Question

I do not understand what Hw 7 problem 1 is asking. Am I suppose to multiply
g = (x y theta)^T
by the se(2) vector
they are both 3x1 matrix.

What is special about the * multiplication?

Answer 1

For #1 I think I should

\[g \prime = g * \xi = \left[\begin{matrix}R(\theta) & d \\ 0 & 1\end{matrix}\right] * \left(\begin{matrix}\xi1 \\ \xi2 \\ \xi3 \end{matrix}\right)\]

Answer 2

Close but not quite...remember last week we defined the transformation*twist vector differently than the regular g.

Answer 3

\[\left[\begin{matrix}\omega & v \\ 0 & 0\end{matrix}\right]\]

where
\[\omega = \left[\begin{matrix}0 & -\xi 3 \\ \xi 3 & 0\end{matrix}\right]\]

Answer 4

No no...what you had was basically right but something is wrong about the g part. One part is there that shouldn't be when you're multiplying g by xi.

Answer 5

\[d = \left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

Answer 6

Yeah d should be 0

Answer 7

why

\[d = \left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

Answer 8

this tells me the derivative of a velocity vector \[\xi \] is just a rotation

Answer 9

Keep in mind we are just transforming a velocity vector...not taking its derivative. We're simply expressing the velocity vector in a different frame. If you think about it, the distance between the two frames should not affect the representation of the velocity but the rotation should.

Answer 10

so the distance between two frames does not effect the velocity but the rotation between the frames will. I do understand this. what does the derivative have to do with this problem? And how do you get a set of differential equations from a rotation?

Answer 11

You can think of the left side as just \[\xi_{new}\]You are just being asked to find it symbolically by multiplying out the right side. g' is just another name for xi.

Answer 12

for #2 \[\omega \]hat
always = 0
so using the Rodregess formula
the exp =1

Answer 13

the magnitude of \[\omega \] hat = 0 not \[\omega \]

Answer 14

How are you calculating \[mag(\omega)?\]

Answer 15

I was taking the magnitude of omega hat, which I now know is incorrect. I need to take the magnitude of omega.

I know omega is a vector but what vector that I can take a magnitude of?

Answer 16

You can extract omega from omega_hat by looking at the structure of omega_hat and how it's defined. You should have in your notes a definition of omega_hat in terms of omega1, omega2, and omega3

Answer 17

Remember in last week's homework you were asked to do that.

Answer 18

[0 -z y]
[z 0 -x]
[-y x 0]

Answer 19

yes so z = omega3 y = omega2 and x = omega1

Answer 20

so the magnitude is square root of x^2+y^2+z^2

Answer 21

correct

Answer 22

is the final answer in the form of a matrix or a number?

Answer 23

What do you think?

Answer 24

the answer should be in the form of a matrix.

Answer 25

Yep

Answer 26

in problem #3 we just take the ln(R) and the answer is a 3x3 matrix.

Answer 27

the sin should be in radians

Answer 28

Yes but to be sure you did it right, you should get something that looks like #2 because the ln is the inverse of the exponential. You should be able to "unhat" it also.

Answer 29

should the angles be in degrees? I say no they should be in radians.

Answer 30

Well in the final answer, it should be angular velocity, not angles. It doesn't matter if it's degrees or radians. Your choice

Answer 31

Actually just stick with radians.

Answer 32

when you take the sin and cos should the argument be in radians?

Answer 33

Yes the R is given in radians for sure.

Answer 34

this applies to #2 as well?

Answer 35

Yes everything is given in radians.

Answer 36

thanks

Answer 37

Np

Answer 38

And also all Taus are assumed to be 1 if not given

Answer 39

problem #3 I use

[r32-r23]
(1/2sin(2tau)) * [r13-r31]
[r21-r12]

Answer 40

I mean

[r32-r23]
(1/2sin(tau)) * [r13-r31]
[r21-r12]

Answer 41

I tried this and omega is a 3x1 vector. do I hat it to get a 3x3 matrix?

Answer 42

You should be multiplying the factor out in front by (R-R^T). R-R^T is a 3x3 matrix so the answer also should be.

Answer 43

for problem #4, in the leftact.m file what does 'type' mean in
function x2 = leftact2(g, x, type) ? I do not understand how it is used.

Answer 44

with the new leftact.m, the program will recognize a 2x1 vector and a 3x1 vector but no other combination.

Answer 45

Yes this last part makes sense. The leftact operation applies to transformations of points. Do you still need me to clear up the first question?

Answer 46

yes. what is type for and how is it used? I entered
g = SE2([x y], theta)
x = [5 6 1]'
leftact(g,x)

my errors Error in SE2/mtimes (line 13)
g.d = g1.d + g1.R * g2.d;

Error in SE2/leftact (line 31)
x2 = g * x;

Answer 47

I can multipy a 3x3 matrix by a 3x1 matrix (velocity) using *, but when I do this in leftact.m I get an error say I have violated mtimes.m. why do I need to use mtimes in leftact?

Answer 48

Ok I'm confused. Let's start from the beginning. did you modify the SE3 file so that it works?

Answer 49

oh hold on we're doing SE2 still

Answer 50

What am I looking for in #4?

Answer 51

>> g = SE2([10 15], pi/3)
0.5000 -0.8660 10.0000
0.8660 0.5000 15.0000
0 0 1.0000

>> x = [5 6 1]'

x =

5
6
1

>> leftact2(g,x)
Error using *
Inner matrix dimensions must agree.

Error in SE2/leftact2 (line 31)
x2 = g.d + g.R * x;

Answer 52

So in this case your g.R is 2x2 and your x is 3x1...that's why you're getting the error. you should have code that accounts for the case where the input is a 3x1.

Answer 53

How do you calculate the new point if the original point is given in homogeneous form?

Answer 54

I do understand that R is a 2x2 and x is a 3x1. Do i code in leftact.m to partition x into \[\hat \omega\] and v?

Answer 55

\[\dot x\] \[\dot y \] and \[\dot \theta \]

Answer 56

\[x = ([\hat \omega ], v)\]

Answer 57

So leftact can take 3 things for x:
1) A 2x1 vector representing a point
2) A 3x1 vector representing a point(x;y;1)
3) A 3x1 vector representing a velocity(vx;vy;0)

In each case it should multiply the g by the x. I think the easiest way to do this for parts 2 and 3 is to make g a homogeneous form matrix(note the difference from SE2 object) and then multiply it by x directly. Then you have a 3x3 times a 3x1 and you'll get a 3x1 homogeneous answer returned.

Answer 58

Should not the vector point be (x;y;z) and the velocity vector be (x,y,theta)? Why do you put a 1 in place of z and a zero in place of theta?

Answer 59

This is SE2...homogeneous representation of a point is [x;y;1] what would the z represent. The velocity vector is just linear velocity in homogeneous form, not including angular velocity.

Answer 60

I see, SE2 is movement or position in 2 space.

Answer 61

Yes SE2 is special euclidian 2-space and SE3 is special euclidian 3-space.

Answer 62

The task is not to take x down to a 2x1 but to take g up to a 3x3.

Answer 63

Yes...in my opinion. The reason is that a star operation would take different forms for point and velocity x. But the matrix operation is the same whether it's a point or velocity.

Answer 64

the result of g*x is the same but how that result is interpreted determines if the product is a point in 2 space or a velocity in 2 space

Answer 65

Yes exactly but it's up to the user to interpret the result...you just need to provide the result.

Answer 66

I will need to figuer out how to turn a 2x2 into a 3x3

Answer 67

figure

Answer 68

You know how to construct a homogeneous matrix in SE2 yes?

Answer 69

That is not fresh in my mind

Answer 70

g in matrix form is \[\left[\begin{matrix}R & d \\ 0 & 1\end{matrix}\right]\]

Answer 71

yes. I meant on the Matlab side

Answer 72

[g.R g.d; 0 0 1]

Answer 73

Also what if I enter x as a point (x,y,1). R = [cos(1) -sin(1)]
{sin(1) cos(1)]

This does not translate well

Answer 74

x is not an SE2 object...x is a vector. x = [x;y;1]

Answer 75

I would be using code similar to SE2.m in creating x.R and x.d. I would need to enter x in as

x = SE2([x y], 1);

this is as if I were entering another g

Answer 76

No. x is a point, not an SE2 object. SE2 objects can be thought of as frames and transformations. g is an SE2 object. x is a vector

Answer 77

x has no rotation. Here's the multiplication you want to do: \[\left[\begin{matrix}g.R & g.d \\ 0 & 1\end{matrix}\right]*\left(\begin{matrix}v1 \\ v2 \\ v3\end{matrix}\right)\]

Answer 78

Where the vector in the right is either \[\left(\begin{matrix}x \\ y \\ 1\end{matrix}\right)\]or \[\left(\begin{matrix}v_x \\ v_y \\ 0\end{matrix}\right)\]

Answer 79

I see, I do not creat an x.R or an x.d

Answer 80

No

Answer 81

My x is a 3x1 vector and it stays a 3x1 vector. Thanks for your patience

Answer 82

No problem

Answer 83

I got in running. Thanks

Answer 84

Cool

Answer 85

I do the same for updating the adjoint.m file

Answer 86

Write out what you plan to multiply to get that adjoint.

Answer 87

if (isa(x,'SE2'))
% if x is a Lie group, then deal with properly.
z = g*x*inv(g);
elseif ( (size(x,1) == 3) && (size(x,2) == 1) )
% if x is vector form of Lie algebra, then deal with properly.
a = [g.R g.d; 0 0 1];
z = g*x*inv(g);
elseif ( (size(x,1) == 3) && (size(x,2) == 3) )
% if x is a homogeneous matrix form of Lie algebra, ...
a = [g.R g.d; 0 0 1];
z = g*x*inv(g);
end

end

Answer 88

You're defining a but never using it. Also in the second case, where x is a 3x1...think about what you're multiplying.

Answer 89

if (isa(x,'SE2'))
% if x is a Lie group, then deal with properly.
z = g*x*inv(g);
elseif ( (size(x,1) == 3) && (size(x,2) == 1) )
% if x is vector form of Lie algebra, then deal with properly.
a = [g.R g.d; 0 0 1];
z = a*x*inv(a);
elseif ( (size(x,1) == 3) && (size(x,2) == 3) )
% if x is a homogeneous matrix form of Lie algebra, ...
a = [g.R g.d; 0 0 1];
z = a*x*inv(a);
end

end

In case 2 and 3, I am multiplying a*x*inv(a). a is a 3x3, x is a 3x1 and inv(a) is a 3x3. That is a bad order to multiply matrices. I will have to do a*inv(a)*x.

Answer 90

a*inv(a) is just the identity. You !cannot! move around matrices arbitrarily. Matrix multiplication is not commutative in general. You need to find another solution.

Answer 91

In part 3 x is a 3x3 so that actually makes sense right.

Answer 92

[0 -z y]
[z 0 -x]
[-y x 0]

Answer 93

x is a transformation. You want to describe a transformation as a 3x3 matrix.

Answer 94

For the second part

Answer 95

\[x = \left[\begin{matrix}R(-\theta) & -R(-\theta)d \\ 0 & 1\end{matrix}\right]\]

Answer 96

why?

Answer 97

You seem to have written the inverse of a transformation. x is just a transformation.

Answer 98

\[x = \left[\begin{matrix}R & d \\ 0 & 1\end{matrix}\right]\]