1 1 4 Lecture Video 1 of 2 Displacement Vectors

00:22:57
https://www.youtube.com/watch?v=JwIPJkS3s24

Resumo

TLDRCette vidéo aborde le concept de vecteurs de déplacement dans le cadre des matrices de transformation homogènes, essentielles pour déterminer la position d'un effecteur terminal d'un manipulateur robotique. Elle explique comment un vecteur est défini en termes de magnitude et de direction, et comment il peut être représenté à l'aide de ses composants dans différentes dimensions. Des exemples pratiques illustrent la détermination des vecteurs de déplacement entre les cadres adjacents d'un manipulateur sphérique et d'un manipulateur Scara, mettant en lumière l'importance des longueurs des bras et des angles d'articulation. Finalement, la vidéo montre comment ces vecteurs sont cruciaux pour les calculs cinématiques dans les systèmes robotiques.

Conclusões

  • 📏 Un vecteur a à la fois une magnitud et une direction.
  • 🔄 La matrice de transformation homogène combine rotation et déplacement.
  • 📐 Les angles des jointures influent sur les déplacements en 3D.
  • 🧮 Les vecteurs de déplacement sont essentiels pour le positionnement des effecteurs terminaux.
  • 🔗 Les longueurs des liens sont cruciales pour les calculs de déplacement.

Linha do tempo

  • 00:00:00 - 00:05:00

    Nous avons été introduits à la matrice de transformation homogène, qui permet de calculer la position de l'effecteur terminal d'un manipulateur robotique en fonction des variables articulaires. Cette matrice se compose d'une matrice de rotation et d'un vecteur de déplacement. Nous avons également défini un vecteur comme une quantité ayant à la fois une magnitude et une direction, et montré comment exprimer le déplacement entre deux points par un vecteur.

  • 00:05:00 - 00:10:00

    Nous avons commencé à établir des vecteurs de déplacement entre les frames adjacentes du manipulateur, en annotant les diagrames cinématiques avec les variables articulaires et les longueurs de lien. Pour chaque paire de frames, nous avons déterminé le vecteur de déplacement dans l'espace tridimensionnel, en tenant compte des valeurs des angles et de l'ampleur des mouvements des joints du robot.

  • 00:10:00 - 00:15:00

    Ensuite, nous avons calculé le vecteur de déplacement de la frame 1 à la frame 2, en examinant l'effet des variations de l'angle Theta 2 sur la direction du déplacement. Nous avons appris que cette direction peut être déterminée à l'aide de relations trigonométriques, en utilisant les longueurs de lien pour exprimer les composantes x et y du déplacement, tout en notant qu'il n'y a pas de déplacement dans la direction z.

  • 00:15:00 - 00:22:57

    Enfin, nous avons analysé le vecteur de déplacement de la frame 2 à la frame 3, qui se situe entièrement dans la direction z, indépendamment de l'impression du joint. Les valeurs sont déterminées par la longueur du lien et l'extension du joint, confirmant le processus de calcul pour tous les vecteurs de déplacement nécessaires entre les frames du manipulateur.

Mostrar mais

Mapa mental

Vídeo de perguntas e respostas

  • Qu'est-ce qu'un vecteur ?

    Un vecteur est une quantité définie par sa magnitude et sa direction.

  • Comment exprimer un vecteur en 2D ?

    Un vecteur 2D est exprimé par ses composants x et y.

  • Quelle est la signification de la matrice de transformation homogène ?

    Elle calcule la position de l'effecteur terminal d'un manipulateur en fonction des variables articulaires.

  • Quelles sont les étapes pour trouver un vecteur de déplacement ?

    Identifier la distance et la direction entre deux points, et exprimer cela en fonction des variables des jointures.

  • Quelle est la relation entre les angles des jointures et les vecteurs de déplacement ?

    Les angles influencent les composants x et y des vecteurs de déplacement en 3D.

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Legendas
en
Rolagem automática:
  • 00:00:01
    previously we have been introduced to a
  • 00:00:04
    kind of Matrix called the homogeneous
  • 00:00:07
    transformation
  • 00:00:09
    matrix which allows us to calculate the
  • 00:00:12
    location of the end Defector of the
  • 00:00:14
    robot manipulator for any values of the
  • 00:00:17
    joint
  • 00:00:20
    variables we learned that a homogeneous
  • 00:00:23
    transformation matrix has two parts a
  • 00:00:27
    rotation Matrix
  • 00:00:30
    and a displacement
  • 00:00:39
    Vector we learned about rotation
  • 00:00:42
    matrices in the last set of videos today
  • 00:00:45
    we will learn how to write displacement
  • 00:00:49
    vectors first let's review what a vector
  • 00:00:53
    is a vector is generally defined as a
  • 00:00:57
    quantity that has both magnitud
  • 00:01:01
    and
  • 00:01:03
    Direction a typical way to express a
  • 00:01:07
    vector is by writing out the vector's
  • 00:01:11
    components for example if we are writing
  • 00:01:14
    a vector in two Dimensions we would
  • 00:01:17
    write a vector with two
  • 00:01:20
    elements the first element is the x
  • 00:01:24
    value and the second element would be
  • 00:01:27
    the Y value
  • 00:01:30
    we can use this approach to write the
  • 00:01:32
    displacement between two
  • 00:01:35
    points suppose that we say that the
  • 00:01:38
    first point is at x = 0 y =
  • 00:01:44
    0 and suppose that we have a second
  • 00:01:48
    point that is at x = 5 and Y =
  • 00:01:55
    8 then we could write the displacement
  • 00:01:58
    from the first point point to the second
  • 00:02:01
    Point as a vector and we would say that
  • 00:02:04
    that Vector is 5
  • 00:02:09
    8 and we would Express that Vector as
  • 00:02:12
    shown here by writing a five in the
  • 00:02:16
    first position and an eight in the
  • 00:02:18
    second
  • 00:02:20
    position can you see how this expression
  • 00:02:23
    has both magnitude and direction we can
  • 00:02:27
    see this more clearly if we picture this
  • 00:02:30
    Vector as a
  • 00:02:32
    triangle like
  • 00:02:39
    this now the direction of this Vector is
  • 00:02:43
    defined by the angle
  • 00:02:46
    here and the magnitude of this Vector is
  • 00:02:50
    defined by the length of this
  • 00:02:55
    hypotenuse now we could write the X and
  • 00:02:59
    Y positions of this Vector
  • 00:03:02
    differently the X position could be
  • 00:03:05
    written as the hypotenuse times the
  • 00:03:09
    cosine of the angle
  • 00:03:13
    Theta and the Y element could be written
  • 00:03:16
    as the hypotenuse times the S of the
  • 00:03:19
    angle
  • 00:03:22
    Theta now here is where we can start to
  • 00:03:26
    see how a vector is really useful for us
  • 00:03:29
    to to find the displacement between one
  • 00:03:32
    frame and the next frame in a kinematic
  • 00:03:37
    diagram in our kinematic diagram this
  • 00:03:41
    angle here will turn out to be the joint
  • 00:03:45
    angle of one of our robot's joints and
  • 00:03:49
    the length of this hypotenuse will turn
  • 00:03:52
    out to be the length of the
  • 00:03:55
    link you'll see how this is more when I
  • 00:03:58
    show you some examples
  • 00:04:02
    for my first example I'm going to show
  • 00:04:05
    you a spherical
  • 00:04:08
    manipulator I also need to draw in all
  • 00:04:11
    of the parts of this kinematic diagram I
  • 00:04:14
    need to label all of the coordinate
  • 00:04:18
    frames
  • 00:04:51
    and I need to draw in my joint variables
  • 00:05:07
    and I need to label all of the
  • 00:05:18
    links now when I find the displacement
  • 00:05:21
    vectors I have to find one displacement
  • 00:05:25
    Vector for each set of adjacent frames
  • 00:05:28
    just like we did with the rotation
  • 00:05:30
    matrices so I should find the
  • 00:05:34
    displacement from frame 0 to frame
  • 00:05:40
    1 I need to find the displacement from
  • 00:05:44
    frame 1 to frame
  • 00:05:47
    2 and I need to find the
  • 00:05:52
    displacement from Frame 2 to frame three
  • 00:05:57
    each of these displacement vectors is
  • 00:06:00
    going to have three elements because
  • 00:06:03
    we're working in threedimensional
  • 00:06:15
    space now let's find each one of these
  • 00:06:18
    vectors one at a
  • 00:06:20
    time the displacement from frame Z to
  • 00:06:24
    frame 1 means what is the displacement
  • 00:06:28
    from the center point of frame 0o to the
  • 00:06:33
    center point of frame
  • 00:06:35
    1 expressed in frame
  • 00:06:39
    zero I need to have an x value a y value
  • 00:06:44
    and a z
  • 00:06:46
    value and the values or the variables
  • 00:06:50
    that I put into this VOR have to be true
  • 00:06:54
    no matter what values we set for the
  • 00:06:57
    joint variables
  • 00:07:00
    finding the displacement from frame 0 to
  • 00:07:03
    frame 1 is fairly easy the displacement
  • 00:07:08
    is right
  • 00:07:09
    here no matter what value we set for
  • 00:07:13
    Theta 1 Theta 2 or
  • 00:07:16
    D3 the displacement between the center
  • 00:07:19
    of frame zero and the center of frame 1
  • 00:07:22
    is always entirely in the z0 direction
  • 00:07:27
    there is no displac M between these two
  • 00:07:30
    frames in either the x0 direction or the
  • 00:07:34
    y0
  • 00:07:36
    direction that means that my X and Y
  • 00:07:40
    values for this displacement will both
  • 00:07:42
    be
  • 00:07:44
    zero what value would I fill in for the
  • 00:07:47
    Z
  • 00:07:48
    displacement here I have to fill in the
  • 00:07:51
    distance between these two frames in the
  • 00:07:54
    Z Direction and here that value is the
  • 00:07:57
    link length A1
  • 00:08:03
    since we're done with the displacement
  • 00:08:05
    from frame 0er to frame 1 let's go on to
  • 00:08:08
    find the displacement from frame 1 to
  • 00:08:11
    frame
  • 00:08:12
    two once again this displacement will
  • 00:08:15
    have to have three elements because this
  • 00:08:18
    is in three-dimensional space we need to
  • 00:08:22
    express the
  • 00:08:24
    displacement from the center of frame
  • 00:08:27
    one to the center of frame frame
  • 00:08:30
    two in the one
  • 00:08:33
    frame and this has to be correct no
  • 00:08:36
    matter what values I set for Theta 1
  • 00:08:40
    Theta 2 and
  • 00:08:41
    D3 finding this displacement Vector is a
  • 00:08:45
    little more tricky than the first
  • 00:08:47
    one right now it appears that the
  • 00:08:50
    displacement from frame 1 to frame 2 is
  • 00:08:54
    a distance of A2 in the X1 Direction
  • 00:09:00
    but it turns out that this is not
  • 00:09:02
    correct the reason why this is not
  • 00:09:05
    correct is because this will not be true
  • 00:09:09
    when Theta 2 becomes a value other than
  • 00:09:13
    zero suppose that Theta 2 was a value of
  • 00:09:19
    90° if Theta 2 was
  • 00:09:23
    90° then our manipulator would not look
  • 00:09:26
    as it's drawn here in that case this
  • 00:09:31
    link A2 would be going straight up in
  • 00:09:36
    the y direction and ending in the joint
  • 00:09:40
    up
  • 00:09:41
    here I can't write the displacement as
  • 00:09:44
    being A2 in the X Direction because that
  • 00:09:48
    won't be true when Theta 2 is not
  • 00:09:52
    zero instead I have to use the triangle
  • 00:09:57
    relationship that we saw a moment ago
  • 00:10:00
    when Theta 2 is some nonzero value the
  • 00:10:04
    center of frame 2 will be up here
  • 00:10:08
    somewhere and frame two will look like
  • 00:10:11
    this this will be
  • 00:10:13
    Z2 and Y2 will be pointing like
  • 00:10:18
    that and X2 will still be into the page
  • 00:10:22
    like this in this case the link will be
  • 00:10:25
    going up like
  • 00:10:27
    this this distance will will still be
  • 00:10:30
    A2 but what will be the displacement in
  • 00:10:34
    the X Direction and the displacement in
  • 00:10:38
    the y
  • 00:10:39
    direction the displacement in the X
  • 00:10:42
    direction will be equal to the
  • 00:10:44
    hypotenuse of the triangle which is
  • 00:10:49
    A2 times the
  • 00:10:52
    cosine of whatever value Theta 2
  • 00:10:57
    has simp similarly the Y displacement
  • 00:11:02
    will be equal to the hypotenuse A2 * the
  • 00:11:06
    S of theta
  • 00:11:09
    2 no matter what value we set for Theta
  • 00:11:13
    2 there will never be any displacement
  • 00:11:15
    between these two frames in the Z
  • 00:11:18
    Direction which is into and out of the
  • 00:11:20
    page so that value will be
  • 00:11:25
    zero now we're finished with the
  • 00:11:28
    displacement vector from frame 1 to
  • 00:11:30
    frame two let's find the displacement
  • 00:11:33
    Vector from frame two to frame
  • 00:11:43
    three the displacement from frame two to
  • 00:11:46
    frame three is Right
  • 00:11:55
    Here Right Now the displacement between
  • 00:11:58
    Frame 2 and frame frame three is
  • 00:12:00
    entirely in the Z direction that is the
  • 00:12:03
    Z2 direction will that be true no matter
  • 00:12:07
    what value we have for the joint
  • 00:12:10
    variable
  • 00:12:12
    D3 it will be true no matter what the
  • 00:12:15
    value is for D3 when D3 becomes nonzero
  • 00:12:20
    frame 3 will slide to the right entirely
  • 00:12:24
    in the Z2
  • 00:12:26
    Direction no matter what value we set
  • 00:12:29
    for D3 there will never be any
  • 00:12:32
    displacement between these two frames in
  • 00:12:34
    the X direction or in the y
  • 00:12:38
    direction so for this displacement
  • 00:12:41
    Vector we can fill in zero for the X and
  • 00:12:44
    the Y
  • 00:12:46
    values now what value should we set for
  • 00:12:50
    the Z
  • 00:12:51
    direction we always draw a kinematic
  • 00:12:54
    diagram to show what the manipulator
  • 00:12:56
    looks like when all of the joint variabl
  • 00:12:59
    are zero so in this diagram D3 is set to
  • 00:13:05
    a value of zero right now the
  • 00:13:08
    displacement between these two frames in
  • 00:13:11
    the Z direction is equal to the link
  • 00:13:13
    length
  • 00:13:15
    A3 but as soon as D3 becomes non zero
  • 00:13:20
    frame 3 will slide along the X Direction
  • 00:13:23
    and the distance between these frames
  • 00:13:25
    will
  • 00:13:27
    increase so the actual ual displacement
  • 00:13:30
    between these frames in the Z direction
  • 00:13:32
    is equal to the link length
  • 00:13:35
    A3 plus however much The Joint variable
  • 00:13:39
    D3 has
  • 00:13:42
    moved so when D3 is equal to Zer the Z
  • 00:13:48
    Direction displacement between these
  • 00:13:50
    frames is just equal to
  • 00:13:52
    A3 but as soon as D3 begins to extend
  • 00:13:57
    and the frame three slid
  • 00:13:59
    in the Z Direction the Z displacement
  • 00:14:02
    between these two frames becomes
  • 00:14:05
    bigger we've now found one displacement
  • 00:14:08
    Vector for each pair of adjacent frames
  • 00:14:11
    so we're done with the displacement
  • 00:14:13
    vectors for this
  • 00:14:15
    manipulator let's take a look at another
  • 00:14:20
    example do you recognize this standard
  • 00:14:23
    form of
  • 00:14:24
    manipulator this is a Scara manipulator
  • 00:14:29
    before I can start to find the
  • 00:14:31
    displacement vectors I need to start by
  • 00:14:34
    filling in everything I need in this
  • 00:14:36
    kinematic diagram starting with the
  • 00:14:39
    coordinate
  • 00:14:47
    frames I'll put the Z direction as the
  • 00:14:52
    axis of rotation for each of my revolute
  • 00:14:55
    joints
  • 00:14:59
    and I'll fill in X and Y so that they
  • 00:15:02
    follow the right hand
  • 00:15:06
    rule Z for a prismatic joint is the
  • 00:15:09
    direction of
  • 00:15:10
    motion and once again I fill in X and Y
  • 00:15:14
    so that they follow the right hand
  • 00:15:18
    rule the end Defector frame just copies
  • 00:15:22
    the frame before it
  • 00:15:29
    next I have to label the positive
  • 00:15:31
    direction of the joint variables for
  • 00:15:34
    this I use the right-and rule
  • 00:15:45
    also lastly I have to label all of the
  • 00:15:48
    link
  • 00:15:56
    lengths now let's start by finding ing
  • 00:15:59
    the displacement from frame 0 to frame
  • 00:16:05
    1 the displacement between frame zero
  • 00:16:09
    and frame 1 is this Vector right
  • 00:16:14
    here remember that when I write that
  • 00:16:17
    Vector I have to write it so that what I
  • 00:16:20
    write will be true no matter what the
  • 00:16:23
    value is for the joint variable Theta
  • 00:16:27
    1 right now it looks like the
  • 00:16:30
    displacement from frame 0 to frame 1 is
  • 00:16:33
    equal to A1 in the Z Direction and A2 in
  • 00:16:38
    the X
  • 00:16:40
    Direction but consider what will happen
  • 00:16:43
    when Theta 1 is not
  • 00:16:46
    zero when Theta 1 is not zero frame 1
  • 00:16:50
    will rotate and move to a new location
  • 00:16:54
    in the XY
  • 00:16:56
    plane to help you see this let's imagine
  • 00:17:00
    that we're looking down at this
  • 00:17:01
    manipulator from the top view I'm going
  • 00:17:05
    to draw what this manipulator would look
  • 00:17:07
    like from the
  • 00:17:09
    top from the top we would see a circle
  • 00:17:12
    showing the first joint and we would see
  • 00:17:15
    the X AIS coming off to the
  • 00:17:18
    right and we would see the Y AIS
  • 00:17:21
    pointing
  • 00:17:23
    up I'll scroll up a little bit so you
  • 00:17:26
    can see what this joint looked like from
  • 00:17:28
    the angled view now when Theta 1 is
  • 00:17:33
    equal to0 we will see the link A2 coming
  • 00:17:38
    off directly along the X Direction
  • 00:17:42
    leading to this second joint over
  • 00:17:45
    here but what will happen when Theta 1
  • 00:17:48
    is let's say
  • 00:17:51
    45° in that case we would see joint two
  • 00:17:55
    having moved up to about here
  • 00:17:59
    and the link A2 would be seen like
  • 00:18:03
    this in this case the displacement from
  • 00:18:07
    frame 0 to frame 1 is no longer entirely
  • 00:18:11
    in the X Direction but instead it has an
  • 00:18:15
    X component and a y component and we can
  • 00:18:19
    see that from this triangle right
  • 00:18:23
    here this angle of the triangle is the
  • 00:18:27
    rotation of the angle Theta
  • 00:18:30
    1 and this distance right here is the
  • 00:18:34
    same as this link length
  • 00:18:38
    A2 from this we can get the X and Y
  • 00:18:41
    displacements using the S and the
  • 00:18:45
    cosine so let's write the displacement
  • 00:18:48
    from frame 0 to frame
  • 00:18:51
    1 the X Direction displacement is equal
  • 00:18:54
    to the length length
  • 00:18:56
    A2 time the cosine of the angle Theta
  • 00:19:01
    1 the Y displacement is equal to the
  • 00:19:05
    length length A2 * the S of theta
  • 00:19:10
    1 now here the Z displacement remember
  • 00:19:14
    is not zero the Z displacement can be
  • 00:19:18
    seen from our original
  • 00:19:20
    view the Z displacement between frame 0
  • 00:19:24
    and frame 1 is a distance of the length
  • 00:19:27
    length A1
  • 00:19:32
    and here we're finished with the
  • 00:19:33
    displacement Vector from frame 0 to
  • 00:19:36
    frame 1 I'm going to set this value
  • 00:19:39
    aside while we find the displacement
  • 00:19:42
    from frame 1 to frame
  • 00:19:44
    two the displacement from frame 1 to
  • 00:19:48
    frame two can be seen right
  • 00:19:54
    here once again right now it looks like
  • 00:19:57
    this display is entirely in the X1
  • 00:20:01
    Direction but we can see that that will
  • 00:20:04
    not be true as soon as Theta 2 is no
  • 00:20:06
    longer
  • 00:20:08
    zero in fact the way that the
  • 00:20:11
    displacement will change as Theta 2
  • 00:20:14
    becomes non zero will be the same for
  • 00:20:17
    these two frames as it was for the zero
  • 00:20:20
    and one
  • 00:20:22
    frame the displacement from frame 1 to
  • 00:20:25
    frame two will also have signs and
  • 00:20:28
    cosiness in the X and Y
  • 00:20:33
    positions the X displacement from frame
  • 00:20:36
    one to frame 2 will be equal to the link
  • 00:20:40
    length A3 which is the hypotenuse in
  • 00:20:43
    this case times the cosine of theta
  • 00:20:48
    2 similarly the Y displacement will be
  • 00:20:52
    the length A3 * the S of theta
  • 00:20:56
    2 here we have we have no displacement
  • 00:21:00
    between frame 1 and Frame 2 in the Z
  • 00:21:03
    Direction and this will be true no
  • 00:21:05
    matter what the value of theta 2 is so
  • 00:21:09
    we can fill in a zero for the Z
  • 00:21:11
    component of the
  • 00:21:15
    displacement now we're done with the
  • 00:21:18
    displacement from frame 1 to frame two
  • 00:21:21
    let's set this aside will we find the
  • 00:21:24
    displacement from Frame 2 to frame three
  • 00:21:27
    the displacement from Frame 2 to frame
  • 00:21:30
    three can be seen right
  • 00:21:35
    here this appears to be a displacement
  • 00:21:38
    entirely in the Z2 direction will that
  • 00:21:42
    be true even when D3 becomes non zero we
  • 00:21:47
    can see that that will continue to be
  • 00:21:50
    true no matter what the value of D3
  • 00:21:53
    is right now the displacement from Frame
  • 00:21:57
    2 to frame three is equal to the link
  • 00:21:59
    length A4 in the Z Direction but this
  • 00:22:03
    distance is going to increase as D3
  • 00:22:07
    becomes large and this Prismatic joint
  • 00:22:11
    extends so we can write the displacement
  • 00:22:14
    between frames 2 and 3 like this the
  • 00:22:18
    displacement in the X and Y directions
  • 00:22:20
    are both zero because no matter what
  • 00:22:24
    value we have for
  • 00:22:26
    D3 there will never be any displacement
  • 00:22:29
    between these two points in X or in
  • 00:22:33
    y the displacement in the Z direction
  • 00:22:36
    will be equal to A4 the distance between
  • 00:22:40
    these two frames when D3 is zero plus
  • 00:22:45
    however much this Prismatic joint has
  • 00:22:49
    extended and this gives us the complete
  • 00:22:52
    displacement from frame two to frame
  • 00:22:55
    three
Etiquetas
  • matrice de transformation homogène
  • vecteur
  • déplacement
  • robotique
  • kinématique
  • sphérique
  • Scara
  • jointures
  • angle
  • magnitude