1. 2.007 Robot Concepts AdamPaxson Adam Paxson

2. Functional Requirements 1.Drive and maneuver 2.Score pucks + balls 3.Move arrow

3. 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 F f = µ s mg –= 0.268*4.5*(1/4)*9.8 = 2.954 N Movement requirement: –t = √(4*m*x/(F D -F f )) ; x = 2m –t = 3.17 sec m=4.5k µ w = 1.810 µ s = 0.268 Γ = 2.6e-3 N*m D=5e-2m FdFd FdFd FfFf FfFf

4. FR1: Drive and Maneuver Front vs Rear wheel drive: steering stability Robot is displaced relative to driving direction Sliders provide restoring force: –2 F f sin(Θ) –F f = µm s g –F f = (0.268)(1.1kg)(9.8) –F restoring ≈ 5.25 sin(Θ) N Θ m=4.5k m s =m/4 m s =1.1kg

5. FR1: Drive and Maneuver Slider vs tracked steering Sum of moments about Center of Mass: –Γ total = Γ d - Γ f Γ t = 2F d r D cos(Θ) –Θ = arctan(2d/w 1 ) –r D = √(d 2 + w 1 /2 2 ) –F D = 2µF n Γ/D = 8.3*(0.4 – d) –F n = m(d/l) = 4.5*(0.4-d)/0.4 Γ f = 2F f r F –r f = √(d-l) 2 + w 2 /2 2 ) –F f = µF n = 0,268*4.5*d/0,4 –Γ total = 2*8.3*(0.4-d)* √(d 2 +0.04)*cos(arctan(d/0.2)) - 2*0.268*4.5*d/0.4* √((d-0.4) 2 +0.36)N*m cm rDrD rDrD rFrF rFrF FfFf FfFf FdFd FdFd d W1W1 W2W2 l m=4.5kg l=0.4m w 1 =0.4m w 2 =0.35m µ=0.268 Γ=2.6e-3 N*m D=5e-2m

6. FR1: Drive and Maneuver Slider vs tracked steering Sum of moments about Center of Mass: –Γ total = Γ d - Γ f Γ t = 2F d r D –r D = w/2 –F d = 2µF n Γ/D = –F n = mg/2 = 4.5*9.8/2 –Γ t = µmgΓw/D Γ f = 2F F r D –r D = l/2-x –F F = ∫F f (x)dx –F f (x) = (µMdx/x) –Γ f = µgml 3 /48 –Γ total = µ*4.5*9.8*2.6e-3*0.4/5e-2 - µ*9.8*4.5*0.4 3 /48 cm rDrD F f (x) FdFd FdFd W l m=4.5kg l=0.4m w=0.4m µ=0.00 Γ=2.6e-3 N*m x

7. FR1: Drive and Maneuver 2WD Front2WD Rear4WD Tracked Drive wheels get stuck in gap Unstable driving dynamics High turning load Most stable steering dynamics Less chance of puck and ball blockage Better handling over obstacles/gaps

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. FR2: Score Pucks and Balls Feasibility of lifting balls and pucks: time to lift 2 axes of motion: –clamp + rotate Γ motor = m load x r arm –= (0.1+0.2)* √((h bin /2) 2 + (d+2r ball ) 2 ) –= 0.3* √((0.0156 + (d+0.75) 2 ) Constraining dimensions: –r arm > h bin /2 (d=0) –r arm = √((h bin /2) 2 + (d+2r ball ) 2 ) W motor = 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 h bin r arm d r balls m ball =0.1kg h bin =0.25m r ball =0.037n

10. FR2: Score Pucks and Balls Feasibility of dropping pucks: center of gravity BLE complication: balls must be deposited at an angle CM y = ∑m i r i / ∑ m i –m base =m max -m tray –m tray =5m b + 0.5 = 1kg –m base =3.5kg –h tray =h bin + l/2*sin(Θ) = 0.25 + 0.0325 = 0.2825 –3.5*0.03 + 1*0.2825 = 0.0861m m b sin(Θ)>µm b g –Θ = arcsin(µ) = 10° φ h bin h tray m b =0.2kg h bin =0.250m h base =3e-2m l=0.375m µ rolling =0.176 Θ h base l

11. FR2: Score Pucks and Balls Feasibility of pushing pucks: frictional force F D = 2µ w F n Γ/D –= 2*1.81*4.5*(3/4)*9.8*2.6e-3/5e-2 = 6.226*(gear ration*e) N F f = µ s mg + µ p m p g –= 0.268*4.5*(1/4)*9.8 = 2.954 N –= 0.466*(0.200*5)*9.8 = 4.567 N –F f = 2.954 + 4.567 = 7.5208 N Motors will have sufficient power m=4.5k m p = 0.200kg µ w = 1.810 µ s = 0.268 µ p = 0.466 Γ = 2.6e-3 N*m D=5e-2m FdFd FdFd FfFf FfFf FfFf

12. Lifting Arm + PushTipping Bin +Push Adaptable to different table layouts Motors have required wattage to lift quickly Very fast scoring method, good countermeasure for molestabots Simple, few moving parts, no motor requirement No extra motors to use for arrow mechanism More time-consuming, vulnerable to opponent Missing the bin would ruin strategy No versatility, only one funtion FR2: Score Pucks and Balls

13. FR3: Move Arrow Extensible ArmMotorized Minibot Spring Minibot

14. FR3: Move Arrow Feasibility of moving arrow: –Required torque = F x r a F = 0.150 * 9.8 = 1.47N r a = 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 F rara haha F = 150g r a = 0.273m h a = 0.539m

15. FR3: Move Arrow Feasibility of mechanism placement: moment, CG Γ arm = ∑m i x i = m arm a/2 –= 0.30*0.310/2 = 0.05N*m CM x = ∑m i x i / ∑m i –= (4.5-m arm )(l/2) + m arm *(d+a)/2 / 4.5 –= (0.84 + (d+0.310)/2/4.5 Arm masses yield appropriate CM locations d a haha Arrow axis m arm = 0.3kg m total = 4.5kg a = 0.310m h a = 0.539m l = 0.4m l

16. FR3: Move Arrow Extremely difficult to accurately maneuver extensible arm Must have access to arrow at the end of the round Difficult to make reliable wiring that detaches Increased weight to lift to 18 inches Springs may not provide sufficient torque to lift arrow Springs may trigger at the wrong time Arm can be used for other functions (opening doors, hindering opponent) Arrow mechanism is activated at precise time and speed Requires no extra motors Leave and forget