ch07 - Solutions to Chapter 7 Exercise Problems Problem 7.1...

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Unformatted text preview: Solutions to Chapter 7 Exercise Problems Problem 7.1 A coupler curve has the approximate straight-line section shown in the figure below. Design a four-bar linkage that will generate the portion of the curve shown. Describe the linkage in sufficient detail that it can be manufactured. Y xB xA Ay A B yB X xA = 6 yA = 3 xB = 18 yB = 12 Solution: To solve the problem, we need to find a coupler curve with an approximate straight-line section. We can use any of the mechanisms identified in section 7.1.2. We will use the Watt straight-line mechanism shown in Fig. 7.1. Assuming the crank is the driver, the coupler curve is shown in Fig. 7.1.1 for the coupler length equal to 5/3 times the crank length. This drawing is based on the results from the program fourbar_analysis.m included on the disk with this book. The linkage is a Grashof, Type 2 double rocker. The straight line part of the coupler curve is given by segment SU, and this distance is approximately equal to the length of the cranks. The line SU is also bisected by the line of centers PQ, and vice versa. Also, line SU is inclined at an angle of approximately -50.2˚ to the line PQ. The length of the line of centers can be computed using the Pythagorean theorem when the coupler and cranks are at right angles to each other as shown in Fig. 7.1.2. Using a crank length of 3 and a coupler length of 5, the center distance is found to be 2(3.9051) = 7.8102. V U 50.2˚ P T S R Fig. 7.1.1: Coupler curve for Watt's mechanism when the coupler length is 5/3 times the length of the cranks. Q - 277 - V 2.5 50.2˚ T 3 2.5 3 P 3.9051 Q R Fig. 7.1.2: Determination of the distance between centers. To use this linkage for the given problem, it is necessary to map point S to point A and point U to point B. The length of the line AB is given by L = (xB xA)2 + (y B yA)2 = (18 6)2 + (12 3)2 = 15 and the angle of inclination of the line AB is given by = tan 1 [ yx yx ] = tan [18 6] = tan [12 ] = 53.13˚ 12 3 9 B B A A 1 1 Therefore, to use the Watt's linkage to generate the curve, we need only scale the linkage so that the length of SU is 15 and rotate the linkage so that the line SU is at and angle of 53.13˚ to the horizontal. The rotation required is 53.13˚ - 50.20˚ or 2.93˚. The length of line SU in Fig. 7.1.1 is 3. Therefore, the scaling factor for the linkage is (15/3) or 7. The final link lengths then are: r =5 1 r2 = 5 r3 = 5 r4 = 5 7.8102 = 39.051 3 = 15.000 5 = 25.000 3 = 15.000 The coupler point T which generates the coupler curve is located midway between the two ends of the coupler. In the position corresponding to Fig. 7.1.2, the point T has the following x and y coordinates: xT = ( xB + xA) / 2 = (18 + 6) / 2 = 12 yT = ( yB + y A) / 2 = (12 + 3) / 2 = 7.5 The coordinates of point P are given by: xP = x T PT cos = 12 (39.051/ 2)cos(2.93) = 12 19.5255cos(2.93) = 7.500 yP = y T PT sin = 7.5 (39.051 / 2)sin(2.93) = 7.5 19.5255sin(2.93) = 6.502 And the coordinates of point Q are given by: - 278 - xQ = xT + TQ cos = 12 + (39.051 / 2)cos(2.93) = 12 + 19.5255cos(2.93) = 31.500 yQ = yT + TQsin = 7.5 + (39.051 / 2)sin(2.93) = 7.5 + 19.5255sin(2.93) = 8.498 The final solution linkage is shown in Fig. 7.1.3. V 15.000 in U (31.500, 8.498) (-7.500, 6.502) P 19.5255 in T 50.2˚ 2.93˚ Q 15.000 in R S Fig. 7.1.3: Final solution linkage. Problem 7.2 Resolve Problem 7.1 if xA = 3 , yA = 3 , xB = 20 , and yB = 15 . Solution: To solve the problem, we need to find a coupler curve with an approximate straight-line section. We can use any of the mechanisms identified in section 7.1.2. We will use the Watt straight-line mechanism shown in Fig. 7.1. Assuming the crank is the driver, the coupler curve is shown in Fig. 7.2.1 for the coupler length equal to 5/3 times the crank length. This drawing is based on the results from the program fourbar_analysis.m included on the disk with this book. The linkage is a Grashof, Type 2 double rocker. The straight line part of the coupler curve is given by segment SU, and this distance is approximately equal to the length of the cranks. The line SU is also bisected by the line of centers PQ, and vice versa. Also, line SU is inclined at an angle of approximately -50.2˚ to the line PQ. The length of the line of centers can be computed using the Pythagorean theorem when the coupler and cranks are at right angles to each other as shown in Fig. 7.2.2. Using a crank length of 3 and a coupler length of 5, the center distance is found to be 2(3.9051) = 7.8102. To use this linkage for the given problem, it is necessary to map point S to point A and point U to point B. The length of the line AB is given by L = (xB xA)2 + (y B yA)2 = (20 3)2 + (15 3)2 = 20.809 - 279 - V U 50.2˚ P T S R Fig. 7.2.1: Coupler curve for Watt's mechanism when the coupler length is 5/3 times the length of the cranks. V 2.5 50.2˚ T 3 2.5 3 Q P 3.9051 Q R Fig. 7.2.2: Determination of the distance between centers. and the angle of inclination of the line AB is given by = tan 1 [ yx yx ] = tan [20 33] = tan [17 ] = 54.78˚ 15 12 B B A A 1 1 Therefore, to use the Watt's linkage to generate the curve, we need only scale the linkage so that the length of SU is 15 and rotate the linkage so that the line SU is at and angle of 53.13˚ to the horizontal. The rotation required is 54.78˚ - 50.20˚ or 4.58˚. The length of line SU in Fig. 7.2.1 is 3. Therefore, the scaling factor for the linkage is (20.809/3) or 6.936. The final link lengths then are: r = 6.936 1 r2 = 6.936 r3 = 6.936 r4 = 6.936 7.8102 = 54.1715 3 = 20.808 5 = 34.680 3 = 20.808 - 280 - The coupler point T which generates the coupler curve is located midway between the two ends of the coupler. In the position corresponding to Fig. 7.2.2, the point T has the following x and y coordinates: xT = ( xB + xA) / 2 = (20 + 3) / 2 = 11.500 yT = ( yB + y A) / 2 = (15 + 3) / 2 = 9.000 The coordinates of point P are given by: xP = x T PT cos = 11.500 (54.1715 / 2)cos(4.58) = 11.500 27.086cos(4.58) = 15.499 yP = y T PT sin = 9.000 (54.1715 / 2)sin(4.58) = 9.000 27.086sin(4.58) = 6.837 And the coordinates of point Q are given by: xQ = xT + TQ cos = 11.500 + (54.1715/ 2)cos(4.58) = 11.500 + 27.086cos(4.58) = 38.500 yQ = yT + TQsin = 9.000 + (54.1715/ 2)sin(4.58) = 9.000 + 27.086sin(4.58) = 11.163 The final solution linkage is shown in Fig. 7.2.3. V 20.808 U (38.500, 11.163) 50.2˚ (-15.499, 6.837) P S 20.808 R Fig. 7.2.3: Final solution linkage. 27.087 T 4.58˚ Q - 281 - Problem 7.3 Determine the cognate linkages that will trace the same coupler curve as that traced by point C in the figure below. C B 30˚ D A 3.25 in E 1.0 in AB = 1.5 in BC = 1.25 in BD = 2.0 in BC = 3.0 in DE = 1.25 in Solution Use Roberts' linkage in Fig. 7.12 as a guide to construct the cognates. First locate pivot O by recognising that triangle MQO is similar to ABC. Then complete the parallagrams indicated in Fig. 7.12. Also, refer to Fig. 4.59. The cognates are shown in the following figure. C B A D A E This figure can be checked using the program cognates.m. The coupler curve and cognates are given in the following. Note that this linkage is a type 2, double rocker. - 282 - Problem 7.4 Determine the cognate linkages that will trace the same coupler curve as that traced by point C in the figure below. A 1.75 in 2.375 in 30˚ D 90˚ E B C DE = 1.125 in BD = 2.25 in BC = 1.0 in Solution Use Roberts' linkage in Fig. 7.12 as a guide to construct the cognates. First locate pivot O by recognising that "triangle" EAO is similar to DBC. Then complete the parallagrams indicated in Fig. 7.12. Also, refer to Fig. 4.59. The cognates are shown in the following figure. - 283 - O A B D C E This figure can be checked using the program cognates.m. The coupler curve and cognates are given in the following. Note that this linkage is a type 2, double rocker. - 284 - Problem 7.5 A spherical four-bar linkage is shown in the figure below. If the angular velocity of link 2 is 100 rad/s (constant), find the angular velocity and angular acceleration of link 4 as a function of the rotation of link 2. Plot the angular velocity and angular acceleration of link 4 for a full rotation of link 2. Make the calculations for the assembly mode shown in the figure. R4 R2 14 ω R1 1ω 2 α4 2 α1 α3 3 4 1 α2 1 R3 α 1 = 110˚ α 2 = 90˚ α 3 = 90˚ α 4 = 90˚ Solution We must solve the problem for and R in Eq. (7.3). 1 from 0˚ to 360˚. For each value of 1, we must solve for P, Q, P = - cos 1cos 1sin 2sin 4 + sin 1sin 2cos 4 Q = sin 1sin 2sin 4 R = cos 1sin 1cos 2sin 4 + cos 1cos2cos 4 - cos 3 Hence referring to Table 3.1, we can obtain values for P2 + Q2 R2 RP where = ±1 is a sign variable, and 2 = 2 tan-1(t). t= Q+ 2 (7.3) 1 given from: (7.4) (7.5) The two solutions correspond to the two solutions obtained in the solution of the position problem of a planar four-bar, and have the same source in the reflection of the driven-crank and coupler about the plane of the moving joint axis of the driving-crank and the fixed joint axis of the driven-crank. By checking both solutions for 2 = 90˚ , we can determine that the solution corresponding to the figure in the problem statement is for = -1. We can also develop relationships between the angular velocities about joints 1 and 2 by differentiation of Eq. (7.1). Differentiation of Eq. (7.1) with respect to time gives, after rearrangement: ˙2 = ˙1 sin 4(cos 1 sin 2 sin sin 2(sin 1 cos 2 sin 2 + sin 1 cos 2 cos 1sin 2 4 + cos 1 sin 2 cos 1 sin 4 sin 1 sin 1 cos sin 2 sin 1 cos 2) 4) (7.6) - 285 - We can simplify this expression using cos and sin 2 2 = cos 3 = cos 4 =0 = sin 3 = sin 4 =1 However, it is relatively easy to write a general MATLAB program to solve Eq. (7.6) directly, so the simplification is not necessary. An expression for the acceleration is given by Eqs. (7.7) and (7.8). These are ˙˙2 = ˙˙ ˙2 B 1 ˙ ˙ C ˙2 D ˙2 ˙1 + A 1 2 + A 1 + A 2 1cos 2sin 4 + cos 1sin 2cos 1sin 4 - sin 2sin 1cos 4) 4(sin 1sin 2cos 1 - cos 1cos 2) 1sin 2sin 2 -cos 1cos 2cos 1sin 2 + cos 1sin 1cos 2) 1sin 2sin 4 - cos 1cos 2cos 1sin 4 + cos 2sin 1cos 4) (7.7) where A = sin 2(sin B = 2sin 2sin C = sin 4(sin D = sin 2(sin (7.8) These equations can also be easily programmed using MATLAB. Once the values are computed, they can be plotted as a function of the rotation of link 2. The resulting program is given in the following: % MATLAB program for problem 7.5 %Program clf; clear all; ans='y'; disp(' ') disp(' Universal Joint Analysis Program') %Input the linkage variables alpha1=110 alpha2=90 alpha3=90 alpha4=90 phi1dot=100 phi1ddot=0 fact=pi/180; al1=alpha1*fact; al2=alpha2*fact; al3=alpha3*fact; al4=alpha4*fact; phi1i=0; phi1f=360; phii=phi1i*fact; - 286 - phif=phi1f*fact; dphi=(phif-phii)/360; i=0 for phi=phii:dphi:phif i=i+1; % Position calculations P=-cos(phi)*cos(al1)*sin(al2)*sin(al4)+sin(al1)*sin(al2)*cos(al4); Q=sin(phi)*sin(al2)*sin(al4); R=cos(phi)*sin(al1)*cos(al2)*sin(al4)+cos(al1)*cos(al2)*cos(al4)cos(al3); arg=P*P+Q*Q-R*R; p1=0; p2=0; if arg>=0 den=R-P; if den==0; den=0.1*10^-16; end t2=(-Q-sqrt(arg))/(den); p2=2*atan(t2); end ph1(i)=phi/fact; temp=p2/fact; if temp<0; temp=temp+360; end ph2(i)=temp; % Velocity calculations num2=sin(al4)*(cos(phi)*sin(p2)*sin(al2)+... sin(phi)*cos(p2)*cos(al1)*sin(al2)-sin(phi)*sin(al1)*cos(al2)); den2=sin(al2)*(sin(phi)*cos(p2)*sin(al4)+... cos(phi)*sin(p2)*cos(al1)*sin(al4)-sin(p2)*sin(al1)*cos(al4)); phi2dot2=-phi1dot*(num2/den2); phi2d(i)=phi2dot2; % Acceleration calculations A2=sin(al2)*(sin(phi)*cos(p2)*sin(al4)+... cos(phi)*sin(p2)*cos(al1)*sin(al4)-sin(p2)*sin(al1)*cos(al4)); B2=2*sin(al2)*sin(al4)*(sin(phi)*sin(p2)*cos(al1)-cos(phi)*cos(p1)); C2=sin(al4)*(sin(phi)*sin(p2)*sin(al2)-... cos(phi)*cos(p2)*cos(al1)*sin(al2)+cos(phi)*sin(al1)*cos(al2)); D2=sin(al2)*(sin(phi)*sin(p2)*sin(al4)-... cos(phi)*cos(p2)*cos(al1)*sin(al4)+cos(p2)*sin(al1)*cos(al4)); phi2ddot2=(phi1ddot*phi2dot2)/phi1dot+(B2/A2)*phi1dot*phi2dot2+... (C2/A2)*phi1dot^2+(D2/A2)*phi2dot2^2; phi2dd(i)=phi2ddot2; end % plot the results. clf; height=0.5; Start with the velocity curve - 287 - width=0.5; xmin=0; xmax=360; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',10); xlabel('Link 2 rotation, deg','color', 'k', 'fontname',... 'times', 'fontsize', 10); ylabel('Link 4 Angular Velocity, rad/s','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') yminv=min(phi2d); ymaxv=max(phi2d); axis([xmin xmax yminv ymaxv]); % Set up the vectors needed to draw the velocity curve1=line('xdata', , 'ydata', ,... 'linewidth', 1.5, 'erasemode', 'xor', 'color', 'r'); set(curve1,'xdata', ph1, 'ydata' ,phi2d); pause clf % Draw the acceleraton curve axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',10); xlabel('Link 2 rotation, deg','color', 'k', 'fontname',... 'times', 'fontsize', 10); ylabel('Link 4 Angular Acceleration, rad/s^2','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') ymina=min(phi2dd); ymaxa=max(phi2dd); axis([xmin xmax ymina ymaxa]); % Set up the vectors needed to draw the acceleraton curve1=line('xdata', , 'ydata', ,... 'linewidth', 1.5, 'erasemode', 'xor', 'color', 'r'); set(curve1,'xdata', ph1, 'ydata' ,phi2dd); The velocity and acceleration plots areshown in Figs. 7.5.1 and 7.5.2. - 288 - Fig. 7.5.1: Velocity plot for problem 7.5. Fig. 7.5.2: Acceleration plot for problem 7.5. Problem 7.6 Resolve Problem 7.4 if 1 = 150˚ but all other data remain the same. - 289 - Solution We must solve the problem for and R in Eq. (7.3). 1 from 0˚ to 360˚. For each value of 1, we must solve for P, Q, P = - cos 1cos 1sin 2sin 4 + sin 1sin 2cos 4 Q = sin 1sin 2sin 4 R = cos 1sin 1cos 2sin 4 + cos 1cos2cos 4 - cos 3 Hence referring to Table 3.1, we can obtain values for P2 + Q2 R2 RP where = ±1 is a sign variable, and 2 = 2 tan-1(t). t= Q+ 2 (7.3) 1 given from: (7.4) (7.5) The two solutions correspond to the two solutions obtained in the solution of the position problem of a planar four-bar, and have the same source in the reflection of the driven-crank and coupler about the plane of the moving joint axis of the driving-crank and the fixed joint axis of the driven-crank. By checking both solutions for 2 = 90˚ , we can determine that the solution corresponding to the figure in the problem statement is for = -1. We can also develop relationships between the angular velocities about joints 1 and 2 by differentiation of Eq. (7.1). Differentiation of Eq. (7.1) with respect to time gives, after rearrangement: ˙2 = ˙1 sin 4(cos 1 sin 2 sin sin 2(sin 1 cos 2 sin 2 + sin 1 cos 2 cos 1sin 2 4 + cos 1 sin 2 cos 1 sin 4 sin 1 sin 1 cos sin 2 sin 1 cos 2) 4) (7.6) We can simplify this expression using cos and sin 2 2 = cos 3 = cos 4 =0 = sin 3 = sin 4 =1 However, it is relatively easy to write a general MATLAB program to solve Eq. (7.6) directly, so the simplification is not necessary. An expression for the acceleration is given by Eqs. (7.7) and (7.8). These are ˙˙2 = ˙˙ ˙2 B 1 ˙ ˙ C ˙2 D ˙2 ˙1 + A 1 2 + A 1 + A 2 1cos 2sin 4 + cos 1sin 2cos 1sin 4 - sin 2sin 1cos 4) 4(sin 1sin 2cos 1 - cos 1cos 2) 1sin 2sin 2 -cos 1cos 2cos 1sin 2 + cos 1sin 1cos 2) 1sin 2sin 4 - cos 1cos 2cos 1sin 4 + cos 2sin 1cos 4) (7.7) where A = sin 2(sin B = 2sin 2sin C = sin 4(sin D = sin 2(sin (7.8) - 290 - These equations can also be easily programmed using MATLAB. Once the values are computed, they can be plotted as a function of the rotation of link 2. The resulting program is given in the following: % MATLAB program for problem 7.6 %Program clf; clear all; ans='y'; disp(' ') disp(' Universal Joint Analysis Program') %Input the linkage variables alpha1=150 alpha2=90 alpha3=90 alpha4=90 phi1dot=100 phi1ddot=0 fact=pi/180; al1=alpha1*fact; al2=alpha2*fact; al3=alpha3*fact; al4=alpha4*fact; phi1i=0; phi1f=360; phii=phi1i*fact; phif=phi1f*fact; dphi=(phif-phii)/360; i=0 for phi=phii:dphi:phif i=i+1; % Position calculations P=-cos(phi)*cos(al1)*sin(al2)*sin(al4)+sin(al1)*sin(al2)*cos(al4); Q=sin(phi)*sin(al2)*sin(al4); R=cos(phi)*sin(al1)*cos(al2)*sin(al4)+cos(al1)*cos(al2)*cos(al4)cos(al3); arg=P*P+Q*Q-R*R; p1=0; p2=0; if arg>=0 den=R-P; if den==0; den=0.1*10^-16; end t2=(-Q-sqrt(arg))/(den); p2=2*atan(t2); end ph1(i)=phi/fact; temp=p2/fact; if temp<0; temp=temp+360; end ph2(i)=temp; - 291 - % Velocity calculations num2=sin(al4)*(cos(phi)*sin(p2)*sin(al2)+... sin(phi)*cos(p2)*cos(al1)*sin(al2)-sin(phi)*sin(al1)*cos(al2)); den2=sin(al2)*(sin(phi)*cos(p2)*sin(al4)+... cos(phi)*sin(p2)*cos(al1)*sin(al4)-sin(p2)*sin(al1)*cos(al4)); phi2dot2=-phi1dot*(num2/den2); phi2d(i)=phi2dot2; % Acceleration calculations A2=sin(al2)*(sin(phi)*cos(p2)*sin(al4)+... cos(phi)*sin(p2)*cos(al1)*sin(al4)-sin(p2)*sin(al1)*cos(al4)); B2=2*sin(al2)*sin(al4)*(sin(phi)*sin(p2)*cos(al1)-cos(phi)*cos(p1)); C2=sin(al4)*(sin(phi)*sin(p2)*sin(al2)-... cos(phi)*cos(p2)*cos(al1)*sin(al2)+cos(phi)*sin(al1)*cos(al2)); D2=sin(al2)*(sin(phi)*sin(p2)*sin(al4)-... cos(phi)*cos(p2)*cos(al1)*sin(al4)+cos(p2)*sin(al1)*cos(al4)); phi2ddot2=(phi1ddot*phi2dot2)/phi1dot+(B2/A2)*phi1dot*phi2dot2+... (C2/A2)*phi1dot^2+(D2/A2)*phi2dot2^2; phi2dd(i)=phi2ddot2; end % plot the results. Start with the velocity curve clf; height=0.5; width=0.5; xmin=0; xmax=360; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',10); xlabel('Link 2 rotation, deg','color', 'k', 'fontname',... 'times', 'fontsize', 10); ylabel('Link 4 Angular Velocity, rad/s','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') yminv=min(phi2d); ymaxv=max(phi2d); axis([xmin xmax yminv ymaxv]); % Set up the vectors needed to draw the velocity curve1=line('xdata', , 'ydata', ,... 'linewidth', 1.5, 'erasemode', 'xor', 'color', 'r'); set(curve1,'xdata', ph1, 'ydata' ,phi2d); pause clf - 292 - % Draw the acceleraton curve axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',10); xlabel('Link 2 rotation, deg','color', 'k', 'fontname',... 'times', 'fontsize', 10); ylabel('Link 4 Angular Acceleration, rad/s^2','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') ymina=min(phi2dd); ymaxa=max(phi2dd); axis([xmin xmax ymina ymaxa]); % Set up the vectors needed to draw the acceleraton curve1=line('xdata', , 'ydata', ,... 'linewidth', 1.5, 'erasemode', 'xor', 'color', 'r'); set(curve1,'xdata', ph1, 'ydata' ,phi2dd); The velocity and acceleration plots areshown in Figs. 7.6.1 and 7.6.2. Fig. 7.6.1: Velocity plot for problem 7.6. - 293 - Fig. 7.5.2: Acceleration plot for problem 7.5. Problem 7.7 The mechanism shown is used for a steering linkage for an automobile. The wheel base is 110 in, and link OFF is driven by the steering column. The toe-in angle ( ) is 9˚. If the link dimensions are given as shown, determine the y error in the Ackermann steering condition (see Figs. 7.23 and 7.26) for a 10˚ CCW rotation of O FF. Recall that the linkage OEEFOF is a parallelogram. OB D B F OF 11 in h C OA EA OE O BOA = 50 in OBB = O AA = 3 in OF F = O EE = 3 in BD = AC = 12 in DC = 26 in OF OE = 28 in h = 8 in Solution The linkage can be analyzed by first considering the parallelogram linkage OFFEOE and then analyzing the two dyads acting in parallel. Because loop O FFEOE is a parallelogram, we can calculate the positions of points D and C as shown in Fig. 7.7.1. Let, r = OFOE = FE = 28 in 1 r2 = OFF = OE E = 5 in r3 = FD = EC = 1 in - 294 - OB y OA 30 in 28 in A F D θ 1 in B E 8 in C θ OF x OE 11 in Fig. 7.7.1: Dimensions for linkage When the wheels are in the neutral position, = 100˚ . Then, xD = r2 cos + r3 = 5cos(100) + 1 = 0.1317 in yD = r2 sin = 5sin(100) = 4.9240 in and = 90˚ . Therefore, when OFF rotates 10˚ CCW, xC = r + r2 cos r3 = 30 + 5cos(100) 1 = 28.1317 in 1 yC = r2 sin = 5sin(100) = 4.9240 in The coordinates of OB and OA are xOB = 11 yOB = 8 and xOA = 11 + 50 = 39 yOA = 8 We can now analyze the dyads. Let the left hand dyad be as shown in Fig. 7.7.2. Then, rOB/D = d1 + d2 or xOB yOB x D = d1 cos 1 + d2 cos 2 = yD = d1sin 1 + d2 sin 2 = 1, = 11 0.1317 = 11.1317 in y = 8 4.9240 = 3.0760 in x 2 To solve for isolate the terms involving on one side of the equation. Then, - 295 - β3 OB d2 β2 B d1 β1 D α1 Fig. 7.7.2: Left hand dyad d2 cos d2 sin = 2= 2 x y d1 cos 1 d1 sin 1 (1) Now square both sides of both equations and add. The result, after simplifying, is 0= 2 x + 2 y 2 + d1 2 d2 2 x d1 cos 1 2 y d1sin 1 Collecting terms and using Eqs. (3.33) and (3.34), a = 2 x d1 = 2( 11.1317)(12) = 267.1622 b = 2 yd1 = 2(3.0760)(12) = 73.8231 c = 2x + 2 + d12 d22 = ( 11.1317)2 + (3.0760)2 + (12)2 (3)2 = 268.3776 y 2 2 2 t = b ± b2 c2 + a2 = 73.8231 + ( 73.8231) (268.3776) + (267.1622) = 117.7391 ca 268.3776 267.1622 In the expression for t , we use the plus sign on the square root. This corresponds to to configuration shown in Fig. 7.7.2. The corresponding value for 1 is 1 = 2tan 1 t = 2tan 1(117.7391) = 179.0268˚ From Eq. (1), tan The angle 2 2 = 2 y x d1 sin d1 cos 1 = 3.0760 12sin(179.0268˚) = 2.8722 = 3.3147 11.1317 12cos(179.0268˚) 0.8665 1 2 is in the first quadrant. Therefore, is given by = tan 1(3.3147) = 73.212˚ - 296 - The wheel stub shaft angle 3= 2 3 shown in Fig. 7.7.2 is given by + 270˚ 1 = 73.212˚+270˚ 9˚= 334.212˚ Then, the equation for the centerline of the wheel is given by y = yOB + (x xOB )tan 3 (2) The right-hand dyad is shown in Fig. 7.7.3. From that figure, rOA /C = d3 + d4 or xOA yOA xC = d3 cos 4 + d4 cos 5 = x = 39 26.1318 = 12.8682 yC = d3 sin 4 + d4 sin 5 = y = yOA yC = 8 4.9240 = 3.076 4, To solve for isolate the terms involving 5 on one side of the equation. Then, β6 d4 OA β5 A d3 C α2 β4 Fig. 7.7.3: Right-hand dyad d4 cos d4 sin 5= 5= x y d3 cos d3 sin 4 4 (1) Now square both sides of both equations and add. The result, after simplifying, is 0= 2 x+ 2 y 2 + d3 2 d4 2 x d3 cos 4 2 y d3 sin 4 Collecting terms and using Eqs. (3.33) and (3.34), e = 2 x d3 = 2(12.8682)(12) = 308.8378 f = 2 yd3 = 2(3.076)(12) = 73.8231 2 2 g = x + y + d32 d2 = (12.8682)2 + (3.076)2 + (12)2 (3)2 = 310.0532 4 f f 2 g2 + e2 73.8231 + ( 73.8231)2 (310.0532)2 + ( 308.8378)2 t= = = 0.0085 ge 310.0532 + 308.8378 - 297 - In the expression for t , we use the plus sign on the square root. This corresponds to to configuration shown in Fig. 7.7.3. The corresponding value for 4 is 4 = 2tan 1 t = 2tan 1(0.0085) = 0.9783˚ Then from Eq. (1), tan The angle 5 5= y x d3 sin d3 cos 4 4 = 3.076 12sin0.9783˚ = 2.8711 = 3.3003 12.8682 12cos0.9783˚ 0.8699 5 5 is in the first quadrant. Therefore, is given by = tan 1(3.3003) = 73.143˚ The angle for the wheel stub shaft is given by 6 = 5 + 270˚+ 2 = 73.143˚+270˚+9˚= 352.143˚ Then, the equation for the centerline of the wheel is given by y = yOA + (x xOA)tan 6 (2) The intersection of the two axes can be found by solving Eqs. (1) and (2) simulaneously. Then, yOA + ( x xOA)tan Solving for x, x= yOB yOA + xOA tan 6 xOB tan tan 6 tan 3 3 6 = yOB + (x xOB )tan 3 = 8 8 + 39tan(352.143˚) + 11tan(334.212) = 30.9923 tan(352.143 ) tan(334.212) ˚ Backsubstituting into Eq. (2) will give the corresponding value for y. Then, y = yOA + (x xOA)tan 6 = 8 + ( 30.9923 39)tan(352.143˚) = 17.659 in In the ideal case, the value for y will be given by yi = wheel base + 8 = 110 + 8 = 102 in Therefore, the error in the linkage is given by y = yi Problem 7.8 Write a computer program to analyze the steering linkage shown in Problem 7.7. If only h can change, determine the optimum value for h that will give the least error in y for the Ackermann steering condition for a ± 15˚ rotation of OFF. y = 102 17.659 = 119.659 in - 298 - Solution The linkage geometry is shown in Fig. 7.8.1. Assume that the wheel base is the same (110 in) as that given in Problem 7.7. OB D B F OF w h C OA EA OE w O BOA = 50 in OBB = O AA = 3 in BD = AC = 12 in DC =26 in OBO A = 30 in Fig. 7.8.1: Basic linkage for Problem 7.8. The independent variables which must be determined are: (toe-in angle), DC, and h. We will first generate the analysis equations and then conduct the optimization. First let r = OFOE = FE 1 r2 = OFF = OE E r3 = FD = CE = (FE DC) / 2 w = (OAOB OFOE ) / 2 The linkage can be analyzed by first considering the parallelogram linkage OFFEOE and then analyzing the two dyads acting in parallel. Because loop O FFEOE is a parallelogram, we can develop equations for the positions of points D and C as shown in Fig. 7.8.2. When the wheels are in the neutral position, = 90˚ . Therefore, when O FF rotates ±15˚, then, 105˚ . The points on the parallogram can be computed from, 75˚ xD = r2 cos + r3 yD = r2 sin and xC = r + r2 cos 1 yC = r2 sin r3 - 299 - OB y OA B F D E h C OE A θ OF θ x w Fig. 7.8.2: Dimensions for linkage The coordinates of OB and OA are xOB = w = (OBOA yOB = h and xOA = OAOB w yOA = h We can now analyze the dyads. Let the left hand dyad be as shown in Fig. 7.8.3. Then, rOB/D = d1 + d2 or xOB yOB x D = d1 cos 1 + d2 cos 2 = yD = d1sin 1 + d2 sin 2 = 1, x y 2 FE) / 2 To solve for d2 cos d2 sin 2 isolate the terms involving x y on one side of the equation. Then, (1) = 2= d1 cos 1 d1 sin 1 Now square both sides of both equations and add. The result, after simplifying, is 0= 2 x + 2 y 2 + d1 2 d2 2 x d1 cos 1 2 y d1sin 1 - 300 - β3 OB d2 β2 B d1 β1 D α1 Fig. 7.8.3: Left hand dyad Collecting terms and using Eqs. (3.33) and (3.34), a = 2 x d1 b = 2 yd1 c = 2x + 2 + d12 d22 y b + b2 c2 + a2 t= ca In the expression for t , we use the plus sign on the square root. This corresponds to to configuration shown in Fig. 7.8.3. The corresponding value for 1 is 1 = 2tan 1 t From Eq. (1), tan The angle 2 2 = 2 y x d1 sin d1 cos 1 1 2 is in the first quadrant. Therefore, 1 y x is given by = tan d1sin d1 cos 1 1 3 The wheel stub shaft angle 3= 2 shown in Fig. 7.8.3 is given by + 270˚ Then, the equation for the centerline of the wheel is given by - 301 - y = yOB + (x xOB )tan 3 (2) The right-hand dyad is shown in Fig. 7.8.4. From that figure, rOA /C = d3 + d4 or xOA yOA xC = d3 cos 4 + d4 cos 5 = x yC = d3 sin 4 + d4 sin 5 = y 4, To solve for isolate the terms involving 5 on one side of the equation. Then, β6 d4 OA β5 A d3 C α2 β4 Fig. 7.8.4: Right-hand dyad d4 cos d4 sin 5= 5= x y d3 cos d3 sin 4 4 (3) Now square both sides of both equations and add. The result, after simplifying, is 0= 2 x+ 2 y 2 + d3 2 d4 2 x d3 cos 4 2 y d3 sin 4 Collecting terms and using Eqs. (3.33) and (3.34), e = 2 x d3 f = 2 yd3 2 2 g = x + y + d32 d2 4 f f 2 g2 + e2 t= ge In the expression for t , we use the plus sign on the square root. This corresponds to to configuration shown in Fig. 7.8.4. The corresponding value for 4 is 4 = 2tan 1 t Then from Eq. (3), - 302 - tan 5= y x d3 sin d3 cos 4 4 The angle for the wheel stub shaft is given by 6 = 5 + 270˚+ Then, the equation for the centerline of the wheel is given by y = yOA + (x xOA)tan 6 (4) The intersection of the two axes can be found by solving Eqs. (3) and (4) simulaneously. Then, yOA + ( x xOA)tan Solving for x, x= yOB yOA + xOA tan 6 xOB tan tan 6 tan 3 3 6 = yOB + (x xOB )tan 3 Backsubstituting into Eq. (2) will give the corresponding value for y. Then, y = yOA + (x xOA)tan 6 In the ideal case, the value for y will be given by yi = wheel base + h Therefore, the error in the linkage is given by y = yi y To optimize the system, we need to determine the value for h that will minimize y We will do this using MATLAB and a simple grid search. For this we will vary h between reasonable limits using equal increments. Once the optimum location is determined approximately, we can refine the grid and redo the grid search with smaller bounds for each of the design variables if desired. For the search, we will use the limits 5< h <10 and 1000 equal increments within this range. For each interation of h, we will increment between 75˚ and 105 and identify the maximum error in the range. The objective will be to minimize the maximum error for all values of and h. The equations which must be solved for each iteration are summarized in the following. r = OFOE = FE 1 r2 = OFF = OE E r3 = FD = CE = (FE DC) / 2 w = (OAOB OFOE ) / 2 xD = r2 cos + r3 yD = r2 sin - 303 - xC = r + r2 cos 1 yC = r2 sin r3 FE) / 2 xOB = w = (OBOA yOB = h xOA = OAOB w yOA = h a = 2 x d1 b = 2 yd1 c = 2x + 2 + d12 d22 y 2 2 2 t = b+ b c +a ca 1 = 2tan 1 t 2 tan = 1 y x d1 sin d1 cos y x 1 1 1 1 2 = tan 2 d1sin d1 cos 3= + 270˚ e = 2 x d3 f = 2 yd3 2 2 g = x + y + d32 d2 4 f f 2 g2 + e2 t= ge 4 = 2tan 1 t 5= y x tan x= d3 sin d3 cos 4 4 3 yOB yOA + xOA tan 6 xOB tan tan 6 tan 3 6 y = yOA + (x xOA)tan yi = wheel base + h y = yi y The MATLAB program for conducting the simple optimization is summarized in the following. - 304 - % MATLAB program for problem 7.8 %Program clf; clear all; ans='y'; disp(' ') disp(' Akermann Steering Linkage Analysis Program') %Input the linkage variables DC=28; FE=28 r3=(FE-DC)/2; OBOA=50 w=(OBOA-FE)/2 OBB=3 OFF=8 DB=12 WB=110 emax=0 alphas=0 emin=10^10 h1=5 h2=10 dh=(h2-h1)/1000 hs=0; r1=FE; r2=5; d1=12; d2=3; d3=12; d4=3; xOB=-w; xOA=OBOA-w; r3=(FE-DC)/2; alp=9; for h=h1:dh:h2 i=0; emax=0; flag1=0; for theta=75:5:105 i=i+1; alpha=alp; fact=pi/180; td=theta*fact; ad1=alpha*fact; xd=r2*cos(td)+r3; yd=r2*sin(td); xc=r1+r2*cos(td)-r3; yc=yd; - 305 - yOB=h; yOA=h; % Analyze the first dyad. deltax=xOB-xd; deltay=yOB-yd; a=-2*deltax*d1; b=-2*deltay*d1; c=deltax^2+deltay^2+d1^2-d2^2; arg1=b*b-c*c+a*a; if arg1<0; flag1=1; end if c==a; c=1.0000000001*a; end t=(-b+sqrt(arg1))/(c-a); beta1=2*atan(t); beta1d=beta1/fact; beta2=atan2(deltay-d1*sin(beta1), deltax-d1*cos(beta1)); beta2d=beta2/fact; beta3=beta2+(3*pi)/2-ad1; beta3d=beta3/fact; % Analyze the second dyad deltax=xOA-xc; deltay=yOA-yc; a=-2*deltax*d3; b=-2*deltay*d3; c=deltax^2+deltay^2+d3^2-d4^2; arg1=b*b-c*c+a*a; if arg1<0; flag1=1; end if c==a; c=1.0000000001*a; end t=(-b-sqrt(arg1))/(c-a); beta4=2*atan(t); betad4=beta4/fact; beta5=atan2(deltay-d3*sin(beta4), deltax-d3*cos(beta4)); betad5=beta5/fact; beta6=beta5+(3*pi)/2+ad1; if beta6>2*pi; beta6=beta6-2*pi; end betad6=beta6/fact; if beta6==beta3; beta6=1.0000000001*beta3; end x=(yOB-yOA+xOA*tan(beta6)-xOB*tan(beta3))/(tan(beta6)-tan(beta3)); y=yOA+(x-xOA)*tan(beta6); yi=-WB+h; % Find the maximum error for the range of theta error=yi-y; if abs(error)>emax & flag1==0 emax=abs(error); end end % Determine if the maximum error is less than emin. if abs(emax)<emin & flag1==0 - 306 - emin=abs(emax); hsave=h; end end emin h=hsave % plot the results. i=0; for theta=75:1:105 i=i+1; alpha=alp; fact=pi/180; td=theta*fact; ad1=alpha*fact; xd=r2*cos(td)+r3; yd=r2*sin(td); xc=r1+r2*cos(td)-r3; yc=yd; xOB=-11; yOB=8; xOA=39; yOA=8; % Analyze the first dyad. deltax=xOB-xd; deltay=yOB-yd; a=-2*deltax*d1; b=-2*deltay*d1; c=deltax^2+deltay^2+d1^2-d2^2; arg1=b*b-c*c+a*a; t=(-b+sqrt(arg1))/(c-a); beta1=2*atan(t); beta1d=beta1/fact; beta2=atan2(deltay-d1*sin(beta1), deltax-d1*cos(beta1)); beta2d=beta2/fact; beta3=beta2+(3*pi)/2-ad1; beta3d=beta3/fact; % Analyze the second dyad deltax=xOA-xc; deltay=yOA-yc; a=-2*deltax*d3; b=-2*deltay*d3; c=deltax^2+deltay^2+d3^2-d4^2; arg1=b*b-c*c+a*a; t=(-b-sqrt(arg1))/(c-a); beta4=2*atan(t); Start with the velocity curve - 307 - betad4=beta4/fact; beta5=atan2(deltay-d3*sin(beta4), deltax-d3*cos(beta4)); betad5=beta5/fact; beta6=beta5+(3*pi)/2+ad1; if beta6>2*pi; beta6=beta6-2*pi; end betad6=beta6/fact; if beta6==beta3; beta6=1.0000000001*beta3; end x=(yOB-yOA+xOA*tan(beta6)-xOB*tan(beta3))/(tan(beta6)-tan(beta3)); y=yOA+(x-xOA)*tan(beta6); yi=-WB+h; error=yi-y; % Store position information in matrices. xx(i)=x; yy(i)=y; thet(i)=theta; err(i)=error; end % plot the results height=0.3; width=0.35; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',8); xlabel('theta (deg)','color', 'k', 'fontname',... 'times', 'fontsize', 8); ylabel('x, y, err (in)','color', 'k', 'fontname',... 'times', 'fontsize', 8); set(gcf,'color', 'w') xmin=75; xmax=105; ymin=min([xx, yy, err]); ymax=max([xx, yy, err]); axis([xmin xmax ymin ymax]); % Set up the vectors needed to draw the linkage curvex = line('xdata', thet, 'ydata', xx,... 'linewidth', 1, 'erasemode', 'xor', 'color', 'b', 'linestyle', ''); hold on; curvey = line('xdata', thet, 'ydata', yy,... 'linewidth', 1, 'erasemode', 'xor', 'color', 'r', 'linestyle', '.'); hold on; curvez = line('xdata', thet, 'ydata', err,... 'linewidth', 1, 'erasemode', 'xor', 'color', 'k', 'linestyle', '.'); legend([curvex, curvey, curvez],'x','y', 'err'); The result from the interation is that the optimum h = 7.81 in. A plot of x, y, and the error is given in Fig. 7.8.5. - 308 - Fig. 7.8.5: Plot of x, y, and error for optimum linkage Problem 7.9 In the rack and pinion mechanism shown in Fig. 7.25, the wheel base is 125 in. If the link dimensions are as given below, plot the y error in the Ackermann steering condition as a function of the displacement u (see Fig. 7.23) for a ±1.5-in displacement of u. p = 55 in a = 3.5 in = 10˚ Solution For the solution, we can follow the procedure given in Example 7.5. We will plot the x and y coordinates of the intersection of the front wheel axes for small increments of the rack displacement, u , in the range -1.5 < u 1.5 in, where the reference frame has its origin at the middle of the rear axle, as shown. The x coordinate can be interpreted as the radius of curvature of the path followed by the vehicle, and the y coordinate is the error from perfect Ackermann geometry. The linkage can be analyzed as two slider crank linkages acting in parallel with a common input, u, applied to the sliders. Figure 7.25 is redrawn as Fig. 7.9.1 below. For the right hand linkage, we can resolve the vector components in the x and y directions, respectively. a cos + b cos µ = r + u a sin + b sin µ = s where µ is the tie rod angle as shown in Fig. 7.9.1. Similarly, for the left side a cos + b cos = r - u (7.15) (7.13) (7.14) b = 12 in r = 11in s = 6.0 in - 309 - y δ aφ b s p θa b µ γ ν u α r u r q i x O Fig. 7.9.1: The rack-and-pinion steering linkage geometry analyzed in Problem 7.9 a sin + b sin = s (7.16) µ may be eliminated from Eqs. (7.13) and (7.14) by segregating the µ terms on one side of each equation, squaring both sides of both equations, and adding to give b2 = (r + u - a cos )2 + (s - a sin )2 or b2 = r 2 + s2 + a2 + u2 + 2 r u - 2 a u cos This equation has the form P cos + Q sin where P = 2 a (u + r) Q=2as R = b2 - a2 - r2 - s2 - u2 - 2 r u +R=0 (7.18) - 2 a r cos - 2 a s sin (7.17) (7.19) Hence the standard solution of Table 3.1 may be applied to obtain values of corresponding to given values of u . Two values of are obtained for each value of u , one positive and one negative. Only the negative value is consistent with the configuration shown in Fig. 7.9.1, so the positive values are discarded. - 310 - Similarly, elimination of from Eqs. (7.15) and (7.16) gives b2 = (r - u - a cos )2 + (s - a sin )2 or b2 = r 2 + s2 + a2 + u2 - 2 r u + 2 a u cos This equation has the form P' cos + Q' sin + R' = 0 where P' = 2 a (r - u) Q=2as R = b2 - a2 - r2 - s2 - u2 + 2 r u - 2 a r cos - 2 a s sin (7.20) (7.21) (7.22) for which the solution is also given by Table 3.1. Values of for incremental values of u throughout the specified range can be calculated. As was the case for , two values of are obtained for each value of u , one positive and one negative. Only the negative solution is consistent with the configuration drawn in Fig. 7.9.1, so the positive solutions are discarded. Now = / 2 , and = + / 2 , where and are the steering angles of the inner and outer front wheels, as shown in Fig. 7.9.1, and values of and may now be calculated. The resulting values of and throughout the range of values of u are listed in Table 7.9.1. Also, and determine the location of the intersection, i, of the axes of the wheels: tan = qy qy , tan = x p2 x+ p 2 (7.23) Hence, (x - p/2) tan = (x + p/2) tan which, when solved for x, gives x= p tan + tan tan 2 tan (7.24) Substitution for x into either of Eqs. (7.23) allows solution for y to give y=q p tan tan tan tan The analysis equations for this problem can be easily solved using MATLAB. A program for doing this is given in the following. % MATLAB program for Problem 7.9 clf; clear all; q=125; p=55; - 311 - a=3.5; alpha=10; alphar=alpha*pi/180; b=12; r=11; s=6; % Loop to analyze the joint du=0.1; i=0; for u = -1.5:du:1.5; i=i+1; ut=u; P=2*a*(u+r); Q=2*a*s; R=b*b-a*a-r*r-s*s-u*u-2*r*u; theta=2*atan((-Q-sqrt(Q*Q-R*R+P*P))/(R-P)) A=2*a*(r-u); B=2*a*s; C=b*b-a*a-r*r-s*s-u*u+2*r*u; phi=2*atan((-B-sqrt(B*B-C*C+A*A))/(C-A)); gamma=pi/2-theta-alphar; delta=phi+alphar-pi/2; tg=tan(gamma); td=tan(delta); x=(p/2)*(tg+td)/(tg-td); y=q-p*tg*td/(tg-td); % Store position information in matrices. xx(i)=x; yy(i)=y; uu(i)=u; fprintf('%10.3f %10.3f %10.3f %10.3f %10.3f\n',u, delta, gamma, x,y) end % plot the results clf; height=0.4; width=0.5; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',10); xlabel('u (in)','color', 'k', 'fontname',... 'times', 'fontsize', 10); ylabel('x, y (in)','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') xmin=-1.75; xmax=1.75; ymin=min([xx,yy]); ymax=max([xx,yy]); axis([xmin xmax ymin ymax]); % Set up the vectors needed to draw the linkage ; - 312 - curvex = line('xdata', uu, 'ydata', xx,... linewidth', 1, 'erasemode', 'xor', 'color', 'b', 'linestyle', '' '); hold on; curvey = line('xdata', uu, 'ydata', yy,... 'linewidth', 1, 'erasemode', 'xor', 'color', 'r', 'linestyle', '.'); legend([curvex, curvey],'x','y'); The results are plotted in Fig. 7.9.2 for the given values of the design parameters (increment for u is 0.1). A comparison of these results with those in Example 7.5 will indicate that the y error for the dimensions in this problem are quite large. It is therefore important to choose the dimensions for the Ackermann drive carefully. Table 7.9.1: Numerical values obtained by solution of Problem 7.9 u -1.500 -1.400 -1.300 -1.200 -1.100 -1.000 -0.900 -0.800 -0.700 -0.600 -0.500 -0.400 -0.300 -0.200 -0.100 0.000 0.100 0.200 0.300 0.400 0.500 0.600 0.700 0.800 0.900 1.000 1.100 1.200 1.300 1.400 1.500 gamma -0.051 -0.020 0.010 0.040 0.069 0.099 0.128 0.156 0.185 0.213 0.242 0.270 0.298 0.326 0.354 0.382 0.410 0.438 0.466 0.494 0.522 0.550 0.578 0.606 0.634 0.663 0.692 0.720 0.749 0.778 0.808 delta -0.808 -0.778 -0.749 -0.720 -0.692 -0.663 -0.634 -0.606 -0.578 -0.550 -0.522 -0.494 -0.466 -0.438 -0.410 -0.382 -0.354 -0.326 -0.298 -0.270 -0.242 -0.213 -0.185 -0.156 -0.128 -0.099 -0.069 -0.040 -0.010 0.020 0.051 x 30.319 28.656 26.919 25.112 23.242 21.314 19.333 17.304 15.234 13.126 10.987 8.821 6.634 4.431 2.218 -0.000 -2.218 -4.431 -6.634 -8.821 -10.987 -13.126 -15.234 -17.304 -19.333 -21.314 -23.242 -25.112 -26.919 -28.656 -30.319 y 127.949 126.140 124.459 122.904 121.475 120.171 118.990 117.934 117.003 116.195 115.511 114.952 114.516 114.205 114.019 113.957 114.019 114.205 114.516 114.952 115.511 116.195 117.003 117.934 118.990 120.171 121.475 122.904 124.459 126.140 127.949 - 313 - Fig. 7.9.2: The coordinates of the intersection of the front wheel axes, i , plotted against the rack displacement, u . x approximates the radius of curvature of the vehicle's path, and y is the error in location of the intersection relative to the rear axle axis. That is, y is the deviation from the Ackermann condition. Note the size of the error compared to that shown in Fig. 7.26 in the text. Problem 7.10 A new subcompact automobile is being designed for rack and pinion steering. Assume that the wheel base is 90 in. Determine the other dimensions such that the error in the Ackermann steering condition is as small as possible for a ±1.5-in displacement of the rack. Solution For the solution, we can follow the procedure given in Example 7.5 to derive the analysis equations. We will determine the values for the x and y coordinates of the intersection of the front wheel axes for small increments of the rack displacement, u, in the range -1.5 < u 1.5 in, where the reference frame has its origin at the middle of the rear axle, as shown. The x coordinate can be interpreted as the radius of curvature of the path followed by the vehicle, and the y coordinate is the error from perfect Ackermann geometry. The linkage can be analyzed as two slider crank linkages acting in parallel with a common input, u, applied to the sliders. Figure 7.25 is redrawn as Fig. 7.10.1 below. For the right hand linkage, we can resolve the vector components in the x and y directions, respectively. a cos + b cos µ = r + u a sin + b sin µ = s where µ is the tie rod angle as shown in Fig. 7.9.1. Similarly, for the left side a cos + b cos = r - u (7.15) (7.13) (7.14) - 314 - y δ aφ b s p θa b µ γ ν u α r u r q i x O Fig. 7.10.1: The rack-and-pinion steering linkage geometry analyzed in Problem 7.9 a sin + b sin = s (7.16) µ may be eliminated from Eqs. (7.13) and (7.14) by segregating the µ terms on one side of each equation, squaring both sides of both equations, and adding to give b2 = (r + u - a cos )2 + (s - a sin )2 or b2 = r 2 + s2 + a2 + u2 + 2 r u - 2 a u cos This equation has the form P cos + Q sin where P = 2 a (u + r) Q=2as R = b2 - a2 - r2 - s2 - u2 - 2 r u +R=0 (7.18) - 2 a r cos - 2 a s sin (7.17) (7.19) Hence the standard solution of Table 3.1 may be applied to obtain values of corresponding to given values of u . Two values of are obtained for each value of u , one positive and one negative. Only the negative value is consistent with the configuration shown in Fig. 7.9.1, so the positive values are discarded. - 315 - Similarly, elimination of from Eqs. (7.15) and (7.16) gives b2 = (r - u - a cos )2 + (s - a sin )2 or b2 = r 2 + s2 + a2 + u2 - 2 r u + 2 a u cos This equation has the form P' cos + Q' sin + R' = 0 where P' = 2 a (r - u) Q=2as R = b2 - a2 - r2 - s2 - u2 + 2 r u - 2 a r cos - 2 a s sin (7.20) (7.21) (7.22) for which the solution is also given by Table 3.1. Values of for incremental values of u throughout the specified range can be calculated. As was the case for , two values of are obtained for each value of u , one positive and one negative. Only the negative solution is consistent with the configuration drawn in Fig. 7.9.1, so the positive solutions are discarded. Now = / 2 , and = + / 2 , where and are the steering angles of the inner and outer front wheels, as shown in Fig. 7.9.1, and values of and may now be calculated. Also, and determine the location of the intersection, i, of the axes of the wheels: tan = qy qy , tan = x p2 x+ p 2 (7.23) Hence, (x - p/2) tan = (x + p/2) tan which, when solved for x, gives x= p tan + tan tan 2 tan (7.24) Substitution for x into either of Eqs. (7.23) allows solution for y to give y=q p tan tan tan tan We now have the equations necessary to analyze the error for any choice of values for p, b, a, r, , and s. The design procedure will be to increment the design variables between reasonable limits and to compute the error in y for -1.5 < u 1.5 in for each choice of the values for the design variables. For each interation, we will increment u between -1.5 and 1.5 and identify the maximum error in the range. The objective will be to minimize the maximum error for all values of the design varialbes.. This is tedius to do by hand, but relatively easy to do using MATLAB. For the analysis, we can start with some relatively large limits for the individual design variables and then reduce the - 316 - limits as the solution location is identified. To begin the iterations, assume that limits for the design variables are 45 < p 5<r 50 12 8<b 0< 12 12˚ 2.5 < a 3<s 3.5 8 A MATLAB program for making the calculations is given in the following. In this program, the limits have been reduced to the following in order to refine the calculations. 45 < p 10 < r 49 12 8<b 6˚ < 12 10˚ 2.5 < a 6<s 3.0 8 % MATLAB program for Problem 7.10 clf; clear all; q=90; % set the limits for the design variables p1=46; p2=49; dp=(p2-p1)/5; a1=2.5; a2=3.0; da=(a2-a1)/5; alpha1=6; alpha2=10; dalpha=(alpha2-alpha1)/5; b1=10; b2=12; db=(b2-b1)/3; r1=6; r2=10; dr=(r2-r1)/3; s1=6; s2=8; ds=(s2-s1)/3; error=10^10; for p=p1:dp:p2 for a=a1:da:a2 for alpha=alpha1:dalpha:alpha2 alphar=alpha*pi/180; for b=b1:db:b2 for r=r1:dr:r2 for s=s1:ds:s2 % Loop to analyze the motion - 317 - du=(3)/10; i=0; err=0; for u = -1.5:du:1.5; i=i+1; ut=u; P=2*a*(u+r); Q=2*a*s; R=b*b-a*a-r*r-s*s-u*u-2*r*u; theta=2*atan((-Q-sqrt(Q*Q-R*R+P*P))/(R-P)) A=2*a*(r-u); B=2*a*s; C=b*b-a*a-r*r-s*s-u*u+2*r*u; phi=2*atan((-B-sqrt(B*B-C*C+A*A))/(C-A)); gamma=pi/2-theta-alphar; delta=phi+alphar-pi/2; tg=tan(gamma); td=tan(delta); x=(p/2)*(tg+td)/(tg-td); y=q-p*tg*td/(tg-td); % Store position information in matrices. xx(i)=x; yy(i)=y; uu(i)=u; % Compute the maximum error for the given iteration if abs(y)>err; err=abs(y); end end % If the maximum error is less than the current value, store the maximum error if abs(err)<error; error=abs(err); psave=p; ssave=s; alphasave=alpha; rsave=r; bsave=b; asave=a; end end end end end end end fprintf('%10.3f %10.3f %10.3f %10.3f %10.3f %10.3f %10.3f %10.3f\n',... error, psave, ssave, alphasave, rsave, bsave, asave) % Analyze the linkage for the optimum values ; - 318 - p=psave; s=ssave; alpha=alphasave; r=rsave; b=bsave; a=asave; du=0.1; i=0; for u = -1.5:du:1.5; i=i+1; ut=u; P=2*a*(u+r); Q=2*a*s; R=b*b-a*a-r*r-s*s-u*u-2*r*u; theta=2*atan((-Q-sqrt(Q*Q-R*R+P*P))/(R-P)) A=2*a*(r-u); B=2*a*s; C=b*b-a*a-r*r-s*s-u*u+2*r*u; phi=2*atan((-B-sqrt(B*B-C*C+A*A))/(C-A)); gamma=pi/2-theta-alphar; delta=phi+alphar-pi/2; tg=tan(gamma); td=tan(delta); x=(p/2)*(tg+td)/(tg-td); y=q-p*tg*td/(tg-td); % Store position information in matrices. xx(i)=x; yy(i)=y; uu(i)=u; end % plot the results clf; height=0.4; width=0.5; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',10); xlabel('u (in)','color', 'k', 'fontname',... 'times', 'fontsize', 10); ylabel('x, y (in)','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') xmin=-1.75; xmax=1.75; ymin=min([xx,yy]); ymax=max([xx,yy]); axis([xmin xmax ymin ymax]); % Set up the vectors needed to draw the linkage curvex = line('xdata', uu, 'ydata', xx,... ; - 319 - linewidth', 1, 'erasemode', 'xor', 'color', 'b', 'linestyle', '' '); hold on; curvey = line('xdata', uu, 'ydata', yy,... 'linewidth', 1, 'erasemode', 'xor', 'color', 'r', 'linestyle', '.'); legend([curvex, curvey],'x','y'); The results from the analysis are: p = 47.8 in a = 2.8 in = 9.2˚ b = 11.33 in r = 7.33 in s = 6.67 in The error in y is plotted in Fig. 7.10.2 for the given values of the design variables. A comparison of these results with those in Example 7.5 will indicate that the y error is quite small for all values of u. Fig. 7.10.2: The coordinates of the intersection of the front wheel axes, i , plotted against the rack displacement, u. x approximates the radius of curvature of the vehicle's path, and y is the error in location of the intersection relative to the rear axle axis. That is, y is the deviation from the Ackermann condition. - 320 - Problem 7.11 The center distance between the driver and follower of a Geneva mechanism is to be 3 in. The driver is to rotate five revolutions for each rotation of the follower. The driving pin is to enter the slot tangentially so that there will be no impact load. Do the following: 1) Design the Geneva mechanism and draw it. 2) Determine the angular velocity and acceleration of the Geneva wheel for one fifth of a revolution if the angular velocity of the driver is 100 rpm CCW. Plot the results. Solution The mechanism must have five stations (N=5). The angle = 360˚/2 N = 360˚/(2 5) = 36˚ The radius to the center of the pin is given by r = C sin = 3sin36˚= 1.763 in where C is the center distance between the driver and the follower of the Geneva mechanism. Also from Fig. 7.31, the radius of the pin at the lowest point on the slot is given by L = C(1 sin ) = 3(1 sin36˚) = 1.237 in The resulting system is shown in Fig. 7.11.1. The position, velocity, and acceleration of the Geneva mechanism are given by Eqs. (7.30), (7.31), and (7.32). The equations are in Fig. 7.31 is given by (1 sinin sinos ) s c cos sin ˙ = ˙ sin ( 1 + sin 2 sin = tan 1 2 (7.30) cos ) (7.31) (7.32) ˙˙ = ˙2 sin cos2 sin (1 + sin2 2sin cos )2 In this problem, = 100 rpm = 100(2 ) / 60 = 10.47 rad / s . The calculations can be done using MATLAB. A program for making the calculations is given in the following. The results are plotted in Fig. 7.11.2. In Fig. 7.11.2, the velocity and acceleration curves are normalized by dividing the velocity and acceleration values by the maximum velocity and maximum acceleration, respectively. % MATLAB program for Problem 7.11 clf; clear all; - 321 - % Loop to analyze the joint i=0; alpha=36; alpr=alpha*pi/180; thdot=100*2*pi/60; Lxt=-30; Lyt2=-0.5; Lyt3=-0.7; for theta = -36:1:36; i=i+1; th=theta*pi/180; phi=atan(sin(alpr)*sin(th))/(1-sin(alpr)*cos(th)); dphi=thdot*sin(alpr)*(cos(th)-sin(alpr))/(1+(sin(alpr))^2-... 2*sin(alpr)*cos(th)); ddphi=-thdot^2*(sin(alpr)*cos(alpr)^2*sin(th))/((1+(sin(alpr))^2-... 2*sin(alpr)*cos(th))^2); % Store position information in matrices. x(i) = theta; angle(i)=phi*180/pi; velocity(i)=dphi; acceleration(i)=ddphi; zro(i)=0; end maxv=max(velocity); minv=min(velocity); normv=max([abs(maxv), abs(minv)]); maxa=max(acceleration); mina=min(acceleration); norma=max([abs(maxa), abs(mina)]); vel=velocity/normv; acc=acceleration/norma; % plot the results clf; height=0.4; width=0.6; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',9); ylabel('Normalized Wheel Velocity, Acceleration','color', 'k', 'fontname',... 'times', 'fontsize', 10); xlabel('Driver Angle (deg)','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') xmin=-36; xmax=36; ymin=min([vel, acc]); ymax=max([vel, acc]); axis([xmin xmax ymin ymax]); % Set up the vectors needed to draw the linkage zeroline = line('xdata', x, 'ydata',zro, 'linewidth', .5, 'erasemode', ... 'xor', 'color', 'k', 'linestyle', '-'); - 322 - hold on; vel = line('xdata', x, 'ydata', vel, 'linewidth', 1, 'erasemode',... 'xor', 'color', 'r', 'linestyle', '--'); hold on; acc = line('xdata', x, 'ydata',acc, 'linewidth', 1, 'erasemode',... 'xor', 'color', 'b', 'linestyle', '-.'); hold on; legend([vel, acc], 'Velocity','Acceleration'); text(Lxt,Lyt2,['Max vel. value is: ', num2str(normv)],'color', 'r'); text(Lxt,Lyt3,['Max acc. value is: ', num2str(norma)], 'color', 'b'); 1.763in 3.00 in 2.427 in 1.237 in Fig. 7.11.1: Geneva mechanism for Problem 7.11 - 323 - Fig. 7.11.2: Plot of normalized velocity, and acceleration for Problem 7.11 Problem 7.12 Resolve Problem 7.11 if the input link rotates three revolutions for each rotation of the follower. Conduct the velocity and acceleration analysis for one third of a rotation. Solution The mechanism must have three stations (N=3). The angle = 360˚/2 N = 360˚/(2 3) = 60˚ The radius to the center of the pin is given by r = C sin = 3sin60˚= 2.598 in where C is the center distance between the driver and the follower of the Geneva mechanism. Also from Fig. 7.31, the radius of the pin at the lowest point on the slot is given by L = C(1 sin ) = 3(1 sin60˚) = 0.402 in The resulting system is shown in Fig. 7.12.1. The position, velocity, and acceleration of the Geneva mechanism are given by Eqs. (7.30), (7.31), and (7.32). The equations are = tan 1 in Fig. 7.31 is given by (1 sinin sinos ) s c - 324 - (7.30) ˙ = ˙ sin ˙˙ = ˙2 (1 + sincos 2 sin 2 sin cos ) (7.31) (7.32) sin cos2 sin (1 + sin2 2sin cos )2 In this problem, = 100 rpm = 100(2 ) / 60 = 10.47 rad / s . The calculations can be done using MATLAB. A program for making the calculations is given in the following. The results are plotted in Fig. 7.12.2. In Fig. 7.12.2, the velocity and acceleration curves are normalized by dividing the velocity and acceleration values by the maximum velocity and maximum acceleration, respectively. % MATLAB program for Problem 7.12 clf; clear all; % Loop to analyze the joint i=0; alpha=60; alpr=alpha*pi/180; thdot=100*2*pi/60; Lxt=-55; Lyt2=-0.5; Lyt3=-0.7; for theta = -60:1:60; i=i+1; th=theta*pi/180; phi=atan(sin(alpr)*sin(th))/(1-sin(alpr)*cos(th)); dphi=thdot*sin(alpr)*(cos(th)-sin(alpr))/(1+(sin(alpr))^2-... 2*sin(alpr)*cos(th)); ddphi=-thdot^2*(sin(alpr)*cos(alpr)^2*sin(th))/((1+(sin(alpr))^2-... 2*sin(alpr)*cos(th))^2); % Store position information in matrices. x(i) = theta; angle(i)=phi*180/pi; velocity(i)=dphi; acceleration(i)=ddphi; zro(i)=0; end maxv=max(velocity); minv=min(velocity); normv=max([abs(maxv), abs(minv)]); maxa=max(acceleration); mina=min(acceleration); norma=max([abs(maxa), abs(mina)]); vel=velocity/normv; acc=acceleration/norma; % plot the results clf; - 325 - height=0.4; width=0.6; axes('position',[.2 .2 width height],'box','on','xcolor','k','ycolor','k',... 'fontname', 'times', 'fontsize',9); ylabel('Normalized Wheel Velocity, Acceleration','color', 'k', 'fontname',... 'times', 'fontsize', 10); xlabel('Driver Angle (deg)','color', 'k', 'fontname',... 'times', 'fontsize', 10); set(gcf,'color', 'w') xmin=-60; xmax=60; ymin=min([vel, acc]); ymax=max([vel, acc]); axis([xmin xmax ymin ymax]); % Set up the vectors needed to draw the linkage zeroline = line('xdata', x, 'ydata',zro,... 'linewidth', .5, 'erasemode', 'xor', 'color', 'k', 'linestyle', ''); hold on; vel = line('xdata', x, 'ydata', vel, 'linewidth', 1, 'erasemode',... 'xor', 'color', 'r', 'linestyle', '--'); hold on; acc = line('xdata', x, 'ydata',acc, 'linewidth', 1, 'erasemode',... 'xor', 'color', 'b', 'linestyle', '-.'); hold on; legend([vel, acc], 'Velocity','Acceleration'); text(Lxt,Lyt2,['Max vel. value is: ', num2str(normv)],'color', 'r'); text(Lxt,Lyt3,['Max acc. value is: ', num2str(norma)], 'color', 'b'); - 326 - 2.598 in 3.00 in 1.5 in 0.402 in Fig. 7.12.1: Geneva mechanism for Problem 7.11 Fig. 7.12.2: Plot of normalized velocity, and acceleration for Problem 7.11 - 327 - ...
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This note was uploaded on 02/20/2011 for the course MEC 411 taught by Professor Shudong during the Winter '11 term at Ryerson.

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