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Thread: Rugby Pitch Problem

  1. #1
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    Rugby Pitch Problem

    If you would have to kick the ball from anywhere on the line (the sides between the black squares in the map below), between the goal posts and over the bar, where on the line should it be best to take the kick in order to maximize the chance of succeeding to score a try? (Let's say that the vertical length of the pitch it 2a and the the length between the goalposts as 2b. Ignore the lengths given in the map below.)
    https://s-media-cache-ak0.pinimg.com...9cdf1ed09e.gif


    I've thought about this and came up with a vague answer: wouldn't it be better if the person kicked the ball at an angle like 45 30 or 60? meaning that it would be easier to aim... But this seems to be too simple... Anyone have an idea?
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  2. #2
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    Re: Rugby Pitch Problem

    To answer this you would have to know the trade-off between the probability of success based on (a) the apparent angle between the goal posts as seen from the point of the kick - the wider that angle the better, and (b) the distance from the point of the kick to the goal posts (the shorter the better). Without that it's hard to say. However, if we can assume that the probability of success is directly related to the apparent width between goal posts( as seen from the point of the kick), then we can derive an expressio as follows:

    Given kicking point along the line at distance x from the goal line, half-width of the pitch = a, and width between goal posts of w, the apparemt angle between the goal posts is:

     \theta = tan^{-1} (\frac {a + \frac w 2 } x ) - tan^{-1} (  \frac {a - \frac w 2 } x )

    To find the maximum value of apparent width you differentiate this with respect to x, set the result to zero, and solve for x. Good luck with that!

    Another, slightly less accurate approach, is to model the apparent width as the w times the cosine of the angle to the midpoint of the goal posts, divided by the distance from the point of the kick to the mid-point of the goal, which is as  d = \sqrt{x^2 + a^2}. This gives:

    Apparent width =  \frac {w \cos \alpha} d = \frac {wx}{x^2 + a^2}

    Take the derivative with respect to x, set to zero, and you get x=a. In other words, the space between the goal posts appears largest when x = a, meaning when the kick is at 45 degrees. This should be reasonably accurate as long as the value of w is small relative to a.
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