MODELING SINGLE LINK MOTION AND MUSCLE BEHAVIOR WITH A MODIFIED PENDULUM EQUATION

Transcription

1 MODELING SINGLE LINK MOTION AND MUSCLE BEHAVIOR WITH A MODIFIED PENDULUM EQUATION By ALLISON SUTHERLIN A SENIOR RESEARCH PAPER PRESENTED TO THE DEPARTMENT OF MATHEMATICS AND COMPUTER SCIENCE OF STETSON UNIVERSITY IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF BACHELOR OF SCIENCE STETSON UNIVERSITY 2015

2 ACKNOWLEDGMENTS Guidance and direction along any research experience is greatly appreciated. Thus, I thank my senior research advisor, Dr. Miles, for your expertise and many laughs you have provided during this research experience. Additionally, I thank the entire Math and Computer Science department for being not only great professors, but good people and role models as well. I have gained valuable insight in not only mathematics, but the meaning of life as well through cherished conversations outside of the classroom. Furthermore, I thank Dr. Matthew Schrager for his assistance in our understanding some of the more complicated geometry, anatomy, and physics components to my research. I also thank the Stetson Undergraduate Research Experience program for giving me the opportunity to receive a SURE grant. I was able to live on campus during the summer and work exclusively with Dr. Miles without outside obligations. This experience taught me a valuable life lesson -that is the only way to count yourself out of a race is to not sign up. Therefore, I am thankful I applied and was chosen as a recipient. Moreover, I thank the softball coaching staff at Stetson University. Without them, I would not have been able to attend Stetson. Lastly, I thank my family, softball family, and the friends at Stetson that have become family for their continued support and encouragement throughout my college career. 2

3 TABLE OF CONTENTS ACKNOWLEDGEMENTS LIST OF FIGURES ABSTRACT CHAPTERS 1. INTRODUCTION 1.1. Background and Objective Physics Background Torque and the Human Body METHODS TO ANALYZE HUMAN MOTION 2.1. Overview Static Analysis Single Link Static Analysis Two Links Dynamic Analysis Single Link Dynamic Analysis Two Links DEVELOPMENT OF SINGLE LINK MODEL 3.1. Seated Leg Raise Exercise Geometric and Force Interpretation of Leg Extension Torque at Knee Joint Relationship Between Torque and Angular Acceleration Simple Pendulum Comparison MODELING MUSCLE BEHAVIOR 4.1 Hooke s Law Background Hooke s Law and Muscle Behavior Muscle Resistance and Damping Term ENERGY EXPENDITURE 5.1 Nonlinear Conservative System Energy Expended in Seated Leg Raise Exercise FUTURE WORK AND APPLICATIONS 6.1. Two Link Motion Compare to Double Pendulum Description of Bench Press Technique Geometric Interpretation of the Bench Press Lift Wide Versus Narrow Grip Comparison Energy and Fatigue Function Conclusion

4 APPENDIX REFERENCES BIOGRAPHICAL SKETCH

5 LIST OF FIGURES FIGURE 1. Graphical representation of torque components First (a), second (b), and third (c) class lever Single link static analysis Two link static analysis Single link dynamic analysis Two link dynamic analysis Position of the lower leg at a fixed time Geometry of the lower leg motion in seated leg extension exercise Path of lower leg as the leg reaches extension Geometry for calculating torque from muscle force Geometry for calculating torque due to the weight of the limb Magnitude of torque over a range of θ values from -π/2 to Limb weight force under examination in the differential equation Geometry of the lower leg motion to show arc length relationship Differential equation results with lower leg extension conditions Simple pendulum diagram Pendulum differential equation results Muscle experiment to prove spring functionality Finding length of muscle using Law of Cosines Graph of the muscle length as joint angle increases through the range of motion 21. Single link motion with muscle component Proper upper arm placement Geometry for bench press lift A simulation of the bench press motion

6 ABSTRACT MODELING SINGLE LINK MOTION AND MUSCLE BEHAVIOR WITH A MODIFIED PENDULUM EQUATION By Allison Sutherlin May 2015 Advisors: Dr. William Miles Department: Mathematics and Computer Science A model that can analyze weight lifting motions would be beneficial to players, coaches, and strength and conditioning instructors. To represent the human body mathematically, the body is broken down into 5 links consisting of the forearm, upper arm, trunk, upper leg, and lower leg. A force and torque analysis is performed for simple, single link motion -motion where one link rotates about a joint- that was motivated by the bench press exercise. The link and corresponding joint for the seated leg raise exercise are considered in the analysis. A relationship between torque and angular acceleration is established using arc length. Applying the relationship, a homogenous, nonlinear, second order differential equation is derived that accounts for the limb weight force and is solved for in MATLAB. The motion of a simple pendulum and resulting differential equations were studied to compare to our lifting model. A muscle forcing term is added to the equation to find the muscles contribution to the exercise by treating the muscle as a spring that behaves according to Hooke s law. Moreover, since muscles act in pairs, a damping term is included to represent a resisting muscle force. The analysis is presented and the total energy expended in the leg raise exercise is calculated. 6

7 1.1 BACKGROUND AND OBECTIVE CHAPTER 1 INTRODUCTION Collegiate athletes are constantly pushed to improve in every aspect of training for their respective sport. In the collegiate game, as well as other elite levels of athletics, strength training makes an enormous impact on performance. Moreover, there is a wide range of exercises an athletes may perform to improve their performance. One exercise that many athletes find particularly challenging is the bench press lift because of the way it tests the lifter s mental and physical capabilities. It is interesting that two athletes of similar build could have very different results with regard to this exercise. While the initial goal in this research project was to model the bench press lift with differential equations, we have revised our approach to modeling simple lifting motions and incorporating a muscle component to the analysis. In addition, we laid the ground work for proposed future work in analyzing two link motion with a differential equation and for building an energy function. 1.2 PHYSICS BACKGROUND In order to develop a relationship that can be used to establish a differential equation, it is necessary to be familiar with properties of forces and torque. An important distinction to make in the analysis is the difference between mass and weight. Weight is a force, which implies weight has vector qualities of magnitude, direction, line of action, and point of application, while mass only has a magnitude since it is not a force [A5]. 7

8 The weight an individual is holding or lifting will be referred to as the load to differentiate between the weights of the body segments. Both magnitude and direction determine the force, which are represented as vectors. The direction of a vector is equivalent to the direction a force acts. Furthermore, the length of the vector is equal to the magnitude of the force [A7]. The line of action is an infinitely extended line in the direction of the force. Linear movement occurs when the force tries to moves the body along its line of action. Otherwise, the line of action of a force does not cross through the body and the force tries to rotate the body [B6]. The point of application of a force is at the exact location at which a force acts on the body. For instance, the point of application of a force on a human body segment is at the center of mass of the segment, which is point on the segment where the mass is properly distributed. [A5]. Published data will be used for the average sized man for the center of mass and lengths of each body segment studied in the lifting motion [A4] TORQUE AND THE HUMAN BODY As previously mentioned in Section 1.3, a body attempts to rotate when the line of action of the force does not extend through the body. When rotation occurs, torque, the tendency of a body to rotate about an axis, is formed [A7]. The equation for torque is the cross product of, τ = r F (1) where τ is torque, r is the distance from the axis of rotation to where the force acts, and F denotes the force that is producing rotation about r. The result of the cross product is, 8

9 τ = r F sin(θ) (2) where θ is the angle between r and the line of action of the force, and rsin(θ) is the length of the lever arm. Figure 1 displays a graphical representation of the result from equation (2). Figure 1 Graphical representation of torque components. In the human body, bones and muscles represented as levers that produce motion. Consider joints in the human body as the axis of rotation or fulcrum. Muscles generate forces that rotate limbs. There are three types of levers that classify rotation. A first class lever is a lever in which the axis is located between the weight and the force. Second class levers are described as a lever in which the weight is between the axis and the force. Third class levers, which represent a majority of the joint systems in the human body, occur when the force is applied between the axis and the weight [B3]. Figure 2 displays each type of lever. 9

10 Figure 2 First (a), second (b), and third (c) class lever. The green vector represents force, red vector is the weight or resistance, and the fulcrum or axis of rotation is the blue triangle. 10

11 CHAPTER 2 METHODS TO ANALYZE HUMAN MOTION 2.1. OVERVIEW Biomechanical researchers have already developed a model that evaluates or predicts the effect of lifting a certain amount of weight, the positioning of the weight, body posture, and the strength capability of the body. One model, created by Chaffin and Anderson, is the foundation for biomechanical research because it transforms human characteristics into mathematical properties [A5]. Body segments are converted into a model of solid links that represent the human body [A1]. Researchers tend to utilize a model with no more than five body links when analyzing full body movement because having more links than necessary greatly increases the complexity of solving the model [A1]. Additionally, each link has a specific length, mass, and moment of inertia that are used to calculate torque, velocity, acceleration, force production and numerous other properties that describe physical motion. The moment of inertia is defined as the segment s resistance to rotation, denoted by M, M = I cm a (3) where I cm is the location of the center of mass of the link, or the point of application in which a gravitational force would act on the segment, and a is the angular acceleration [A4]. The formula for the center of mass is, I cm = r 2 dm = mr 2 (4) 11

12 Where r is the perpendicular distance each mass is located from the axis of rotation of a particular link, and m is the mass of the link. Therefore, the following equation is also equivalent for the moment of inertia, M = mr 2 θ (5) To adjust this model for weight lifting, the model is simplified into a two dimensional sagittal plane. This means that the body is split vertically into left and right halves. Therefore, we can analyze the motion of certain lifts, like the bench press, on one half of the body and assume symmetry applies to the other halve. However, we need to determine if the individual is holding a load while remaining stationary, which indicates a static analysis should be performed, or if the individual is moving a load, indicating we need to perform a dynamic analysis [A5]. The following sections give an example of each type of analysis for a single and a two link model, along with a description of what each analysis entails. 2.2 STATIC ANALYSIS FOR SINGLE LINK A static analysis for a single link is utilized when an individual holding a load but is not motion. The first step in a static analysis is determining what external forces are acting upon the body. Consider a stationary individual who is standing upright with their arms extended downward while holding a load, as seen in Figure 3. 12

13 Figure 3 Single link static analysis of arm holding a load Now we will determine what external forces are acting on the body. If the individual were to drop the load they were holding in their hand, then the load would accelerate downward due to the force of gravity. Thus, the gravitational attraction created by the mass of the load and the mass of the earth creates the weight force of the load held by the person. By applying Newton s second law of motion, which states that the acceleration of a body is directly proportional to the net force acting on the body and inversely proportional to the mass of the body and is in the direction of the net force, we can determine the weight of the load [A3]. Using the equation for Newton s second law of motion, F = ma (6) Where F denotes the force, m is mass, and a is acceleration, we can define the weight force generated by the mass of the load by W = mg (7) Where W denotes the force from weight, m is the mass, and g is the gravitational acceleration ( m/s^2). 13

14 Since the person is not in motion, there exists another external force that is canceling out the weight force that allows the individual to hold the load. This implies the individual achieves a static equilibrium, meaning that the sum of the external forces acting on the mass equal zero. This is the first condition for equilibrium and arises from Newton s third law of motion, which states that whenever one body exerts a force on a second body, the second body exerts an equal and opposite force on the first [A3]. This means that as the weight force acts downward from the load, there exists a reaction that is equal in magnitude and opposite in direction. The reactive force must act along the same line of action of the weight force. Otherwise, the body would rotate. Given the first condition for equilibrium, we can continue our analysis by calculating the reactive forces and actuating torques. Torque is a force, and therefore has the same vector properties, as discussed in Section 1.5, that forces do. Moreover, when the summation of the torque is zero, the second condition of equilibrium is achieved [A5]. 2.3 STATIC ANALYSIS FOR TWO LINKS In the previous section, the arm was assumed to be one link. Now, the arm is analyzed as two links, upper arm and the fore are as seen in Figure 4, while only holding a load. 14

15 Figure 4 Two link static analysis illustrating location of the elbow, the joint that connects the upper arm and forearm links. First we will determine when static equilibrium for two links is achieved. Each link is treated separately. All the external forces and torques on the link furthest away from load are examined first. In Figure 4 above, the forces would be examined as follows: the weight of the upper arm, the resultant elbow force, torque at the elbow, and the weight of the forearm link. When two or more forces acting in a straight line are acting on a body, the forces can be added together. For the elbow, the force from the upper arm and forearm can be added together to create the resultant elbow force [A5]. The static equilibrium equations for the shoulder are solved by summing the forces and then summing the torques. Note that the direction of each force and torque must be taken into consideration. We will assume that negative vectors act downward and yield clockwise rotation. 2.4 DYNAMIC ANALYSIS FOR SINGLE LINK In a dynamic analysis for a single link, an individual is in motion while holding or lifting a load. Another component is adding the analysis. Not only is gravity acting, but now inertial forces are acting on the body. The inertial forces are decomposed into two 15

16 orthogonal forces- a tangential force and radial force. The tangential force is the force that is tangent, or perpendicular, to the link s motion. The equation for tangential force is, F t = mrθ (8) Where m is the mass of the link, r is the distance from the joint to the center of mass of the link, and θ is the instantaneous angular acceleration of the link. The radial force acts along the radius of the arc of motion and is denoted by, F R = mrθ 2 (9) where F R is the radial force of rotation away from the joint center, m is the mass of the link, θ is the angular velocity of the segment. Since the radial force acts along the radius of the arc of motion, its line of action passes through the joint, or the axis of rotation. This creates a reactive force that counteracts the radial force to keep the joint together while the body undergoes movement. This reactive force is called the centripetal force [A5]. Consider a dynamic analysis on the forearm segment with an initial position as seen in Figure 4. The individual will flex, or bend, their arm until the forearm is perpendicular to the upper arm, shown in Figure 5. 16

17 Figure 5 Single link dynamic analysis of forearm. Thus, only the forearm link is rotating about the elbow joint. A torque analysis at the elbow will involve summing the torques of the static effect of gravity, or the weight of the body, tangential force from equation (7), and the moment of inertia of the forearm to get the total torque [A5]. This is given by the following equation, T E = mgsin(θ)r + mr 2 θ + I cm θ (10) where T E denotes the torque at the elbow and mgsin(θ)r is the effect of gravity from the limb in motion. Recall that the moment of inertia can be substituted with equation (9). 2.5 DYNAMIC ANALYSIS FOR TWO LINKS Similar to the approach for a two link static analysis, the dynamic analysis for a two link system consists of adding the reactive forces and torques caused by an adjacent link to the kinetics of the link being added to the analysis. The term kinetics refers to the forces associated with motion and the forces that result from motion [A5]. In a two link dynamic analysis, both links can move linearly and rotate. For example, consider an individual who is lifting his arms above his head, as depicted in Figure 6. 17

18 Figure 6 Two link dynamic analysis of the forearm and upper arm. The initial position is (a) and (b) is the position of the links at end of the motion. Thus, the inertial forces acting on the elbow have two components: a rotational effect about the elbow and an effect of the forearm motion about shoulder. These two components correspond to the tangential and radial forces acting at the center of mass of the forearm link. Coriolis, a French mathematician, found that in addition to tangential and radial forces acting on a link, the difference between the two rotational accelerations, θ 1and θ 2, creates a separate inertial force, called the Coriolis force. The difference for two link motion is, 2θ 1θ 2r 1 m 1 (11) where θ 1 and θ 2are the angular velocities for each respective rotational acceleration, r 1 is the length of the forearm, and m 1 is the mass of the forearm. A Coriolis force arises when one link rotates at different speed than its connecting link [A5]. 18

19 Now we transition into analyzing simple lifting motions to apply the methodology s described in Chapter 2. 19

20 CHAPTER 3 DEVELOPMENT OF SINGLE LINK MOTION MODEL 3.1. SEATED LEG RAISE EXERCISE We focus our attention on a single link dynamic approach to analyze the seated leg raise exercise. In this exercise, a seated individual raises his lower leg, the part of the leg from the knee to the ankle, to maximum extension. Note that extension occurs when the limb is in a straight line, while flexion is when the limb is bending [A2]. Figure 7 shows the position the individual is in while in the process of extending their leg. Figure 7 Position of the lower leg at a fixed time [A2]. 3.2 GEOMETRIC AND FORCE ANALYSIS OF LEG EXTENSTION Now we will analyze the geometry and forces involved in the leg extension motion. A crucial measure is the angle of pull, δ, the angle between where the muscle attaches to the bone and the direction the muscle force, or the bone on which movement occurs [A2]. The angle of pull stays relatively constant throughout the motion, meaning that the angle of pull will rotate with the segment as the segment rotates, but the orientation of the angle of pull will remain the same. 20

21 The lifter s leg starts rising at the vertical axis of π, indicated by the purple 2 dashed line in Figure 8. Realistically, the initial position of the leg for the exercise can begin at any angle with respect to the vertical axis. For simplicity, we assume the lifter initial positon is at the vertical axis. Figure 8 Geometry of lower leg motion in seated leg extension exercise. The motion of the lower leg rising to extension is activated by the angle of pull. The muscle exerts a force, known as the muscle force that must be greater than the limb weight in order to for rotation to occur. Since the vertical axis and limb weight are parallel, the angles are the same. As the lower leg is raised to maximum extension; the path the segment travels follows circular motion in a counter clockwise direction. From this motion, we can decompose all forces into tangential and radial components, as discussed in Section 2.4. However, the line of action of the radial force passes through the knee joint, which is the 21

22 axis of rotation for leg extension. Therefore, when calculating the torque at the knee, the radial forces sum to zero and only the tangential force contributes to the torque, as seen in Figure 9. Figure 9 Path of the lower leg as the leg reaches extension. The radius of the circle is r, the length of the lower leg with the knee as the axis of rotation, and the blue vectors indicate the tangential force. 3.3 TORQUE AT KNEE JOINT To gain an understanding of how torques interact to cause rotation, the torque acting on the knee joint during the leg raise motion is computed using published mean values for the angle of pull, forces acting on the link, center of mass of the link, and the length of the leg [A2]. Consider an individual performing the leg extension exercise whose lower leg is 0.36 m, exerts a muscle force of 1000 N, a body weight of 700 N, and an angle of pull of 0.25 rad. The torque of the muscle and limb weight force are calculated separately for simplicity and then summed together to find the net torque acting on the knee joint. Figure 10 shows the geometry of the force and angles involves for the muscle force. 22

23 Figure 10 Geometry for calculating torque from muscle force where δ is the angle of pull (0.25 rad), d is the distance from the joint to where the muscle force vector attaches to the lower leg (0.05m), a is the lever arm, and F m denotes the muscle force. By equation (1), the torque from muscle force is calculated as follows: τ m = F m a (12) τ m = 1000 a (13) using right triangle trigonometry to find a, sin(0.25) = a 0.05 (14) a = 0.05 sin (0.25) (15) a = m. (16) Therefore, the torque due to muscle force is, τ m = (17) τ m = 12.4 N m. (18) The calculation for torque due to the limb weight is more complicated than the muscle force because we need to know the center of mass of the limb, which is at 43.4% of lower leg length, and b, the length of the lever arm, must be calculated before computing the 23

24 torque. Recall that center of mass indicates where the point of application of the limb weight force is on the lower leg. The calculation is as follows: d = (19) d = m. (20) Thus, the distance from the knee joint to the point of application of the limb weight force is m. Figure 11 shows the geometry and the force due to the weight of the limb. Figure 11 Geometry for calculating torque due to the weight of the limb. Now, δ, the angle between the lower leg and b, is 0.5 rad, and b is distance from the knee joint to the line of action of the force due to limb weight. The force due to limb weight for a man with 0.36 m lower leg is given as, F w = 37.8 N. (21) With this information and equation (1), the torque of the force due to limb weight is, τ w = F w b (22) right triangle trigonometry is used to solve for b, cos(0.5) = b d (23) b = cos (0.5) (24) b = 0.14 m. (25) 24

25 Thus, the distance of the lever arm to center of mass of the limb is 0.14m. Substituting back into equation (22) τ w = (26) τ w = 5.2 N m. (27) The net torque at the knee joint is found by summing the magnitude and taking into account the direction of each torque. Since the force due to limb weight acts downward, the quantity is subtracted from the muscle force, τ K = ( F m a) ( F w b) (28) τ K = (29) τ K = 7.2 N m (30) The torque at the knee joint has a magnitude of 7.2 N m. This implies the torque acts in a clockwise direction. With this geometry and force understanding, we can calculate the torque at the knee joint in MATLAB for a range of θ values. Figure 12 displays the results of our code; which indicates the magnitude of torque is the greatest when the lower leg is at the vertical axis and approaches zero as the lifter s leg reaches extension. Notice the magnitude of the torque is 7.2 N m, is the same as the result in (30). A copy of the code is in Appendix B. 25

26 Figure 12 Magnitude of torque over a range of θ, the change in knee angle with respect to the vertical axis, values from -π/2 to RELATIONSHIP BETWEEN TORQUE AND ANGULAR ACCELERATION The changes in θ, the knee angle, essentially dictate the motion of the leg. Here, we aim to describe how the tangential component of rotational force, torque, is related to the angular acceleration. The setting has the subject performing a seated leg raise exercise. No weights, other than the weight of the leg are included in the analysis. The graphical representation is given in Figure 13 below. 26

27 Figure 13. Limb weight force, mg, under examination in the differential equation. To find the torque of the limb weight, the lever arm is found by drawing a perpendicular line with the line of action of the force and the axis of rotation. The length of this line is calculated using trigonometry and the torque is the lever arm times the force. The generalized equation for the torque caused by the limb weight force is, τ K = mg sin(θ)r (31) Where mg is the force from the weight of the limb, r is the length from the axis of rotation to the point of application of the limb weight force on the link, and θ is the angle between the force applied and the axis of rotation. If more forces were to be considered, their torques would be calculated and added to equation (14) along with an additional tangential and moment of inertia components to find the total torque acting on the knee joint. To begin, we relate θ to θ by examining distance at which the lower leg moves following the circular path of motion, as shown in Figure

28 Figure 14 Geometry of the lower leg motion to show arc length relationship Where r is the length of the lower leg, θ is the angle created by the starting position of the leg and the ending position, and the arc length, or length of the arc, is denoted as s. For constant r, the rate of change of s with respect to time, or the velocity is, v = rθ. (32) The second derivative yields the acceleration as, a = rθ. (33) Thus, the velocity in the tangent direction is rθ and the acceleration in the tangent direction is rθ. Notice substituting (35) into equation (8) gives, F = mrθ (34) which is the tangential component of force as discussed in Section 2.4. The cross product of equation (34) and r is torque, τ = mr 2 θ. (35) To find the net torque acting on the knee joint we will sum the torques as mentioned in Section 2.4 equation (14), τ K = mg sin(θ)r +mr 2 θ +mr 2 θ. (36) 28

29 We substitute (36) into τ K to set the equation for torque equal to the sum of the components that contribute to torque, mr 2 θ = mg sin(θ)r +mr 2 θ +mr 2 θ. (37) Simplifying the equation gives, θ = g sin(θ) (38) r where m, r, and g are constants representing mass, distance from the axis of rotation to the point of application of the force, and the acceleration due to gravity respectively. Given the complex nature of this homogenous, non-linear, second order ordinary differential equation, MATLAB ODE45 is used to obtain an approximate solution. The initial conditions, stated below, are based on the seated leg raise exercise, θ(0) = π 2 (39) θ (0) = 0. (40) Appendix C contains the code of the program. The results from Figure 15 indicate that as θ, the knee angle, approaches zero, the angular speed of the lower leg reaches its maximum. Provided θ is not allowed out of certain ranges, we conclude that this differential equation describes the motion due to the limb weight force of the leg raise exercise. 29

30 Figure 15 Differential equation results with leg extension conditions. 3.5 SIMPLE PENDULUM EQUATION COMPARISON The leg extension motion described in the previous section is similar to the motion of a pendulum. Thus, the differential equation for pendulum motion is examined to compare to our single link motion model. Figure 16 illustrates the motion of the pendulum system. A mass m hangs from a rod of length L. The string is released at an initial angle and rotates periodically about the pivot point P. 30

31 Figure 16 Simple pendulum with mass m hanging from a rod of length L with pivot point P. Using Newton s second law, the resulting equation of motion for the pendulum is obtained by setting the sum of the torques acting on the string equal to the torque equation, as we did for single link motion. Recall the equation for torque is found by multiplying the moment of inertia, ml 2, times the angular acceleration,θ, mg sin(θ) L = ml 2 θ (41) Simplifying (*) yields, θ + g sin(θ) = 0 (42) L As means to validate our model, we compare results from the pendulum equation of motion to our model for single link motion. MATLAB is used to solve (42) and g and L are set to be one for simplicity because those terms vary depending on the system. 31

32 Figure 17 Pendulum differential equation solved for in MATLAB. Notice that the results in Figure 17 from the pendulum equation are the same as our single link motion that only considers the limb weight. This observation propels us into the next section where we modify the pendulum equation to accommodate for muscle force. 32

33 CHAPTER 4 MODELING MUSCLE BEHAVIOR 4.1 HOOKE S LAW BACKGROUND To improve our model for single link motion, a muscle forcing term is added to the differential equation that accounts for the muscle s contribution to motion. We speculate that this term behaves according to Hooke s law because the muscle force varies as the muscle is extended and contracted. A system is set in motion by either moving the mass from its equilibrium position and releasing the spring with an initial velocity, or imposing a nonzero initial velocity on the spring that causes the mass to move from its equilibrium [A8]. In our study, we release the spring with an initial velocity of zero. Hooke s law states that force required to put the spring in motion is proportional to the change in the position of the spring. This is denoted as, F = kx (43) Where F is the magnitude of the force, x is amount of elongation or compression from equilibrium, and k is the spring constant. 4.2 HOOKE S LAW AND MUSCLE BEHAVIOR To apply Hooke s law to muscles, we first need to confirm that a muscle behaves like a spring. Researchers performed an experiment to investigate if a muscle had spring like functionality. A muscle was attached to a bar that moved to simulate the compression, or shortening, of a muscle, as seen in Figure 18A. One end of the bar has a weight inside a basket that pulls on the muscle by a force T when the catch, located on 33

34 the other side of the bar, is released. To begin the experiment, the muscle is configured to its maximally extended position. Releasing the catch with the muscle maximally extended results in a force, T 0, produced by the muscle. Figure 18 A muscle is stimulated when the catch mechanism is released (A). The muscle produces more force than the weight in the basket which causes the muscles to decreases rapidly and then level out as time passes. The behavior of the muscle force and length from (B) and (C) indicate that the muscle acts like a spring [A9]. In Figure 18B, the force in the muscle decreased from T 0 to T. Figure 18C shows how the length of the muscle shortened drastically when the catch was released but as time passes the length of the muscle remains at a constant length. The fact that the muscle immediately shortened in length and reduces force productions indicates that the muscle acted like a spring [A9]. Therefore, the structure of a muscle consists of elastic like properties whose behavior can be modeled mathematically as a spring. 34

35 The lengths of the links and the joint angle in the leg raise exercise are analyzed to generate an expression for the x term from Hooke s law. The lengths of the upper and lower leg are denoted below in Figure 19 as b and c, are assumed to be 0.45 m and 0.5 m respectively. Applying law of cosines, we can relate the lengths of the links to the joint angle, θ, to determine the length of the muscle, x. Figure 19 The lengths of the upper and lower leg are b and c respectively. The joint angle, θ, is used in the law of cosines formula to find the length of the muscle. The muscle length is given as, x = b 2 + c 2 2bc cos (θ) (44) The muscle elongates through the motion of the leg extension exercise. The change in muscle length for a range of θ values from 0 to pi/2 is computed in MATLAB, as seen in Figure 20. The code is located in appendix D. The figure indicates that when the joint angle reaches its maximum range of motion, the length of the muscle also reaches its maximum elongation. 35

36 Figure 20 Graph of the muscle length as joint angle increases through the range of motion. For modeling purposes, a value for k, the spring constant, is chosen based on an estimated force production by the muscle. This value is found from published data in [A4]. We set the k*x term equal to the estimated force produced by the muscle and solve for the spring constant to get a value for k. From this equation we can determine information about the speed and position of human motion. Restrictions on theta, the joint angle, must also be considered. The muscle forcing term added to our simple motion model is, mrθ + mgsin(θ) + k b 2 + c 2 2bc cos (θ) = 0. (45) Or similarly, 36

37 θ = g r sin(θ) k mr b2 + c 2 2bc cos (θ) (46) 4.3 MUSCLE RESISTANCE AND DAMPING TERM Adding a damping term to (46) allows for the accommodation for an internal resistance force produced by the body that affects lifting motions. In investigating muscle functions, we found that the human body contains three types of muscles. Cardiac muscles are muscles in the heart. Smooth muscles line the walls of hollow organs, like in the digestive system, that help move substances throughout. Lastly, there are skeletal muscles that attach to skeletal bones. When these muscles contacts, there is another muscle that is paired with it that relax. Since skeletal muscles only pull in one direction, its counterpart reacts and pulls in the opposite direction to straighten the joint out again. Without this physical arrangement, it would not be possible to straighten your legs when you walk or bend your fingers to grip something [B7]. The muscle that works to counteract the movement of its pair is represented as the damping term in our model. The muscle force is estimated and is used to back fit the equation as we did for the muscle forcing term. The model with the damping term included is, θ = g r sin(θ) d mr θ k mr b2 + c 2 2bc cos (θ) (47) Where d represents the force produced by the counteracting muscle and θ is the change in speed of the muscle. MATLAB ODE45 is used to solve (47). Figure 21 displays the results. 37

38 38

39 39

40 Figure 21 Results of modified pendulum equation with muscle forcing and damping terms. The figure is decomposed into two separate graphs due to differences into the vertical axis for the functions. Comparing the results from MATLAB to the differential equation that only considered the limb weight force (42) to the limb weight and muscle forcing terms (47), a noticeable decrease in lifting time when the muscle forcing terms is observed. In (42), the time that elapsed for leg to pass through the range of motion was 2 seconds, while in (47), the lifting motion takes approximately seconds to complete. This value is found in MATLAB by calculating where θ is at its maximum, which is roughly The index in the array where the maximum θ occurs is returned and then used to calculate the corresponding time value. 40

41 Therefore, the magnitudes of the muscle and damping force have a great effect on time the motion takes to complete the motion. To generalize this observation for more single link motions, the amount of time elapsed during the lifting motion varies depending on magnitude of the muscle and damping forces. 41

42 CHAPTER 5 ENERGY EXPENDITURE 5.1 NONLINEAR CONSERVATIVE SYSTEM Expanding on proposed future work to build an energy and fatigue component, the total energy of the leg extension motion was calculated. However, due to time constraints, an energy and fatigue function could not be established. Given the complexity of our model, pendulum equations and applications to energy were studied to gain insight into energy expenditure and how the kinetic and potential energy components function. Kinetic energy is energy a body possesses by being in motion, 1 2 mv2, where m is mass and v is velocity. Potential energy is the energy that is stored in a body because of its positon. The law of conservation of energy, which states that the sum of the kinetic and potential energy is a constant, implies that the system is conservative. Therefore, we first consider a basic nonlinear conservative system and then apply the process to our model of single link motion and muscle behavior. Recall the differential equation of a simple pendulum, d 2 x = g sin(x) (48) dt 2 L Where g is the acceleration due to gravity and L is the length of the body in motion. Letting g and L equal one for simplicity, d 2 x dt 2 = sin(θ). (49) 42

43 The law of conservation of energy states that the total energy of a system is the sum of the kinetic and potential energies. Differential equations where this law is applied follow the form, m d2 x dt2 = F(x) (50) Where F(x) represents a function that is assumed to be differentiable for all values of x. Converting the differential equation (50) to a nonlinear autonomous system, that is a system of differential equations that do not explicitly depend on the independent variable, time, gives, dx dt = y dy = F(x). (51) dt m To solve the differential equation, dt is eliminated, dy = F(x). (52) dx my Using the separation of variables technique, we obtain my dy = F(x) dx. (53) Suppose we let x be x 0 and y from (51) be y 0 when t is t 0. Then integrating the left side from y 0 to y and the right side from x 0 to x yields, 1 x 2 my2 1 my = F(x)dx (54) x 0 43

44 Which is equivalent to splitting the integral from x 0 to x into the integral from 0 to x minus the integral of 0 to x 0, 1 x 2 my2 F(x)dx = 1 my F(x)dx. (55) 0 x 0 Notice that 1 2 m(dx dt )2 = 1 2 my2 is the kinetic energy and x V(x) = F(x)dx 0 (56) Is the potential energy of the system. Simplifying (55) gives 1 2 my2 + V(x) = h (57) Where the constant h represents the total energy of the system and is denoted by, h = 1 2 my 0 2 x 0 0 F(x)dx. (58) This approach of calculating energy of a nonlinear conservative system was provided by [A8] and will be used in the following section to demonstrate how the total energy of the leg extension motion is calculated. 44

45 5.2 ENERGY EXPENDED IN SEATED LEG RAISE EXERCISE To determine the total amount of energy expended in the leg extension motion, we follow a similar approach to finding the total energy in a pendulum system. Notice that the mass term has already been divided through this equation. Thus, the differential equation of motion is, θ = g r sin(θ) d mr θ k mr b2 + c 2 2bc cos (θ). (59) MATLAB was used to find the velocity and θ functions of (59). Refer to Appendix E for the code. Recall that the kinetic energy is calculated by 1 2 mv2. The velocity function was approximated using trapezoidal rule, a technique in approximating definite integrals by finding the area under the curve using trapezoids. MATLAB ODE45 returns a vector of 69 values approximating the solution θ and θ on the interval of 0 to seconds. The interval was chosen based on the range of motion of the lift. The range of motion cannot exceed π, and the corresponding time value when the motion is at its maximum is However, the values MATLAB returns are not uniformly spaced over the interval. To use the trapezoidal rule, uniformly spaced points were constructed and then fit with the not uniform points located inside of the uniformly spaced points with a piecewise linear function. Then, we use the constructed function to get evenly spaced points for numerical integration. The code is also located appendix E. Kinetic energy was found to be Joules. For the potential energy, we let F(x) = g r sin(θ) d mr θ k mr b2 + c 2 2bc cos (θ). (60) 45

46 The integral of negative F(x), which gives the potential energy, was also approximated using the trapezoidal rule. A similar construction is used to create uniformly spaced points in MATLAB to calculate potential energy, which is Joules. Therefore, the total energy expended during the seated leg raise exercise is Joules by the law of conservation. 46

47 CHAPTER 6 FUTURE WORK AND APPLICATIONS 6.1 TWO LINK MOTION AND MUSCLE BEHAVIOR MODELED WITH A DOUBLE PENDULUM Just as we used a simple pendulum for single link motion to validate our model, we hypothesize that for two link analysis, a double pendulum system could be compared to two link motion. Furthermore, adding muscle terms to a two link system would involve a complex arrangement and understanding of multiple muscle springs and damping, or resistance, forces interacting. We suspect that an additional link adds another muscle pair to the model, meaning another muscle forcing and damping term is added to the equation. In addition, initial conditions for the muscle, damping, and limb weight forces would need to be established and back fitted for the model. Examining the angle of pull in the leg extension motion propels further investigation into a force analysis on the bench press motion. This will involve additional research on a dynamic analysis for a two link body motion and determining the relationship between two angle of pulls, load from the barbell, the weight of two links, and other extraneous forces not yet considered. We hypothesize this force analysis will allow us to adjust our differential equation to describe the motion of the bench press lift. Furthermore, initial conditions based on the bench press motion will have to be determined for the differential equation. Essentially, we laid the foundation for such an examination to occur through studying the geometry of a two link lifting motion, the bench press lift. We also consider an analysis that can be performed once a model for two 47

48 link motion is set by altering the initial grip of the lift. These concepts are discussed in more detail in the following sections DESCRIPTION OF BENCH PRESS LIFT The bench press is performed with the lifter initially in a horizontal position, lying face up on a workout bench with a barbell overhead. The lifter presses the barbell upward to remove it from the rack and holds the barbell directly overhead with their arms fully extended, hence forming a right angle between the body and the extended arms. We will define this position as the initial position. For this project, the lifter will have a neutral grip, meaning the lifter s grip on the barbell is roughly shoulder width apart. The bench press motion can be decomposed into two phases: the down phase and the up phase. To complete the down phase, the lifter starts at the initial position and brings the barbell down in a vertical direction to their mammillary line, or even with the chest. In this position, the forearm is perpendicular to the body. Pressing the barbell from the mammillary line to the initial position completes the up phase of the lift. It is important to note that proper placement of the upper arm is crucial to preventing shoulder injuries. Figure 22 below illustrates the proper placement of the upper arm to ensure the elbow is in the correct position for the lift. The likelihood of injury increases when the upper arm placement forms a line that is perpendicular to the shoulder [B1]. 48

49 Figure 22 Proper upper arm placement (red) for bench press lift when looking down on the lifter on the bench. The black lines represent incorrect positions of the upper arm GEOMETRIC INTPRETATION OF BENCH PRESS MOTION Before we delve into the geometric interpretation of the bench press lift, it is important to note that we did not consider any extraneous external or internal forces acting upon the body. However, we did impose physiological constraints on the model. For instance, the length of the upper arm and the length of the forearm must preserve the same length throughout the duration of the lift and must function in ways that are consistent to human motion. This means that a joint should not surpass its maximum range of motion, or the angle that a segment can move while remaining attached to the joint [B2]. Surpassing this angle would result in an injury to the lifter. Additionally, we analyzed the geometry for the right arm and assumed by symmetry that the left arm behaves in a similar manner. Given the constraints on the model, the analysis starts by examining the linear and rotational movement produced as a result of the change in shoulder angle, θ. Note that the shoulder is fixed at the origin and will only produce rotation in a vertical plane during the lift. As the upper arm rotates about the shoulder, the angle at the elbow, β, adjusts to allow the hand to move along the line L in Figure

50 Figure 23 Geometry for bench press lift at a fixed point of time in the motion. This creates a right triangle with the shoulder at the bottom vertex that allows us to calculate all variables related to the position of the arms during the lift using right triangle trigonometry. The calculation is as follows: cos(θ) = X. (61) A Solving for X yields, X = A cos(θ). (62) This gives us the x coordinate for the position of the elbow. Following a similar calculation for the y coordinate yields, Y = A sin(θ). (63) Thus, the position of the elbow is at (A cos(θ), A sin (θ)). Now we can solve for θ using inverse trigonometry, θ = sin 1 ( B2 Z 2 ). (64) A+B 50

51 These equations were used to create a program in MATLAB that illustrates the geometry and motion of the bench press lift, as seen in Figure 24. The code is located in appendix A. Figure 24 A simulation of the bench press motion 51