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Apple Pi Robotics Gaining a Mechanical Advantage.

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Presentation on theme: "Apple Pi Robotics Gaining a Mechanical Advantage."— Presentation transcript:

1 Apple Pi Robotics Gaining a Mechanical Advantage

2 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)

3 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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5 Mechanics 101 – Gaining an Advantage Levers Belts/Pulleys – Chain/Sprockets Gears Let's talk Forces!

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

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

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

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

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

11 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

12 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

13 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

14 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

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

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

17 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

18 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

19 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

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

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

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

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

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

25 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

26 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

27 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

28 2 nd Order Lever – L1 F1 (applied force) F2 (load)‏ - Wheel Barrow - Nutcracker - Bottle Opener - Crowbar L2

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

30 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

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

32 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

33 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

34 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”

35 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

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

37 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

38 Resolution of Force Y X F FxFx FyFy

39 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

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

41 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

42 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

43 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

44 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

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

46 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

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

48 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)‏

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

50 Belts-Sheaves and Chain-Sprockets

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

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

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

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

55 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

56 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

57 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

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

59 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

60 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

61 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”

62 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

63 “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

64 “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)

65 That's All Folks!


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