The (x, y) coordinates of the hand at the top of a robot arm are given by: x = L, cos(8,) + L2 cos(8, + 0,) y = L, sin(8,) + L,sin(8, + 02) Where L1 and L2 are the distances of the parts of the arm from the shoulder to he elbow and from the elbow to the hand, Đ1 and 62 are the angles that these arts make with the x-axis. These angles can be expressed as: 0,(t) = a,t³ + a,t“ + azt³ 02(t) = b,t5 + bzt“ + bąt³ vhere t is the time. We assume that the arm starts to move at t = 0s and stops t tf=2s. The conditions that the angular position 61(t), angular velocity and ngular acceleration should meet at the final time tf= 2 (in sec) are expressed n the following matrix form and allow us to obtain the coefficients a1, az and 3: [a 3x22||a2 园- [01f – 010] 24 4x23 20x23 12x22 6x2 Jlaz] 25 23 5x24 vith 0lf= 437/180, 010 = -197/180. a) Use MATLAB left division to solve the above system and get the oefficients a1, az and az ) Similarly the coefficients b1, b2 and b; can be obtained by solving a similar ystem: 25 24 23 [82f – 0201 5x24 20x23 12x22 6x2 3x22 ||b2 = 4x23 [b3. vith 02f=1517/180, 020 = 447/180. Obtain the coefficients bi bɔ and b; using MATLAB matrix inversion

Database System Concepts
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Chapter1: Introduction
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The (x, y) coordinates of the hand at the top of a robot arm are given by:
x = L, cos(8,) + L, cos(8, + 0,)
y = L, sin(0,) + L,sin(8, + 02)
Where Li and L2 are the distances of the parts of the arm from the shoulder to
the elbow and from the elbow to the hand, 01 and 02 are the angles that these
parts make with the x-axis. These angles can be expressed as:
0,(t) = a,t5 + a,t“ + azt³
02(t) = bịt5 + b;t + b3t³
where t is the time. We assume that the arm starts to move at t= 0s and stops
at tf=2s. The conditions that the angular position G1(t), angular velocity and
angular acceleration should meet at the final time tf = 2 (in sec) are expressed
in the following matrix form and allow us to obtain the coefficients a1, az and
a3:
25
24
23
3x22||a2 =
[az.
[01f – 010]
5x24
4x23
| 20x23 12x22 6x2
with 0lf=43x/180, 610 = -197/180.
a) Use MATLAB left division to solve the above system and get the
coefficients a1, az and as
b) Similarly the coefficients b1, bɔ and b3 can be obtained by solving a similar
system:
25
24
23
[82f – 0201
5x24
4x2
3x22||b2=
| 20x23 12x22 6x2 ]lb,]
with 02f=151/180, 620 = 447/180.
Obtain the coefficients b1, b2 and b3 using MATLAB matrix inversion.
Transcribed Image Text:The (x, y) coordinates of the hand at the top of a robot arm are given by: x = L, cos(8,) + L, cos(8, + 0,) y = L, sin(0,) + L,sin(8, + 02) Where Li and L2 are the distances of the parts of the arm from the shoulder to the elbow and from the elbow to the hand, 01 and 02 are the angles that these parts make with the x-axis. These angles can be expressed as: 0,(t) = a,t5 + a,t“ + azt³ 02(t) = bịt5 + b;t + b3t³ where t is the time. We assume that the arm starts to move at t= 0s and stops at tf=2s. The conditions that the angular position G1(t), angular velocity and angular acceleration should meet at the final time tf = 2 (in sec) are expressed in the following matrix form and allow us to obtain the coefficients a1, az and a3: 25 24 23 3x22||a2 = [az. [01f – 010] 5x24 4x23 | 20x23 12x22 6x2 with 0lf=43x/180, 610 = -197/180. a) Use MATLAB left division to solve the above system and get the coefficients a1, az and as b) Similarly the coefficients b1, bɔ and b3 can be obtained by solving a similar system: 25 24 23 [82f – 0201 5x24 4x2 3x22||b2= | 20x23 12x22 6x2 ]lb,] with 02f=151/180, 620 = 447/180. Obtain the coefficients b1, b2 and b3 using MATLAB matrix inversion.
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