2.007 Robot Concepts AdamPaxson Adam Paxson

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

FR1: Drive and Maneuver Feasibility of 2WD: –Tractive force –Frictional force –Assume distribution F D = 2µ w F n Γ/D –= 2*1.81*4.5*(3/4)*9.8*2.6e-3/5e-2 = N F f = µ s mg –= 0.268*4.5*(1/4)*9.8 = N Movement requirement: –t = √(4*m*x/(F D -F f )) ; x = 2m –t = 3.17 sec m=4.5k µ w = µ s = Γ = 2.6e-3 N*m D=5e-2m FdFd FdFd FfFf FfFf

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

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 )*cos(arctan(d/0.2)) - 2*0.268*4.5*d/0.4* √((d-0.4) )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

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

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

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

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 –= ( )* √((h bin /2) 2 + (d+2r ball ) 2 ) –= 0.3* √(( (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 100% eff: 6W 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

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 = 1kg –m base =3.5kg –h tray =h bin + l/2*sin(Θ) = = –3.5* * = m 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

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 = N –= 0.466*(0.200*5)*9.8 = N –F f = = N Motors will have sufficient power m=4.5k m p = 0.200kg µ w = µ s = µ p = Γ = 2.6e-3 N*m D=5e-2m FdFd FdFd FfFf FfFf FfFf

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

FR3: Move Arrow Extensible ArmMotorized Minibot Spring Minibot

FR3: Move Arrow Feasibility of moving arrow: –Required torque = F x r a F = * 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

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 –= ( (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

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