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2.007 Robot Concepts

2.007 Robot Concepts. Adam Paxson. Functional Requirements. Drive and maneuver Score pucks + balls Move arrow. FR1: Drive and Maneuver. Feasibility of 2WD: Tractive force Frictional force Assume 25-75 distribution F D = 2µ w F n Γ/D = 2*1.81*4.5*(3/4)*9.8*2.6e-3/5e-2 = 6.226 N

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2.007 Robot Concepts

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  1. 2.007 Robot Concepts AdamPaxson

  2. Functional Requirements • Drive and maneuver • Score pucks + balls • Move arrow

  3. FR1: Drive and Maneuver • Feasibility of 2WD: • Tractive force • Frictional force • Assume 25-75 distribution • FD = 2µwFnΓ/D • = 2*1.81*4.5*(3/4)*9.8*2.6e-3/5e-2 = 6.226 N • Ff = µsmg • = 0.268*4.5*(1/4)*9.8 = 2.954 N • Movement requirement: • t = √(4*m*x/(FD-Ff)) ; x = 2m • t = 3.17 sec Fd Fd m=4.5k µw = 1.810 µs = 0.268 Γ = 2.6e-3 N*m D=5e-2m Ff Ff

  4. Θ FR1: Drive and Maneuver • Front vs Rear wheel drive: steering stability • Robot is displaced relative to driving direction • Sliders provide restoring force: • 2 Ff sin(Θ) • Ff = µmsg • Ff = (0.268)(1.1kg)(9.8) • Frestoring ≈ 5.25 sin(Θ) N m=4.5k ms=m/4 ms=1.1kg

  5. FR1: Drive and Maneuver W1 • Slider vs tracked steering • Sum of moments about Center of Mass: • Γtotal = Γd - Γf • Γt = 2FdrDcos(Θ) • Θ = arctan(2d/w1) • rD = √(d2 + w1/22) • FD = 2µFnΓ/D = 8.3*(0.4 – d) • Fn = m(d/l) = 4.5*(0.4-d)/0.4 • Γf = 2FfrF • rf = √(d-l)2 + w2/22) • Ff = µFn = 0,268*4.5*d/0,4 • Γtotal = 2*8.3*(0.4-d)* √(d2+0.04)*cos(arctan(d/0.2)) - 2*0.268*4.5*d/0.4* √((d-0.4)2+0.36)N*m d rD rD m=4.5kg l=0.4m w1=0.4m w2=0.35m µ=0.268 Γ=2.6e-3 N*m D=5e-2m cm Fd Fd l Ff Ff rF rF W2

  6. FR1: Drive and Maneuver W Ff(x) x • Slider vs tracked steering • Sum of moments about Center of Mass: • Γtotal= Γd - Γf • Γt = 2FdrD • rD = w/2 • Fd = 2µFnΓ/D = • Fn = mg/2 = 4.5*9.8/2 • Γt = µmgΓw/D • Γf = 2FFrD • rD = l/2-x • FF = ∫Ff(x)dx • Ff(x) = (µMdx/x) • Γf = µgml3/48 • Γtotal = µ*4.5*9.8*2.6e-3*0.4/5e-2 - µ*9.8*4.5*0.43/48 Fd m=4.5kg l=0.4m w=0.4m µ=0.00 Γ=2.6e-3 N*m cm Fd l rD

  7. FR1: Drive and Maneuver

  8. FR2: Score Pucks and Balls • Concept 1: drop 5 balls via tilted tray, push pucks and remaining balls into lower bin • Concept 2: lift 5 balls, pucks and remaining balls via 2-axis arm

  9. Feasibility of lifting balls and pucks: time to lift 2 axes of motion: clamp + rotate Γmotor = mloadxrarm = (0.1+0.2)*√((hbin/2)2 + (d+2rball)2) = 0.3* √((0.0156 + (d+0.75)2) Constraining dimensions: rarm > hbin/2 (d=0) rarm = √((hbin/2)2 + (d+2rball)2) Wmotor= E/t = mgh/t = 0.3*9.8*0.25/t Available watts @ 100% eff: 6W Watts @ 50% / 4motors = 0.75W Expected lift time: 1 sec * n balls FR2: Score Pucks and Balls mball=0.1kg hbin=0.25m rball=0.037n rarm hbin d rballs

  10. FR2: Score Pucks and Balls φ • Feasibility of dropping pucks: center of gravity • BLE complication: balls must be deposited at an angle • CMy = ∑miri/ ∑ mi • mbase=mmax-mtray • mtray=5mb + 0.5 • = 1kg • mbase=3.5kg • htray=hbin + l/2*sin(Θ) • = 0.25 + 0.0325 = 0.2825 • 3.5*0.03 + 1*0.2825 = 0.0861m • mbsin(Θ)>µmbg • Θ = arcsin(µ) = 10° mb=0.2kg hbin=0.250m hbase=3e-2m l=0.375m µrolling=0.176 l htray Θ hbin hbase

  11. FR2: Score Pucks and Balls • Feasibility of pushing pucks: frictional force • FD = 2µwFnΓ/D • = 2*1.81*4.5*(3/4)*9.8*2.6e-3/5e-2 = 6.226*(gear ration*e) N • Ff = µsmg + µpmpg • = 0.268*4.5*(1/4)*9.8 = 2.954 N • = 0.466*(0.200*5)*9.8 = 4.567 N • Ff= 2.954 + 4.567 = 7.5208 N • Motors will have sufficient power Fd Fd Ff m=4.5k mp = 0.200kg µw = 1.810 µs = 0.268 µp = 0.466 Γ = 2.6e-3 N*m D=5e-2m Ff Ff

  12. FR2: Score Pucks and Balls

  13. FR3: Move Arrow

  14. FR3: Move Arrow F • Feasibility of moving arrow: • Required torque = F x ra • F = 0.150 * 9.8 = 1.47N • ra = 0.273m • Γ = 1.47*0.273 = 0.4N*m • Required work = ΓΘ/t • = 0.4*pi/2t • High wattage required • Available torque from torsion spring: 0.1N*m ra F = 150g ra = 0.273m ha = 0.539m ha

  15. FR3: Move Arrow Arrow axis • Feasibility of mechanism placement: moment, CG • Γarm = ∑mixi = marma/2 • = 0.30*0.310/2 = 0.05N*m • CMx = ∑mixi/ ∑mi • = (4.5-marm)(l/2) + marm*(d+a)/2 / 4.5 • = (0.84 + (d+0.310)/2/4.5 • Arm masses yield appropriate CM locations marm = 0.3kg mtotal = 4.5kg a= 0.310m ha = 0.539m l = 0.4m ha a d l

  16. FR3: Move Arrow

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