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
7. • Driving gear(back) convert torque
into speed, and the speed again
convert into torque
• Not efficient
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
10. • A third, larger type of gear
• Greater ratio
• Size constraints
11. “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)
12. • 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
