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enter image description here

So, I'm making a game. The game has archer in it. Arrows are simulated in 2D, $(x,y)$

An archer acquires a target and needs to know at what angle he needs to tilt his bow for the arrow to land at the target.

variables:

$t =$ time
$a =$ angle

The following constants are known:

$V_{0} =$ velocity, the initial velocity of the projectile.
$H =$ the initial height of the archer
$G =$ Gravity
$L =$ the distance to the target
$F =$ A simplified air friction constant.
$V$ (velocity) $=$ $V_{0} - F t$

I'm looking for $a$, the angle of the bow.

I know that $V_{x} = \cos(a)$ and $V_{y} = \sin(a)$

I know that: $V_{x} \cdot t - 0.5 \cdot F \cdot t^{2} = L$

I know that: $t = \frac{V_{y} + \sqrt{V_{y}^{2} + 2 \cdot G H}}{G}$

Which gives me these functions:

public static double length(double height, double angle, double velocity) {
        
    double vz = Math.sin(angle*2*Math.PI)*velocity;
    double v = Math.cos(angle*2*Math.PI)*velocity;
    double t = getTime(height, vz);
    return getLength(v, t);
}
    
public static double getTime(double height, double vz) {
    return (vz + Math.sqrt(vz*vz + 2*G*height))/G;
}
    
public static double getLength(double v, double time) {
    return time*v - 0.5*FRICTION*time*time;
}

This allows me for testing angles, and they work. In fact, my only solution right now is to test for the correct angle by binary search. It usually get a solution with 10-15 iterations.

There are many solutions for this without air friction, but I'm looking for one with it included.

However, I can't manage to substitute t and solve for $a$.

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  • $\begingroup$ Wiki is your friend: en.wikipedia.org/wiki/… $\endgroup$
    – Gert
    Commented Feb 7, 2022 at 10:56
  • $\begingroup$ Does not answer the question. It does not have a solution for an angle with air friction. It has some reasoning with air friction, but this is advanced friction, while I am fine to simplify mine to a constant $\endgroup$
    – Jake
    Commented Feb 7, 2022 at 12:22

1 Answer 1

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Starting with :

$$m\,\ddot x=-F\\ m\,\ddot y=-m\,g$$

and $$x(0)=0~,\dot x(0)=v\,\cos(\alpha)\\ y(0)=h~,\dot y(0)=v\,\sin(\alpha)$$

you obtain the solutions

$$x(t)=-\frac 12\,{\frac {F{t}^{2}}{m}}+v\cos \left( \alpha \right) t\tag 1$$ $$y(t)=-\frac 12\,g{t}^{2}+v\sin \left( \alpha \right) t+h\tag 2$$

with $~x(t)=L~$ you obtain the time $~t_L~$ that the projectile reach the distance L. substitute this time in equation (2)
$$y(t=t_L) =0=f(\alpha)$$ from the solution $~f(\alpha)=0~$ you obtain $~\alpha~$


$$f(\alpha)=-\frac 12\,{\frac {g \left( v\cos \left( \alpha \right) m-\sqrt {{v}^{2} \left( \cos \left( \alpha \right) \right) ^{2}{m}^{2}-2\,FLm} \right) ^{2}}{{F}^{2}}}+{\frac {v\sin \left( \alpha \right) \left( v \cos \left( \alpha \right) m-\sqrt {{v}^{2} \left( \cos \left( \alpha \right) \right) ^{2}{m}^{2}-2\,FLm} \right) }{F}}+h $$

Remarks:

the start velocity $~v~$ must be greater then $ ~{\frac {\sqrt {2}\sqrt {F\,L\,m}}{m}}~$ otherwise you don't obtain real solution for the angle

Edit

numerical solution $~f(\alpha)=0~$with Newton method

$$\alpha_{n+1}=\alpha_n-\frac{f(\alpha_n}{f'(\alpha_n)}$$

Example

Data=[F= 0.1,L=20,m=1,g=10,v=15,h=2]

$$n=0,\alpha_1=0+\frac{f(0)}{f'(0)}=0.346~,f(\alpha_1)=-0.888\\\\ n=1~,\alpha_2=0.404~,f(\alpha_2)=-0.02\\ n=2~,\alpha_3=0.406~,f(\alpha_3)=-0.00001$$

so you need 2 interation to obtain the solution

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  • $\begingroup$ Thank you Eli. Forgive my ignorance, but I need to solve this for α. I.E α =equation of other stuff, so α must be factored out and put on the left hand side. I tried doing this with wolframalpha and it couldn't do it. $\endgroup$
    – Jake
    Commented Feb 7, 2022 at 14:51
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    $\begingroup$ The analytical solution is not useful because you have more then one solution , instead put all the data und solve for $\alpha$ I use Maple fsolve $\endgroup$
    – Eli
    Commented Feb 7, 2022 at 15:38

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