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Math Help - robotics and simple ik

  1. #1
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    robotics and simple ik

    Question: how do we solve SYSTEM 2 (see bellow) to prove SYSTEM 3?
    -----------------------------------------------------------
    hello,
    im having a little trouble with robotics, inverse kinematics and its equations
    if you do have heard of robotics IK, then this should be fairly easy
    its the simplest occasion (
    planar 2-link/2-dof/2r manipulator) from the introduction to kinematics (robotics)
    -----------------------------------------------------------
    [L1, L2] = length of each link, [1, 2] = angles, [cos12] = cos(1 + 2), [sin12] = sin(1+2)
    -----------------------------------------------------------
    SYSTEM 1 (forward kinematics equations)
    -----------------------------------------------------------
    px = L1cos1 + L2cos12

    py = L1sin1 + L2sin12
    -----------------------------------------------------------
    SYSTEM 2 (rearrangement of system 1)
    -----------------------------------------------------------
    px = (L1 + L2cos2)cos1 + (-L2sin2)sin1

    py = (L2sin2)cos1 + (L1 + L2cos2)sin1
    -----------------------------------------------------------
    SYSTEM 3 (system 2 solved for sin1, cos1)
    -----------------------------------------------------------
    sin1 = px (-L2sin2) + py (L1 + L2cos2)

    cos1 = px (L1 + L2cos2) + py (L2sin2)
    -----------------------------------------------------------

    Question: How do we get from SYSTEM 2 to SYSTEM 3?

    of course i dont expect an analytical answer, i wont bother you that much
    i just hope for some internet directions since i dont know the correct english terms to run a proper search

    the problem is that i have found different answers/ solutions for sin1, cos1 in different books but not one has the proof
    -----------------------------------------------------------
    thank you
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  2. #2
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    Quote Originally Posted by sklnner View Post
    Question: how do we solve SYSTEM 2 (see bellow) to prove SYSTEM 3?
    -----------------------------------------------------------
    hello,
    im having a little trouble with robotics, inverse kinematics and its equations
    if you do have heard of robotics IK, then this should be fairly easy
    its the simplest occasion (
    planar 2-link/2-dof/2r manipulator) from the introduction to kinematics (robotics)
    -----------------------------------------------------------
    [L1, L2] = length of each link, [1, 2] = angles, [cos12] = cos(1 + 2), [sin12] = sin(1+2)
    -----------------------------------------------------------
    SYSTEM 1 (forward kinematics equations)
    -----------------------------------------------------------
    px = L1cos1 + L2cos12

    py = L1sin1 + L2sin12
    -----------------------------------------------------------
    SYSTEM 2 (rearrangement of system 1)
    -----------------------------------------------------------
    px = (L1 + L2cos2)cos1 + (-L2sin2)sin1

    py = (L2sin2)cos1 + (L1 + L2cos2)sin1
    -----------------------------------------------------------
    SYSTEM 3 (system 2 solved for sin1, cos1)
    -----------------------------------------------------------
    sin1 = px (-L2sin2) + py (L1 + L2cos2)

    cos1 = px (L1 + L2cos2) + py (L2sin2)
    -----------------------------------------------------------

    Question: How do we get from SYSTEM 2 to SYSTEM 3?

    of course i dont expect an analytical answer, i wont bother you that much
    i just hope for some internet directions since i dont know the correct english terms to run a proper search

    the problem is that i have found different answers/ solutions for sin1, cos1 in different books but not one has the proof
    -----------------------------------------------------------
    thank you
    Do you know how to solve two linear equations in two unknowns, say by elimination or substitution or using matrices? If not, try this Google search on

    solving 2 linear equations in 2 unknowns elimination.

    System 2 is equations of this form

    \begin{matrix} p_x = a x + by \\ p_y = cx + dy \end{matrix}<br />

    where x = \cos 1,\ y = \sin 1,\ a = L_1 + L_2 \cos 2, etc.

    In matrix form

    \begin{bmatrix} p_x \\ p_y \end{bmatrix} =<br />
\begin{bmatrix} a & b \\ c & d \end{bmatrix}<br />
\begin{bmatrix} x \\ y \end{bmatrix}.<br />

    The solution is

    \begin{bmatrix} x \\ y \end{bmatrix} =<br />
{\begin{bmatrix} a & b \\ c & d \end{bmatrix}}^{-1}<br />
\begin{bmatrix} p_x \\ p_y \end{bmatrix}<br />
=\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}<br />
\begin{bmatrix} p_x \\ p_y \end{bmatrix}<br />
= \begin{bmatrix} d p_x - b p_y \\ -c p_x + a p_y \end{bmatrix}.<br />

    Back to non-matrix form, System 3 is

    \begin{array}{l}x = d p_x - b p_y \\ y = -c p_x + a p_y \end{array}.

    But note the order of the equations is reversed as you have written it:

    \begin{array}{l}y = -c p_x + a p_y \\ x = d p_x - b p_y \end{array}.
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  3. #3
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    Jun 2007
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    A - thnx a lot
    -----------------------------------
    B - i tried it with matrices
    but isnt the inverse matrix of this:
    ---
    a b
    c d
    ---
    this?:
    ---
    1/(ad-bc) *

    d -c
    -b a
    ---
    that was my problem, i dont see how (ad-bc) could be 1 in order for the inverse matrix to be without [1/(ad-bc)]
    but i must be saying something wrong here(?)
    -----------------------------------
    C
    if you visit here (its not spam, just another site):

    Protein Inverse Kinematics and the Loop Closure Problem

    press ctrl+f (browser search) and enter this phrase: "click for solution"
    you can see that the solution for sin1 / sin(θ1) (last line of the solution text) is the same but with a denominator which is equal to this:
    px^2 + py^2
    which is different than 1, which means one of the two solutions is wrong(?)
    -----------------------------------
    D
    i cant ask for anything more from you JakeD after the long answer you bothered to write, so i ll practise it some more to find where my silly mistakes are, and i ll get back
    -----------------------------------
    E
    can i too generate these images? i tried but couldnt get it to work - is there any guide in the forum for this?
    must i change the script lang somehow?
    -----------------------------------
    again thnx
    edit: its late here, so if anyone replies know that i ll probably check in again tomorrow
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  4. #4
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    Quote Originally Posted by sklnner View Post
    A - thnx a lot
    -----------------------------------
    B - i tried it with matrices
    but isnt the inverse matrix of this:
    ---
    a b
    c d
    ---
    this?:
    ---
    1/(ad-bc) *

    d -c
    -b a
    ---
    that was my problem, i dont see how (ad-bc) could be 1 in order for the inverse matrix to be without [1/(ad-bc)]
    but i must be saying something wrong here(?)
    -----------------------------------
    C
    if you visit here (its not spam, just another site):

    Protein Inverse Kinematics and the Loop Closure Problem

    press ctrl+f (browser search) and enter this phrase: "click for solution"
    you can see that the solution for sin1 / sin(θ1) (last line of the solution text) is the same but with a denominator which is equal to this:
    px^2 + py^2
    which is different than 1, which means one of the two solutions is wrong(?)
    -----------------------------------
    D
    i cant ask for anything more from you JakeD after the long answer you bothered to write, so i ll practise it some more to find where my silly mistakes are, and i ll get back
    -----------------------------------
    E
    can i too generate these images? i tried but couldnt get it to work - is there any guide in the forum for this?
    must i change the script lang somehow?
    -----------------------------------
    again thnx
    edit: its late here, so if anyone replies know that i ll probably check in again tomorrow
    You are right about the missing factor 1/(ad-bc) in the inverse. I was looking at System 3 and trying to justify it.

    Now

    ad-bc = (L_1 + L_2 \cos \theta_2)^2 + (L_2 \sin \theta_2)^2 = L_1^2 + 2 L_1 L_2 \cos \theta_2 + L_2^2

    using \sin^2 \theta_2 + \cos^2 \theta_2 = 1.

    Looking at



    in the page you provided, I think I see why ad-bc might equal 1. Let X_1, X_2 be the coordinate vectors for the links. Then from the picture, the angle between these vectors is angle \theta_2. So

    \cos \theta_2 = X_1^T X_2/L_1 L_2

    where L_i = || X_i || = \sqrt{ X_i^T X_i} and thus

    L_1^2 + 2 L_1 L_2 \cos \theta_2 + L_2^2 = X_1^T X_1 + 2 X_1^T X_2 + X_2^T X_2 = (X_1 + X_2)^T (X_1 + X_2) = || X_1 + X_2 ||^2.

    But the vector X_1 + X_2 = (x,y) in the picture. If this vector is assumed to have length || X_1 + X_2 || = 1, then ad - bc = 1.

    Apparently (x,y) = (p_x,p_y) in your equations, so that ad -bc = ||p||^2 = p_x^2 + p_y^2, which explains why this would be in the denominator if (p_x,p_y) is not assumed to have length 1.

    For the mathematical script, see the http://www.mathhelpforum.com/math-he...-tutorial.html.
    Last edited by JakeD; June 13th 2007 at 10:01 PM.
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  5. #5
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    Joined
    Jun 2007
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    sorry for abandoning my own topic, yesterday was really busy
    i didnt have the chance to read the tutorial yet but i will
    ---
    as for the problem, we agree
    the 2 solutions are the actually the same if in the first one the quantity "px^2 + py^2" (or the hypotenuse/length from (0,0) to x,y) is assumed to be 1
    ---
    one problem is im pretty sure its not 1 and now i have found enough sources to say that the denominator should be there expressed in either of the two forms: ~(px, py) or ~(L1, L2)

    i ll be examined on this book soon and every solved problem in it solves the 2x2 systems as if 1/(ad-bc) was 1 and im talking about totally different manipulators

    so i dont know what to do, most sources in the internet use another, 3rd way splitting the angle θ1 and solving for two more angles but i cant understand this yet because i dont know anything about producing the atan2 function which is needed
    ---
    anyway thanx a lot JakeD, i ll keep trying to figure this out
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