The problem:
I want to calculate at which angle I have to shoot an arrow in order to hit a target which is x meters away.
What I know:
- Force of the bow
- Weight of the arrow
- Distance to the target
- Gravity in meters/sec (don't know if this is even needed)
For simplicity (and because I'm absolutely bad at math), let's say the target is at the same height as me and there's no wind.
What I want to know:
Are there any other variables involved in order to perform the calculation?
What's the actual formula?
I already did some googling and found some formulas and even calculators, but they never show the actual degree of the shot.
Best Answer
You might have come across a formula like this:
$$ d = \frac{v_0^2 \sin (2\theta_0)}{g},$$
where $d$ is the distance in meters at which the projectile returns to the same height from which it was fired, $v_0$ is the initial velocity of the projectile in meters per second, $\theta_0$ is the initial angle from horizontal at which the projectile is launched, and $g$ is the acceleration of gravity in meters per second per second.
You can of course do this in different units as long as you keep them consistent.
This is an idealized formula that does not account for the projectile slowing down due to air resistance or for other aerodynamic effects. I would expect it to work poorly for high initial angles.
The information you gave did not include initial velocity, but you can estimate that by integrating the force over the distance the arrow travels while still in contact with the string to get the total work done on the arrow, and then setting that quantity equal to $\frac12 mv_0^2,$ where $m$ is the mass of the arrow.
(If you're working in pounds or ounces the distinction between mass and weight can have a big impact on the calculations; if you are working in metric units, usually when someone says an arrow weighs $w$ grams they actually mean the mass is $w$ grams; the gram is not a unit of force.)
If you are using a modern compound bow, I believe these usually maintain a nearly constant draw strength for most of the length of the draw, dropping to a much lower force only just before the arrow is fully drawn. In that case I think you could get a good approximation of the work done on the arrow by multiplying the force by the length of the draw.
So supposing you have the force $F$ measured in Newtons, the draw length $L$ measured in meters, and the arrow's mass measured in kilograms, you would have
$$ \frac12 mv_0^2 = FL. $$
Solving for $v_0^2$ (you don't actually need $v_0$ itself for the other formula!),
$$ v_0^2 = \frac{2FL}{m}. $$
Now you can plug this into the first formula to get the distance at which to place the target if you shoot at an angle $\theta_0,$ or take a given distance $d$ and solve the formula for $\theta_0$:
$$ \theta_0 = \frac12 \arcsin\left(\frac{gd}{v_0^2}\right) = \frac12 \arcsin\left(\frac{mgd}{2FL}\right) . $$
The function $\arcsin(\cdot)$ is the inverse sine function, also called $\sin^{-1}$ and available on most scientific calculators. Just make sure you have the calculator set to return degrees if you want degrees, or do the conversion from radians to degrees yourself if you need to.
Of course to use this you now have not only the concern about the possible sources of error in this idealized calculation, but also the difficulty of being sure about the angle at which the arrow leaves the bow.
The derivation of that last line of formulas, in case you had trouble working it out:
We start with the well-known formula $$ \frac{v_0^2 \sin (2\theta_0)}{g} = d.$$
Multiply by $g/v_0^2$ on both sides: $$ \sin (2\theta_0) = \frac{gd}{v_0^2 }.$$
Take the inverse sine function of both sides: $$ 2\theta_0 = \arcsin\left(\frac{gd}{v_0^2}\right) . $$
Divide both sides by $2.$ $$ \theta_0 = \frac12 \arcsin\left(\frac{gd}{v_0^2}\right) .$$
That's the first version of the formula for $\theta_0$. To get the other version, replace $v_0^2$ by $\frac{2FL}{m}$: $$ \theta_0 = \frac12 \arcsin\left(\frac{gd}{2FL/m}\right) .$$
Multiply both top and bottom of the fraction on the right by $m$: $$ \theta_0 = \frac12 \arcsin\left(\frac{mgd}{2FL}\right) .$$