I need some help with calculations. Are there any equations that showed the range at which your arrow would still be hitting target like the picture shown below?


NOT like the picture below



I just need to know this for my theoretical bow. I probably have every variable ready; I just need to know the equation. Many people have told me that they can shoot up to 70-80 meters without any slight angle change in the angle of the arrow. However, I want to be more concise than that.


Thank you very much for your help.

Well when I get home I could find the equation of motion for a for a thrown object. That assumes no air friction. Which may not be accurate at distance due to the drag from the fletching. Feathers fletching are lighter than plastic vanes but have higher friction. Giving more control and greater initial speed but slows down faster.

Little searching turned up [url] http://en.wikipedia.org/wiki/Trajectory[/url] I guess if you were to find the equation of motion section and plot the Y= versus x distance. That would give a rough path of travel for the arrow. But as I mentioned above the arrow drag isn't really negligible

Drag equation: http://en.wikipedia.org/wiki/Drag_%28physics%29 unfortunately figuring out the drag coefficient isn't really trivial and requires actual testing.
http://en.wikipedia.org/wiki/Drag_coefficient

There may be some free ballistic calculators out there for your PC that have been tuned for archery. I'd recommend checking out the freeware sights.

I have looked into this in some detail and posted the calculations on this webpage about 4 years ago:

http://mysite.verizon.net/tsafa1/longbow/longbow.htm

The calculation is straight forward in a vacuum. If the arrow is shot at a 45 degree angle and there is no drag (air resistance). Gravity is the only factor and the formula for range is

range = v^2/g

Kevin, you said you want to hit that target while it is a the top of its arch. I would expect that means dividing the calculated range above by 2. The arrow should be completely horizontal at the half way point where it is at the top of its arch.

Once you factor in drag it becomes very difficult. The drag is subject to variations in the condition of the air.
range = v^2/g ( 1 + cv^2/mg)^-0.74

c = a constant number specific to each individual arrow that allows for the effect of drag.

Kevin, again divide the range by 2 to figure where the top of the arch is.

To calculate Exit Velocity.

v = (eFx/m)^1/2

v = The exit velocity.
e = The efficiency (all force applied in the same direction). We can assume it will be 70% for most medieval bows. Modern bows are close to 1. If the value is 1 is has no effect in the equation.
F= The force to draw the bow fully, commonly known as draw-weight. Keep if mind that for every inch you underdraw a 50 lb bow, you loose 3 lbs of force or 6%.
m= Mass of the arrow. This will vary based on the arrowhead and length.
x= This is the distance that the string will travel from its rest position to its fully drawn position. This is equally important to the draw-weight. If you underdraw you arrow this factor will deminish. Also remember that for every inch you underdraw a 50 lb bow, you loose 3 lbs of force or 6%.

Note: The ^1/2 in the formula is the same as taking the square root of (eFx/m).

So I was looking at a site that had similar information.

Force is not a constant. it is dependent on draw length. for a long bow the draw weight (force) may quite likely be linearly dependent on the draw distance. This conveniently makes the force draw curve a triangle so the equation should look like:
v = (eW/m)^1/2
where W = Fx/2 Being a triangle the area under the curve is base x height / 2


One last note to calculate x = draw lenght - brace height.
That would be the simple way to say what was put in the last post

I guess the other question is how much of an angle is acceptable?