Question

for #1 part b of hw 11, when I find

Answer 1

for #1 part b of hw 11, when I find \[g _{\omega}=g _{e}g _{7}^{-1}\]

but when I go to find \[g _{h}= g _{w}^{-1}g _{e}g _{7}^{-1}\] I always get the 4x4 identity matrix

Answer 2

\[g _{h}\] is suppose to be a pue rotation matrix. What am I doing wrong?

Answer 3

How do you find gw?

Answer 4

\[g _{w} = g _{e}g _{7}^{-1} = g _{i}g _{7}^{-1}\] I am given that l3 = 0.5 so \[g _{7}=\left[\begin{matrix}1 & 0 & 0.5\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right]\] and so \[g _{7}^{-1}=\left[\begin{matrix}1 & 0 & -0.5\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right]\] I am given \[g _{i}\] then I multiply them together to get \[g _{w}\]

Answer 5

Ah ok you missed an important point... pos(gw) = pos(ge*g7^-1), but gw does not equal ge*g7^-1. You need to get gw by first getting alpha1, alpha2, and alpha3

Answer 6

I thought Pos(gw) was (xw, yw, zw) and I need alpha 1, 2 and 3 to get xw, yw and zw or visa versa

Answer 7

Pos(gw) is (xw, yw, zw). Just like you did last week, you use this position to solve for the angles before the wrist. Pos(gw) can be figured out from known quantities.

Answer 8

I am to use \[\beta\] \[\gamma \] \[\delta \] and r to find \[\alpha _{1}\] and \[\alpha _{2}\]

Answer 9

I have to go for now be back later. You need to use geometry. It should be in the notes.

Answer 10

thanks
later

Answer 11

Were you able to move along?

Answer 12

no, I looked at the geometry section of the elbow manipulator in my notes. All it says is to solve (x,y) projection problem to get alpha1 and the (r,z) projection problem to get alpha2 and alpha3.

Answer 13

I know what a projection onto a plane means but I do not see how I can get the needed information without having the angle or the x y projections.

Answer 14

Well you have xw, yw, and zw right?

Answer 15

I have the \[d _{i}\] which comes from the \[g _{i}\] that is given. I need the \[g _{\omega} \] to get the \[d _{\omega} \] which is \[\left(\begin{matrix}x _{\omega} \\ y _{\omega} \\ z _{\omega}\end{matrix}\right) \] I know that the way I was getting the \[g _{\omega} \] is not right but I do not know of another way.

Answer 16

How are you getting gw?

Answer 17

Not to sound short but I thought this would be quicker. I explained it in my forth posting.

Answer 18

I know now from what you stated that Pos(gw) = Pos(ge*g7^-1) does not mean gw = ge * g7^-1

Answer 19

Ok let me make it clear.

\[pos(g_w) = pos(g_e*g_7^{-1})\]\[g_e = \left[\begin{matrix}R_e & d_e \\ 0 & 1\end{matrix}\right]\]\[g_7 = \left[\begin{matrix}I & \left(\begin{matrix}0 \\ l_3 \\ 0\end{matrix}\right) \\ 0 & 1\end{matrix}\right]\]Therefore\[d_w = \left(\begin{matrix}x_w \\ y_w\\z_w\end{matrix}\right) = pos(g_e*g_7^{-1})\]

Answer 20

This is exactly the same thing as you did last week but in 3D.

Answer 21

I see a great similarity in this to what I did. Your saying I do not know ge and that is why you have written ge in its general form. I am thinking that I do know ge because it is gi. I have put l3 in the wrong place. I put it in the x position, I did not see the lettering on the axis in the problem. I do see were l3 should be in the y position.

Answer 22

I'm not saying ge is unknown I'm just giving you the general solution. In 1b, gi = ge for the first case and gf=ge in the second case.

Answer 23

In inverse kinematics problem ge is always known. It doesn't make sense to calculate joint angles if you don't know where your end effector is.

Answer 24

I was looking back at last weeks hw and saw something that looked like what I have in this problem. That I can find gw by \[g _{\omega}=g _{1}(\alpha _{1})g _{2}(\alpha _{2})g _{3}(\alpha _{3})g _{4}\]

Answer 25

After you get alpha1, alpha2, and alpha3, you can use this to get gw yes.

Answer 26

That is what I need to finish that problem.

Answer 27

I did move to the last problem and I have a question.

Answer 28

Ok

Answer 29

I am asked to design a splined trajectory from alpha0 to alpha1 and the movement time is to be 8 seconds. I looked in my notes and I see \[A = X ^{-1}\alpha \] but I am not sure how to code it to move in a curved path.

Answer 30

Do you know how to write a function that would give you the angle as a function of time?

Answer 31

a(1) = q0;
a(2) = qv0;
a(3) = (3 * (q1-q0) - t * (2 * qv0 + qv1)) / t^2;
a(4) = (-2 * (q1-q0) + t * (qv0 + qv1)) / t^3;

q0 is initial position
q1 is final position
qv0 is initial velocity
qv1 is final velocit
t is time

I took the equations from the notes and solved for alpha

Answer 32

As for a matlab function, I am not sure.

Answer 33

Ok I have to go for now. I'll do an example and talk a bit about the matlab in class.

Answer 34

Ok thanks

Answer 35

When I calculate \[g _{h} = g _{\omega}^{-1}g _{e}g _{7}^{-1} = g _{\omega}^{-1}g _{i}g _{7}^{-1} \] I am suppose to get a pure rotational matrix

Answer 36

Yeah

Answer 37

I did not

Answer 38

My guess is you calculated something incorrectly.

Answer 39

I am going back over them

Answer 40

\[g _{7}=\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0.5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{matrix}\right]\] which would make \[g _{7}^{-1}=\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & -0.5 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{matrix}\right]\] is that correct?

Answer 41

Yes I believe so.

Answer 42

In the notes \[r=\sqrt{x _{\omega}^{2}+y _{\omega}^{2}}\] Should I change r to \[r=\sqrt{x _{\omega}^{2}+y _{\omega}^{2}+z _{\omega}^{2}}\]

Answer 43

Maybe if it makes sense to you. It depends what it's being used for.

Answer 44

When I find \[\omega \] I use the quadratic formula, which gives me two values for \[\omega \] I do not know which one to use.

Answer 45

well both are possible solutions

Answer 46

In the notes it says just pick one. I took it to mean the as long as \[\omega \] is not complex then both will give you a correct answer. This means that I will only need to do the rest of the calculations with the one I pick and not do it again because the one I picked did not give me the right answer.

Answer 47

Yes I think you're right. You can just pick one. How are you getting alpha1?

Answer 48

when I have \[\omega \] and I go to find \[g _{\omega}=g _{1}(\alpha _{1})g _{2}(\alpha _{2})g _{3}(\alpha _{3})g _{4}\] say \[\alpha _{1}\] rotates about the z axis this would mean \[g _{1}(\alpha _{1})=\left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & \cos (\alpha _{1}) & -\sin (\alpha _{1}) & 0 \\ 0 & \sin (\alpha _{1}) & \cos (\alpha _{1}) & 0 \\ 0 & 0 & 0 & 1\end{matrix}\right]\] and if \[\alpha _{2}\] rotated about the x axis then \[g _{2}(\alpha _{2})=\left[\begin{matrix}\cos (\alpha _{2}) & -\sin (\alpha _{2}) & 0 & 0 \\ \sin (\alpha _{2}) & \cos (\alpha _{2}) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{matrix}\right] \] is this the right idea because Ithe manipulator is in x y and z directions

Answer 49

Yes it's the right idea but you seem to be leaving out the displacements

Answer 50

\[d _{1}=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] \[d _{2}=\left(\begin{matrix}0 \\ 0 \\ l _{0}\end{matrix}\right)\] \[d _{3}=\left(\begin{matrix}0 \\ l _{1} \\ 0\end{matrix}\right)\] \[d _{4}=\left(\begin{matrix}0 \\ l _{2} \\ 0\end{matrix}\right)\] but \[g _{4} = \left[\begin{matrix}1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{matrix}\right]\] because it is just a translation

Answer 51

why do you say that? g4 is a rotation about the x by alpha4

Answer 52

Do you say that because you only take the dispalcement part of g4 as part of gw?

Answer 53

when I find \[g _{\omega} \] using Pieper's approach I break \[g _{4}(\alpha _{4})\] into a translational part \[g _{4}\] and a rotational part

Answer 54

Yes that's fine. That doesn't mean g_4 is only a translation. It just means you only use the translation part for gw.

Answer 55

I have class I have to go

Answer 56

Alright I'll check back tonight.

Answer 57

thanks

Answer 58

back at 6:30

Answer 59

don't think i'll be on then but i'll check your questions later. Just go over your formulas again. Make sure everything makes sense to you

Answer 60

I am on #2, I have the polynomials and I need to plot the manipulator movement. Does that mean plot the alpha functions?

Answer 61

No you should plot the actual manipulator. There was a plot function distributed with homework 2 that can achieve this for you.

Answer 62

Are you refering to the parametric plot?

Answer 63

No there is something like planarR3plot or plotPlanarR3

Answer 64

I found it. It states that I need to plot 5-7 points along the trajectory.

Answer 65

But it is just a display. Would I show many plots on the same display?

Answer 66

Yes.

Answer 67

thanks later

Answer 68

Thanks. I am on the last question. The demo. We have the spline code made. No errors in running the code. The manipulator starts in the sleep position and moves the home position. We press enter and the screen shows almost a perfect unit matrix with a 17 in the z position (3 4 position). but the manipulator does not move until it finishes and moves back to the sleep position. We have check the code now for about an hour and still can not find anything. Is there a special something we should be lokking for?

Answer 69

How are you telling the manipulator to follow the trajectory?

Answer 70

I'm not going to debug your code for you guys. That's your job. My usual technique for debugging is to look at different values throughout the program. Make sure your alpha_traj looks like you would expect it to look etc. The answer to your question is no there is nothing special that you need to look for. There is just a programming error somewhere.

Answer 71

When this does work, it will move from the home position to the first position. Then only slightly to the next positions until the final position is reached. Then it will move to the sleep position. Is that correct?

Answer 72

So all we need is the forward kinematics.

Answer 73

I'm not so sure what forward kinematics has to do with this. You need a video of your manipulator going through a spline trajectory. It should start at the angles specified by alpha_i and end at the angles specified by alpha_f. It will start out moving slowly, speed up in the middle, and slow down toward the end if it is following the proper spline trajectory.

Answer 74

We had six unknowns and we needed six equations to solve this problem like was done in class. We took the second derivative to come up with the other two equations. Was that correct?

Answer 75

We used the same method as in the first #2 of this homework to solve for the alphas. how does the manipulator know to speed up and slow down? Is it built into the equations?

Answer 76

You seem to be confused about what you're doing. You are designing a trajectory (position vs. time) for each alpha. Each alpha has its own cubic trajectory that it should be following. The coefficients of the cubic equation depend on alphai, alphaf, and tf. The purpose of having them follow cubic trajectories is that they slowly speed up at the beginning and slow down at the end. So yes it is built into the equation that the speed is low at the beginning and end but fast in the middle.