question on effort

les3002

i have to design a vehicle that will accelerate up a 1/8 gradient hill to 11.11m/s in 15s

i need the effort required and wheel torque.

mass = 170kg
rolling resistance = 50N/t (0.05N/kg)
air resistance = 13.6N
g = 9.81m/s^2

i have calculated:

a = 11.11*(1/15) = 0.74m/s^2
theta = 7.18 deg
sin theta = 0.125 deg

so i have the effort as:

E = ma+mgsin_theta+Froll+Fair
E = 125.8+(170*9.81*0.125)+0.05+13.6
E = 347.9125N



is E the torque i need at the wheels?
if not how would you go about working it out.

i tried with a motor of 40N, wheel radius of 0.2m and gear ratio of 10:1

i worked it out like this.
axle torque = 40*10 = 400N

so to find the torque at the wheel do i divide the axle torque by wheel radius

400/0.2 = 2000 that seems wrong to me??

andrevdh

You need to multiply the rolling resistance with the mass:

[tex]E = m(a + R + g/8) + 13.6 = 357\ N[/tex]

which makes only a slight difference.

To find the torque at the wheel, [tex]\Gamma _w[/tex], you need to multiply the effort by your wheel radius. The effort will be shared by the four wheels (if all four a driving) reducing the torque to a quarter at each wheel.

In order to choose an appropiate motor you need to consider the torque of the motor,[tex]\Gamma _m[/tex], and gear ratio [tex]G[/tex] such that

[tex]\Gamma _m \times G = \Gamma _w[/tex],

http://www.blueink.com/CLASS/physcom1/gear.htm

Torque, [tex]\Gamma[/tex], is the turning effect of a force [tex]F[/tex]. If the force is applied with a longer lever arm [tex]r[/tex] the turning effect will be greater:

[tex]\Gamma = F \times r[/tex]

the S.I. units of torque is newtons meter.