Question

Hw 12b, states I should 'break up the trajectory into two segments with the same time duration'. Does that mean I find the greatest distance traveled, find the max velocity based on that distance, split that duration of travel into two even parts and create an initial and final set of inverse kinematic vector equations for both the initial alpha and the final alpha? Meaning a total of four sets of three alpha vector equations, two sets for the initial alpha(0) and two sets for the final alpha(5)?

Answer 1

You know the total time is 5 seconds for the trajectory. Two time segmets of equal duration would be 2.5 seconds each. You should find the end effector configuration at T=0, 2.5, and 5. Then figure out the joint angles at each time and use that to spline the positions together.

Answer 2

Ok

Answer 3

because of 1a, I have the end configuration of the initial and the final configurations (T=0 and T=5). So I need to find the end configuration at T = 2.5

Answer 4

Right

Answer 5

Once I have all the end effector configurations, I will want to find the inverse kinematics so I will know all the joint angles at the three positions.

Answer 6

Yeah but again you have the joint angles at T=0 and t=5 from the first part.

Answer 7

Since I am to find the straight line trajectory and that means use the equation given.

Answer 8

That will give me the midpoint end effector configuration.

Answer 9

or what ever other time (t) I should want

Answer 10

Right

Answer 11

Finding the inverse kinematics of (x y theta)T sounded easier than I thought

Answer 12

For the three link planar manipulator you solved that problem in the homework 2 weeks ago. You should have formulas for each joint angle.

Answer 13

Yes, yes I did. I know x and y so I can find r. I know the lengths so that gives me the beta and gamma and delta. That gives me alpha1 and 2, and those give me alpha3. What is meant by the warning 'be careful of multiple solutions'. I know I will have multiple solutions for alpha1 and 2 but why the warning?

Answer 14

r is actually a function of xw and yw dont forget that. I think the warning just means you should pick the solution that makes the most sense. In this case, I would imagine the alphas at the waypoint should be between those at T=0 and T=5. It's conceivable that they wouldn't be for one of the solutions.

Answer 15

I think I did forget that about x and y because I am trying to figure out why I got a complex delta.

Answer 16

The solutions are posted on t-square from the previous homeworks so you can get your formulas there.

Answer 17

Gotta go for now.

Answer 18

Ok thanks, later

Answer 19

for 1b, I found the midpoint angles, formed the spline trajectory between initial and midpoint ( 3 functions) and the spline trajectory between midpoint and final (3 more functions). Is that all 1b is asking for? What does it mean by 'spline the desirded joint angles together'?

Answer 20

This is essentially a spline with 1 waypoint(the midpoint). You should use what I talked about last thursday to spline all the angles together.

Answer 21

I see the material we went over but I do not see how it it applies here.

Answer 22

I am reading the notes but I do not follow how this applies.

Answer 23

You have alphaf, alphai, alphav, tf, ti, and tv. You can make a splined trajectory with 1 waypoint.

Answer 24

Will I have a p(t) and a q(t) for each set of alphas?

Answer 25

Yes

Answer 26

how do you find the inverse kinematics of \[\xi \]

Answer 27

In 1c, I am to do the same thing as in 1b but with \[\xi \] but how do I find the inverse of \[\xi \]

Answer 28

Problem c is the same as problem b, except that instead of a linear end effector trajectory, the end effector trajectory is now defined by xi.

Answer 29

this is SE(2), \[\omega \] is the sum of the main diagonal but I do not know g so I do not know \[\omega \] so I do not see how I can solve. This is what I have \[g _{e}\left( t \right)=g _{e}\left( 0 \right)e ^{\xi t}\] \[g _{e}^{-1}(0)g _{e}(t)=g _{e}^{-1}(0)g _{e}(0)e ^{\xi t}\] \[e ^{\xi t}=g _{e}^{-1}(0)g _{e}(t)\]

Answer 30

You have ge(5), ge(0), and t. Isn't that all you need to solve for xi?

Answer 31

I do know ge(0) and ge(5) and t=2.5 at the midpoint.

Answer 32

if I made \[e ^{2.5 \xi }=g _{e}^{-1}(0)g _{e}(2.5)\] what would \[g _{e}(2.5) \] be

Answer 33

Ok in part b you had a formula for the trajectory. You plugged in t=2.5 to get the ge at 2. correct?

Answer 34

Am I not thinking about this in the right way?

Answer 35

No you are not. Do you agree with my previous statement?

Answer 36

Yes. I got \[g _{e}(2.5)\] by using the vector form of the problem. In part c I am instructed 'Instead of vectorizing' which tells me not to use the ge(2.5) from part b.

Answer 37

Yes I just want you to agree that you got ge(2.5) by plugging in t=2.5 into a formula for the trajectory of ge.

Answer 38

Yes.

Answer 39

Ok. In part c you also need a formula for the trajectory of the end effector. Once you get this formula, you can just plug in t = 2.5 to get ge(2.5). The only difference is this time there is a different formula for the trajectory of ge.

Answer 40

\[g _{e}(t)=g _{e}(0)e ^{\xi t}\]

Answer 41

Yes that's correct. What's the only unknown in that formula?

Answer 42

\[\xi \]

Answer 43

Ok. You should know how to get xi. It's been done in homeworks and you had to do it on the test.

Answer 44

Yes, for SE(3). This is SE(2) and I am unsure as how to find \[\omega \]

Answer 45

I have \[\omega = \left( \frac{ 1 }{ \tau } \right) atan(R _{21},R _{11})\] for \[R _{21} and R _{11}\] what rotation matrix do I create?

Answer 46

Check homework 8 problem 3 solutions. You've done this in SE2

Answer 47

To figure out xi, you first need g, the transformation from gi to gf.

Answer 48

My mistake, I see that I did an SE(2) to find \[\xi \] I see how to find \[\xi \] now.

Answer 49

I solve \[g _{e}(2.5) = g _{e}(0)e ^{\xi *2.5}\] and find the joint angles for \[g _{e}(2.5)\] then do what I did in part b

Answer 50

Yep

Answer 51

thank you for your help, goodnight

Answer 52

Good night you're welcome.

Answer 53

If \[\xi = \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)\] would \[e ^{\xi t} = e ^{\left(\begin{matrix}x \\ y \\ z\end{matrix}\right) t}=\left(\begin{matrix}e ^{xt} \\ e ^{yt} \\ e ^{zt}\end{matrix}\right)\]

Answer 54

No. You need to take the matrix exponential. This has also been done numerous times before, including on the test.

Answer 55

\[g _{e}(midway) = e ^{\xi \frac{ \tau }{ 2}} = \left[\begin{matrix}R(\omega \frac{ \tau }{ 2 }) & (-\frac{ 1 }{ \omega })(unit matrix - R(\omega \frac{ \tau }{ 2 }) )Jv \\ 0 & 1\end{matrix}\right]\] \[\xi = \left(\begin{matrix}v \\ \omega\end{matrix}\right) \] has been found

Answer 56

Seems reasonable except you need to multiply by ge(0) on the left.

Answer 57

I got the above equation from the notes. I believe I see why I would multiply it by ge(0) though. It is because of the trajectory we are trying to follow given in the problem

Answer 58

Yes as before \[g_f = g_ie^{\xi*\tau}\]

Answer 59

Is it \[g = g _{f}g _{i}^{-1}\] or \[g = g _{i}^{-1}g _{f}\] I have it in my notes both ways. I have done homework both ways and on my test one way

Answer 60

I say it is \[g = g _{i}^{-1}g _{f}\]

Answer 61

That's fine.

Answer 62

1d asks to plot. I am not sure how to plot this.

Answer 63

You've plotted trajectories before. What is different about this situation?

Answer 64

I am just not sure of how to handle the p(t) and the q(t).

Answer 65

With the planarR3 from the last homework, I need alpha 1, 2 and 3 but I have p(t) and q(t).

Answer 66

Well first of all you don'e need to plot the planarR3, you just need to plot the frame of the end effector over time. You should have p(t) and q(t) for each alpha so for any time t, you can figure out alpha1, alpha2, and alpha3.

Answer 67

are the p(t) and q(t) the alphas?

Answer 68

I plot my end effector path from 0 to 2.5 using the p(t) equation to give me alpha1, 2 and 3, and from 2.5 to 5 using mthe q(t) to give me alpha1, 2 and 3. I use planarR3 and plug in the alpha's. My angles at p(2.5) do not = my angles at q(2.5) so I have a large displacement in the middle of my graph. Am I thinking about this right?

Answer 69

Is 1c suppose to supply alpha values that stay between the initial and final configurations when plotted in 1d?

Answer 70

Ok let's just do one question at a time. p(0) should equal q(0). The spline from T=2.5 to T=5 is equal to q(t-tv). So it should go from 0 to 2.5 just like p(t) does.

Answer 71

p(2.5) should equal q(0) is what I meant. apologies

Answer 72

I do understand and agree that p(2.5) should = q(0). I am plotting q(t) from 2.5 to 5, I think this might be my problem but I need to check to see if my p(2.5) = q(0).

Answer 73

Yeah you should plot q(t-tv) from 2.5 to 5. That would be the issue.

Answer 74

My p(2.5) does not = q(0). When I print p(1) I get three numbers but I thought I would only get one

Answer 75

A problem the same as this, with three joints, was done in class. I wrote down a 8x3 matrix that was multiplied by a 3x1 alpha matrix the result being a 8x1 coeff matrix. I double check that I copied this down correctly and put it into matlab right. I use these coeff to complete the p(t) and q(t) functions that make the alphas. Is that much right in steps.

Answer 76

Yes all that makes sense.

Answer 77

Once you have the coefficients all you have to do is plot them.

Answer 78

plot the trajectories that is.

Answer 79

so plot p(t) for 0<t<2.5 and plot q(t-tv) for 2.5<t<5

Answer 80

Ok I have to go for now. I'll check back later.

Answer 81

do you think you will be back tonight?

Answer 82

ok thanks

Answer 83

% p(t) for 0<t<2.5

alpha1p = -0.524 + 0*(2.5) + 0.0522*(2.5)^2 + -0.078*(2.5)^3

% q(t-tv) for 2.5<t<5

alpha1q = -0.284 + 0.1571*(2.5 - 2.5) +...
0.0107*(2.5 - 2.5)^2 + -0.0112*(2.5 - 2.5)^3

This is matlab code. The coeff are what I calculated in matlab. This is just alpha1 for p(2.5) and q(2.5 - 2.5) = q(t - tv) = q(0), they do not equal. None of the alphas at this point equals.

Answer 84

for part b and part c should I have found g wrist before we found alpha1, 2 and 3 since each joint has only one axis of rotation?

Answer 85

For parts b and c, you should find inverse kinematics exactly the same way you did 2 homeworks ago. First find xw,yw...calculcate alpha1 and alpha2, and then calculate alpha3.

It's exactly the same problem as 2 homeworks ago.

Answer 86

What does it mean when I get beta and delta as complex numbers?

Answer 87

and my p(2.5) does not = q(0)

Answer 88

For any of the alphas?

Answer 89

No alphas equal between p(2.5) and q(0)

Answer 90

I looked at your answers. I think all you had to do was subtract or add 2*pi to your angles from q.