Calculating Wheel Torque to Accelerate Up a Hill

[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:

$$E = m(a + R + g/8) + 13.6 = 357\ N$$

which makes only a slight difference.

To find the torque at the wheel, $$\Gamma _w$$, 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,$$\Gamma _m$$, and gear ratio $$G$$ such that

$$\Gamma _m \times G = \Gamma _w$$,

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

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

$$\Gamma = F \times r$$

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

 

[m2003]

I would approach this problem by first breaking down the forces acting on the vehicle. The main force required to accelerate the vehicle up the hill is the force of gravity, which is dependent on the mass of the vehicle and the slope of the hill. In this case, the force of gravity is 170kg*9.81m/s^2*sin(7.18 degrees) = 125.8N.

Next, we need to consider the forces that oppose the motion of the vehicle, such as rolling resistance and air resistance. Rolling resistance is dependent on the weight of the vehicle, so we can calculate it by multiplying the weight (170kg*9.81m/s^2) by the rolling resistance coefficient (0.05N/kg). This gives us a rolling resistance force of 85.5N. Air resistance is a bit more complex, as it is dependent on the speed of the vehicle and the aerodynamic properties of the vehicle.

To calculate the effort required, we can use the equation E = ma + mg*sin(theta) + Froll + Fair, where m is the mass of the vehicle, a is the acceleration, g is the acceleration due to gravity, theta is the slope of the hill, Froll is the rolling resistance force, and Fair is the air resistance force. Plugging in the values given in the problem, we get E = 170kg*0.74m/s^2 + 125.8N + 85.5N + 13.6N = 347.9N. This is the total effort required to accelerate the vehicle up the hill.

To calculate the wheel torque, we need to take into account the gear ratio and the wheel radius. The gear ratio tells us how much the motor torque is multiplied before it reaches the wheels. In this case, with a gear ratio of 10:1, the torque at the wheels will be 10 times larger than the motor torque. We also need to consider the wheel radius, as torque is equal to force multiplied by radius. So, to calculate the torque at the wheels, we divide the total effort required by the gear ratio and the wheel radius. In this case, it would be 347.9N/(10*0.2m) = 1739.5N*m.

It is important to note that this calculation assumes that the motor is able to provide a constant torque throughout the acceleration