CSCE 452 Intro to Robotics

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CSCE 452 Intro to Robotics
Inverse Kinematics
1
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
Forward/Direct Kinematics:
CSCE 452 Intro to Robotics
X
Given x οƒ  a unique q
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
Workspace
CSCE 452 Intro to Robotics
• Reachable Workspace
• Dexterous Workspace
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
X
CSCE 452 Intro to Robotics
𝛾
𝛾
Given X
Find q=(1, 2, 3)
Algebraic Solution
CSCE 452 Intro to Robotics
The kinematics of the example seen before
are:
c123 ο€­ s123 0 l1c1  l2 c12 οƒΉ
οƒͺs
123
B
0
οƒͺ
T
ο€½
T
ο€½
W
3
οƒͺ 0
οƒͺ
 0
c123
0
0
0 l1s1  l2 s12 οƒΊοƒΊ
οƒΊ
1
0
οƒΊ
0
1

Assume goal point is specified by 3
numbers:
𝑐𝛾
𝑠𝛾
𝐡
π‘Šπ‘‡ =
0
0
−𝑠𝛾
𝑐𝛾
0
0
0 π‘₯
0 𝑦
1 0
0 1
Algebraic Solution (cont.)
By comparison, we get the four equations:
c ο€½ c123
s ο€½ s123
CSCE 452 Intro to Robotics
x ο€½ l1c1  l2 c12
y ο€½ l1s1  l2 s12
Summing the square of the last 2
equations: x 2  y 2 ο€½ l 2  l 2  2l l c
1
2
1 2 2
From here we get an expression for c2
x  y ο€­l ο€­l
c2 ο€½
2l1l2
2
2
2
1
2
2
Algebraic Solution (III)
CSCE 452 Intro to Robotics
• When does a solution exist?
• What is the physical meaning if no solution
exists?
• Two solutions for 2 are possible. Why?
Using c12=c1c2-s1s2 and s12= c1s2-c2s1:
x ο€½ k1c1 ο€­ k 2 s1
y ο€½ k1s1  k 2 c1
where k1=l1+l2c2 and k2=l2s2. To solve these eqs,
set π‘Ÿ = π‘˜12 + π‘˜22 and φ= 𝐴𝑇𝐴𝑁2(π‘˜2 , π‘˜1 )
Algebraic Solution (IV)
k1
CSCE 452 Intro to Robotics
φ
l2
2
k2
l1
Then: k1=r cos φ , k2=r sin φ , and we can write:
x/r= cos φ cos 1 - sin φ sin 1
y/r= cos φ cos 1 - sin φ sin 1
or: cos(φ +1) = x/r, sin(φ +1) =y/r
CSCE 452 Intro to Robotics
Algebraic Solution (IV)
Therefore:
φ+1 = Atan2(y/r,x/r) = Atan2(y,x)
and so:
1 = Atan2(y,x) - Atan2(k2,k1)
Finally, 3 can be solved from:
1+ 2+ 3 = γ
Geometric Solution
IDEA: Decompose spatial geometry into
several plane geometry problems
CSCE 452 Intro to Robotics
y
L2
L1
x
Applying the “law of cosines.”
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
CSCE 452 Intro to Robotics
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