Linkage Based Compound Bow Force Analysis Jeremy Prince Cj M

# Introduction

With today’s technology being unlimited, the variations in bows are endless. You have the option to choose between a recurve, compound, or crossbow. Each of these options has their own different variations to choose between. With all this being said, the main technological advances in the bow market today are directed towards the compound bow. For example, the compound bow’s variations include the number of cams, the size and weight of the cams, and the endless limitations for the speed and accuracy of the arrow. We chose our main focus for a force analysis problem to be a linkage based compound bow.

# Problem Statement

A linkage based compound bow is a very unique type of bow. It replaces the traditional style cams with a ternary linkage. The reason for using ternary links is to maximize let off without sacrificing arrow speed. The linkage is substituted in to the bow to create the desired let off for the operator. The let off is defined as the decreased force after reaching the peak pull force, allowing one to retain the bow string at its full drawn position more effortless. A desired let off is between 65% and 80% of the maximum pull force. Other than the ternary links the compound bow is very similar to other styles.

# Assumptions and Parameters

To start with, we had to make simplifying assumptions and set constant parameters. The simplified assumptions we used were, a constant string length, the string pull is level and straight, constant limb cross-sectional area, constant linkage cross-sectional area, the limb deflection causes increased velocity of the arrow. The constant parameters include, limb length of 12.5 inches, bow frame of 24 inches, short side linkage length of 2inches and long side linkage length of 5 inches, limb to link string of 41.85 inches, link to pull point string of 23.25 inches, and the angle between the short and long side of the linkage is 98 degrees.

%* Constant Bow Dimensions ***%
r1a = 12; r1b = r1a; %Bow Frame
r2 = 12.5; r12 = r2; %Length of Limb
r3a = 2; r3b = 5 ; r13a = r3a; r13b = r3b; %Bow Links
r4 = r1a + r2*.9; r14 = r4; %Link to Pull Point String
r6 = 1.8*r4; r16 = r6; %Limb to Link Strings
%Changing the lenghts of r4 and r6 controls the force response time

%* Constant Bow Angles ***%
theta1a = 90*dtr; theta1b = -90*dtr;
alpha3 = 98*dtr; alpha13 = -alpha3;
theta5 = 180*dtr;
%Changing alpha3 effects the toggle position

constants = [r1a,r1b,r2,r12,r3a,r3b,r13a,r13b,r4,r14,r6,r16, theta1a,theta1b,alpha3,alpha13,theta5];

%
% Initial Input for Bow Kinematics Target Function
%

r5 = 14;
input = [r5];

%
% Defining Initial Guess for the Bow Kinematics Target Function
%

theta2 = 55*dtr; theta12 = -theta2;
theta3a = 15*dtr; theta13a = -theta3a;
theta4 = -75*dtr; theta14 = -theta4;
theta6 = -88*dtr; theta16 = -theta6;
target_o = [theta2,theta12,theta3a,theta13a,theta4,theta14,theta6,theta16];
target = target_o;

# Position, Velocity, and Acceleration Equations

We started solving the force analysis by setting up and solving a simplified schematic of the linkage based compound bow. We solved for the mobility of the bow and created a vector model. We were able to create four loop equations and two constraint equations to solve for ten unknowns. We proceeded to solve for the velocity and acceleration of the model. Unfortunately, we were unable to get a closed form solution for the position problem, which in turn forced us to use an interpretation function in Matlab. Using the interpretation method we were able to find an acceptable solution. By plotting the vector model and using our initial guesses and constant parameters we were able to verify our solution was valid.
After the solving the position problem, we took the derivative of the positions loop and constraint equations. We then separated the derived equations into X and Y equations. Separating the equations allowed us to plug them into matrix form where it becomes a linear problem. From here we are able to take the inverse of the matrix containing unknown terms and multiplying them by another matrix containing the known terms. We took a second derivative of the position equations and repeated the process for the acceleration equations.

%
% Solving Velocity Problem:
%

%* Defining Internal Variables for Matrix Size Reduction **
c2 = cos(theta2); s2 = sin(theta2);
c12 = cos(theta12); s12 = sin(theta12);
c3a = cos(theta3a); s3a=sin(theta3a);
c13a = cos(theta13a); s13a = sin(theta13a);
c3b = cos(theta3b); s3b = sin(theta3b);
c13b = cos(theta13b); s13b = sin(theta13b);
c4 = cos(theta4); s4 = sin(theta4);
c14 = cos(theta14); s14 = sin(theta14);
c5 =cos(theta5); s5 = sin(theta5);
c6 = cos(theta6); s6 = sin(theta6);
c16 = cos(theta16); s16 = sin(theta16);

%* Defining A Matrix For Velocity and Acceleration Calculations ***
A = [r2*s2, r3a*s3a, 0, r4*s4, 0, 0, 0, 0, 0, 0
r2*c2, r3a*c3a, 0, r4*c4, 0, 0, 0, 0, 0, 0
0, 0, 0, 0, 0,r12*s12, r13a*s13a, 0, r14*s14, 0
0, 0, 0, 0, 0,r12*c12, r13a*c13, 0, r14*c14, 0
r2*s2, 0, r3b*s3b, 0, r6*s6, -r12*s12, 0, 0, 0, 0
r2*c2, 0, r3b*c3b, 0, r6*c6, -r12*c12, 0, 0, 0, 0
-r2*s2, 0, 0, 0, 0, r12*s12, 0, r13b*s13b,0, r16*s16
-r2*c2, 0, 0, 0, 0, r12*c12, 0, r13b*c13b,0, r16*c16
0, -1, 1, 0, 0, 0, 0, 0,0, 0
0, 0, 0, 0, 0, 0, -1, 1,0, 0];

%* Defining b Matrix For Velocity *
b_v = [c5;-s5;c5;-s5;0;0;0;0;0;0]*r5_dot;

%* Velocity Calculation ***
v = inv(A)*b_v;

%* Defining Velocity to Corresponding Variables ***
theta2_dot = v(1); theta12_dot = v(6);
theta3a_dot = v(2); theta13a_dot = v(7);
theta3b_dot = v(3); theta13b_dot = v(8);
theta4_dot = v(4); theta14_dot = v(9);
theta6_dot = v(5); theta16_dot = v(10);

%
% Solving Acceleration Problem:
%

%* Defining b Matrixs For Acceration **
b_a1 = [-r2*theta2_dot^2*c2-r3a*theta3a_dot^2*c3a-r4*theta4_dot^2*c4;
r2*theta2_dot^2*s2+r3a*theta3a_dot^2*s3a+r4*theta4_dot^2*s4;
-r12*theta12_dot^2*c12-r13a*theta13a_dot^2*c13a-r14*theta14_dot^2*c14;
r12*theta12_dot^2*s12+r13a*theta13a_dot^2*s13a+r14*theta14_dot^2*s14;
-r2*theta2_dot^2*c2-r3b*theta3b_dot^2*c3b-r6*theta6_dot^2*c6;
r2*theta2_dot^2*s2+r3b*theta3b_dot^2*s3b+r6*theta6_dot^2*s6;
-r12*theta12_dot^2*c12-r3b*theta3b_dot^2*c3b- r16*theta16_dot^2*c16;
r12*theta12_dot^2*s12+r3b*theta3b_dot^2*s3b+r16*theta16_dot^2*s16;
0; 0];

b_a2 = [c5;-s5;c5;-s5;0;0;0;0;0;0]*r5_ddot;
b_a = b_a1 + b_a2;

%* Acceleration Calculation ***
a = inv(A)*b_a;

%* Defining Acceleration to Corresponding Variables ***
theta2_ddot = a(1); theta12_ddot = a(6);
theta3a_ddot = a(2); theta13a_ddot = a(7);
theta3b_ddot = a(3); theta13b_ddot = a(8);
theta4_ddot = a(4); theta14_ddot = a(9);
theta6_ddot = a(5); theta16_ddot = a(10);

# Virtual Work Force Analysis

Virtual work force analysis is defined as the total work done by external forces added together to equal zero. Our external forces included the pull force and the bow limbs acting as torsional springs. We assumed the spring constant to be 4.5 lbs/rad. The torsional spring torque is equal to the spring constant, K multiplied by the change in the limb angle. We took the spring torque and multiplied by the angular velocity of the limbs. We took this result and divide by the pull velocity to obtain our pull force.

%
% Solving for Virtual Work
%

T2 = k*(theta2_us-theta2); T12 = k*(theta12_us-theta12);% Limb Torque Calculations

%* Force Calculation **
F5 = -((T2*theta2_dot)+(T12*theta12_dot))/r5_dot;%Pound Force
F5_store(j) = F5;%Pound Force

# Results

We plotted our results with a graph representing Pull Force (lb) vs. Draw Length (in). Our plot shows us that one applies minimal force towards the bow string until it reaches a draw length of 14in. Once 14in. is reached the pull force drastically increases and reaches its peak pull force of 65lbs. around a draw length of 24in. Once one achieves a draw length of 24in. the pull force is reduced, this is what is called the let off. This result is comparable to a realistic outcome.

## Ideal Results:

### Note:

The area under the curve of the Ideal Force Curve is greater than the results we received. That implies there is more energy available from the compound bow with cams instead of a linkage based bow.

# Problems Experienced

Some of the problems we encountered while trying to achieve our result were our initial constant parameters. These include the string lengths, the limb links, the frame links, the linkage links, and linkage angles. These problems were overcome by trial and error manipulation, and the generosity of Dr. Steven Canfield. We achieved acceptable results but with a greater time window and a better understanding of linkage based compound bows, more desirable results can be found.

# References:

Dr. Stephen Canifield

http://webzoom.freewebs.com/mayflowerarchers/News,%20Articles,%20Documents/Modern%20Archery%20-%20Compound%20Bow.pdf

page revision: 7, last edited: 11 Dec 2010 23:16