Archery trajectory computation

[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?

 

[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.

 

[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!

 

[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.

 

[PJC01]

I would approach this problem by first defining all the variables involved. These would include the initial velocity of the arrow, the weight of the arrow, the angle of the bow upon release, and the distance between the sight aperture and the eye. From there, I would use the principles of projectile motion to calculate the trajectory of the arrow.

Next, I would use the given information about the sight bar setting at 9 meters and 18 meters to determine the change in angle of the bow. This change in angle, combined with the known distance between the sight aperture and the eye, can be used to calculate the angle needed for a target at 90 meters.

Then, I would use the calculated angle and the initial velocity of the arrow to determine the height of the sight aperture needed to hit the target at 90 meters. This would involve solving for the height in the equation for projectile motion, taking into account the change in angle and the distance between the sight aperture and the eye.

Finally, I would use the calculated height to determine the necessary change in the sight bar setting to achieve the desired angle and hit the target at 90 meters.

It is important to note that this calculation would be an approximation, as factors such as drag and wind resistance would affect the trajectory of the arrow. However, with the given information and assumptions, a rough estimate of the necessary sight bar setting could be determined.