Author 
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Kevin S.

Posted: Mon 07 Dec, 2009 1:12 pm Post subject: Need Some Help In Calculating Arrow Flight 


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 7080 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.




Joel Minturn

Posted: Mon 07 Dec, 2009 3:27 pm Post subject: 


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




Matthew Fedele
Location: Auburn, NY USA Joined: 21 Jul 2005
Posts: 64

Posted: Mon 07 Dec, 2009 3:54 pm Post subject: 


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.




Bill Tsafa
Location: Brooklyn, NY Joined: 20 May 2004
Posts: 599

Posted: Tue 08 Dec, 2009 8:00 am Post subject: 


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 drawweight. 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 drawweight. 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).
No athlete/youth can fight tenaciously who has never received any blows: he must see his blood flow and hear his teeth crack... then he will be ready for battle.
Roger of Hoveden, 11741201
www.poconoshooting.com
www.poconogym.com




Joel Minturn

Posted: Tue 08 Dec, 2009 9:13 am Post subject: 


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?




Kevin S.

Posted: Tue 08 Dec, 2009 2:56 pm Post subject: 


Bill Tsafa wrote:  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 drawweight. 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 drawweight. 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). 
Wait, so the top arch would be the maximum straight shooting range? If that's the case, then thank you very much ( I have already calculated the range using the equation that you have given me)




Bill Tsafa
Location: Brooklyn, NY Joined: 20 May 2004
Posts: 599

Posted: Tue 08 Dec, 2009 5:13 pm Post subject: 


Kevin Sanguanlosit wrote: 
Wait, so the top arch would be the maximum straight shooting range? If that's the case, then thank you very much ( I have already calculated the range using the equation that you have given me) 
Yes Kevin. Top of the arch is most likely where the arrow will be most horizontal. That should be at half the range. I would expect some difference in practice unless you test under perfect lab conditions. This formula assumes a 45 degree angle of release.
Joel you are right about reducing the length by the brace height. I took it for granted that everyone would think to do so, but I think not. There are a number of ways to try to calculate velocity. I had considered one that takes into account the elasticity of the wood but its just complicates things beyond the point of basic understanding. There is no easy way to measure elasticity, other then to solve for it, so I left that alone.
Your adjustment to the formula seems to use an average draw weight. Problem is that the force does not go up consistently as you draw the bow. On a 100 lb bow, half draw is not 50 lbs. It is more like 30 lbs. It goes up a lot in the last few inches. I believe a formula that takes into account elasticity factors this in properly.
No athlete/youth can fight tenaciously who has never received any blows: he must see his blood flow and hear his teeth crack... then he will be ready for battle.
Roger of Hoveden, 11741201
www.poconoshooting.com
www.poconogym.com




Glennan Carnie

Posted: Wed 09 Dec, 2009 12:39 am Post subject: 


Quote:  Your adjustment to the formula seems to use an average draw weight. Problem is that the force does not go up consistently as you draw the bow. On a 100 lb bow, half draw is not 50 lbs. It is more like 30 lbs. It goes up a lot in the last few inches. I believe a formula that takes into account elasticity factors this in properly. 
It depends on the bow. On an MR warbow the drawforce curve is almost the opposite  the force climbs dramatically early in the draw, then lets off as you approach full draw. Replica MR bows (same wood, same ringcount, same crosssection, same tiller) made by Master Bowyer Steve Stratton demonstrate this property. My warbow draws 28lb/inch from brace; this drops off to 15lb/inch at full draw.
Although I'm no expert I believe this drawforce property is also true of Scythian / Mongolian bows, too.
A Victorian design target bow is most likely to have a linear drawforce curve; and is also most likely to stack (increase weight toward the end of the draw)




Lafayette C Curtis

Posted: Wed 09 Dec, 2009 7:23 am Post subject: 


Glennan Carnie wrote:  Although I'm no expert I believe this drawforce property is also true of Scythian / Mongolian bows, too. 
I don't know about Scythian bows (they're so unusual in many respects), but it's definitely true with staticrecurve bows like the ones popular among medieval Mongols. In fact, that's the whole point of having their bows recurved in the first place!




