StephanieCalculations092709

Stephanie 9/27/09:

Given: Weight(W)=30lbs mass=30/32.2=0.931677lbs velocity=8in/sec= 2/3ft/sec Assumptions: wheel radius=4inches=1/3ft

rotational velocity: v=rw 2/3ft=(1/3ft)w w=(2rev/s)(60)=120rpm

__**1: Flat ground**__

F=CrrFn F=(0.15)(0.931677)(32.2)= Torque: T=m*r T=(30/32.2)(1/3ft)=0.31 ftlbs
 * For two motor drives, each motor torque=0.15528ftlbs

F=T/r: 0.31/(1/3ft)=0.93lbs a=F/m=(0.93lbs)/(0.931677)=1 ft/s^2

Power: P=[T(2pi)(w)]/33000= P(hp)= [(0.31)(2pi)(120rpm)]/33000=0.0071hp P(W)=P(hp)*746=(0.0071hp)(746)=5.28W


 * This is a total power of 2.64 Watts per motor
 * Check:

Energy: v=2/3ft/s, course=20ft (20ft)(2/3ft/s)=13.33seconds
 * give double time for obstacles: (13.33)(2)=26.67 seconds
 * Allow time for 3 trials: (26.67)(3)= 80 seconds
 * Allow buffer: T=100 seconds

E=PT=(27.13W)(100s)=2317J I=P/V=(27.13)/36=0.7536A Q=IT=(0.7536A)(100s)=75.36C/3600=.0209Ah Imax<3C: 0.07536<0.0627
 * use 36 volt batteries
 * check: E=VQ=(36V)(75.36)=2713J

__**2: Ramps**__

__For an inclined plane (30 degrees)__ Torque: T=m*g*r*sin(theta) T=(30/32.2)(32.2ft/s^2)(2/3ft)(0.5)=5 ftlbs
 * For two motor drives, each motor torque=2.5lbs

F=T/r: 5/(1/3ft)=15 lbs a=F/m=(15lbs)/(0.931677)=13.9752 ft/s^2

Power: P=[T(2pi)(w)]/33000= P(hp)= [(5)(2pi)(120rpm)]/33000=0.11424hp P(W)=P(hp)*746=(0.11424hp)(746)=85.22W


 * This is a total power of 42.61Watts per motor
 * Check:

Fdrag= rho(Cd)(FrontalArea)V^2+(RollingResistance)(Fnormal)

Summary:


 * **Obstacle** || **Power per Motor** || **Total Power** || **Torque Per Motor** || **Total Torque** ||
 * Flat Ground || 2.64Watts || 5.28Watts || 0.15528ftlbs || 0.31ftlbs ||
 * 30degree downhill || 42.61Watts || 85.22Watts || +2.5ftlbs || +5ftlbs ||
 * 30degree uphill || 42.61Watts || 85.22Watts || -2.5ftlbs || -5ftlbs ||