] = 150 ft --> [itex]y_{R}[/itex] = 16.7 ft;
[itex]x[/itex] = 200 ft --> [itex]y_{R}[/itex] = 17.3 ft;
[itex]x[/itex] = 250 ft --> [itex]y_{R}[/itex] = 17.6 ft;
[itex]x[/itex] = 300 ft --> [itex]y_{R}[/itex] = 17.9 ft;
[itex]x[/itex] = 350 ft --> [itex]y_{R}[/itex] = 18.0 ft;
[itex]x[/itex] = 400 ft --> [itex]y_{R}[/itex] = 18.1 ft;
[itex]x[/itex] = [itex]\infty[/itex] --> [itex]y_{R}[/itex] = 19 ft;

I did not try for other angles to see how different are the results.
 
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  • #5


I would suggest that the right field fence should also be 4 feet high in order to maintain equal difficulty in hitting a home run to both fields. By keeping the height consistent, other factors such as the distance and angle of the fences can be adjusted to create a balanced playing field. Additionally, it is important to consider other variables such as wind patterns and the strength of the players in determining the overall difficulty of hitting a home run to either field. Ultimately, maintaining consistency in the height of the fences can help ensure fairness and competitiveness in the game.
 

FAQ: Two different fences i need them to be about the same in a baseball field

How far apart should the two fences be in a baseball field?

The two fences in a baseball field should be at least 300 feet apart, with a recommended distance of 345 feet. This allows for a fair and challenging playing field for hitters and fielders.

What is the standard height for a baseball field fence?

The standard height for a baseball field fence is 8-10 feet, with a recommended height of 10 feet. This height is tall enough to prevent most home runs, but still allows for exciting plays at the wall.

Can the two fences be made of different materials?

Yes, the two fences in a baseball field can be made of different materials as long as they meet the standard height and distance requirements. Some common materials used for fences include chain link, wood, and vinyl.

How do the two fences affect the game of baseball?

The two fences in a baseball field can greatly impact the game. A shorter distance between the fences can lead to more home runs, while a taller fence can make it more difficult for players to hit those home runs. The distance and height of the fences also affect the strategy and positioning of fielders.

Are there any regulations for the design of the two fences in a baseball field?

Yes, there are regulations for the design of the two fences in a baseball field. These include the height and distance requirements, as well as any safety measures such as padding on the fence or breakaway sections. The design of the fences must also comply with any local building codes.

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