2.007 robot concepts adampaxson adam paxson. functional requirements 1.drive and maneuver 2.score...
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![Page 1: 2.007 Robot Concepts AdamPaxson Adam Paxson. Functional Requirements 1.Drive and maneuver 2.Score pucks + balls 3.Move arrow](https://reader035.vdocuments.site/reader035/viewer/2022062421/56649d595503460f94a38d75/html5/thumbnails/1.jpg)
2.007 Robot Concepts2.007 Robot Concepts
AdamAdam PaxsonPaxson
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Functional RequirementsFunctional Requirements
1. Drive and maneuver2. Score pucks + balls3. Move arrow
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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=4.5k
µw = 1.810
µs = 0.268
Γ = 2.6e-3 N*m
D=5e-2m
Fd
Fd
Ff Ff
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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) NΘ
m=4.5k
ms=m/4
ms=1.1kg
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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=4.5kg
l=0.4m
w1=0.4m
w2=0.35m
µ=0.268
Γ=2.6e-3 N*m
D=5e-2m
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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=4.5kg
l=0.4m
w=0.4m
µ=0.00
Γ=2.6e-3 N*m
x
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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
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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
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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=0.1kg
hbin=0.25m
rball=0.037n
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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=0.2kg
hbin=0.250m
hbase=3e-2m
l=0.375m
µrolling=0.176
Θ
hbase
l
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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=4.5k
mp = 0.200kg
µw = 1.810
µs = 0.268
µp = 0.466
Γ = 2.6e-3 N*m
D=5e-2m
Fd
Fd
Ff Ff
Ff
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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
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FR3: Move ArrowFR3: Move Arrow
Extensible Arm
Motorized Minibot
Spring Minibot
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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 = 150g
ra = 0.273m
ha = 0.539m
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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 = 0.3kg
mtotal = 4.5kg
a = 0.310m
ha = 0.539m
l = 0.4m
l
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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