V2 slideshow initial
- 1. ‘Creative Journey’ Group XYZ Rhys Morrison, Bryan Sito, Kevin Huynh, Samuel Warren
- 2. Understanding the Problem Performance ‘𝑝’ = (𝑚𝑔ℎ/𝑡𝑉𝐼)*(𝑚𝑔ℎ/𝑡D3) Weight (m)(kilograms): 8 -> ??? Kg Height (h)(metres): 0.5 -> 1m Time (t)(seconds): As few as possible Max. Axial Length (D)(metres): Minimize as much as possible g = 9.8m/s^2 V = 6V I (nominal) = 2.6A Motor Torque (nominal) = 80g.cm
- 3. Find output torque required on 6mm shaft to lift minimum 8kg weight T = F.R = mg*0.003m = 8*9.81*0.003 = 0.23544N.m = 2400.82g.cm Compare to input torque to find required conversion for gearbox (Specification sheet) 80g.cm = 0.00784532N.m input torque 0.23544/0.00784532 = 30.010 conversion needed Design a gearbox using standard gears to achieve required ratio Create iterative solutions to achieve maximal performance balance of weight and lift speed without overloading motor or building to excessive size
- 4. Derive gear sizes, pitch circle diameters (hence effective radii) and ratios from construction sketches
- 5. 18mm/8mm = 2.25 Gear ratio 2.25^? = 31.01 ? = 4.52 gear-downs required to lift minimum weight at listed motor performance
- 6. Large axial length Loss in power Not efficient
- 7. • Driving gear(back) convert torque into speed, and the speed again convert into torque • Not efficient • A third, larger type of gear • Greater ratio • Size constraints
- 8. • Driving gear(back) convert speed into torque • The smaller(front) gear which move along with the big gear then drive the output gear • Not efficient
- 9. Planetary gear system
- 10. “Final” Initial design • Overlapping gear structure, highly compact • Diagonal placement to reduce axial length • Simple, repeating design. Easy to modify • Cubic shape gives maximum volume with minimum side length • 63*63mm, including side tabs • (Motor mounting not finalized)
- 11. • Top-down view of gearbox showing alignment of gears. • Significant room to play around with • Slight separation of each gear to avoid friction. Include small washer or spacer