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|-+  General Category - Industrial Robot Forum
| |-+  Robot Geometry, Linear Algebra, Forward and Inverse Kinematics
| | |-+  Calculating position
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Author Topic: Calculating position  (Read 5245 times)
asimo
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« on: November 18, 2008, 03:31:11 AM »

Hi,

I've started this thread to continue on the one here:
http://www.robot-forum.com/robotforum/kuka_robots/krl_undocumented_features-t1174.0.html

There was a question on matrices and I think that it has become too much off topic from the original: KRL Undocumented Features. Because I would like this thread to continue for the undocumented features, I suggest that we continue the discussion of calculation here. I hope that I will not offend people on this thread!

Continuing on mookie's question:
Quote
you guys make my head hurt..

can anybody confirm or deny jims assessment of it? its drawn in pretty little pictures that help me think a little bit.. next to breasts, pictures of homogenous transformations is the next best thing.


...and Jim's post:
Quote
here's how (I think) I understand it:
                                                                                       Rx(C)         Ry(B)        Rz(A)         TRANS       
                                                                                 |   (60*PI/180)        0              0               100  |
POS here {x 100,y 200,z 300,a 30,b 45,c 60}  =    |           0          (45*PI/180)    0               200  |
                                                                                 |           0                0           (30*PI/180)  300  |
                                                                                 |           0                0              0                 1    | 

each column is a vector. Last value of each vector is a "scale factor" or unit vector.
Angles are in radians. Transpose does not matter as long as you stick to the same unit.
...


I would deny it (no offense Jim icon_smile)

I think that the matrix should look like this one instead:
Code:
|C(b)C(a)  S(c)S(b)C(a)-C(c)S(a)   C(c)S(b)C(a)+S(c)S(a)   x|
|C(b)S(a)  C(c)C(a)+S(c)S(b)S(a)  C(c)S(b)S(a)-S(c)C(a)    y|
|-S(b)      S(c)C(b)                       C(c)C(b)                      z|
|0            0                                0                                 1|

...where C=Cos and S=Sin. The first 3 columns represents the i,j,k unit vectors for the resulting axis system and the last represent the point coordinates.

Hope this helps.
« Last Edit: November 18, 2008, 03:34:19 AM by asimo » Logged
Jim Tyrer
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« Reply #1 on: November 18, 2008, 02:35:28 PM »

None taken, Asimo. I'm as intrigued as everyone else and never had time to learn to do this properly.
How about a new board dedicated to robot kinematics/geometry? (Not just Kuka)
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asimo
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« Reply #2 on: November 19, 2008, 05:24:03 AM »

Jim: Yes, nice suggestion.

I have recalculated the matrix with a software and I confirm that the result is good. This is the "magical mystery matrix". As is, you can find the orientation of any coordinates.
« Last Edit: November 19, 2008, 05:29:21 AM by asimo » Logged
SkyeFire
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« Reply #3 on: November 20, 2008, 02:21:56 PM »

That's interesting -- I flipped Asimo's matrix into Excel just to play around with, and I notice when the input XYZABC is all zeros, the output of the matrix is an Identity matrix.  Which I'm sure means something of Great Significance to those people who understand Matrix Algebra better than I do...    17
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asimo
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« Reply #4 on: November 21, 2008, 02:35:25 AM »

Hi SkyeFire,

Yeah, this is really funny. In the reality, it means that there is no rotation (all axes vectors are on themselves) and that there is no translations (it's all zero's).

But it's even better than that:
- Invert that matrix (M^-1) and you have the inverted position (INV_POS(FRAME or POS)).
- Multiply it by itself (M1 x M2) and you have the geometric operator (:)

That's why I called it the "magical mystery matrix"!
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bert
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« Reply #5 on: November 21, 2008, 02:42:34 PM »

My favourite web resource is the "Matrix and Quaternion FAQ"

www.j3d.org/matrix_faq/matrfaq_latest.html

It's a pretty no-nonsense webpage that actualy explains things quite nicely
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SkyeFire
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« Reply #6 on: November 21, 2008, 03:04:13 PM »

...wow.  I think I just learned more about how matrix math works in robot positioning in the last five minutes than I have in the last few months of on-and-off research.  I'm going to have to dig out my old Matrix Algebra textbooks in order to fully *grasp* what this means, but I think I can see just how *major* this is.   grinser043

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doug1880
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« Reply #7 on: December 02, 2008, 04:49:11 AM »

If you like matrices and vectors you may like to work thru Prof Gilbert Strang's course in Linear Algebra (MIT). Go to http://ocw.mit.edu/OcwWeb/Mathematics/18-06Spring-2005/VideoLectures/index.htm. All free.

You need RealPlayer for the videos - it's free.

You can download and save the video lectures for future viewing or storage on a DVD or CD.  See http://ocw.mit.edu/OcwWeb/web/help/faq4/index.htm#t3 for how to edit the URL in order to do it.

There's a great course in Electricity and Magnetism (2nd year physics) by MIT's GREAT(!) Prof Walter Lewin at http://ocw.mit.edu/OcwWeb/Physics/8-02Electricity-and-MagnetismSpring2002/VideoAndCaptions/index.htm. Ya should have a year or so of differential and integral calculus under your belt to really follow the material.  You shouldn't need mutivariable or vector calculus though (Div, Grad, Curl, Laplacian, etc), or diff equations either.

doug
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