Tie rack calculations

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Summary of Geartrain Calculation Results

Calculations proved that the gear train in the rotating tie rack is used to decrease speed and increase torque provided by the motor. The mechanical advantages of these ratios are that the ties rotate at a speed slow enough to be easily visible to the user, yet with enough torque that rotation occurs regardless of the weight of the ties or interferences with the belt from closet items.


In conclusion, it was determined that the output gained from the input is optimized to meet design goals for the rotating tie rack, and any changes to the mechanisms of the system would be altering the size and number of gears. However, this is also optimized in that there are no mechanical, cost-savings or material advantages to changing the components of the gear train.


The values below summarize the results calculated using the following analysis:


Motor Speed: ω1 = 51314.84 RPM

Motor Torque: τ1 = 0.0284 N-m

Stall Torque: τs = 0.0142


Output Speed: ω10 = 70.26 RPM

Output Torque: τ10 = 138.28 N-m

Diagrams for Geartrain Calculations

Gear Train

The figure below represents the gear train located in the rotating tie rack. The numbers represent each step of the gear reduction from the motor input (1) to the belt gear output (10) which drives the rotating belt for hanging ties. These numbers are used as subscripts in the speed and torque calculations to differentiate each step, and the arrows indicate the direction of rotation for each shaft. Thus, the rotational velocities calculated are in these corresponding directions.


Image:24-441 07 Team06 GearT.jpg


Free Body Diagram

The following free body diagram is also useful, while it is not necessary for the speed and torque values desired for the design. The green arrows again indicate the direction of rotation for each shaft (dotted line), and input (τ1) and output (τ10) torque are the important values which were analyzed. The blue arrows indicate reaction forces, including forces on the gear shafts, from meshing gear teeth and between the gear belt and pulleys; the double arrows indicate normal forces and the single arrows indicate tangential forces between the gears and tension forces between the pulleys. These forces were not calculated as there are not relevant to the analysis and design considerations being pursued.

Image:24-441 07 Team06 FBD3.jpg

Geartrain Calculations

Variables:

ω = rotational velocity

τ = torque

τs = stall torque

α = constant (stall torque/speed)

r = radius

N = number of teeth


Known:

Ngears = 52

Npinions = 10


Speed of Output

The gear train creates a decrease of speed and increase of torque from the actual motor output. The final speed is slow enough to measure with a stopwatch. This was done by marking a point on the rotating belt, and measuring the time that it took for one rotation (the point returns to where it started). The marking is shown in the figure below:


Image:24-441 07 Team06 Speed.jpg


Results:


Time for one rotation of belt gear = 5s : 38ms

Circumference of belt gear = 0.126 m


Speed = length/time (m/s)

Angular speed = ω = 2π/T (radians/s)


ω10 = length / [r10 * time] = (0.126 m) / [(0.02m)(5.38 s)] = 1.17 rotations/sec

ω10 = 70.26 RPM = output speed


Gear Ratios/Calculations (Relating input to output)

ω1 r1 = ω2 r2 and ω2 = ω3

ω2 = ω4 = [r1 / r2] ω1 (1)


ω3 r3 = ω7 r7 and ω7 = ω4

ω7 = ω3 = [r3 / r7] ω3 (2)


substitute (1) into (2)

ω7 = ω4 = [r3 r1 / r7 r2] ω1 (3)


ω4 r4 = ω8 r8 and ω8 = ω5

ω8 = ω5 = [r4 / r8] ω34 (4)


substitute (3) into (4)

ω3 = ω5 = [r4 r3 r1 / r8 r7 r2] ω1 (5)


ω5 r5 = ω9 r9 and ω9 = ω6

ω9 = ω6 = [r5 / r9] ω5 (6)


substitute (5) into (6)

ω9 = ω6 = [r5 r4 r3 r1 / r9 r8 r7 r2] ω1 (7)


substitute ω10 = ω6 into (7)


ω10 = [r5 r4 r3 r1 / r9 r8 r7 r2] ω1 (8)


ω10/ ω1 = [r5 r4 r3 r1 / r9 r8 r7 r2] (9)


ω1/ ω10 = [r9 r8 r7 r2 / r5 r4 r3 r1] (10)


ω1 = [(0.019m)( 0.019m)(0.019m)(0.0115m) / (0.003m)(0.003m) (0.003m)(0.004m)] * 70.26 RPM

ω1 = 51314.84 RPM = motor speed


Stall Torque

The stall torque of the motor was calculated by finding an equilibrium point between an opposing torque and motor output torque. This was done by turning the motor sideways, powering it, and hanging weights on the pulley attached to the motor shaft until weight suspension was reached (τ1 = τ2). The string is of negligible mass and was tightly affixed to the pulley to prevent slipping by both knotting it with a rubber band. The setup used is shown in the figure below:


Image:24-441 07 Team06 Torque.jpg


Results:

Motor stalled when 0.8 lb weight was creating a torque. Thus:


0.8 lbs = 0.363 kg

Radius of motor pulley = 0.004 m


τs = F*r = (0.363 kg)*(9.81 m/s2)*(0.004 m)

τs = 0.0142 Nm = 14.24 mN-m


Engine Torque

A relationship between torque and speed can provide an equation to calculate the torque output from the motor when it runs in the device. Assuming ideal motor function, thus a linear distribution:


τ = α ω + τs

α = τs / speed = 0.0142 mN-m / 51314.84 RPM = 2.77 x 10-7


τ1 = α ω1 + τs = (2.77 x 10-7) (51314.84 RPM) + 0.0142 N-m


τ1 = 0.0284 N-m = motor torque


Applying torque ratios:


τ6 = (ω1/ ω6) τ1


τ10 = (r10 / r6 )*(ω1/ ω6)* τ1


τ10 = (0.02m/0.003m)*( 51314.84 RPM / 70.26 RPM) * 0.0284 N-m


τ10 = 138.28 N-m = output torque

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