Question

on hw#5, I am not clear as to what is being asked for in problem 1a.

Answer 1

in hw#1 the hammer was at (6,0) and rotated (in frame B) -pi. in hw#3 the hammer was at (1,0) and rotated (in frame B) -pi. nowhere was the hammer rotated -pi/2.

Answer 2

I'll answer your question this evening.

Answer 3

Ok thanks

Answer 4

Dr. Vela is defining the relationship between the base of the hammer and the hammer head in the problem description. This hasn't really been done before. He is only saying that the transformation from the base to the head has a d of [1;0] and a rotation by -pi/2. Does that help?

Answer 5

Am I to reference any prior homework to solve 1a?

Answer 6

Is the problem asking me to rotate and translate the world frame, O, to the frame A (from earlier homework #3) and then to translate and rotate that frame to the location described in the homework and call that frame A'?

Answer 7

In homework 3, you are given the transformations from O to A and B, which are each the base of the hammer. Now in this homework you are given the transformation from the base of the hammer to the head. Question 1A is asking you to compute the overall transformation from O to the hammer head in A and from O to the hammer head in B.

Answer 8

1a has nothing to do with 'adjoint'

Answer 9

Nope not really

Answer 10

thanks I will try to continue on the rest this evening

Answer 11

Sounds good

Answer 12

I took gOA, I calculated in Hw3, and multiplied it with the hammer head point and rotation

gOA*gHAMMER


gHAMMER = ({1 0 }, R)

and got a different answer than

gHAMMER*gOA

this tells me I am doing something incorrect in Hw5 1a

Answer 13

I am thinking of gHAMMER as a frame

Answer 14

never mind I got it

Answer 15

Yeah gOA*gHammer should not equal gHammer*gOA

Answer 16

in 2a is the response

qpOA = ({0 0}, R(theta)) ?

Answer 17

or

gAB = ({0 0}, R(theta))

Answer 18

gAB = ({x y}, R(theta))

where x and y are constants for all values of theta

Answer 19

none of that worked, how should I be approach 2a?

Answer 20

Ok so in 2A, you want to describe the following statement: A transformation of a certain point returns that same point.

Answer 21

the pole is the point in the plane whose
coordinates do not change when the point is rigidly transformed by g

Answer 22

all I can think of is

g = ({0 0}, and the rotation is any angle)

Answer 23

Hmmm ok start by writing out the equation of a transformation of a point by g.

Answer 24

Conceptually, we're not trying to figure out g...we're trying to figure out qp for a given g. So you want to write an equation taht relates g and qp, not a specific g

Answer 25

qp = gAB*qp

Answer 26

Right...now the only other thing we have to say is that the coordinates of the point do not change after the transformation...

Answer 27

Well actually I suppose you've already done that in your equation...so that's basically it

Answer 28

2b I have

qp = d +R*qp

but what does it mean by what frame was it computed in?

Answer 29

If you write superscripts and subscripts on the variables, the frame it's calculated in is the superscript over qp. However, problem 2 asks to solve for qp, meaning you can not use qp as part of it's definition. In other words, you have to do some algebra to isolate qp

Answer 30

qp = gAB*qBB

the B's cancel leaving

qp = qAB

and the frame is A

Answer 31

What you've essentially written is qAp = gAB*qBp and that qAp = qBp...we already knew that. What you want to do is define qAp=qBp in terms of R and d of g. As the problem states, you should first write out your previous equations in terms of d and R.

Answer 32

You've done that here: qp = d +R*qp. Now just express qp only in terms of d and R...not in terms of itself.

Answer 33

if I do this

qp = gAB*qp
= d + R*qp
qp - R*qp = d
qp = d/(1 - R)

now qp is in terms of d and R

Answer 34

Close but remember that you're dealing with matrices. For one thing, you can't divide matrices (you can only multiply by the inverse). And the other thing is that the side you multiply on is important so that inv(A)*A*B = B but A*B*inv(A) does not equal B.

Answer 35

how do you subtract a matrix from 1 as in (1 - R)?

Answer 36

The 1 in this case would be the identity matrix, with 1s on the diagonal and 0s everywhere else.

Answer 37

qp = d*inv(1 - R)

Answer 38

Almost there but when you factored out qp on the left side, I think you factored it out on the wrong side.

Answer 39

qp = -d*inv(R - 1)

Answer 40

Nope... what did you get when you factored out qp on the left side of the equation?

Answer 41

qp*(1-R)

Answer 42

qp = inv(1-R)*d

Answer 43

qp = gAB*qp
= d + R*qp
qp - R*qp = d
(1 - R)*qp = d
inv(1 - R)*(1 - R)*qp = inv(1 - R)*d
qp = inv(1 - R)*d

Answer 44

yep that looks good to me

Answer 45

where 1 = I

Answer 46

what frame does it mean? in what frame ? since qp is the same in all frames, does a frame apply?

Answer 47

It's not the same in all frames, it's only the same in the two frames which the g relates. I think Dr. Vela is trying to make you see taht the point you found is in the frame A or frame B, but not in the observer frame.

Answer 48

for 3a do I multiply the vector by the rotation matrix to get the vector in relation to the frame?

Answer 49

Yes that's correct.