Archery trajectory computation

AI Thread Summary
The discussion focuses on calculating the necessary adjustment of a bow sight to hit a target at 90 meters, given previous settings for 9 and 18 meters. The user has established equations relating angle adjustments to distance, neglecting drag for simplicity. They recognize that while neglecting air resistance simplifies the math, it significantly impacts accuracy, especially at longer ranges. The calculations suggest a sight adjustment of approximately 9 cm, but the user anticipates needing a higher adjustment in practice. They plan to test the calculations outdoors when conditions allow and report back on the results.
tedzpony
Messages
2
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
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!
 
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.
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top