HW7sol

# HW7sol - ME 576 Homework#7 1 Bollinger 8.7 Δθ 2 error(E...

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Unformatted text preview: ME 576 Homework #7 1. Bollinger 8.7 Δθ 2 error (E) Δθ R 2 The error, E, is ⎛ ⎝ E=R ⎜ 1 − cos θ⎞ ⎟ 2⎠ Since R is 5” and max. error should be 0.001” E⎞ ⎛ Δθ = 2 cos −1 ⎜ 1 − ⎟ R⎠ ⎝ ⎛ 0.001 ⎞ =2cos-1 ⎜ 1⎟ = 0.04rad = 2.29° 5⎠ ⎝ The length of the straight line segment is L = 2R sin Δθ = 2 × 5 × sin(0.02) = 0.200" 2 Time required to traverse is Δt = L υ = 0.200" 60(s) = 0.040sec 300" (min) B.C. for the first three segments θi x (ti ) = R cos(θ i ) 0 0.00 π 1 0.04 π − 0.04 2 2 0.08 3 ti 0.12 2 π − 0.08 2 π − 0.12 2 Equations for linear interpolation x (t ) = x (t i −1 ) + [t - ti -1 ] x (ti ) − x (ti −1 ) ti - ti −1 y (ti ) − y (ti −1 ) y (t ) = y (t i −1 ) + [t - ti -1 ] ti - ti −1 y (ti ) = R sin(θ i ) 0.000 i 5.000 0.200 4.997 0.400 4.985 0.599 4.965 2. Bollinger 8.10 ⎡ ⎛ dy ⎞2 ⎤ ⎛ dx ⎞2 x + y = υ ⇒ ⎢1 + ⎜ ⎟ ⎥ ⎜ ⎟ = υ 2 ⎢ ⎝ dx ⎠ ⎥ ⎝ dt ⎠ ⎣ ⎦ 2 2 2 dy = 1.5 dx t =0 dy = 1.17 dx t =4 R = ⎡r (ti ) r (ti ) r (t f ) r (t f ) ⎤ ⎣ ⎦ r (t) = A ⋅ b(t) Segment 1. ⎡1 R1 = ⎢ ⎣1 2.77 4.16 ⎡0 ⎢ 0 B1 = ⎢ ⎢0 ⎢ ⎢1 ⎣ 0 0 1 0 5⎤ 0⎥ ⎦ 4 4 (1.33)3 (1.33) 1.33 1 2 3(1.33) 2 ⎤ ⎡0 ⎥⎢ 2(1.33) ⎥ ⎢0 = 1 ⎥ ⎢0 ⎥⎢ 0 ⎥ ⎣1 ⎦ 0 0 1 0 2.37 1.78 1.33 1 5.33 ⎤ 2.67 ⎥ ⎥ 1⎥ ⎥ 0⎦ ⎡0 0 2.37 5.33 ⎤ ⎢ ⎥ ⎡1 2.77 4 5 ⎤ ⎢0 0 1.78 2.67 ⎥ A 1 = R ⋅ B −1 = ⎢ ⎥ 1⎥ ⎣1 4.16 4 0⎦ ⎢0 1 1.33 ⎢ ⎥ 1 0⎦ ⎣1 0 ⎡1.8413 -2.8478 2.7735 1 ⎤ =⎢ ⎥ ⎣-0.1911 -1.1779 4.1603 1⎦ −1 Segment 2. ⎡4 R2 = ⎢ ⎣4 5 0 5⎤ 0⎥ ⎦ 7 2 ⎡(1.33)3 ⎢ (1.33) 2 B2 = ⎢ ⎢ 1.33 ⎢ ⎢1 ⎣ ⎡ 2.37 ⎢1.78 =⎢ ⎢1.33 ⎢ ⎣1 ⎡3.0938 A2 = ⎢ ⎣1.6875 3(1.33) 2 (2.66)3 2(1.33) 1 0 (2.66) 2 2.66 1 5.33 18.96 2.67 1 0 7.11 2.67 1 -18.5625 -10.1250 3(2.66) 2 ⎤ ⎥ 2(2.66) ⎥ 1⎥ ⎥ 0⎥ ⎦ 21.33⎤ 5.33 ⎥ ⎥ 1⎥ ⎥ 0⎦ 38 18 -21⎤ -6 ⎥ ⎦ Segment 3 ⎡7 R3 = ⎢ ⎣2 5 0 ⎡18.96 ⎢ 7.11 B3 = ⎢ ⎢ 2.67 ⎢ ⎢1 ⎣ ⎡ 2.1086 A3 = ⎢ ⎣ 0.0286 10 3.25 ⎤ 4.5 3.8 ⎥ ⎦ 21.33 43 5.33 1 0 42 4 1 -21.7427 1.1392 3(4) 2 ⎤ ⎡18.96 ⎥⎢ 2(4) ⎥ ⎢ 7.11 = 1 ⎥ ⎢ 2.67 ⎥⎢ 0 ⎥⎣1 ⎦ 75.9776 -6.6872 21.33 5.33 1 0 -80.9776 ⎤ 11.1862 ⎥ ⎦ 64 16 4 1 48 ⎤ 8⎥ ⎥ 1⎥ ⎥ 0⎦ 3. Bollinger 8.10 ⎡ ⎛ dy ⎞2 ⎤ ⎛ dx ⎞2 x + y = υ ⇒ ⎢1 + ⎜ ⎟ ⎥ ⎜ ⎟ = υ 2 ⎢ ⎝ dx ⎠ ⎥ ⎝ dt ⎠ ⎣ ⎦ 2 2 2 dy = 1.5 dx t =0 dy = 1.17 dx t =4 R* = [r * (0) r * (0) r * (1) r * (1)] r * (ui ) = A * ⋅b * (ui ) Segment 1. ⎡1 * R1 = ⎢ ⎣1 2.77(1.33) 4.16(1.33) ⎡ 2 -3 ⎢1 -2 * −1 B1 = ⎢ ⎢ -2 3 ⎢ -1 ⎣1 0 1 0 0 ⎡1 * * * A1 = R1 ⋅ B1 −1 = ⎢ ⎣1 4 4 5(1.33) ⎤ ⎥ 0 ⎦ 1⎤ 0⎥ ⎥ 0⎥ ⎥ 0⎦ 3.68 5.53 4 4 ⎡ 2 -3 6.65⎤ ⎢1 -2 ⎢ 0 ⎥ ⎢-2 3 ⎦ ⎢ -1 ⎣1 ⎡ 4.33 -5.03 3.69 =⎢ ⎣-0.47 -2.07 5.53 0 1 0 0 1⎤ 0⎥ ⎥ 0⎥ ⎥ 0⎦ 1⎤ 1⎥ ⎦ Segment 2. ⎡4 R* = ⎢ 2 ⎣4 5(1.33) 0(1.33) 5(1.33) ⎤ ⎥ 0 ⎦ 7 2 ⎡ 7.30 * A* = R * ⋅ B1 −1 = ⎢ 2 2 ⎣ 4.0 -10.95 6.65 -6.00 0 4⎤ 4⎥ ⎦ Segment 3 ⎡7 R* = ⎢ 3 ⎣2 5(1.33) 0(1.33) 10 4.5 ⎡ 4.97 * A * = R * ⋅ B1 −1 = ⎢ 3 3 ⎣0.05 r (t) = A ⋅ b(t) ⎧ui 3 ⎫ ⎪ 2⎪ ⎧ xi (t ) ⎫ * ⎪ui ⎪ ⎨ ⎬ = Ai ⎨ ⎬ ⎩ yi ( t ) ⎭ ⎪ui ⎪ ⎪1 ⎪ ⎩⎭ ⎧3ui 2 ⎫ ⎪ ⎪ ⎧ xi (t ) ⎫ 1 * ⎪ 2ui ⎪ Ai ⎨ ⎨ ⎬= ⎬ ⎩ yi (t ) ⎭ 1.33 ⎪1 ⎪ ⎪0 ⎪ ⎩ ⎭ 3.25(1.33) ⎤ 3.8(1.33) ⎥ ⎦ -8.62 6.65 2.45 0 7⎤ 2⎥ ⎦ 4. Use the constant acceleration method. Δr = 1.697 - 0.693 = 1.004 m Δθ = 45 - (-30) = 75° Δz = 0.4 m Vrmax = 10 m min Ar max = 15 cm VΘ = 2 rev 10 m 60 sec m 2 = 0.15 = sec = 720° sec 2 sec sec AΘ max = 4 rev 2 = 1440° 2 sec sec = 0.1 m VΖ max = 6 m min sec AΖ max = 20 cm 2 = 0.2 m sec sec First calculate the time intervals for three splines of each axis without the consideration to coordinate them. r‐axis 10 = 1.111 sec 60 × 0.15 1.004 × 60 Δx − Δt1 = − 1.111 = 4.913 sec Δt2 = Vr 10 Δt1 = Δt3 = Δt1 = 1.111 sec θ − axis Δt1 = 2 = 0.5sec 4 Vx2 7202 = = 360 However Δx = 75 < Ax 1440 1 ⎛ 0.208 ⎞ 2 ∴ Δt1 = ⎜ ⎟ = 0.228 sec ⎝4⎠ Z‐axis 0.1 = 0.5sec 0.2 0.4 Δt2 = − 0.5 = 3.5sec 0.1 Δt3 = Δt1 = 0.5sec Δt1 = Without coordination, the velocity profiles would be like Modify the velocity profiles according to r‐axis Δt = Δt1 + Δt 2 + Δt 3 = 1.111 + 4.913 + 1.111 = 7.135 seconds For r‐axis, use max. velocity and accelerations For θ ‐axis Δt1 = Vθ = 1.111 Αθ Δt 2 = 75 Δθ − Δt1 = 4.913 ⇒ − 1.111 = 4.913 Vθ Vθ ∴ Vθ = 12.45° = 0.217 rad sec. s Αθ = 11.206° 2 sec For Z‐axis Δt1 = VΖ = 1.111 ΑΖ Δt 2 = 0.4 Δz − Δt1 = 4.913 ⇒ − 1.111 = 4.913 Vz Vz Vz = 0.664 m sec Α z = 0.0598 m 2 sec Boundary conditions r - axis θ − axis θ = −30° t0 r = 0.786 θ = −23.084 r = 0.167 θ = 12.45 r = 1.604 r − 0.167 tf t1 t2 r = 0.693 r=0 r = 1.697 r=0 θ =0 θ = 38.084 θ = 12.45 θ = 45° θ =0 z − axis z=0 z=0 z = 0.037 z = 0.0664 z = 0.363 z = 0.0664 z = 0.4 z=0 A1= 0.0000 0.0750 0 0.6928 ‐0.0000 0.0000 0.1667 0.6002 0.0000 ‐0.0750 .0705 ‐2.1227 0 0.0978 0 ‐0.5236 0.0000 0.0000 0.2172 ‐0.6443 0.0000 ‐0.978 1.3953 ‐4.1936 0 0.0299 0 ‐0.4000 0.0000 0.0000 0.0664 ‐0.4369 0.0000 ‐0.0299 0.4264 ‐1.5215 A2= A3= ...
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## This document was uploaded on 12/28/2011.

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