599-1 ME Homework 2 Solutions
1. a. Define the overall length of the spring as x (which is initially equal to 1 = 150 mm), and the left and right sides of the triangle as 2 = 55 mm and 3 = 130 mm, respectively. Also define the angle between the bottom and ride sides as and the height of the ankle as y ,. Using the law of cosines, we can find x2 + 2 3 = cos 1 2x 3 and hence the height y=
3 2 2
2
(1)
sin =
3
(x 1
+
4x
2 3 2 2 3
2 2 2
)
.
(2)
Solving for the initial case x = 1 , we find the values y = 46.9 mm, = 21.1. The length of the spring as a function of y requires manipulation of equation (2), which yields four possible solutions. Plugging in the known values of y and x for the initial case, we can select the right one: x=
2 2
+
2 3
2y 2 + 2
2 2 2 3
2 2 2
y
2 3
y2 + y4 .
(3)
b. Plugging in y = 46.9 mm, we find x = 161.6 mm, which is a displacement of about 11.6 mm. c. The Jacobian of equation (2) is 4y + J= x = y 2
2 2
2 2 y 2 2 y + 4y 3 2 3 2 2 2 2 2 2 4 2 3 2y 3 y + y
2 2 2 3
+
2 3
2y 2 + 2
2 2 2
y
2 2 3
y + y4
(4)
which, incidentally, is not an equation that humans need to derive. Mathematica can find equation (4) very quickly. Using the approximation x Jy , we find a value of about 14.1 mm, which is about 22% too high. This is not bad. d. Using this same Jacobian, we can use Fankle = J T Fspring to find the approximate force in the spring, 3.2 kN. e. The potential energy for the case of 7 mm vertical displacement, 6.4 kN force is about 22.4 J. There are a number of reasons why this could be in error. First, in reality energy is stored or dissipated in other places in the foot, such as the pad of tissue under the heel, and other soft tissue. Second, a linear spring will store energy than an exponential spring which has relatively low initial stiffness. The actual amount of energy stored in the plantar fascia (the name of the connective tissue in question) has been estimated at between 12 and 17 J. 2. a. While any convention may be used, the use of joint angles can be regarded as the best, if we regard the angles of the joints as the variables being controlled. The kinematics in terms of joint angles are: x= y=
1 1
cos sin
1 1
+ +
2 2
cos(
1
1
+
2
2
)+
3
3
cos(
1
1
+
2
2
+
3
3
)
sin(
+
)+
sin(
+
+
)
where the joint angles are numbered from the base leftward, measuring the counter-clockwise change in segment angles, and x is positive to the left, and y is positive down. The velocity, using time differentiation, is x = 1 sin( 1 ) 1 2 sin( 1 + 2 ) 1 + 2 3 sin( 1 + 2 + 3 ) 1 + 2 + 3 . y = 1 cos( 1 ) 1 + 2 cos( 1 + 2 ) 1 + 2 + 3 cos( 1 + 2 + 3 ) 1 + 2 + 3
(
(
) )
( (
) )
b. The Jacobian is 1s1 2 s12 3 s1 2 3 2 s12 3 s1 2 3 3 s1 2 3 J= 1c1 + 2 c12 + 3c1 2 3 2 c12 + 3 c1 2 3 3 c1 2 3 where s1 = sin 1 , s12 = sin( 1 + 2 ) , s1 2 3 = sin( 1 + 2 + 3 ) , and likewise for cosines. It is easily verified that the Jacobian multiplied by the vector of joint velocities equals the fingertip velocity. c. The vertical force is obviously a function of joint torques, but this question is complicated by the fact that not all sets of joint torques can be applied in the static case. This is because some sets of joint torques will cause the system to move! We can think of this as a constraint acting on the joint torques, which lets us choose any two of the joint torques, and let the constraint fill in the third. We can use elementary statics to solve this problem, but another way is to look at the Jacobian in the equation T = J T F , which is a 3 by 2 matrix relationship: T1 J11 J 21 Fx T2 = J12 J22 F T3 J13 J 23 y This equation seems to tell us that the joint torques are determined by the forces at the tip, and that is indeed true. It is saying that any set of tip forces, in the static case, is associated with one and only one set of joint torques. To solve this problem, we can arbitrarily look at the top two rows of the matrix equation, and use it to give us the fingertip forces: Fx J11 F = J y 12 J21 1 T1 J22 T2
and the third torque is found by using the third row of the matrix equation:
1 J11 J21 T1 T3 = [ J13 J23 ] . J12 J22 T2 We can get the same answer using any other two rows of the equation, or by using simple statics. But the main trick is to realize that all three torques cannot be arbitrarily set if the system is to remain static. Of course, the equations above are also only valid if the torques are chosen so that the finger presses downward on the surface. We have also said nothing of the friction which is needed to keep the tip from sliding horizontally. d. We may regard segments 2 and 3 as a single segment, with length
23
=
2 2
+
2 3
2
2 3
cos165 .
This segment is used to break down the inverse kinematics problem, as shown in the program i n v k i n . m. This program solves for the angle between segment 2 and the combined segment, and between the combined segment and segment 1. This can be used to find the joint angle 2 . Next, the angle between vertical and the line connecting the first joint
e.
f. g.
h. i.
with the is fingertip calculated, along with the angle between this line and segment 1. From this, 1 can be found. We can use the singular values of the Jacobian to tell us about the relationship between joint torques and fingertip forces without having to examine any specific set of torques of forces. Recall that the maximum singular value tells us the maximum vector magnitude of joint torques which can be associated with a unit magnitude force vector. The minimum singular value relates the minimum torque magnitude to a unit force vector. Looked at another way, the minimum singular value tells us when a maximum force can be exerted by a unit torque which satisfies the static condition. The program H W 2 _ 2 . m shows that this minimum is reached when the fingertip is at maximum extension, near 80 mm. Unfortunately, the large force which can be exerted is in the direction of the extension, rather than vertical. An example of a similar situation is when humans lift heavy loads on their shoulderswhen their legs are perfectly straight, they can bear the heaviest loads. To look at just the vertical force, we only need to use the first column of J T . The singular value in this case is actually equal to the magnitude of the column vector, and is plotted by H W 2 _ 2 . m. The minimum is reached at around 25 mm. The manipulability is given by the product of singular values, which is proportional to the volume of the Jacobian ellipsoid. H W 2 _ 2 . m plots this measure of manipulability, which is nearly constant for lateral positions up to about 50 mm, with a slight peak at about 35 mm. It then decreases with increasing extension. So the trade-off of extension, where large forces can be exerted, is that there is poor manipulability. These large forces can also be exerted only in a rather limited set of directions. The columns of the Jacobian are the vector fingertip velocities associated with each joint velocity. The rows of the Jacobian, or the columns of its transpose, correspond to the set of joint torques associated with each fingertip force vector.
% HW 2, problem 2 l1 = 55; l2 = 30; l3 = 25; % define lengths of segments
for i = 1:81, % Step through all 81 positions x = i - 1; % go from 0 to 80 xs(i) = x; th = invkin(x)*pi/180; s1 = sin(th(1)); s2 = sin(th(2)); s3 = sin(th(3)); c1 = cos(th(1)); c2 = cos(th(2)); c3 = cos(th(3)); s12 = sin(th(1)+th(2)); s123 = sin(th(1)+th(2)+th(3)); c12 = cos(th(1)+th(2)); c123 = cos(th(1)+th(2)+th(3)); J = [-l1*s1-l2*s12-l3*s123 -l2*s12-l3*s123 -l3*s123; ... l1*c1+l2*c12+l3*c123 l2*c12+l3*c123 l3*c123]; minsvd(i) = min(svd(J')); prodsvd(i) = prod(svd(J')); minsvd2(i) = min(svd(J(2,:)')); end; subplot(311) % Plot the minimum singular value vs. position x: plot(xs, minsvd); xlabel('Lateral position x'); title('Minimum singular value vs. position'); subplot(312) % Plot the product of singular values vs. position x: plot(xs, prodsvd); xlabel('Lateral position x'); title('Product of singular values vs. position');
subplot(313) % Plot the min singular value of J(2,:) vs. position x: plot(xs, minsvd2); xlabel('Lateral position x');...