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Math Help - Target Interception

  1. #1
    Newbie
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    Jan 2010
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    Target Interception

    Hello,

    it's been a while since I did any real math so I'm having a bit of a hard time with what should be a simple problem. I need to find an equation for intercepting a target (Item A) that's moving at a constant speed by throwing Item B at it from a fixed point. What I'm really interested in the angle at which I should throw Item B. (this is for a video game I'm making by the way)

    I know the speed of Item A, I know the distance between Item A and B and I know that Item B will be thrown at a constant speed... I just need to figure out the angle (or the components x and y).

    The way I went about it was by saying that the distance travelled by Item A at the moment of impact is:

    Xa = Vax * t (Xa being the distance travelled and Vax its speed in x)

    and the angle of Item B is :

    cos th = adj / hip = Yb / (Vb * t)

    substituting t we get :

    Xa = Vax * Yb / (Vb * cos th)

    Knowing that the distance in X between A and B is Xt = Xa + Xb :

    Xt - Xb = Vax * Yb / (Vb * cos th)
    Xt - Yb * tan th = Vax * Yb / (Vb * cos th)

    Which finally gives :

    Xt * cos th - Yb * sin th = Vax * Yb / Vb

    The only thing unknown in this equation is the angle th... any help on how I could solve this, and eventually any corrections on the procedure?

    Thanks,
    i
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  2. #2
    MHF Contributor
    Grandad's Avatar
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    Hello mansiva

    Welcome to Math Help Forum!
    Quote Originally Posted by mansiva View Post
    Hello,

    it's been a while since I did any real math so I'm having a bit of a hard time with what should be a simple problem. I need to find an equation for intercepting a target (Item A) that's moving at a constant speed by throwing Item B at it from a fixed point. What I'm really interested in the angle at which I should throw Item B. (this is for a video game I'm making by the way)

    I know the speed of Item A, I know the distance between Item A and B and I know that Item B will be thrown at a constant speed... I just need to figure out the angle (or the components x and y).

    The way I went about it was by saying that the distance travelled by Item A at the moment of impact is:

    Xa = Vax * t (Xa being the distance travelled and Vax its speed in x)

    and the angle of Item B is :

    cos th = adj / hip = Yb / (Vb * t)

    substituting t we get :

    Xa = Vax * Yb / (Vb * cos th)

    Knowing that the distance in X between A and B is Xt = Xa + Xb :

    Xt - Xb = Vax * Yb / (Vb * cos th)
    Xt - Yb * tan th = Vax * Yb / (Vb * cos th)

    Which finally gives :

    Xt * cos th - Yb * sin th = Vax * Yb / Vb

    The only thing unknown in this equation is the angle th... any help on how I could solve this, and eventually any corrections on the procedure?

    Thanks,
    i
    I can't really comment on the accuracy of your working, since you don't explain clearly enough the initial conditions. For example, I can't see why you have x_t = x_a + x_b. However, I can explain how to solve an equation like the one you've ended up with, and it's this:

    Let x_t\cos\theta - y_b\sin\theta = r \cos(\theta+\alpha)
    =r\cos\theta\cos\alpha -r\sin\theta\sin\alpha
    \Rightarrow \left\{\begin{array}{l }r\cos\alpha=x_t\\r\sin\alpha=y_b\end{array} \right.

    Square and add:
    r^2=x_t^2+y_b^2
    Divide the second equation by the first:
    \tan\alpha = \frac{y_b}{x_t}
    So, with these values of r and \alpha, you now solve:
    r\cos(\theta - \alpha) =\frac{v_{ax}y_b}{v_b}
    which gives:
    \theta=\arccos\left(\frac{v_{ax}y_b}{v_b\sqrt{x_t^  2+y_b^2}}\right)+\arctan\left(\frac{y_b}{x_t}\righ  t)
    Grandad
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