Calculating Rolling Torque for Robot Motors

[Physic_fan]

Hi,
I have a question about one of my project involved 2electrical motors that I have to size to accelerate a Robot on robber wheels.
there are a lot of confusing stuff in the internet that make me give up.:(
so here it is:
the weight of Robot is 10 kg, standing on two legs that has 2 robber wheels 0.02m Radios
under each leg running on Asphalt from standing to reach 20km/h in 2 Sec and continue that speed,
how much torque and RPM I need on the shaft of each motor?
**let say coefficient of friction constant is 0.5 .

my calculation for this is below ,correct me if i am wrong.
Fr=u*mg so Fr=(0.5)*(10kg*9.8)=49 N
T=Fr*R -> T=49 * 0.02m= 0.98 Nm /2 =0.49 Nm on each motor
can anybody help me to finish this calculations?
thanks

 

[R Power]

The torque you calculated is the max torque you can apply to wheels via motor. Exceeding this torque would cause wheel slip.
But I think this isn't what you require! :P

20KM/h in 2 sec...that means an acceleration = 5.5 m/sec^2.
Now , angular acceleration = 5.5 / r = 275 rad/sec^2
Now if you know the inertia of the wheels you are using...you can calculate torque required...
T = I x ang. acc. (I = moment of inertia of wheels)
This T will be the starting torque you require to accelerate the robo. But since you want later a constant speed to be attained then you probably need variable torque motor.

Note: Use wheels which are like discs then you can easily calculate the moment of inertia.

I don't know if I am 100% correct above but this was fun and wanted to help you.

 

[jack action]

The only thing that will stay constant in your powertrain is power (less minor efficiency losses). So the power of your motor (P_m) will be the same as the power of your wheel (P_w) and the same as the power of your robot (P_r). So (all in SI units):

P_m = P_w = P_r

T_mw_m = T_ww_w = F_rv_r

Where: P is power, T is torque, w is rpm, F is the force at the tire contact patch and v is the robot speed.

What you need to know is F_r which cannot exceed the friction force (i.e. the one you calculated in your post). The one you actually need, depends on three things: Aerodynamic drag, rolling resistance and the inertia force due to your robot's acceleration.

Please, read the theory (at the bottom of the page) on http://hpwizard.com/car-performance.html" . It uses cars as examples, but everything applies to your robot.

 

[koolraj09]

Hi R POWER and jack action, but i have a few questions:
1) R-POWER, in the equation T=I*alpha what is I that your going to take. Because it's not the inertia of the wheels that matters here but inertia of the robot mass which is in translational motion.
2) Jack-action. In your case what the power of the robot is F*v. But what F should I take to size my motor? Is it only friction force? neglecting drag, what can inertia force be taken as in this case? If it is F=M*a, then what is "a" here?

 

[R Power]

R-POWER, in the equation T=I*alpha what is I that your going to take. Because it's not the inertia of the wheels that matters here but inertia of the robot mass which is in translational motion.

Initially, you need torque equal to F*r (where F is static friciton) in order to overcome static friction. Once the robo is in motion, the coefficient of dynamic friction will come into account.
So now total torque required to accelerate to desired velocity...
T = I*alpha + (Dynamic Fricition * r)
Here I will be of wheels only...
once you have reached 20 Km/h, the torque you require to maintain constant speed is just = Dynamic friction * r
Thats why you need a variable torque motor.

 

[jack action]

2) Jack-action. In your case what the power of the robot is F*v. But what F should I take to size my motor? Is it only friction force? neglecting drag, what can inertia force be taken as in this case? If it is F=M*a, then what is "a" here?

[PLAIN]http://hpwizard.com/images/accelerating-forces.GIF

These are all the forces acting on you robot.

Weight: $$mg$$
Force at the tire (from your motors): $$F_t = F_{tr} + F_{tf}$$
Rolling Resistance: $$F_r = F_{rr} + F_{rf} = f_r mg$$
Aerodynamic Drag force: $$F_D = 0.5 \rho C_{D} Av^2$$
Inertia force: $$ma$$

For now, ignore the aerodynamic lift ($$F_L = F_{Lr} + F_{Lf}$$).

They are related to each other this way:

$$F_t = F_{r} + F_{D} + ma$$
$$F_t = f_r mg + 0.5 \rho C_DAv^2 + ma$$

$$F_t$$ is the force you are looking for to determine the torque at the wheel.

At very low speed with no acceleration:

$$F_t = f_r mg + 0.5 \rho C_DA(0)^2 + m(0)$$
$$F_t = f_r mg$$

At constant speed:

$$F_t = f_r mg + 0.5 \rho C_DAv^2 + m(0)$$
$$F_t = f_r mg + 0.5 \rho C_DAv^2$$

At low speed with acceleration:

$$F_t = f_r mg + 0.5 \rho C_DA(0)^2 + ma$$
$$F_t = f_r mg + ma$$


In short: $$F_t = f_r mg + 0.5 \rho C_DAv^2 + ma$$

The acceleration and speed depends on your wishes. But, no matter what, there is a maximum on $$F_t$$. Either traction wise or power wise:

Traction wise:
$$F_{t max} = \mu mg$$

Power wise:
$$F_{t max} = \frac{P_{max}}{v}$$

The next image shows those limits (traction limit and max power limit). This gives you the maximum acceleration possible at a given speed. When acceleration reaches 0, then you have attained your maximum speed.

[PLAIN]http://hpwizard.com/images/available-power.GIF

So anything below and to the left of those lines are possible combinations of speed and acceleration of your robot. You can actually plot this graph for your robot on http://hpwizard.com/car-performance.html" . Read the theory at the bottom of the page for detailed info.

once you have reached 20 Km/h, the torque you require to maintain constant speed is just = Dynamic friction * r

This is not true, unless it happens that you need all your friction to fight the drag and rolling resistance ($$F_t = f_r mg + 0.5 \rho C_DAv^2$$).

 

[koolraj09]

hey jack action and r power that really helped a lot. Thanks very much.

 

[Physic_fan]

$$F_r = F_{rr} + F_{rf} = f_r mg$$

 

[Physic_fan]

$$f_r mg$$

thanks for all, but at the end can you explain the components of above formula, and where they come from?

 

[jack action]

$$f_r mg$$

thanks for all, but at the end can you explain the components of above formula, and where they come from?

It's the rolling resistance. $$f_r$$ is the http://hpwizard.com/tire-friction-coefficient.html" .