Transformation Exercises: Denavit- Hartenberg Method Some images - - PowerPoint PPT Presentation
Transformation Exercises: Denavit- Hartenberg Method Some images and exercises from: Introduction to Autonomous Mobile Robots, Siegwart, Nourbakhsh, 2011 Robot Dynamics and Control Second Edition, Spong, Hutchinson, Vidyasagar, 2004 Spacecraft
Some images and exercises from: Introduction to Autonomous Mobile Robots, Siegwart, Nourbakhsh, 2011 Robot Dynamics and Control Second Edition, Spong, Hutchinson, Vidyasagar, 2004 Spacecraft Robot Kinematics Using Dual Quaternions, Valverde, Alfredo & Tsiotras, Panagiotis, 2018
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§ Given two frames of reference, tra
§ A robot sees a thing. Where is the thing in the world? § There’s a thing in the world. Where is it wrt. the robot? § A robot moves around. Where is it in the world?
§ We can do this with translation/rotation matrices,
§ 𝜊" = $
"𝑈 × 𝜊$
§ Or, faster with the Denavit-Hartenberg method.
https://matrix.reshish.com/multiplication.php is nice for running matrix multiplications
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§ z axis is axis of motion
§ Rotation around z for revolute § Translation along z for prismatic
§ xi axis orthogonal to zi an
§ y axis: right-hand rule
§ Fingers point +x § Thumb points +z § Palm faces +y
§ xi axis must
§ Which may mean translating origin
+x +z +y
4 4 § ai-1 : lin
link le length distance Zi-1 ⇆ Zi along Xi
§ αi-1 : lin
link twis ist angle between Zi-1 ⇆ Zi around Xi
§ di : lin
link offset distance Xi-1 to Xi along Zi
§ θi : jo
join int angle le angle between Xi-1 and Xi around Zi
Valverde, Alfredo & Tsiotras, Panagiotis. (2018). Spacecraft Robot Kinematics Using Dual Quaternions. https://www.mdpi.com/2218-6581/7/4/64/htm
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Ri
i−1 =
cosθi −sinθi cosαi,i+1 sinθi sinαi,i+1 ai,i+1cosθi sinθi cosθi cosαi,i+1 −cosθi sinαi,i+1 ai,i+1sinθi sinαi,i+1 cosαi,i+1 di 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥
x y z Tx x y z Ty x y z Tz
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§ Define the axes according to the DH rules. § Then, draw lines and arcs for a and θ.
§ Why not α and d?
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Z axes point out towards us
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§ Give the DH parameters.
Z axes point out towards us
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§ Give the final transformation matrix.
Ri
i−1 =
cosθi −sinθi cosαi,i+1 sinθi sinαi,i+1 ai,i+1cosθi sinθi cosθi cosαi,i+1 −cosθi sinαi,i+1 ai,i+1sinθi sinαi,i+1 cosαi,i+1 di 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥
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§ Give the final transformation matrix. § Plugging in…
Ri
i−1 =
cosθi −sinθi cosαi,i+1 sinθi sinαi,i+1 ai,i+1cosθi sinθi cosθi cosαi,i+1 −cosθi sinαi,i+1 ai,i+1sinθi sinαi,i+1 cosαi,i+1 di 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥
𝑈
( ) =
cos 𝜄( −sin 𝜄( 𝑏),(cos 𝜄( sin 𝜄( cos 𝜄( 𝑏),(sin 𝜄( 1 1 𝑈5
( =
cos 𝜄5 −sin 𝜄5 𝑏(,5cos 𝜄5 sin 𝜄5 cos 𝜄5 𝑏(,5 sin 𝜄5 1 1
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§ Give the final transformation matrix. § Plugging in…
Ri
i−1 =
cosθi −sinθi cosαi,i+1 sinθi sinαi,i+1 ai,i+1cosθi sinθi cosθi cosαi,i+1 −cosθi sinαi,i+1 ai,i+1sinθi sinαi,i+1 cosαi,i+1 di 1 ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥
𝑈5
) = 𝑈 ( )×𝑈5 (=
cos(𝜄( + 𝜄5) −sin(𝜄( + 𝜄5) 𝑏),(cos 𝜄( +𝑏(,5cos(𝜄( + 𝜄5) sin(𝜄( + 𝜄5) cos(𝜄( + 𝜄5) 𝑏),(sin 𝜄( +𝑏(,5sin(𝜄( + 𝜄5) 1 1
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§ This is al
§ Similar derivations can be done for cylindrical
§ This is why we name configurations.
§ If you know these, you can subdivide an arm.
cos(𝜄( + 𝜄5) −sin(𝜄( + 𝜄5) 𝑏),(cos 𝜄( +𝑏(,5cos(𝜄( + 𝜄5) sin(𝜄( + 𝜄5) cos(𝜄( + 𝜄5) 𝑏),(sin 𝜄( +𝑏(,5sin(𝜄( + 𝜄5) 1 1
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§ Could you do the same thing using a sequence
§ How many steps would it take?
§ Could you do the same thing given a word
§ Given numbers? § As a derivation?
§ How many joints would you max out at?
§ Directly? § Using DH parameters?
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1.
a.
What are the frames?
b.
What are the DH parameters?
c.
What are the individual transformation matrices?
d.
What’s the final transformation matrix?
2.
§ This is just a sequence of rotations – see Spong.
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§ What is the complete, derived
a.
What are the frames?
b.
What are the DH parameters?
c.
What are the individual transformation matrices?
d.
What’s the final transformation matrix?
e.
What is the final transformation matrix for this wrist, derived wi withou
using the DH method?
§
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§
§
z0 is along axis of motion
§
Origin 0 (O0) is arbitrary, but makes sense
§
x0 is normal to the page.
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§
§
z1 is along axis of motion
§
Origin 0 (O1) is easy, because z0 and z1 are the same (no origin movement necessary)
§
x1 Is normal to the page currently (but not when joint 0 moves!)
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§
§
z2 is along axis of motion
§
x2 is chosen parallel to x1 so that θ2 is zero.
§
§
Chosen as shown.
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b.
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c.
§ c = cosine, s = sin
0T1 = 1T2 = 2T3 =
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d.
0T3 =
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§ What is the complete, derived transformation
§ This is exercise 3 and exercise 2