# Archery Problem

1. Sep 11, 2006

### GreenLRan

the turkish bow of the 15th and 16th centuries greatly outperformed western bows. the draw force F(x) of the turkish bow versus the bowstring displacement x (for x negative) is approximately represented by a quadrant of the ellipse (F(x)/Fmax)^2+((x+d)/d)^2=1 . Calculate the work done by the bow in accelerating the arrow, taking Fmax=360N , d=.7m and arrow mass m= 34g. assuming that all of the work ends up as arrow kinetic energy, determine the maximum range R of the arrow. (the actual range is about 430m) Compare with the range for a bow that acts like a simple spring force with the same Fmax and d.

First I started by getting the draw force F(x) in terms of x.
F(x)=-Fmax(x^2+2xd)+1
then I did the integral of that function with respect to x from -.7 to 0 to get the work. And then i set the work equal to kinetic energy and solved for v0. I got v0 = 69.88m/s. giving R=v0^2/g=497.81m

The answer in the back of the book is
R=v0^2/g=(pi*Fmax*d)/(2*m*g)=1188m
and the turkish bow is larger by a factor of pi/2

I would like to know what i'm doing wrong, and how they derived that
R=v0^2/g=(pi*Fmax*d)/(2*m*g). Thanks!

2. Sep 11, 2006

### schattenjaeger

(F(x)/Fmax)^2+((x+d)/d)^2=1

Multiply both sides by Fmax^2

F(x)^2+Fmax^2((x+d)/d)^2=Fmax^2

F(x)^2=Fmax^2(1-[(x+d)/d]^2)

sqrt of both sides, F(x)=Fmax*sqrt((1-[(x+d)/d]^2))

looks like you got your expression for F(x) wrong, but it's like 3:15 at night here and I might be wrong, but yours seems too simple

3. Sep 11, 2006

### GreenLRan

haha, awsome, that worked. I guess i should have triple checked it before posting. Well anyway how do they get R=(pi*Fmax*d)/(2*m*g)?