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Apple Pi Robotics Gaining a Mechanical Advantage

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The Design Process Identify a need Establish design criteria and constraints Evaluate alternatives (systems or components) & make decision Build a prototype Test/evaluate prototype against criteria Analyze, “tweak” ( ), redesign ( ), retest Document specifications, drawings to build (communicate design)

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Seeking alternatives: The Creative Process Find an “off-the-shelf” solution that someone else has used to solve a similar problem, or Find a bunch of existing stuff, combine it, and adapt it. Generate a unique, custom-made solution “from scratch” – Innovation!

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Mechanics 101 – Gaining an Advantage Levers Belts/Pulleys – Chain/Sprockets Gears Let's talk Forces!

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Forces – F4 F1 F2 Force has Magnitude & Direction If F1= F2 & F3 = F4 ? F3 M

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Forces – F4 F1 F2 If F1= F2 & F3 < F4 ?? F3 M

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Forces – F4 F1 F2 If F1< F2 & F3 < F4 ?? F3 M

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Forces – F4 F1 F2 If F1< F2 & F3 < F4 F3 M

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Practical Example: Static Model F = mass x a (gravity) = Weight F/2

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Practical Example: Dynamic Model F1 f2 (F1-f2) = mass x acceleration or (F1-f2)/mass = acceleration Provided there is enough reaction force (F2) to support F1 F2 = u * wt/2 (approx) F1 is all about Torque at Wheels (HP is the product of speed and torque) f2 is friction due to windage F2 is reaction force (tire friction) F2 wt

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Practical Example: F=ma (mass of body x gravitational acceleration) Windage is proportional to speed 2 ma = u w V 2 We have reached terminal velocity uwV2uwV2

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Types of Forces F (external) = mass * acceleration F (friction) =u * Wt F (weight) = Wt = mass* acceleration (gravity) F (wind) = windage const. * velocity 2 F (impulse) = mass * velocity change

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Coefficients of Friction (examples) Aluminum Steel.61 Copper Steel.53 Brass Steel.51 Cast Iron Copper 1.05 Concrete (wet) Rubber.30 Concrete (dry) Rubber 1.0 Polyethylene Steel.20 Materials Coefficient

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Mechanical Advantage – The Lever L1L2 F1 F2 F1 x L1 = F2 x L2

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Mechanical Advantage – The Lever 4 ft2 ft 10 lbs F2 F2 = ??

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Mechanical Advantage – The Lever 4 ft2 ft 10 lbs F2 10 lbs x 4 ft = F2 x 2 ft F2 = 10 lbs x 4 ft / 2 ft = 20 lbs

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Mechanical Advantage – The Lever D2 F1 F2 Trade Off - Force vs Distance D1/D2 is Inversely Proportional to F1/F2 D1 2 ft 4 ft

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Mechanical Advantage – The Lever D2 F1 F2 So ……….. If D1 is 1ft, then D2 = ?? If velocity of the left most side of lever is 6 inches/sec then velocity of the right most side is = ?? Examples of real devices?? D1 2 ft 4 ft

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Mechanical Advantage – The Lever D2 F1 F2 -- See Saw -- Catapult -- Oars -- Pry Bar D1 L2 L1

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Mechanical Advantage – The Lever First Order Another Configuration Examples? F1 F2 L2 L1

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Mechanical Advantage – The Lever First Order Crow/Pry Bar Claw Hammer F1 F2 L2 L1

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Balancing Forces – Arm/Boom L1 F1 F2 If F1 = F2 Equilibrium/No Motion - Right? Fulcrum

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Balancing Forces – Arm/Boom L1 F1 F2 Fulcrum M If F1 = F2 Equilibrium/No Motion - Right?

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Balancing Forces/Equilibrium – L1 F1 F2 This is the REAL mechanical system! So F2 must be greater to balance out the effect of boom weight m=Wt/32.2 M

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Balancing Forces/Equilibrium – L1 F1 F2 If (F1 x L1)+ (Wt x L1/2) = F2 x L1, then equilibrium Wt = ma = F w L1 2 Center of Gravity (CG) M

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2 nd Order Lever – L1 F1 (applied force) F2 (load) F1 x L1 = F2 x L2 Equilibrium Mechanical Advantage – If L2 is ½ of L1 Then F2 is how much greater then F1? What common items use 2 nd order lever? L2

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2 nd Order Lever – L1 F1 (applied force) F2 (load) - Wheel Barrow - Nutcracker - Bottle Opener - Crowbar L2

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Balancing Forces/Equilibrium – L1 F1 Wt ? L2/2 L2 How heavy is counterbalance? F2 10 lbs 2 lbs

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Balancing Forces/Equilibrium – L1 F1 2 lbs 10 lbs 44 lbs L2/2 L2 (2 * L2) + (10*L2*2) = Wt *L2/2 Wt = (2*L2)+(10*L2*2)/(L2/2) Wt = 22*L2/(L2/2) = 44 lbs

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Balancing Forces – L1 F1 F2 Remove effects of Wt by counterbalancing How heavy is counterbalance? L2 2*Wt Wt L2/2

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Let's Take a Look at Mechanism on “Little Foot” F1 F2 (F2=F1) 3” 21” F2 x u mass on end of arm Frictional Force Bungee Force Velocity = v

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Let's Take a Look at Mechanism on “Little Foot” F1 F2 3” 21” Weight F2 x u Frictional Force = F2 * u = 15lbs *.70 = 10.5 lbs Estimated Velocity of 4 ft/sec Weight = mass * 32.2 = 6 oz. Approx. Frictional Force Bungee Force Velocity = v

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Gain lever advantage of 7:1 (21” / 3”) Assume 4 ft/sec velocity at end of arm We have accelerated a 6 oz. weight and will be decelerating “very quickly” (.02 secs?) m = 1/3 lb / 32.2 ft/sec 2 Kinetic energy stored = mv – Pulse Energy = m (v1 -v2) = F(t1-t2) For our example this calculates to approx. 15 lbs force Approx. 50% greater than our calculated frictional force Inertial Switch Mechanism on “Little Foot”

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BUT... What happens if for some unknown reason it doesn't work? F1 F2 3” 21” Wt = 6 oz. F2 x u What can we do to effect the outcome? Frictional Force Bungee Force

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Mechanical Advantage – The Lever D2 F1 F2 Demo – Resolution of Forces Where did it go? D1

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Resolution of Forces - L2 F2 F1 All forces applied to the arm can be “resolved” into perpendicular (rotational) and parallel (compression/extension) force components L1

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Resolution of Force Y X F FxFx FyFy

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Resolution of Force How? A Little Math Using SIN, COS & TAN Functions - If any two are known you can solve for any others F FxFx O FyFy

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Resolution of Force How? A Little Math Y X F FxFx FyFy O SOH-CAH-TOA ?

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Resolution of Force How? A Little Math Y X F FxFx FyFy O SOH-CAH-TOA S in (O) = O pposite/ H ypotenus C os(O) = A djacent/ H ypotenus T an(O) = O pposite/ A djacent

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Resolution of Force How? A Little Math Y X F FxFx FyFy O Examples: For angles of 0 – 90 degrees, Sin & Cos vary between 0 – 1, Tan varies between 0 and infinity Sin(30) =.5 Sin(45) =.707 Cos(30)=.866

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Resolution of Force How? A Little Math Y X F FxFx FyFy O Examples: Sin(30) =.5 Sin(45) =.707 Cos(30)=.866 If (O) = 30 degrees and F = 5 lbs, then Sin (30) = Fx/5 lbs or Fx = Sin(30) x 5 lbs = 2.5 lbs

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Resolution of Forces - L2 F2 f2x f2y F1 f1x f1y All forces applied to the arm can be “resolved” into perpendicular and parallel force components MOVIE CLIP – 2008 Robot L1

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Apple Pi 2008 Robot - L1 F1 F2 If f1x x L1 = f2x x L2 Equilibrium Pneumatic Cylinder f1x f1y f2y f2x L2

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Apple Pi 2008 Robot - L1 F1 F2 Take 3 different points to do calculations for – determine worst case (largest F2). Pneumatic Cylinder f1x f1y f2y f2x L2

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Simple Lever vs Compound Lever lbs leveraged to 12,800 lbs

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Design Challenge - Paper Design – First 4 Steps of Process 1. Identify a need - DONE! 2. Establish design criteria and constraints 3. Evaluate alternatives & make decision 4. Build a prototype (In this case a “paper” prototype)

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Design Challenge - Design a solution to satisfy the following problem using lever technology

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Belts-Sheaves and Chain-Sprockets

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Pulley (Block/Tackle) Mechanical Advantage Number of supporting lines MA2 = ? MA1 = ? F=5 lbs F=? lbs F=5 lbs

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Pulley (Block/Tackle) Mechanical Advantage How about this one? MA = ? F=5 lbs F=? lbs

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Pulley (Block/Tackle) Mechanical Advantage Number of supporting lines MA2 = ? MA1 = ? F=5 lbs F=? lbs F=5 lbs

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Pulley (Block/Tackle) Mechanical Advantage How about this one? MA = ? F=5 lbs F=? lbs

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Belts, Sheaves & Sprockets Belt Types V - belt Does slip Depends on tension Flat belt Does slip Depends on tension Cogged generally solidly “coupled” to sprocket Teeth molded into belt Mesh with slots in pulley Flat or V groove No slip/highly efficient

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Pulleys/Sprockets- Configuration of pulleys with belt to gain mechanical advantage Dia. = 1” Dia. = 4” Each pulley has a circumference of (Pi) x D So pulley 1 has a circumference of 3.14” while pulley 2 has a circumference of 12.56”. It will take 4 revolutions of P1 to “feed out” enough belt to allow for 12.56” of the circumference of P2 to be “conveyed”. (assuming no slippage) Pulley #2 will go at ¼ the speed (RPM's) as P #1 and generate 4 x's the torque P#1 P#2

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Pulleys/Sprockets- Another configuration of pulleys with belt Different diameter pulleys Dia. = 1” RPM x1 Dia. = 4” RPM x ¼ Dia. = 2” RPM x ½

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Pulleys/Sprockets- What are speeds if “N” is 10 revs/sec.? Dia. = 1” Dia. = 4” Dia. = 2” N = 10 revs/sec

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Belts/Pulley vs. Chain/Sprocket Whats the Difference? * Belts generally used for lower torque requirements * Belts can slip (could be good?) * Chain can be separated (maintenance) * Belts do not need lubrication

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Gears- Gears behave like “closely coupled” pulleys Differe Dia. = 4” Each gear has a circumference of (Pi) x D So gear #1 has a circumference of 3.14” while gear #2 has a circumference of 12.56”. It will take 4 revolutions of G1 to “circum navigate” the perimeter of G2 to allow for 12.56” of the circumference of G2 (assuming no slippage) Gear #2 will go at ¼ the speed (RPM's) as Gear #1 G# 2 G#1Dia. = 1” Gears behave like “closely coupled” pulleys Differences – G#2 turns opposite direction & no chance for slippage

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Compound Gears- G#2 Dia.4” Previous example showed that G#2 will go at ¼ the speed of G#1. G#3 coupled directly to G#2 so it goes at ¼ the speed also. Then G#4 will go at 1/3 the speed of G#3 since is has 3 times the circumference (or diameter). Total mechanical advantage is 4 x 3 or 12. If G#1 is going 120 RPM's, then G#4 is ? This compound gear set provides 12:1 ratio (1/12 the speed) with 12 x's the torque. G#2 G#1 G#1 & 3 Dia. = 1” G#3 G#4 G#4 = Dia. 3”

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Design Challenge Design gear/sprocket mechanism that will drive “Big Ben II” for Guilford (propose a solution) Motor will provide 120 RPM's & is sufficiently sized to drive the mechanisms. ( Guilford already purchased) Clock to have hour, minute and second hands

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“MANTIS” Drive Train Design Speed to be 12 fps CIM Motors have 5000 rpm top speed Will use 2 speed transmission – High range has 9.4:1 gear ratio – Chain sprockets available are 12, 15, 22, 26, and 30 tooth Drive wheel is 8” diameter Select drive components after transmission

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“MANTIS” Drive Train For 12 fps Wheel circumference of 8” x 3.14 = 25” (approx) 25 in/rev /12 in/ft = 2.08 ft/rev X revs/sec = 12 ft/sec /2.08 ft/rev = 5.77 revs/sec 5000 rpm = 83 rev/sec (5000 rev/min /60 secs/min) Total ratio is 83 rev/sec /5.77 rev/sec = 14.38:1 Ratio of transmission is 9.4:1 so we need additional 14.38/9.4 (1.53:1) ratio Looking at available sprockets (12, 15, 22, 24 and 30) we picked 22 and 15 (1.46:1 ratio)

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That's All Folks!

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