Calculating Torque Requirements for a Mobile Robot

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Discussion Overview

The discussion centers around calculating the torque requirements for a mobile robot, considering factors such as mass, friction, acceleration, and drag forces. Participants explore various equations and models to determine the necessary torque and power for the robot's movement, including both simplified and more complex approaches.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the basic equations for torque and friction force, suggesting that torque can be calculated as Torque = Force * radius.
  • Another participant combines the equations to express torque in terms of the coefficient of rolling resistance and normal force, indicating that torque is needed for both constant speed and acceleration scenarios.
  • A participant questions the assumptions of using arbitrary values for velocity and acceleration in the power equation.
  • One reply clarifies that at constant velocity, acceleration is zero, and discusses how to determine minimum power requirements based on desired performance metrics.
  • A participant inquires about modifying power and torque formulas when a drag force is known, leading to a discussion on how to incorporate drag into the calculations.
  • Another participant emphasizes the need for more information and suggests using an inverse dynamic model for accurate torque determination at the robot's joints.
  • Some participants express skepticism about the simplified equations, arguing they may lead to inadequate results for real-world applications.
  • One participant acknowledges the complexity of the theory but defends the use of simplified methods for their specific mobile robot design.

Areas of Agreement / Disagreement

Participants express a mix of agreement and disagreement regarding the adequacy of simplified equations for calculating torque and power. While some believe these methods are sufficient for a basic mobile robot, others argue that they are overly simplistic and may not yield reliable results.

Contextual Notes

Participants note that the discussion involves various assumptions, such as the neglect of complex dynamics in multi-joint systems and the potential impact of drag forces, which may not be fully addressed in the simplified models presented.

tebrown3
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I am building a robot with a total mass of 835g, using two servo motors conected to the tires made of rubber which have a 3.33cm radius, and i wanted to know how to find out the total torque required to move the robot taking friction into account, the only things i know are:

Torque=Force*radius

Friction Force =Coefficient of Rolling ressistance*Normal force
 
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Mixing both of your equations:

Torque/radius = Coefficient of Rolling resistance * Normal force

Or:

Torque = radius * Coefficient of Rolling resistance * Normal force

And:

Normal force = Robot mass * g

This will be the torque required at constant speed (acceleration = 0). When you will accelerate, the torque needed will be:

Torque = radius * (Coefficient of Rolling resistance * Robot mass * g + Robot mass * acceleration)

Or:

Torque = radius * Robot mass * (Coefficient of Rolling resistance * g + acceleration)

What is more important is the power required by your motors, which is:

Power = Robot velocity * Robot mass * (Coefficient of Rolling resistance * g + acceleration)
 
ok thanks but I am still a little confused
when you said

Power = Robot velocity * Robot mass * (Coefficient of Rolling resistance * g + acceleration)

can we assume a random reasonable velocity and an reasonable acceleration?
 
Last edited:
Well, when you will be «cruising», your velocity is constant, meaning the acceleration is zero. If you put the maximum velocity you wish to achieve, the acceleration is also zero and the equation will give you the minimum power you need from your motor.

You can check http://hpwizard.com/car-performance.html" for more detailed info. It is for cars and it also includes aerodynamic forces, but the theory applies to your robot as well (find the theory explained at the bottom of the page). The graph labeled «MAXIMUM ACCELERATION & SPEED» is basically the equation I gave you for power where the acceleration available drops as velocity increases for a given power.

For example, using the simplified version of the calculator, with a 2 W motor (or motors), a mass of 835 g and a coefficient of friction of 0.8, your robot should be able to reach a top speed of 13-14 km/h and have enough acceleration to reach 10 km/h in about 2-3 s (depending how you connect your motor to the wheels).

The power you need depends only on the performance you want to achieve (top speed and time taken to reach your cruising speed).
 
Last edited by a moderator:
What if a drag force is known. How should power and torque formulas be modefied?
 
deff0 said:
What if a drag force is known. How should power and torque formulas be modefied?

You add the drag force (FD) this way:

Torque = radius * (Robot mass * (Coefficient of Rolling resistance * g + acceleration) + FD)

Power = Robot velocity * (Robot mass * (Coefficient of Rolling resistance * g + acceleration) + FD)

Where FD = 0.5 * air density * drag coefficient * Robot frontal area * (Robot velocity)²
 
You need much more information to be able to determine the required torque at the actuated joints of a robot. Typically, an inverse dynamic model is used to relate the dynamics of an input force
\begin{equation}
f_i(t)
\end{equation}
to the end effector acceleration
\begin{equation}
a_p(t)
\end{equation}
 
Thank you, Jack!
 
  • #10
Hey, Fex32!

Thank you for MIT's links. I'll look them through. But looks like that complicated theory is more about multi-joint assemblies, like robot arms and manipulators.
There are a couple of drives and wheels to move my bot and no plans to land it on Mars. I believe simplified method will do. It's also implemented in Drive Motor Sizing Tool http://www.robotshop.com/dc-motor-selection.html
 
  • #11
Ohhhh, It's a mobile robot. lol.
I thought it was a robot manipulator.
Then I agree with the above.

Cheers,
 

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