Question

Hey,
for HW #5 question #4:
for part (a) are we supposed to use the end-effector configuration that Dr. Vela just went over in class where you have ge(alpha) where alpha is a vector and then we define R(alpha) and d(alpha). In the notes these end up being really big matrices.

also for part (b) would we just have
ge(alpha)=g1(alpha1)g2(alpha2)g3(alpha3)g4(alpha4) ?

Answer 1

I have another question pertaining to the same problem. If class when we went over this Dr. Vela drew the zero-configuration picture and that is what he used to determine his g1(aplha), g2(alpha2), etc. Are we to do the same thing for this problem?

For instance for g2(alpha2) I am getting R(alpha2) = \[\left[\begin{matrix}\cos(\alpha2) & -\sin(\alpha2) \\ \sin(\alpha2) & \cos(\alpha2)\end{matrix}\right]\] and also d(alpha2) =\[\left[ 0 0 l1 \right]^{T}\]

I don't know if that is right or not, but I'm not sure what else to do.

Answer 2

I think you are looking at the more complex example in 3d space, problem 4 is only in 2d space. the d(alpha2) should be pretty simple = [l1, 0]T

Answer 3

Let me try to answer one by one...For part A you should be doing some geometry. It's exactly the same thing as what was worked out in the first week. It should have some sines, cosines, l1, l2, l3, etc. I think what you're thinking of is the Jacobean matrix which does not apply here.

For part B what you have is correct but you should be more specific. you should write out each g matrix, including R and d components. As a hint, the first g has a rotation by alpha1 and no displacement. Your R for alpha2 looks right but the d, as the previous answer pointed out, should be \[\left(\begin{matrix}l2 \\ 0\end{matrix}\right)\]

Does that clear it up any?

Answer 4

Thanks for your help! That does clear things up. I still have a question about the d for alpha2. Why is it \[\left(\begin{matrix}l2 \\ 0\end{matrix}\right)\] and not \[\left(\begin{matrix}l1 \\ 0\end{matrix}\right)\]

Answer 5

My apologies. Your answer is correct.

Answer 6

Oh ok thanks!

Answer 7

Why is d not
[ l1cos(alpha1), l1sin(alpha1)] ?

Answer 8

Because remember that in the first transformation, you are rotating by alpha1. The next transformation should be in relation to the new coordinate axes, so in this case only along the x axis. \[g^0_{\alpha2} = g^0_{\alpha 1}*g^{\alpha 1}_{\alpha 2}\]So the second g in your kinematic chain should be the transformation from alpha1 to alpha2, not all the way from the beginning.

Answer 9

OK that makes sense. Would the rotation at alpha4 would have a d = {0 0} ?

Answer 10

You don't need to take into account the distance to the end effector frame but g(alpha4) does have a displacement.

Answer 11

In the notes, for the 3D configuration, my end effector chain ends in g5

ge(alpha) = g1(alpha1)*g2(alpha2)*g3(alpha3)*g4(alpha4)*g5

g5 was basically an identity matrix with an l3 included to adjust for the length of the end effector arm. Since this, 2D, problem has no end effector arm length, I would have the g5 (equivalent) to be an indentity matrix. Am I thinking ablout this correctly?

Answer 12

I think in the example, g5 is just a length from the final joint to the end effector. In this case, the end effector can rotate and is a joint in itself. You can treat eat as any other joint. The g for alpha4 is analogous to the g for alpha5.

Answer 13

alpha3 is what I meant at the end there

Answer 14

So, if my configuration ends in a final rotation with no length to the end effector then my end effector chain will end in a g(alpha) ?

ge(alpha) = g1(alpha1)*g2(alpha2)*g3(alpha3)

Answer 15

no length from the joint to the end effector

Answer 16

*g4(alpha4). The last joint is the end effector essentially. How did you define the g for alpha3?

Answer 17

I wrote the last ge(alpha) equation as an example. For this problem, my g3(alpha3) is

d = {l2 0}

R = [ cos(alpha3) -sin(alpha3) ]
[ sin(alpha3) cos(alpha3)]

Answer 18

Ok that's right. The last joint is exactly analogous to that right? So you still need to make g4(alpha4) , but you don't have to do a g5 to get to the end effector you can just end at g4.

Answer 19

For 4a, I believe we did something like this in a prior homework. In that homework there was no rotation joint at the end effector therefore no alpha4. Does not alpha4 need to be taken into account for 4a?

Answer 20

Yes it does because there is a rotation at the end effector joint. That's why you need g4(alpha4)

Answer 21

In working through problem 4, do you need to do 4b before 4a?

Answer 22

Oh I see what you're talking about...alpha4 doesn't affect the location of the end effector but it affects the orientation.

Answer 23

yes

Answer 24

Yes so in 4a for your expression of the location of the end effector, alpha4 does not come into play but in your expression for the angle it does.

Answer 25

I would have a d and an R expression for 4a that is the result of 4b?

Answer 26

No you would have an expression for x_e, y_e, and theta_e. It's the same thing that was done the first week of class using geometry.

Answer 27

4a should not use a kinematic chain of Lie groups. It's purely geometry.

Answer 28

I feel comfortable using geometry until I get to alpha4 where no length is involved.

Answer 29

Let me work on it a moment

Answer 30

Alpha4 does not affect x_e or y_e. What is your expression for theta? Hint: don't overthink it.

Answer 31

I do not have a theta.

Answer 32

Just intuitively what should be the angle of the end effector?

Answer 33

I woud say the angle of the end effector should be the sum

alpha1 + alpha2 + alpha3

but the end effector would not close on the object as desired.

Answer 34

That's close by why would alpha4 not get added on to that?

Answer 35

xe = l1cos(alpha1) + l2cos(alpha1 + aplha2) + l3cos(alpha1 + aplha2 + alpha3)

where would I add alpha4?

Answer 36

Again, x_e and y_e are not affected at all by alpha4. It's only a rotation and therefore only affects the angle of the end effector, not the location.

Answer 37

I would not have an R because I am only using geometry not kinematics. x_e and y_e are location values that are arrived at through geometry. I need to use alpha4 for rotation only.

I understand these statements but I do not see how to put the pieces together.

Answer 38

You're really overthinking it....Your x_e is right. You may not have an "R", but you still have an angle of the end effector, which in this case equals alpha1+alpha2+alpha3+alpha4.

Answer 39

I do understand that the answer would involve
theta = alpha1 + alpha2 + alpha3 + alpha4
If I multiply x_e by cos(theta), I know that is not correct
If I add cos(theta) to x_e, I know that makes little sense
I can not add alpha4 to the last cos factor
xe = l1cos(alpha1) + l2cos(alpha1 + aplha2) + l3cos(alpha1 + aplha2 + alpha3 + alpha4)
that would out me at a different location

with xe = l1cos(alpha1) + l2cos(alpha1 + aplha2) + l3cos(alpha1 + aplha2 + alpha3)
I have reached the end effector, and now I need to turn so my next move will be something in addition to my last moves. something that turns the end effector.

Answer 40

Ok let me just summarize from the beginning:

You are looking for the following representation\[\left(\begin{matrix}x_e \\ y_e \\ \theta e\end{matrix}\right)\]x_e and y_e are not affected by alpha4. You wrote the correct equation for x_e above. I'm sure you know how to write the one for y_e also. The only thing alpha4 affects is theta_e. It affects in a simple way - it is added to the other alphas to figure out the final angle of the end effector such that\[\theta_e = \alpha_1+\alpha_2+\alpha_3+\alpha_4\]

That's all there is to it.

Answer 41

I thought I was looking for

\[\left(\begin{matrix}x _{e} \\ y _{e}\end{matrix}\right)\]

not

\[\left(\begin{matrix}x _{e} \\ y _{e} \\ \theta \end{matrix}\right)\]

Answer 42

That is not correct. As with any frame in SE2, it needs a location and a rotation to completely describe it.

Answer 43

Ok thank you for all your help

Answer 44

No problem.

Answer 45

in the comment section of the Adjoint.m code, the adjoint id defined as

Ad_g1 (g2) = g1 * g2 * inverse g1

should it not be

Ad_g1 (g2) = inverse g1 * g2 * g1

Answer 46

http://openstudy.com/study#/updates/5059ca5fe4b0cc1228935a4f

Answer 47

Thanks pvpman