2.007 Robot Concepts2.007 Robot Concepts

AdamAdam PaxsonPaxson

Functional RequirementsFunctional Requirements

1. Drive and maneuver2. Score pucks + balls3. Move arrow

FR1: Drive and ManeuverFR1: 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

m=4.5kµw = 1.810µs = 0.268Γ = 2.6e-3 N*m

D=5e-2m

FdFd

Ff Ff

FR1: Drive and ManeuverFR1: 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

Θ

m=4.5kms=m/4ms=1.1kg

FR1: Drive and ManeuverFR1: Drive and Maneuver• 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

cm

rD rD

rF rF

Ff Ff

Fd Fd

d

W1

W2

l

m=4.5kgl=0.4mw1=0.4mw2=0.35mµ=0.268Γ=2.6e-3 N*mD=5e-2m

FR1: Drive and ManeuverFR1: Drive and Maneuver• 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

cm

rD

Ff(x)

Fd

Fd

W

l

m=4.5kgl=0.4mw=0.4mµ=0.00Γ=2.6e-3 N*m

x

FR1: Drive and ManeuverFR1: Drive and Maneuver

2WD Front 2WD Rear 4WD 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 FR2: Score Pucks and BallsBalls

• 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 FR2: Score Pucks and BallsBalls

• 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

hbin

rarm

drballs

mball=0.1kghbin=0.25mrball=0.037n

FR2: Score Pucks and FR2: Score Pucks and BallsBalls

• 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°

φ

hbin

htray

mb=0.2kghbin=0.250mhbase=3e-2ml=0.375mµrolling=0.176

Θ

hbase

l

FR2: Score Pucks and FR2: Score Pucks and BallsBalls

• 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

m=4.5kmp = 0.200kgµw = 1.810µs = 0.268µp = 0.466Γ = 2.6e-3 N*m

D=5e-2m

FdFd

Ff Ff

Ff

Lifting Arm + Push Tipping 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 FR2: Score Pucks and BallsBalls

FR3: Move ArrowFR3: Move Arrow

Extensible Arm

Motorized Minibot

Spring Minibot

FR3: Move ArrowFR3: Move Arrow• 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

F

ra

ha

F = 150gra = 0.273mha = 0.539m

FR3: Move ArrowFR3: Move Arrow• 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

d

a

ha

Arrow axismarm = 0.3kgmtotal = 4.5kga = 0.310mha = 0.539ml = 0.4m

l

FR3: Move ArrowFR3: 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