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Unformatted text preview: ME577 Homework #6 Problem 1 Here is a simplified diagram of the pitchers arm: S H W E The following notation will be used for the solution: hx , hy , hz : w x , wy , wz : fax , fay , faz : hux , huy , huz : Position of the hand in the x,y, and z directions Position of the wrist in the x,y, and z directions Position of the elbow in the x,y, and z directions Position of the shoulder in the x,y, and z directions Position of the center of mass of the forearm in x, y, and z directions Position of the center of mass of the Humerus in x, y, and z directions ￿h , ￿h , ￿ h : rva Position, Velocity, and Acceleration vectors of the hand ￿w , ￿w , ￿ w : rva ￿e , ￿e , ￿ e : rva ￿s , ￿s , ￿ s : rva ￿fa , ￿fa , ￿ fa : rva ￿hu , ￿hu , ￿ hu : rva Position, Velocity, and Acceleration vectors of the wrist Position, Velocity, and Acceleration vectors of the elbow Position, Velocity, and Acceleration vectors of the shoulder Position, Velocity, and Acceleration vectors of the center of mass of the forearm Position, Velocity, and Acceleration vectors of the center of mass of the Humerus ￿ , ￿, ￿ : xyz Unit vectors in the x, y, and z directions e x , e y , ez : s x , sy , sz : A) To determine the velocity of the center of masses of the hand, forearm and humerus, the velocity at each of the measured points (hand, wrist, elbow, and shoulder) must first be calculated. For the hand, the position vector from the fixed origin to the 3rd knuckle is given by the following equation: ￿h = hx ￿ + hy ￿ + hz ￿ r x y z To calculate the velocity at this point, differentiate the position vector with respect to time. This gives the equation for the velocity of the hand. ˙x ˙y ˙z ￿ h = hx ￿ + hy ￿ + hz ￿ v Using the provided position data for the hand, the first derivative of the position can be approximated using the forward difference equation: r ) rh ˙ ￿h (t1 ) = ￿h (t1 ) ≈ ￿h (t22 −￿1 (t1 ) v r t −t This calculation is then repeated for each time interval. N ME577 Homework #6 Problem 1 v Using the same calculation, but substituting the position data for the wrist, elbow, and shoulder, ￿w , ￿e, and ￿s can be found. v v Using the assumption that the position data given for the 3rd knuckle is roughly the position of the v center of mass. Then ￿h is the velocity vector for the hands center of mass. To compute the velocity of the center of mass of the forearm, the vector from the wrist and the elbow must be determined. It can be calculated with the following equation: ￿e = ￿w + ￿e/w r r r Rearranging to solve for the vector from the wrist to the elbow yields: ￿e/w = ￿e − ￿w r r r If we assume the center of mass of the forearm is half way between the wrist and the elbow, then the position vector for the center of mass is: ￿fa = ￿w + 1 ￿e/w r r 2r Which simplifies to: ￿fa = 1 (￿e + ￿w ) r r 2r Taking derivative of this equation with respect to time yields the velocity of the forearm. ￿fa = 1 (￿e + ￿w ) v v 2v Finally, the velocity of the center of mass of the Humerus can be determined by first calculating the vector from the elbow to the shoulder, and then following a similar calculation as was done for the forearm. It was assumed that the center of mass of the Humerus was half way along it’s length. This leads to the following equations. ￿hu = 1 (￿s + ￿e ) r r 2r ￿hu = 1 (￿s + ￿e ) v v 2v The total velocity at each time step is then given by the magnitude of the velocity vector. ￿ ˙ ˙ ˙ |￿h | = h2 + h2 + h2 v x y z ￿ ˙2 ˙2 ˙2 |￿fa | = fa x + fa y + fa z v ￿ ˙2 ˙2 ˙2 |￿hu | = hu x + hu y + hu z v Figure 1 shows these magnitudes plotted versus time. O ME577 Homework #6 Problem 1 25 Hand Forearm Humerus Velocity (m/s) 20 15 10 5 0 0 0.5 1 1.5 2 2.5 Time (s) 3 3.5 4 4.5 5 Figure 1: Velocities of the center of masses of the hand, forearm and Humerus B) To calculate the accelerations of the center of masses of the hand, forearm, and Humerus, a similar process must be taken. The position vectors of the hand, wrist, elbow and shoulder need to be differentiated with respect to time. For the hand, this acceleration vector will take the form of: ˙ ¨x ¨y ¨z ￿ h = ￿ h = hx ￿ + hy ￿ + hz ￿ a v Once again, a forward difference approximation at each time interval will be used to calculate this derivative. ˙ ￿ h ( t 1 ) = ￿ h ( t1 ) ≈ a v ￿h (t2 )−￿h (t1 ) v v t 2 − t1 aa a Similarly, ￿ w , ￿ e , and ￿ s can then be calculated. To find the acceleration of the center of mass of the forearm and Humerus, the velocity equations for those points must be differentiated with respect to time. This results in the following equations: ￿ fa = 1 (￿ e + ￿ w ) a a 2a ￿ hu = 1 (￿ s + ￿ e ) a a 2a aa a These vectors can be found directly by substituting in the previously found ￿ w, ￿ e, and ￿ s vectors. P ME577 Homework #6 Problem 1 The magnitude of the acceleration vectors is then calculated similarly to the magnitude of the velocity vectors. These magnitudes are plotted versus time in Figure 2. 1500 Hand Forearm Humerus Acceleration (m/s2) 1000 500 0 0 0.5 1 1.5 2 2.5 Time (s) 3 3.5 4 4.5 5 Figure 2: Velocities of the center of masses of the hand, forearm and Humerus C) To determine the magnitude of the distraction force at the shoulder, the simplified arm has to be split into separate free body diagrams. One for the arm and wrist, one for the forearm and one for the Humerus. Figures 3-5 show these free body diagrams. e Fw t Mw (mb + mh )g Figure 3: FBD of the wrist and hand Q ME577 Homework #6 Problem 1 t Fe Mw Fw Me b mfa g Figure 4: FBD of the forearm Fs p Ms b mhu g Me Fe Figure 5: FBD of the Humerus The governing equation to solve for the forces in each link is Euler’s First Law. ￿￿ F = m￿ cg a Writing out this equation for the hand and the wrist yields: ￿ Fw + (mb + mh )g (−￿ ) = (mb + mh )￿ h y a ￿ Solving for Fw yields: ￿ Fw = (mb + mh )￿ h + (mb + mh )g￿ a y Next, Euler’s First Law is used on the forearm. ￿ ￿ Fe − Fw + mfa g (−￿ ) = mfa ￿ fa y a ￿ Solving for Fe gives: ￿ ￿ Fe = mfa ￿ fa + Fw + mfa g￿ a y Finally, summing forces at the shoulder: ￿ ￿ Fs − Fe + mhu g (−￿ ) = mhu ￿ hu y a R ME577 Homework #6 Problem 1 Then the shoulder force can be solved for. ￿ ￿ Fs = mhu ￿ hu + Fe + mhu g￿ a y To determine the distraction force at the shoulder, a unit vector pointing from the shoulder to the elbow along the humerus is needed. First, the position vector from the shoulder to the elbow is calculated. ￿e = ￿s + ￿e /s r r r r Solving for ￿e /s : ￿e /s = ￿e − ￿s r r r e If we define ￿r as the unit vector pointing from the shoulder to the elbow along the Humerus, then ￿e /s can also be written as the following: r ￿e /s = l￿r r e r Where l is the distance between the shoulder and elbow. This distance can is the magnitude of ￿e /s . l = |￿e − ￿s | r r e Combining the previous three equations and solving for ￿r yields: ￿r = e Written in component form, this is: ￿r = e ex − s x ￿ x l + ￿e −￿s rr |￿e −￿s | rr ey − s y ￿ y l + ez − s z ￿ z l The distraction force then is simply the dot product of the shoulder force and this unit vector. ￿ ￿e Fdist = Fs · ￿r To calculate the various forces, the following masses and constants were used: máíÅÜÉê=j~ëë m TU=âÖ _~ää=j~ëë mb KNQR=âÖ e~åÇ=j~ëë mh KMMSm coêÉ~êã=j~ëë mfa KMNSm eìãÉêìë=j~ëë mhu KMOUm g VKU=ãLëO dê~îáí~íáoå~ä=^ÅÅÉäÉê~íáoå Figures 6 and 7 show the magnitude of these forces versus time. S ME577 Homework #6 Problem 1 1000 Distraction Force at Shoulder (N) 500 0 500 1000 1500 0 0.5 1 1.5 2 2.5 Time (s) 3 3.5 4 4.5 5 Figure 6: Magnitude of the distraction force at Shoulder 3000 Wrist Elbow Shoulder 2500 Magnitude of Force (N) 2000 1500 1000 500 0 0 0.5 1 1.5 2 2.5 Time (s) 3 3.5 4 4.5 5 Figure 7: Magnitude of total forces at Wrist, Elbow, and Shoulder T Problem 3 From the data file, I found that vary from -0.99 to 0.39. But based on the figure, it cannot be positive, or the knee bended to the wrong direction. I believe the figure has an error. The angle2 should be between the vertical direction and the tibia. But I did not correct it. Based on the anthropometric data and H=1.9 m, we take 0.481, 0.447, Therefore, cos sin cos sin cos sin sin cos 0.09 -0.78 -0.79 Xo -0.8 -0.81 -0.82 -0.83 -0.84 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 0.9 1 0.7 0.8 0.9 1 Figure 1 Xo vs. time 0.5 0.4 0.3 Angle1 0.2 0.1 0 -0.1 -0.2 -0.3 0 0.1 0.2 0.3 0.4 0.5 t 0.6 Figure 2 Angle1 vs. time 0.4 0.2 0 Angle2 -0.2 -0.4 -0.6 -0.8 -1 -1.2 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 0.9 1 0.8 0.9 1 Figure 3 Angle2 vs. time 0.5 0.4 0.3 Angle3 0.2 0.1 0 -0.1 -0.2 -0.3 -0.4 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 Figure 4 Angle3 vs. time -0.3 -0.31 -0.32 -0.33 Xa -0.34 -0.35 -0.36 -0.37 -0.38 -0.39 -0.4 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 0.9 1 0.7 0.8 0.9 1 Figure 5 Xa vs. time 0.25 0.2 0.15 Ya 0.1 0.05 0 -0.05 -0.1 -0.15 0 0.1 0.2 0.3 0.4 0.5 t 0.6 Figure 6 Ya vs. time 0.15 0.1 Xb 0.05 0 -0.05 -0.1 -0.15 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 0.9 1 0.7 0.8 0.9 1 Figure 7 Xb vs. time 0.6 0.4 Yb 0.2 0 -0.2 -0.4 -0.6 0 0.1 0.2 0.3 0.4 0.5 t 0.6 Figure 8 Yb vs. time 0.2 0.15 Xc 0.1 0.05 0 -0.05 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 0.9 1 0.7 0.8 0.9 1 Figure 9 Xc vs. time 0.5 0.4 0.3 0.2 Yc 0.1 0 -0.1 -0.2 -0.3 -0.4 -0.5 0 0.1 0.2 0.3 0.4 0.5 t 0.6 Figure 10 Yc vs. time 0.2 0.15 Xd 0.1 0.05 0 -0.05 0 0.1 0.2 0.3 0.4 0.5 t 0.6 0.7 0.8 0.9 1 0.7 0.8 0.9 1 Figure 11 Xd vs. time 0.7 0.6 0.5 0.4 Yd 0.3 0.2 0.1 0 -0.1 -0.2 -0.3 0 0.1 0.2 0.3 0.4 0.5 t 0.6 Figure 12 Yd vs. time Code of Matlab: clear load walking_data.txt -ascii disp('Start mapping process') t=walking_data(:,1); Xo=walking_data(:,2); anglea=walking_data(:,3); angleb=walking_data(:,4); anglec=walking_data(:,5); Xa=Xo+0.481*cos(anglea); Ya=0.481*sin(anglea); Xb=Xa+0.447*cos(anglea+angleb); Yb=Ya+0.447*sin(anglea+angleb); Xc=Xb+0.09*cos(anglec); Yc=Yb+0.09*sin(anglec); Xd=Xc-0.228*sin(anglea); Yd=Yc+0.228*cos(anglea); plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, plot(t, Xo) anglea) angleb) anglec) Xa) Ya) Xb) Yb) Xc) Yc) Xd) Yd) ...
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This document was uploaded on 12/21/2011.

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