AME 352
VECTORS
2. VECTORS
Vector algebra forms the mathematical foundation for kinematics and dynamics. Geometry of motion is at the heart of both the kinematics and dynamics of mechanical systems. Vector analysis is the time-honored tool for describing geometry. A Vector can be described either geometrically or algebraically. Vectors
Geometric presentation Vectors are denoted by bold-face characters such as θ R R, V, etc. The magnitude of a vector, such as R, is V denoted as: θ | R | , R , or R The angle of a vector is denoted as θ which is measured θ θ positively cou nter-clockwise (CC W) with respect t o a well-defined axis. It is common to consider the positive x-axis as the reference axis. In kinematics and dynamics a vector may represent F A position, velo city, acceleration, or force/moment. Note : Since we cannot write by hand in boldface, we denote a vector with an over-score arrow or an under-score line, for example R or R. Algebraic presentation A vector can be projected onto the x- and y-axes of a y Cartesian frame in order to form its analytical representation. R y R ⎧ R x ⎫ ⎧ R cosθ ⎫ ⎧cosθ ⎫ θ R ⎨ R ⎬ ⎨ ⎬ R ⎨ ⎬ ⎩ sinθ ⎭ ⎩ y ⎭ ⎩ R sinθ ⎭ x This representation remains valid regardless of whether the angle is in the first, second, third, or fourth quadrant, R x as long as the angle is measured CCW with respect to the positive x-axis.
=
=
=
Rotated Vector
If a vector such as R is rotated 90 CCW, it will be denoted as R . The rotated vector will have the same magnitude as R but its x-y components will be different: ⎧ R cos(θ + π 2 )⎫ ⎧− R sin θ ⎫ ⎧− sin θ ⎫ ⎧− R y ⎫ R=⎨ = ⎨ = R ⎨ ⎬ ⎬ ⎬= ⎨ ⎬ π + 2 ) ⎭ ⎩ R cos θ ⎭ ⎩ cos θ ⎭ ⎩ R x ⎭ ⎩ R sin(θ +
y R y
R
θ
R y
R
x
R x
P.E. Nikravesh
R x
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VECTORS
Position vectors In general a position vector describes the position of one point with respect to another point. Either of these points could be moving or stationary. When a point moves, the orientation and/or the magnitude of its position vector change as well. Examples: R BA connects two points that are defined
on two separate moving links;
R CO
A
B
R BA
C
R CO
connects point O on the
O
ground (stationary) to point C on a moving link. Note: A point depicted as a small black circle in a figure is assumed stationary (defined on the ground). Index of A Vector In kinematic analysis of mechanisms, it is helpful to assign indices to position vectors. Most commonly, an index refers to the end points of the vector. The first letter in an index indicates the head of the vector (the arrow) and the second letter refers to the tail B of the vector. For example vectors R BC , which reads position of B R BC
relative to C , and
R
that reads position of C relative to O. CO
In some problems, for notational simplification, the index may carry a single number. For example vector R 3 in the diagram is the same as vector
R
BC
R BA dt
=
d
R CO dt
=
R BA
.
Q B A
≡
R CO
V BA (velocity
VCO
≡
of B relative to A)
R CO
O
Linear Velocity and Acceleration The time derivative of a position vector represents the velocity of the point the position vector represents. For example: d
C
R3
RCQ
R BA
C
(velocity of C relative to O)
This velocity is also called linear velocity.
R CO
O
Note: Linear
velocity is defined for a point (not for a link or a vector). When the reference (the tail) point of a position vector is stationary, that point’s index could be dropped from the index of the velocity vector. For example: d
R CO dt
=
VCO
=
VC
(a)
If we consider another stationary point such as Q, we get d
(b) R CQ VCQ VC dt This means that the velocity of a point with respect to the ground is independent of the defined reference point on the ground. This vector represents the velocity of C relative to the ground or the absolute velocity of C . The second time derivative of a position vector, or the time derivative of a velocity vector, denotes the acceleration of the point the position vector represents. For example, acceleration of B relative to A is denoted as =
=
2
d
2
dt
P.E. Nikravesh
R BA
=
R BA
d ≡
dt
V BA
=
V BA
≡
A BA
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VECTORS
The absolute acceleration of C (acceleration of C with respect to the ground) is denoted as 2
d
2
dt
R CO
=
R CO
d ≡
dt
VC
=
≡A V C C
Note: Linear acceleration is defined for a point (not for a link or a vector). Angular Velocity and Acceleration The time rate of change in the orientation of a vector, or a link, is defined as the angular velocity of that vector or link. In planar systems, the angular velocity is the time rate of rotation about the z-axis and it is denoted as d
θ dt
=
R
θ
≡ ω θ
We consider a counter-clockwise (CCW) rotation as positive and a clockwise (CW) rotation as negative. Note: Angular velocity is defined for a vector or a link, not for a point. Note: Vectors that are defined on the same body experience the same angular velocity. For example, R BA , R BC , and R CA have the same angular velocity since they are defined between the points of the same body.
B
R BA R BC A C
RCA
The time rate of change of the angular velocity of a vector, or a link, is defined as the angular acceleration of that vector or link. In planar systems, the angular acceleration is a vector along the z-axis and it is denoted as 2
d
2
θ
=
dt
θ
d
≡
ω dt
=
≡ α ω
We consider a counter-clockwise (CCW) rotation as positive and a clockwise (CW) rotation as negative. Position, Velocity, and Acceleration Vectors in Mechanisms Position vectors must be constructed between well-defined points of a mechanism. The magnitude and the angle of a position vector must reveal certain information about the position and orientation of a link with respect to another link or with respect to the ground. Depending on the type of a position vector, the corresponding velocity and acceleration vectors may be decomposed differently. The following examples show typical position vectors, and their corresponding velocity and acceleration vectors that appear in kinematic analysis of mechanisms. Constant Magnitude, Constant Angle The position vector is defined between two non-moving points. Q R QO
=
⎧cos ψ ⎫ ⎬, ⎩ sinψ ⎭
LQO ⎨
VQO
=
0
,
A QO
=
0
where LQO is the constant magnitude and ψ is the constant angle of
R QO
O
ψ
the vector. Zero vectors for the velocity and acceleration should be obvious since both ends of the position vector are fixed to the ground. Variable Magnitude, Constant Angle Two links form a sliding joint. One of the links (the rod) is fixed to the ground and the other (the block) slides on the rod. The position vector is defined between a point on the ground and a point on the block parallel to the sliding axis. This vector does not rotate since neither link can rotate. P.E. Nikravesh
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VECTORS
R BO
V BO
⎧cosψ ⎫ ⎨ ⎬ ⎩ sinψ ⎭ ⎧cosψ ⎫ ⎨ R ⎬≡V , sin ψ ⎭ ⎩
=
s
=
A
R BO
BO
BO
s
B
A BO
=
Vs
⎧cosψ ⎫ ⎨ ⎬≡A sin ψ ⎭ ⎩
R BO
R BO
s
BO
Vs
O
ψ
As
(It is assumed that the slip components are positive.) The velocity and acceleration vectors are along the axis of R BO —they are called slip velocity and slip acceleration. Constant Magnitude, Variable Angle The position vector is defined between two points on a link. The link is free rotate. R BA
where
=
⎧cosθ ⎫ ⎨ ⎬ ⎩ sinθ ⎭
t
A
B
n
R BA A
Vt
θ
is the constant length of the vector.
L BA
V BA
L BA
A
⎧− sinθ ⎫ ⎨ ⎬≡V cosθ ⎭ ⎩
ω L BA
=
(It is assumed that the angular velocity and acceleration are positive.)
t
BA
Vt ω R =
This vector is perpendicular to A BA
=
R BA —it
is called tangential velocity.
⎧cosθ ⎫ ⎧− sinθ ⎫ − ω L ⎨ ⎬ + α L ⎨ ⎬≡A sinθ ⎭ cosθ ⎭ ⎩ ⎩ 2
BA
BA
A n = −ω 2R
n
BA
+
t
A BA
At =α R
R BA
The acceleration vector contains two components: one in the opposite direction of normal , and one perpendicular to
R BA
called tangential acceleration.
Variable Magnitude, Variable Angle A block and a rod form a sliding joint. Both the rod and the block rotate together; i.e., the two links have the same angular velocity and the same angular acceleration. R BA
V BA
R BA
BA
BA
t
+
c
s
A
n
t
B Vt
R BA
s
V BA
A
θ Vs
s
The velocity vector can be decomposed into two components: a tangential velocity perpendicular to and a slip velocity along the axis of =
A
A
BA
V
Vt =ω R
A BA
A
⎧cosθ ⎫
⎨ ⎬ ⎩ sinθ ⎭ ⎧− sinθ ⎫ ⎧cosθ ⎫ = ω R ⎨ ⎬ + R ⎨ ⎬≡V θ cos sin θ ⎭ ⎩ ⎭ ⎩ =
called
R BA
R BA
,
(It is assumed that the slip components, and angular velocity and acceleration are positive.)
.
⎧cosθ ⎫ ⎧− sinθ ⎫ ⎧cosθ ⎫ ⎧− sinθ ⎫ − ω R ⎨ ⎬ + α R ⎨ ⎬ + R ⎨ ⎬ + 2 ω R ⎨ ⎬ sinθ ⎭ cosθ ⎭ sinθ ⎭ cosθ ⎭ ⎩ ⎩ ⎩ ⎩ 2
BA
BA
A n = −ω 2R
≡
n
A BA
+
t
A BA
+
BA
At =α R s
A BA
+
BA
As
Ac = 2ω V s
c
A BA
The acceleration vector is decomposed into four components: a normal component in the
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VECTORS
opposite direction of the axis of
R BA
R BA
R BA
, a tangential component perpendicular to
, and a Coriolis component perpendicular to
R BA
, a slip component along
.
Coordinates, velocity, and acceleration of A Point Another type of position vector describes the coordinates of a point with respect to the origin of a Cartesian reference frame. For example, R CO describes the coordinates of point C with respect to the origin O. When the reference point is the origin, for notational simplification, it may be dropped from the index; e.g., we may say R C instead of R CO .
B
y
C RCO
O
x
One common step in kinematic analysis is to determine the velocity and acceleration of a point based on some other known velocities and accelerations. For this purpose we start with a position vector equation and then take its time derivatives. Here we consider two typical examples. In the first example, two points are defined on the same B y VC link. The velocity of point C is given and we want to AC R BC determine the velocity of point B. We start from the following position vector equation: C R BO (a) R BO = RCO + R BC The time derivative of this equation yields V BO
=
VCO
V B
=
VC
+
+
V BC t
V BC
=
VC
+ ω R BC
A B
=
A C
+
A BC
n
A BC
A C
=
A C − ω R BC
2
+
A BC
t
=
+
x
(b)
Assuming a known CCW angular velocity, t he vector summation in (b) can be constructed graphically as shown. Acceleration of B is obtained from the time derivative of (b): n
RCO
O
V BC
AC
VC
V B
t
A BC A B
t
A BC
+ α R BC
(c)
Assuming known acceleration of C and known CCW angular velocity and acceleration of the link, the acceleration expression in (c) can be constructed graphically as shown. Equations (a), (b) and (c) can also be evaluated algebraically as: ⎧ x B ⎫ ⎧ xC ⎫ ⎧cosθ ⎫ (d) R B = ⎨ ⎬ = ⎨ ⎬ + L BC ⎨ ⎬ y y θ sin ⎩ B ⎭ ⎩ C ⎭ ⎩ ⎭ V B
A B
⎧ x B ⎫ ⎧ xC ⎫ ⎧− sinθ ⎫ (e) ⎨ ⎬ = ⎨ ⎬ + ω L BC ⎨ ⎬ ⎩ y B ⎭ ⎩ yC ⎭ ⎩ cosθ ⎭ ⎧ x B ⎫ ⎧ xC ⎫ ⎧cosθ ⎫ ⎧− sinθ ⎫ = ⎨ ⎬ = ⎨ ⎬ − ω L BC ⎨ ⎬ + α L BC ⎨ ⎬ (f) y y θ θ sin cos ⎩ B ⎭ ⎩ C ⎭ ⎩ ⎭ ⎩ ⎭
=
y
2
In the second example, two points are defined on two links of a sliding joint. The velocity and acceleration of point C are given and we want to determine the velocity and acceleration of point B. We start with the position vector equation that is identical to (a). The corresponding velocity equation: t s s V B = VC + V BC + V BC = VC + ω R BC + V BC (g)
VC
B
R BC
C AC
R BO
RCO
x O
If the angular velocity (both links have the same angular
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VECTORS
velocity) and the slip velocity are given, then
V B can
be
s
V BC
constructed graphically as shown. The acceleration expression is written as A B
=
A C + A BC
= =
n
t
s
c
A C + A BC + A BC + A BC + A BC
2
VC
s (h)
s
A C − ω R BC + α R BC + A BC + 2ω V BC
V B
Assuming that the acceleration of C , the angular velocity and acceleration of the link, and the slip velocity and accelerations are given, then the acceleration expression in (h) can be constructed graphically as shown. Equations (a), (g) and (h) can also be solved algebraically: ⎧ x B ⎫ ⎧ xC ⎫ ⎧cosθ ⎫ (i) R B = ⎨ ⎬ = ⎨ ⎬ + R BC ⎨ ⎬ y y θ sin ⎩ B ⎭ ⎩ C ⎭ ⎩ ⎭
V B
A B
t
V BC
A B c
A BC s
A BC
AC t
A BC
n
A BC
⎧ x B ⎫ ⎧ xC ⎫ ⎧− sinθ ⎫ ⎧cosθ ⎫ (j) ⎨ ⎬ = ⎨ ⎬ + ω R BC ⎨ ⎬ + R BC ⎨ ⎬ y y θ θ cos sin ⎩ B ⎭ ⎩ C ⎭ ⎩ ⎭ ⎩ ⎭ − sinθ ⎫ ⎧ x B ⎫ ⎧ xC ⎫ ⎧cosθ ⎫ ⎧− sinθ ⎫ ⎧cosθ ⎫ BC ⎧⎨ = ⎨ ⎬ = ⎨ ⎬ − ω R BC ⎨ ⎬ + α R BC ⎨ ⎬ + R BC ⎨ ⎬ + 2ω R ⎬ ⎩ y B ⎭ ⎩ yC ⎭ ⎩ sinθ ⎭ ⎩ cosθ ⎭ ⎩ sinθ ⎭ ⎩ cos θ ⎭
=
2
(k)
Vector Loop The position vectors that are defined for kinematic analysis of a mechanism should form one or more kinematic loops (also called closed chains). As an example the vectors that are defined for the four-bar in (a) form a loop. These vectors may be directed differently to form a loop as shown in (b) If we navigate in a loop from vector to vector, the vector that is navigated from tail-to-head is considered positive and a vector that is navigated from head-to-tail is considered negative. For example, for the four-bar in (a), if we navigate through the loop in the following fashion O2 → A → B → O4 → O2 , vectors R AO and R BA are navigated positively and R BO and R O O 2
4
4
2
are navigated negatively. R BA
B
A
A R BO 4
R AO 2
O2
RO4 O2
R AB
RO B
R AO 2
O4
4
R O4 O2
O2
B
O4
(a) (b) A kinematic loop can be expressed as a vector equation. For example, the vectors in the four bar (a) form the vector-loop equation R AO
2
+
R BA − R BO
4
− RO O 4
2
= 0
Similarly, the vectors in (b) form the vector-loop equation R AO
2
− R AB + RO B − RO O 4
4
2
= 0
Obviously, there are many other possible scenarios. Any of these vector loop equations can be used for kinematic analysis of the mechanism. Examples Several examples of commonly used planar mechanisms are presented here. For each mechanism a set of vectors are defined to form a vector-loop equation(s). P.E. Nikravesh
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VECTORS
(a) Four-bar R AO
R BA − R BO
+
2
4
− RO O 4
= 0
2
B
R BA
A
R BO 4
R AO 2 R O4 O2
O2
O4
(b) Slider-crank (inversion 1) R AO
R BA − R BO
+
2
2
(c) Slider-crank (inversion 2)
= 0
R AO
− RO O − R AO
2
4
2
=
0
4
A A R BA
R AO 2
R AO
B O2
(d) Slider-crank (inversion 3) R O A − R O
+
2
4 O2
4
A
2
O2
R BO 2
R AO
R AO 4
RO
(e) Slider-crank (inversion 4)
= 0
R AO
+
2
R BA − R BO
R BA
4
R AO 2
2
O2
B
O4
RO
4 O2
2
− RO O − R BO − R AB 4
2
R BO 2
O2
(f) Slider-crank (inversion 2 - variation) R AO
=
(g) Slider-crank (inversion 3 - variation)
0
R AO
4
2
− R O O − R BO − R AB 4
A
R AB
R AO
R BO 4
R AO 2
4
independent vector loop equations can be constructed:
R AO
− R AC − R CO − RO O
P.E. Nikravesh
R BA − R BO
2
4
= 0
(a)
4
2
=
0
O4
A R BA
R CO
+
4
O4
. Therefore two
R AO
R BO
R O4 O2
(h) Six-bar mechanism (a four-bar and a slider-crank) This mechanism contains two R AC C independent loops: O4 CAO2O4 and
2
B
R O4 O2
O2
0
2
O4
R AO 2
=
4
O2
A
2
2
R BA
B
O2 ABO2
= 0
2
A
R O A
R AO
O4
4 O2
RO
4 O2
O2
R BO
B 2
(b)
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A third loop may be visualized as
O4 CABO2O4 .
However, this is a redundant loop—its vector
loop equation can be obtained from subtracting the second equation from the first: R BA − R BO
+
2
R AC + R CO
+
4
RO O 4
= 0
2
If we combine the two ground vectors, R CO
4
+
R AC + R BA − R BO
= 0
4
R BO and R O O , into one vector we 2
4
2
(c)
We only need two of these three equations. (i) Six-bar mechanism (two slider-cranks) This mechanism contains two independent loops: R AO − R AO − R O O 0 (a)
R CO
R BO
4
4
+
4
R CB − R CO
1
O1
2
− RO O 1
2
+
RO
4 O2
= 0
1
1
R CB
(b)
A third dependent loop can be constructed as R AO + R BA + R CB − R CO − R O O = 0 (c) 2
C
1
=
2
get:
2
We only need two of these three equations.
RO O 1
B
R BO
2
4
R AO
2
A
O2
R AO
RO
4 O2
4
O4
Note: In
examples containing a sliding joint, the vector that connects the two links must be defined parallel to the axis of the sliding joint. This is a rule that we must follow, otherwise we can make a simple problem too difficult to solve. For example, for this slider-crank, we should not define a vector directly from O4 to A. Instead, we should define two vectors, axis and
Note:
R AB
R BO
4
B
R AB
R BO 4
A
R AO
O4
4
which is parallel to the sliding
perpendicular to the sliding axis.
If the ground vector is not parallel to the x-axis, we may decompose it into two vectors—one parallel and one perpendicular to the x-axis. For example, R O O may be replaced by R QO + R O Q . 4
2
2
4
O4
R O4 O2
RO Q 4
O2
P.E. Nikravesh
R QO
Q
2
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