# Archery trajectory computation

1. Jan 15, 2012

### tedzpony

Okay, this is not a homework problem, and I think if I really played with it long enough I could work something out. But really, I just want an answer, so I'm hoping someone on here who really loves solving problems like this can just knock it out. It would be much appreciated. I'm wondering about a sight bar setting for my bow (archery).

Here's what I know. I set the sight in a certain position to shoot dead center at 9 meters. I set my sight precisely 1 cm lower to hit the center of the target at 18 meters. This raises the angle of the bow upon arrow release. The sight aperture (aim point) is 99.2 cm from my eye. On the two separate shots, the bow and arrows are the same, therefore same force, same initial velocity, same projectile weight.

I don't know the initial velocity, or arrow weight. I would neglect drag, though I know that makes this slightly less accurate. I also recognize that's not a lot of information, but my gut tells me that knowing the difference between two separate sight settings while everything else remains constant is probably enough to solve it for someone who knows how.

Now the question. How much would I have to lower the sight aperture (therefore raising the bow angle) to hit a target at 90 meters?

2. Jan 15, 2012

### Simon Bridge

The big decider would be drag - which is non-linear.
Neglecting drag makes it a lot less accurate. Air resistance is significant for real projectiles. But neglecting drag makes the math simpler:

$d=v^2\sin(2\theta)/g$ ... relates the angle to the range if the target is about the same height as the arrow starts... since v is always the same, I can just put k=v2/g and proceed...

Sight is set for 9m
$9=k\sin2\theta_0$ ...(1)

1cm higher (over 99.2cm - gives the angle change as $\delta_1=\arctan(1/99.2)$ gets you to 18m
$18 = k\sin2(\theta_0+\delta_1)$ ...(2)

which is kinda neat because this gives two equations and two unknowns.

$$k=\frac{9}{\sin2\theta_0} \qquad \text{...(3)}$$$$\theta_0 = \frac{1}{2}\arctan{\bigg [ \frac{\sin 2\delta_1}{2-\cos 2\delta_1} \bigg ] } \qquad \text{...(4)}$$

(you want to check equation (4))
Now remains only to plug the numbers into:

$90 = k\sin2(\theta_0 + \delta_2)$ ...(5)

Which we can use to find $\delta_2$ and get the elevation in cm at $e=99.2\tan(\delta_2)$ ... (6)

Which is:
$$e=(99.2)\tan \bigg [ \frac{1}{2}\arcsin \big ( \frac{90}{k} \big ) - \theta_0 \bigg ] \qquad \text{...(7)}$$

I figure you can plug the numbers in.
... someone should check my algebra.

Caveat: all this assumes zero drag. It will be interesting to see how close it is IRL.

3. Jan 15, 2012

### tedzpony

Simon Bridge,

Thank you very much. Yes, I recognize that drag will be significant, especially over 90 meters, but I also knew that it would make the question impossible, because I couldn't possibly provide the data that would be required to factor that in. It certainly will be intersting to see how it pans out in the real world. Unfortunately, there aren't any indoor ranges that I know of that reach out to 90 meters, and it's still way too cold right now to shoot outdoors. That said, although it will be a wait, I will do my best to remember to come back here and let you know the results whenever it warms up and I can "check your math" with bow in hand! Again, thanks a bunch!

4. Jan 15, 2012

### Simon Bridge

I get:
$\delta_1 = 0.0101^\circ$
$\theta_0=0.0101^\circ$
$k = 445.8 \text{m/kg}$
... which suggests (reality check) you are releasing the arrows faster than 200kmph and you have a maximum range a bit under 400m (45degrees) ... sound right?

$e \approx 9cm$

I expect you'll need higher than that.