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

  1. Jun 16, 2011 #1
    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
  2. jcsd
  3. Jun 16, 2011 #2


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    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?
  4. Jun 16, 2011 #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)?
  5. Jun 17, 2011 #4

    jack action

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    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" [Broken], you have the following equations:



    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.
    Last edited by a moderator: May 5, 2017
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