MS5 Schadenfreude

MS5 Schadenfreude

• 1.
  TEAM 33: DESIGN REVIEW ShadaeBoakye-Yiadom, Jonathan Bruns, Jarred McDuffey, Jolene Ng, Julia Roth
• 2.
  OVERVIEW ZONE 4
• 3.
  FUNCTIONAL REQUIREMENTS & SPECIFICATIONS Ableto lift up three cubes at a time (207.36 grams) (3* 69.12 g where 69.12 grams is the weight of one cube) Able to travel back and forth from the border between zones 3 and 4 to the goal in 1 minute (2.7 inches/sec) Able to reach the goal height to score points (12 inches) Able to push the largest block F =2μsN = 2.5 lbs
• 4.
  DESIGN CONCEPTS COMPARISON
• 5.
  DESIGN CONCEPTS COMPARISON (continued)
• 6.
  FINAL DESIGN
• 7.
  FINAL DESIGN SUBSYSTEMS
• 8.
  CAD Final Design:
• 9.
  CAD Subsystem 1:Scissor Lifting Mechanism
• 10.
  CAD Subsystem 2:Dustpan Mechanism
• 11.
  CAD Subsystem 2:Dustpan Mechanism
• 12.
  ANALYSIS Subsystem 1: Movingthe blocks out of the way Assume that the gear ratio is the highest gear ratio FD = FF/2 = 5.56 N TD= ƒs*r*FD = (1.5)(5.56)(.0381) = .5889 Nm Planetary Gearbox: At 4.5 V: At 6 V: TS=.0013Nm TS=.0173 Nm N0=10200 rpm NO= 13600 rmp TR= .2*ɣ*TS = .2*.15*.0173 = 5.19*10-4Nm MR= TD/TS =.3178/ 5.19*10-4 = 1134.7 TD≤ ɣ*M*TS .15*400*.0173 = 1.038 and .5889 ≤ 1.038 TSg= 1.038 M = 400 N0g=1/M*13600 = 34 T = -.0305*n+1.038 n = 14.72 rpm %n = 14.72/34 *100 = 43.3 % Not within the maximum operating efficiency while pushing but will provide us enough torque to be able to push the blocks out of the way.
• 13.
  Analysis (cont) Total weightof Scissor lift and rack = 4.52 N The radius of the pulley is .5 inches. Torque needed to pull the scissor lift up with safety factor of 1.5 = .0861 Nm We found that a planetary gear motor with an 100:1 gear ratio can lift the weight with 34% of the total torque (approximately 4% away from the range of maximum efficiency). T_D = 1.5*4.52* .0127 = .0861 Nm T_sg = .15*100*.0173 = .2595 Nm T_sg > T_D Using the equations for an operation line: T= - (.2595/136)n+.2595 Plugging in T_D for the torque we get that n = 90, which is 66% of the no- load speed. Subsystem: Scissor lift
• 14.
  Analysis (Cont.) Required weightto lift = 1.13 N 24 tooth gear connected to pinion. 1 inch pulley connected to the shaft, belt and other pulley( .5 inches) connecting this shaft to the motor. T_r = 1.5*1.13*.0127 = .02153 Nm T_m = T_r*(d_m/d_r) = .0127*(.02153/.0254) = .010765 Nm Using the metal gear box we would use 6% of the total torque which is not within the range of maximum efficiency. T_metal = 1.13(.15) = .1695 T_m/ T_metal = .0635 Using the planetary gear motor with a 20:1 gear ratio we use 20.7% of the total torque which is within the maximum efficiency of the motor. T_max = .2*.15*.0173 = 5.19 * 10^(-4) Nm M_max = T_m/T_max = .010765/5.19*10^(-4) = 20.7 T_D = .15*20*.0173 = .0519 > .010765 = T_m T_m/T_D = 20.7% which is within maximum efficiency of the motor, therefore we will be try to use the planetary gear motor in order to raise the shovel up to its required height. Gears and motor to lift the shovel
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  SUMMARY EXPECTED CHALLENGES: Stability throughoutthe RMP especially with so much weight on the front Staying within the size limitations ensuring the blocks do not fall of the table when we push them Getting both moving parts to work together RISKS: turning the dustpan to drop the cube without the use of a motor Pushing the blocks into an area that we will later need to travel Getting our RMP to go in between both blocks surrounding the goal without damaging our RMP
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  Assumptions and Procedure Assumptions: Wehave assumed that everything is going to be cut exactly to the correct size. We assumed that the density of the Aluminum in the lab is also 2.7 grams/cm^3 We will be able to fit the cost of extra planetary gear motors into the budget. There will be 1 inch and half inch pulleys available for the belt we will use to transfer torque to our gears on the rack. The weight of the pinions is negligible compared to the weight that the rack is putting on the scissor lift. Procedure: Used the density and dimensions of all the parts to approximate all the weights of our parts that we will be lifting. Found the torque necessary to lift the weights of the parts and checked every single motor to see which ones we could possibly use to generate the necessary torque. Once we had the motors narrowed down we used further analysis to find which motor and gear ratio would be the most efficient. Made sure that the motors we have chosen will fit into our design without any spacing issues.
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  ANALYSIS CONTINUED Double gearbox: At6 V: TS=.0031 Nm NO= 11500 rpm TR= .2*ɣ*TS = .2*.15*.0031 = 9.3*10-5Nm MR=.3178/ 9.3*10-5 = 3417 M = 344 TD≤ ɣ*M*TS .15*344*.0031 = .15996 and .3178 ≮ .15996 Therefore it will not provide enough torque to push blocks Analysis Showing the double gearbox and metal gearbox do not provide enough torque to push blocks: Metal gearbox: At 6 V: TS=1.13 Nm NO= 100 rpm TD≤ ɣ*M*TS .15*1*1.13 = .1695 and .3178 ≮ .1695 Therefore it will not provide enough torque to push blocks
• 18.
  ANALYSIS (continued) Weight Analysis Massof the shovel:V*D = (4.2)*(5.5)*(1/16)*(44.23) = 63.86 grams Mass of cubes: 3*17 grams = 51 grams axle is placed at (⅓)*(2.25) = .75 from the left side L1 = 2.25” L2 = 0.75” Taking moments about the axle is mshovel and cubes *((l1/2)- l2 ) = mW*l2 mW ≥ 31.93 grams l2 l1