1. Advanced Design Applications Power and Energy © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications Teacher Resource Learning Cycle 3

2. Machine Components: Gears © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

3. Types of Machine Components Machines are useful tools that engineers have designed to do a certain task. There are many components of machines that manipulate motion: Threads Springs Gears Bearings Pulleys Cables, chains sprockets Clutches and brakes © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

4. Gear Types Spur axes are parallel teeth are parallel Helical axes are parallel teeth are not parallel © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

5. Gear Types Bevel Plain: axes intersect on center line teeth are straight Spiral: axes intersect on center line teeth are curved © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

6. Gear Types Hypoid: offset shafts axes do not intersect Worm: used for large speed ratios and significant sliding © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

7. Gear Types Rack and Pinion: Special Spur gear set The rack translates rectilinearly © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

8. Gearing Calculations Gear geometry (tooth) Gear train (multiple gears working together) Gearing loads (tangential and separating) © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

9. Gear Train Calculations Gears can be used to manipulate motion: Change speed Change torque Change rotational motion to a different axis or a straight line Multiple reductions can be used to make a larger change in speed and/or torque. With gears that are spinning on the same shaft, their speed and torque will be the same, but their number of teeth may not be. © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

10. Gear Train Calculations To change speed or torque, you need to use ratios and cross multiplication: Use variables and numbers to denote gears in the gear train. Speed and torque have an inverse relationship: S 1 /S 2 = T 2 /T 1 Speed and number of teeth have an inverse relationship: S 1 /S 2 = N 2 /N 1 Teeth and torque have a direct relationship: N 1 /N 2 = T 1 /T 2 © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

11. Gear Reductions Each gearset is called a “reduction.” In some cases, an engineer cannot accomplish a speed or torque that they want. This is usually because of space (the gears would become too big). In this case, you will need to use multiple reductions… © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

12. Gear Reductions A single reduction looks like this when looking at the axes of the shafts: A double reduction looks like this when looking at the axes of the shafts: © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

13. Gear Reductions The gears should be numbered in order that the power is transferred through.  1 is the starting gear (usually a pinion on a motor)  2 is the mating gear to pinion #1  3 is a second pinion on the same shaft as gear 2  4 is a second gear that mates with #3 Please note: Since #2 and #3 are on the same shaft, their speed and torque will be the same. © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

14. Single Gear Reduction Sample Calculation 1: N = number of teeth, S = speed, T = torque S 2 /S 1 = N 1 /N 2 S 2 /11000 = 13/47 S 2 /11000 = 0.277S 2 = 11000(0.277) = 3042 T 2 /T 1 = N 2 /N 1 T 2 /4 = 47/13 T 2 /4 = 3.62T 2 = 3.62(4) = 14.46 © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications N 1 = 13S 1 = 11000 RPMT 1 = 4 in lb N 2 = 47S 2 = ____ RPMT 2 = ___ in lb

15. Single Gear Reduction Sample Calculation 2: N = number of teeth, S = speed, T = torque N 2 /N 1 = S 1 /S 2 N 2 /7 = 10000/4500 N 2 /7 = 2.222N 2 = 2.222(7) = 15.56 But, we cannot have a fraction of a tooth! So we must round and re-calculate the speed, S 2 ! Therefore N 2 = 16 S 2 /S 1 = N 1 /N 2 S 2 /10000 = 7/16 S 2 /10000 = 0.4375T 2 = 10000(0.4375) = 4375 © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications N 1 = 7S 1 = 10000 RPM N 2 = ____S 2 = 4500 RPM

16. Double Gear Reduction Sample Calculation: N = number of teeth, S = speed, T = torque S 2 /S 1 = N 1 /N 2 S 2 /14000 = 13/43 S 2 /14000 = 0.302S 2 = 14000(0.302) = 4233 Because S 2 and S 3 are on the same shaft, they are equal! © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications N 1 = 13S 1 = 14000 RPMT 1 = ____ in lb N 2 = 43S 2 = _____ RPMT 2 = 12 in lb N 3 = 17S 3 = _____ RPMT 3 = ____ in lb N 4 = ____S 4 = 1500 RPMT 4 = ____ in lb

17. Double Gear Reduction N 4 /N 3 = S 3 /S 4 N 4 /17 = 4233/1500 N 4 /17 = 2.822N 4 = 2.822(17) = 47.97 But, we cannot have a fraction of a tooth! So we must round and re-calculate the speed, S 4 ! Therefore N 4 = 48 S 4 /S 3 = N 3 /N 4 S 4 /4233 = 17/48 S 4 /4233 = 0.4375T 2 = 4233(0.4375) = 1852 Now that the first two columns are complete, we can fill in the missing torque values. T 1 /T 2 = N 1 /N 2 T 1 /12 = 13/43 T 1 /12 = 0.302T 1 = 0.302(12) = 3.63 © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications

18. Double Gear Reduction Since gears #2 and #3 are on the same shaft, their torque will be the same. Therefore, we only have the torque for gear 4 left. T 4 /T 3 = N 4 /N 3 T 4 /12 = 48/17 T 4 /12 = 2.82T 2 = 2.82(12) = 33.84 © 2014 International Technology and Engineering Educators Association STEM  Center for Teaching and Learning™ Advanced Design Applications