Question

hw6 problem 1, what does poles have to do with this problem? All I have is a fixed frame of reference , an object's location point and a transformation.

Answer 1

is the object the pole?

Answer 2

Remember the pole is the point about which a transformation can be expressed as a pure rotation. Another way of saying this is that the pole is a point that has the same coordinates in both the initial and final frames of the transformation. In problem 1, you are given a transformation, gBC. You want to find the pole for this transformation.

Answer 3

from hw 5, I worked out the equation of the pole in terms of d and R. knowing the transformation, in hw 6, I can do the same. the pole is located at the same point in all frames. would this not mean that tha answer to part a is the answer to part b? meaning it does not matter which frame I look at the pole from I would get the same result.

Answer 4

The pole is not the same in every frame as I mentioned last time. The pole has the same coordinates in the initial and final frame, in this case frame B and frame C. It DOES NOT have the same coordinates in Frame A or any general frame you might think of.

Answer 5

to find the pole in frame A, I would go from frame A to B times the transformation from frame B to C times the pole in reference to frame C. I would write the equation as I travel from the starting point (frame A) to the ending point (pole in frame C)?

Answer 6

You have the pole in frame B right?

Answer 7

You may want to do part B first.

Answer 8

I thought for the pole in frame C was the same as the pole in frame B

qBp = qCp

Answer 9

Yes that's correct. You have qBp and gAB. It should be trivial to get qAp right? You've been doing that all semester.

Answer 10

yes, thanks

Answer 11

for hw 6 #2, \[\xi\] is given as (d, R). Can the equation

\[\xi O1 = gOA * \xi A1\]

be multiplied the same way as

gO1 = gOA*gA1

Answer 12

Yes, in general what you wrote is true but be careful about how the star operations between a transformation and a twist is defined.

Answer 13

for \[\xi A1 = (2,2,\pi /10)\]

and

\[gOA = (7,3,\pi /2)\]

does \[gOA*\xi A1 = ({5,5}, \left[\begin{matrix}0 & -1 \\ 1 & 0\end{matrix}\right]*\left[\begin{matrix}\cos(\pi/2) & -\sin(\pi/2) \\ \sin(\pi/2) & \cos(\pi/2)\end{matrix}\right])\]

Answer 14

\[g^O_A*\xi^A_1 = \left[\begin{matrix}R^O_A & 0 \\ 0 & 1\end{matrix}\right]*\left(\begin{matrix}v \\w \end{matrix}\right)\]

Answer 15

Hw6 #3, if B has a velocity would not C have the same velocity?

Answer 16

Think about a hammer...if you keep the base still and strike a nail does that base and the head have the same velocity?

Answer 17

earlier this was established: qBp = qCp, is there something else we need to show for part B? I am a little confused what we are starting qp as

Answer 18

I used the formula that was derived for homework 5number 2. where \[qp=(1-R)^-1*d\] and used that for part b in this problem. I'm not sure if that is right but I considered part b to be my qbp and used that in part a to find qap

Answer 19

for Hw6 #3, all I have is gBC and ξbB
so I would think that to find
ξbC
I would do
\[ξbC=gBC ^{-1}∗ξbB ^{-1}\]

but does the inverse of ξbB make sense?

Answer 20

cmcmillan: your approach was correct. pvpman...does that answer question?

Answer 21

I think so, does that mean qp=(1−R)−1∗d using R2 and d2 would be creating qBp? Then as qp is a pole then qCp = gCB*qBp

Answer 22

bebo: close but you need to convert this back into C's frame...so you just need one more factor at the end. This should look like an adjoint operation. Anytime you see two points on the same rigid body and you know the motion of one...think adjoint.

Answer 23

pvpman: correct

Answer 24

Well hold on a second...you want qPA, not qPC. qPC = qPB by definition right

Answer 25

well I was thinking of getting part b first, then using qBp I can find qAp = gAB*qBp = d1R1qBp, right?

Answer 26

Yes that's right...but I think you'll find that applying qCB to qBp will give you the same coordinates back. Remember that the pole is defined as the point which has the same coordinates in the initial and final frames of the transformation.

Answer 27

gCB is what I meant by qCB

Answer 28

oh I see the question asks for it in frame B, which I found earlier to be qBp = (1−R2)−1∗d2, I don't need qCp because that was already proven in the last HW

Answer 29

Well you don't need it but you already have it anyway. qCp = qBp

Answer 30

I am a little confused with question 3 as well, is this what you mean?
\[ξ_{C}^{b}=g_{B}^{C}*(\xi _{B}^{b}) ^{-1}*g_{C}^{B}\]
If so, how do you inverse a vector?

Answer 31

No that is not what I mean...sorry I didn't notice the inverse of the vector. The rest makes sense though right?

Answer 32

Another way to write it that might make it more clear....\[\xi^C_C = g^C_B*\xi^B_B*g^B_C\]That way the letters cancel.

Answer 33

ok yeah that makes more sense, that means the lowercase b(the body) is the frame of both euclideans here (the superscript)?

\[ξ_{C}^{b}=g _{B}^{C} ∗ξ_{B}^{b}∗g _{C}^{B}\]

Answer 34

Well remember that the body velocity of a frame is the velocity of the frame seen from that frame. So another way that to write the body velocity that makes sense is\[\xi^b_C = \xi^C_C\]\[\xi^b_B = \xi^B_B\]

Answer 35

ok, a little confusing but I think I get it now

Answer 36

Yeah the description of velocities in this context takes some time to get used to.

Answer 37

For prob 3, I have ξbC = gcB * ξBB * gbC, that's correct right?

Answer 38

Yes but write each matrix symbolically,,, like R d 0 1. You can use the equation editor

Answer 39

right, gcB = \[\begin{vmatrix} 0.866, -0.5 , 0 \\ 0.5 ,0.866, 0 & \\ 0,0,1 \end{vmatrix}\]

Answer 40

No, symbolically. Like this\[\left[\begin{matrix}R & d \\ 0 & 1\end{matrix}\right]\]

Answer 41

actually never mind, haha I had a neg in the wrong place, now my main diagonal is all zeros, that's correct right?

Answer 42

but my gcB was \[\begin{bmatrix} R(-\pi /6), 0 \\ 0 , 1 \end{bmatrix}\] and gbC \[\begin{bmatrix} R(\pi /6), 0 \\ 0 , 1 \end{bmatrix}\]

Answer 43

Yes...what I was trying to get you to see was that \[\hat \omega = R* \hat \omega * Rinv\]so the omega hat should not change...which should also make sense because the angular velocity would be the same at both points on the rigid body.

Answer 44

I see, so the w_new = w_old, but only the linear velocity changes right?

Answer 45

Yes.... in the 2D case the R and the Rinv cancel...but this is a special situation, That wouldn't normally happen.

Answer 46

kk, yes I think Vela gave us a formula to calc for the 3D case, but I got the ans for this hw problem, thank you!

Answer 47

np