1. Mechanical Concepts 101 Shannon Schnepp Dennis Hughes Anthony Lapp 10/29/05

2. Basic Concepts: Equations  Force = Mass * Acceleration  Torque = Force * Distance = Work  Power = Work/Time  Power = Torque * Angular Velocity

3. The friction coefficient for any given contact with the floor, multiplied by the normal force, equals the maximum tractive force can be applied at the contact area. Tractive force is important! It’s what moves the robot. normal force tractive force torque turning the wheel maximum tractive force normal force friction coefficient = x weight Basic Concepts: Traction

4. Basic Concepts: Traction Equations F friction =  * F normal Experimentally determine  : F normal = Weight * cos(  ) F parallel = Weight * sin(  ) Fnormal Ffriction Weight  Fparallel When F friction = F parallel, no slip F friction =  * Weight * cos(  ) F parallel = Weight * sin(  ) =  * Weight * cos(  )  = sin(  ) / cos(  )  = tan(  )

5. Basic Concepts: Coefficient of Friction  Materials of the robot wheels (or belts)  High Friction Coeff: soft materials, “spongy” materials, “sticky” materials  Low Friction Coeff: hard materials, smooth materials,shiny materials  Shape of the robot wheels (or belts)  Want the wheel (or belt) surface to “interlock” with the floor surface  Material of the floor surface  Surface conditions  Good: clean surfaces, “tacky” surfaces  Bad: dirty surfaces, oily surfaces

6. Basic Concepts: Free Body Diagrams W fAfA NANA fB fB NBNB AB The normal force is the force that the wheels exert on the floor, and is equal and opposite to the force the floor exerts on the wheels. In the simplest case, this is dependent on the weight of the robot. The normal force is divided among the robot features in contact with the ground. The frictional force is dependent of the coefficient of friction and the normal force (f = mu*N).

7. Basic Concepts: Weight Distribution more weight in back due to battery and motors front The weight of the robot is not equally distributed among all the contacts with the floor. Weight distribution is dependent on where the parts are in the robot. This affects the normal force at each wheel. more normal force less normal force less weight in front due to fewer parts in this area EXAMPLEONLY

8. Basic Concepts: Weight Transfer EXAMPLEONLY robot accelerating from 0 mph to 6 mph inertial forces exerted by components on the robot more normal force is exerted on the rear wheels because inertial forces tend to rotate the robot toward the rear less normal force is exerted on the front wheels because inertial forces tend to rotate the robot away from the front In an extreme case (with rear wheel drive), you pull a wheelie In a really extreme case (with rear wheel drive), you tip over!

9. Basic Concepts: Gears  Gears are generally used for one of four different reasons: 1. To reverse the direction of rotation 2. To increase or decrease the speed of rotation (or increase/decrease torque) 3. To move rotational motion to a different axis 4. To keep the rotation of two axes synchronized

10. Basic Concepts: Gears  The Gear Ratio is a function of the number of teeth of the gears  Consecutive gear stages multiply N1N1 N2N2 N3N3 N4N4 Gear Ratio is (N 2 /N 1 ) * (N 4 /N 3 ) Efficiency is.95 *.95 =.90

11. Basic Concepts: Gears N1N1 N2N2 N3N3 N4N4 Gear 4 is attached to the wheel Remember that T = F * R w Also, V =  * R w T 4 = T 1 * N 2 /N 1 * N 4 /N 3 *.95 *.95  4 =  1 * N 1 /N 2 * N 3 /N 4 F = T 4 / R w V =  4 * R w Wheel Diameter - D w D w = R w * 2 F push

12. Lifting/Moving Objects Example 1: A box weighs 130 lbs and must be moved 10 ft. The coefficient of friction between the floor and the box is.25.  How much work must be done??

13. Lifting/Moving Objects  f = mu*N =.25*130  f = 65 lbs  so…  Work = f * dist  Work = 65 * 10 = 650 ft lbs

14. Lifting/Moving Objects  Example 2: The arm weighs 10 lbs and moves 3 ft vertically. The mechanism that contains the balls weighs 5 lbs. The balls weigh 3 lbs. The mechanism and balls move 6 ft vert.  Work = Force 1*Dist 1 + Force 2*Dist 2 = 10 lbs * 3 ft + 8 lbs * 6 ft = 30 + 48 = 78 ft lbs

15. Lifting/Moving Objects  Example 2A:  Desire this motion to be completed in 10 seconds.  Power = 78 ft lbs / 10 seconds *(60sec/1min) *.02259697 = 10.6 Watts  Note: There is only a certain amount of power available.

16. Lifting/Moving Objects  Example 2B:  Desire this motion to be completed in 3 seconds.  Power = 78 ft lbs / 3 seconds *(60sec/1min) *.02259697 = 35.3 Watts

17. Combined Motor Curves

18. Motor Calculations  Motor Power = Power Available = Free Speed / 2 * Stall Torq. / 2 * C.F.  Where:  Free Speed is in rad / min  Stall Torque is in ft lbs  Conversion Factor =.02259697

19. Motor Calculations  Free Speed (rad/min) = RPM * 2 Pi (rad/rev)  Stall Torque (ft*lb) = (in oz)*(1 ft/12 in)*(1 lb/16 oz)

20. Motor Calculations Drill Motor  Free Speed = 20000(rev/min)*2PI(rad/rev) = 125664 rad/min  Stall Torque = 650 (Nmm)*(1 lb/4.45 N)* (1 in/ 25.4mm)*(1 ft/12 in) =.48 ft lbs

21. Motor Calculations Drill Motor  Power = Free Speed / 2 * Stall Torque / 2 *Conv. Factor = 125664 / 2 *.48 / 2 *.02259697 = 340 W

22. Choosing a Motor  Need 78 ft lbs of Torque (ex 2)  Try Globe Motor w/ Gearbox  Working Torque = Stall Torque / 2  = (15 ft lbs @ 12 V) / 2  = 7.5 ft lbs

23. Gear Ratios  Gear Ratio = Torque Needed / Torque Available = 78 ft lbs / 7.5 ft lbs = 10.4 :1  Now time to find the gear train that will work!

24. Choosing a Motor  In Summary:  All motors can lift the same amount (assuming 100% power transfer efficiencies) - they just do it at different rates  BUT, no power transfer mechanisms are 100% efficient  If you do not account for these inefficiencies, your performance will not be what you expected

25. Materials  Steel  High strength  Many types (alloys) available  Heavy, rusts,  Harder to processes with hand tools  Aluminum  Easy to work with for hand fabrication processes  Light weight; many shapes available  Essentially does not rust  Lower strength

26. Material  Lexan  Very tough impact strength  But, lower tensile strength than aluminum  Best material to use when you need transparency  Comes in very limited forms/shapes  PVC  Very easy to work with and assemble prefab shapes  Never rusts, very flexible, bounces back (when new)  Strength is relatively low

27. Structure  Take a look at these two extrusions - both made from same Aluminum alloy:  Which one is stronger?  Which one weighs more? 1.0” 0.8” Hollow w/ 0.1” wallsSolid bar

28. Structure  The solid bar is 78% stronger in tension  The solid bar weighs 78% more  But, the hollow bar is 44% stronger in bending  And is similarly stronger in torsion

29. Structural Equations  It all boils down to 3 equations: Where:  = Bending Stress M = Moment (bending force) I = Moment of Inertia of Section c = distance from Central Axis Where:  = Tensile Stress F tens = Tensile Force A = Area of Section Where: = Shear Stress F shear = Shear Force A = Area of Section BendingTensileShear

30. Stress Example  Let's assume we have a robot arm (Woo hoo!) that's designed to pick up a few heavy weights. The arm is made out of Al-6061, and is 3/8" tall, 1" wide, and 3 feet long. The yield strength is about 40,000 PSI. In the competition they are hoping to to pick up 3 boxes of 15 lbs each. Will this arm be strong enough?