Novint Falcon. Novint Falcon Novint Falcon Novint Falcon. Novint Falcon Daniel J. Block et al., licensee Versita Sp. z o. o.

Size: px

Start display at page:

Download "Novint Falcon. Novint Falcon Novint Falcon Novint Falcon. Novint Falcon Daniel J. Block et al., licensee Versita Sp. z o. o."

Silas Garrett
5 years ago
Views:

1 Novint Falcon 1,, 3, Novint Falcon Novint Falcon Novint Falcon Novint Falcon Novint Falcons 13 Daniel J. Block et al., licensee Versita Sp. z o. o. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs license, which means that the text may be used for non-commercial purposes, provided credit is given to the author. Novint Falcon d-block@illinois.edu mark.michelotti@gmail.com rsree@illinois.edu 18

2 i (x, y, z) O P p A i r E i c a, b, d e A i,b i C i Θ 1i, Θ i Θ 3i (u i,v i,w i ) A i P (u i,v i,w i ) cos φ i sin φ i p i = sin φ i cos φ i p + 1 r φ i i (x, y, z) p i =(p ui p vi p wi ) T p ui,p vi p wi p ui p vi p wi = a cos θ 1i c +(d + e + b sin θ 3i ) cos θ i = b cos θ 3i = a sin θ 1i +(d + e + b sin θ 3i ) sin θ i θ 3i (x, y, z) i i i θ 3i = ± cos 1 p vi b. θ 3i θ i θ i p ui p wi θ i (p ui + c) + p wi + a a(p ui + c) cos θ 1i αp wi sin θ 1i = (d + e) + (d + e)b sin θ 3i + b sin θ 3i. (1) 183

3 t 1i = tan θ 1i sin θ 1i = t 1i 1+t 1i cos θ 1i = 1 t 1i, 1+t1i l i l 1i l i l i t 1i + l 1i t 1i + l i =, = p wi + p ui +cp ui ap ui b sin θ 3i be sin θ 3i bdθ 3i de ac + a + c d e = 4αp wi = p wi + p ui +cp ui ap ui b sin θ 3i be sin θ 3i bdθ 3i de ac + a + c d e t 1i θ 1i θ 3i θ i (x, y, z) C θ 3i θ 1i (θ 11,θ 1,θ 13 ) (x, y, z) Dimension a b c d e r Value 6 mm 13 mm 16.3 mm 1 mm 1 mm 37 mm ϕ ϕ ϕ function [theta1] = inverse_kinematics(x, y, z) r = 37; a = 6; b = 13; c = 16.3; d = 1; e = 1; phi =.5 %radians pu = x*cos(phi) + y*sin(phi) r; pv = x*sin(phi) + y*cos(phi); pw = z; theta3 = acos(pv/b); l = pw^+pu^+*c*pu *a*pu+a^+c^ d^ e^ b^*sin(theta3)^ *b*e*sin(theta3) *b*d*sin(theta3) *d*e *a*c; l1 = 4*a*pw; l = pw^+pu^+*c*pu+*a*pu+a^+c^ d^ e^ b^*sin(theta3)^ *b*e*sin(theta3) *b*d*sin(theta3) *d*e+*a*c; t1 = ( l1 sqrt(l1^ 4*l*l))/(*l); theta1 = atan(t1)*; x k =(x k y k z k ) T k f (x, y, z) (θ 11 θ 1 θ 13 ) T B k k f f 184

4 x = initial guess B = initial Jacobian approximation for k =, 1,,... end Solve Bksk = f(xk) for sk xk+1 = xk + sk yk = f(xk+1) f(xk) Bk+1 = Bk + ((yk Bksk)sk T )/(sk T sk) (x, y, z) x, y, z k p k i k d k p k d k i (x, y, z) k p =,k d =. k i = 1 (x, y, z) =(, 14, 135) (x, y, z) =( 4, 3, 11) D(z) = z 1 T sz D(z) = Ts z 1 T s =

5 Component Description 1 TMS3 Digital Controller Encoder LED emitters and photosensors 3 Supplementary Sensors 4 Motor Leads 5 Controller Buttons 186

6 A B A B SS SCK MISO MOSI ± OUT A OUT B LOAD SUPPLY BRAKE 187

9531446853 1 3 4 5 6 7 8 9 Pixel Distance

7 Actual Distance (mm) y = x x R² = Pixel Distance τ = Iα d x = rα dt I = mr 3 d x = 3g sin dt θ θ d x = 3g dt θ x = 188

8 Θ x Θ y x y ( π ) p ix = p ix d i (1 cos γ) cos + tan 1 m ( π ) p iy = p iy d i (1 cos γ) sin + tan 1 m z p 1,p p 3 x y b x b y ( p iz = d i mb ) x + b y sin γ m +1 d dt x 1 = x d dt x = 3g θ. (x, y, z) OAC ODE m y = mx γ ABC DEF AB BC = DE EF sin θ x sin θ y = sin α cos θ x sin π/ α cos θ y tan θ x sin α = tan θ y cos α m = tan α = tan θ x. tan θ y γ ABC tan γ = AB BC = sin θ x sin α cos θ x = tan θ x sin α θ x = θ y = p 1,p p 3 y = mx 3g/ λ 1, = J ± 4J 1 +J J J = ( 3g 1 θ x 1 3g θ x ) = ( ) 1. J 1 J θ < θ < x 1 x θ = k p x 1 k d x R θ = k p (R x 1 ) k d ( d dt R x ) H(s) = (3g/)k ds + (3g/)k p s + (3g/)k d s + (3g/)k p. d l = mp ix + p iy m +1. k p k d k p > k d > 189

9 m/s m/s k p k d ζ =.6 g =9.81m/s t r ω n = ( ( )) 1 1 ζ π tan 1 =.77 1 ζ ζ ω n =.77rad/s ωn = k p k p =.5 () ζω n = k d k d =.3 (3) d x = 3g sin dt θ θ = sin 1 (u/3g) d x = dt u u θ ωn = k p =7.67 ζω n = k d =3.3 attenuating differentiator G(s) = ωs s+ω ω =5rad/s G(s) G(z) = 5z 5 z.8187 k p =.5 k d =.3 19

10 .1 Position (m) Time (sec) k p =.5 k d =.3 k p =1 k d =1 k p = 7.67 k d = 3.3 k p = 1 k d = 1 Position (m) Time (sec) k p = 3 k d = 1 Position (m) Time (s) 191

11 Position (m) Time (sec).6 / Control Education & Applications files/thesis.m4v y position (m) x position (m) References Proc. Australasian Conference on Robotics and Automation 19

12 193

CHAPTER 40 CARTESIAN AND POLAR COORDINATES

CHAPTER 40 CARTESIAN AND POLAR COORDINATES EXERCISE 169 Page 462 1. Express (3, 5) as polar coordinates, correct to 2 decimal places, in both degrees and in From the diagram, r = 32 + 52 = 5.83 y and 5