Question

In hw#10 problem #1, I am to find the product of exponentials based off of hw#9 problem #4. I am to use the striaght out configuration as reference. All the g(alpha)'s, in hw#9 prob #4, are based on a sitraight up configuration. Would straight out be my initial condition, g0? I would not need to change anything about the g(alphas) from hw#9 #4 because the alphas are generic, would I?

Answer 1

The \[e ^{\xi \alpha}\] would be the \[g(\alpha)\] from hw#9 prob#4

Answer 2

Something like that but I think in your notes he described how to find the "twist" associated with a transformation. You shouldn't just be taking the ln of your gs.

Answer 3

\[\xi = \left(\begin{matrix}\hat g \omega \\ \omega\end{matrix}\right) \]

Answer 4

It's different for different joints in general but for revolute joints it's. \[\left(\begin{matrix}q_i*\omega_i \\ \omega_i\end{matrix}\right)\]

Answer 5

what is/does the twist do for/to the manipulator?

Answer 6

Why is the twist different for different joints?

Answer 7

What is a twist in general?

Answer 8

yes

Answer 9

No I'm asking you

Answer 10

It is a way to discribe a turning at a joint.

Answer 11

the twist does not explain a translation.

Answer 12

Not correct. A twist is nothing more/nothing less than a velocity(linear and angular)\[\xi=\left(\begin{matrix}v_1 \\ v_2\\ v_3\\w_1\\w_2\\w_3\end{matrix}\right)\]For the 3D case

Answer 13

When I think of velocity, I think of an object moving a distance in a period of time (car, baseball, person running ...). When I think of velocity in connection with athe manipulator arm I think of speed of rotation of the arm. This speed of rotation is always angular. I do not grasp the idea of the involvment of the linear velocity. Why do I have a v in the twist?

Answer 14

It's not the velocity of the joint you're describing it's the velocity of the transformation from joint to joint. All the joints on your manipulator are revolute but clearly you can see that there is a linear velocity as well as an angular one of each link.

Answer 15

Do you mean the speed at which any joint moves from that joints last position to its new position?

Answer 16

I'm not following. A revolute joint by definition only revolves. However since it is attahed to a link of length greater than 0 that link both revolves and has a linear displaement(and therefore linear veloity). Just keep your elbow still and turn it. You will see that your hand both rotates and change position.

Answer 17

I understand.

Answer 18

The initial condition for #1 would be \[g _{0} = \left[\begin{matrix}R _{x}(-\pi/2) & \left(\begin{matrix}0 \\ l _{1}+l _{2}+l _{3}+l _{4} \\ l _{0}\end{matrix}\right) \\ 0 & 1\end{matrix}\right]\]

Answer 19

The individual twists would be base off of hw#9 prob#4

Answer 20

What do you mean based off? I don't think there's any reason you have to look at HW9 for this problem.

Answer 21

I thought the wording implied to use last weeks homework.

Answer 22

Well you are doing the same thing as last week's homework but using a different method. So you should get the same results.

Answer 23

I have class now. Thanks for your help. If your around I wiil be back at 12:30. Thanks.

Answer 24

Ok that's fine.

Answer 25

It looks like the joint between l3 and l4 has no rotation. Is that correct?

Answer 26

It rotates about the y. The axis of rotation is coming out of the end effector.

Answer 27

So my product of exponetials is \[g _{e}(\alpha) = e ^{\hat \xi _{1}\alpha _{1}}e ^{\hat \xi _{2}\alpha _{2}}e ^{\hat \xi _{3}\alpha _{3}}e ^{\hat \xi _{4}\alpha _{4}}e ^{\hat \xi _{5}\alpha _{5}}*g _{0} \]

Answer 28

Yes I think that's right.

Answer 29

I find the indivial values of \[\xi \] is what I find

Answer 30

Yes

Answer 31

q a point the represents the joint position

Answer 32

q is a point on the axis of rotation

Answer 33

Is there a difference? The axis of rotation runs through the joint and the joint is all I am interested in. Do I have the understanding correct?

Answer 34

Ah yes there is a slight difference. You may be able to find an "easier" point on the axis of rotation than the joint itself. You also may not be able to. It depends.

Answer 35

In the figure, say I am interested in joint/axis 2. The q for this joint is
\[\sqrt{(l _{0})^{2}+(l _{1})^{2}}\]

Answer 36

q is the straight line distance from the origin to the joint.

Answer 37

No that's not right...that's not the axis of rotation.

Answer 38

The \[\omega \] at alpha2 is \[\left(\begin{matrix}0 \\ l _{1} \\ l _{0}\end{matrix}\right) \]

Answer 39

No that's also not right. w is a unit vector aligned with the axis of rotation in the base frame.

Answer 40

The q at alpha2 is \[\left(\begin{matrix}0 \\ l _{1} \\ l _{0}\end{matrix}\right) \]

Answer 41

Yes that's the easiest one.

Answer 42

I take q and turn it into q-hat

Answer 43

q is a 3x1, q-hat is a 3x3 matrix

Answer 44

How do I come up with omega

Answer 45

how are you going to make a 3x3 matrix out of a 3x1 point?

Answer 46

\[q = \left(\begin{matrix}a \\ b \\ c\end{matrix}\right) \]

\[q-\hat = \left[\begin{matrix}0 & c & -b\\ -c & 0 & a \\ b & -a & 0\end{matrix}\right] \]

Answer 47

I don't think that's quite right. What you're actually doing is qxw. Here is how to make a matrix out of q if that's how you want to compute it: http://en.wikipedia.org/wiki/Cross_product#Conversion_to_matrix_multiplication

w is just a unit vector pointing in the same direction as the axis of rotation.

Answer 48

So if my axis of rotation is y then w = (0 1 0) transpose

Answer 49

Yep

Answer 50

I thought anything with a hat on it was suppose to be some form of even matrix. I do not want to make this harder on myself by doing something that we were not shown. If q is a 3x1 and w is a 3x1 and I do remember to do the cross product to get qw then I am fine.

Answer 51

Yeah that's fine. Things with a hat are matrices yes. There is a technique for finding a cross product which may be easier than whatever you know which is to make that matrix and then just multiply it by the w. It's your choice but it's really just a cross product.

Answer 52

For #2a, I would use the Pieper's Solution.

Answer 53

I think Pieper's solution is just the one that breaks down the inverse kinematics into the hand and the wrist. You can just use geometry for 2

Answer 54

For #2a, I find the angles and create a forward kinematics equation.

Answer 55

You don't need to find the angles. You just need to compute the forward kinematics equation.

Answer 56

\[g _{e}(\alpha) = g _{1}(\alpha _{1})g _{2}(\alpha _{2})\]

Answer 57

Ok sure but write it out interms of ls and cosines and sines.

Answer 58

This is what I have

\[g _{e}(\alpha) = \left[\begin{matrix}\cos (\alpha _{1}) & -\sin (\alpha _{1}) & 0\\ \sin (\alpha _{1}) & \cos (\alpha _{1}) & 0 \\ 0 & 0 & 1\end{matrix}\right]*\left[\begin{matrix}\cos (\alpha _{2}) & -\sin (\alpha _{2}) & l _{1}\\ \sin (\alpha _{2}) & \cos (\alpha _{2}) & 0 \\ 0 & 0 & 1\end{matrix}\right]*\left[\begin{matrix}1 & 0 & l _{2}\\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}\right] \]

Answer 59

I would just write it out in equations. It should look familiar. It's been done a couple times in the past.

Answer 60

I think it was done in HW1 if you look back. In this case the intuitive solution is easier than this product of Lie algebras.

Answer 61

So when it says planar, that is another way of saying 2D forward kinematics?

Answer 62

If that is the case, then when asked for \[d _{e} = \left(\begin{matrix}x _{e} \\ y _{e}\end{matrix}\right)\] xe would be an equation with l's and cos, and ye would have l's and sin

Answer 63

That would mean \[g _{e}(\alpha) = d _{e}(\alpha)\]

Answer 64

in 2D

Answer 65

I'm not following you here. Yes planar means enclosed in a plane or equivalently 2D. Yes xe is ls and cosines and ye is ls and sines. I don't know what you mean by g=d

Answer 66

I meant to write qe not ge.

qe = de

Answer 67

For #2b, I just use the geometric approach to find alpha1 and alpha2.

For #2c, I look at the equations that solve for alpha1 and alpha2 and find the cases where no solution exists.

Am I correct in the approach in solving both questions?

Answer 68

If your approach makes sense to you then you should feel confident in it.

Answer 69

Can you explain #3? I think I am looking for the inverse kinematics but I will be using the geometric method to find the the position and the Pieper's method to find the angles.

How is this manipulator fully controllable as opposed to all the other manipulators we have dealt with?

What does fully controllable have to do with using the Pieper's method?

Answer 70

I'm actually not really sure what is meant by fully controllable and how that relates to pieper's method. What difficulty are you having in understanding the rest of the problem?

Answer 71

#3 starts out asking to 'Work out' the inverse kinematics and tells me to find the solution for the plane using the Pieper's method and find the position part using the geometric method.

The second paragraph talks about solving the problem by spliting the forward kinematics into the gw and gh parts. Giving gw = (ge*)(g4)^-1.

This is as much as I understand the question to be asking.

Answer 72

Yes that is what the question is asking. The main goal is to solve the inverse kinematics using Pieper's method. Your solution from problem 2 should be of use. That's all he's saying.

Answer 73

In #2, I solved the forward kinematics with two equations. xe = l's and cos and ye = l's and sin.

Answer 74

In 2b you solved inverse kinematics

Answer 75

Solving for inverse kinematics is just finding the alphas.

Answer 76

Yes.

Answer 77

Can I not do that in #3 without going through the gw and gh steps?

Answer 78

You can try but I think you will find it's hard.

Answer 79

\[g _{w} = g _{e}^{*}g _{4}^{-1} = g _{1}(\alpha _{1})g _{2}(\alpha _{2})g _{3}\]

Answer 80

Yes.

Answer 81

What does \[g _{e}^{*}\] mean

Answer 82

Desired end effector configuration.

Answer 83

So I am suppose to have some known end effector configuration? I do not have a known end effector configuration unless I work out the forward kinematics. Would I use the form I used in #2a (l's, cos and sin) or would I use the homogen. form?

Answer 84

State the problem of inverse kinematics.

Answer 85

I do not follow.

Answer 86

When someone asks you to work out inverse kinematics, what are they asking?

Answer 87

They want me to find the angles of the manipulator joints.

Answer 88

For a given end effector position.

Answer 89

In problem 2a, you solved forward kinematics(end effector configuration for given alpha) even though you were not actually given the alphas. You solved it symbolically such that any alphas could be plugged in. This is exactly the same thing, except what is given and what is desired is reversed.

Answer 90

I follow what you are saying. I am having difficulty doing it.

Answer 91

You wrote the equation for gw right? Using this equation and the given information you can figure out the position of the wrist.

Answer 92

I start with \[g _{e}^{*}g _{4}^{-1} = g _{1}(\alpha _{1})g _{2}(\alpha _{2})g _{3}\] I know \[g _{e}(\alpha) = g _{1}(\alpha _{1})g _{2}(\alpha _{2})g _{3}(\alpha _{3})g _{4}\]

Answer 93

I don't think you understood what I was saying. ge* is given. g4 is given. You can get gw.

Answer 94

g4 is a translation over the distance l3.

Answer 95

Yes.

Answer 96

If I multiply a translation by the inverse of that translation would I get a unit vector?

Answer 97

as in

g*g^-1 = g1(alpha1)g2(alpha2)g3(alpha3)g4 * (g4)^-1

Answer 98

Yes of course multiplying a matrix by its inverse gives you the identity. What do you mean?