Suppose a wheeled robot is moving on a horizontal platform with a constant speed. Due to constant speed, acceleration of the robot is zero. Hence forces due to traction on wheels should be zero and in that case torque applied by motor is also should be zero. And as per the equation P = M*w
But power is not zero as motor is on. I’m not able to relate it. Please help me to find out whats going wrong in my method.
@Saurabh Samant At constant speed in a vacuum (no air resistance) on a flat surface, no additional power is needed to move an object at constant velocity. However in order to move from a stationary position to a specific velocity, acceleration is needed. In a real environment as well, there is air resistance, and no surface is perfectly flat, so a bit of power will be needed.
Thank you for your effort. I just want to know what is the friction coefficient used ? in my application I want the rover to work in sand with pneumatic tires will this equation work for me ? also you didn’t mention the tire width and its impact ?
@Bla bla An R/C servo is a type of geared motor which allows you to send signals which control the angular position of the output shaft. This is different than a normal DC gear motor where you do not have control over the position.
@Ahmed You can see all equations and assumptions made here: Ahmed
When operating in sand, there will be slipping, which is much more difficult to factor in.
These equations are meant to provide some values to use as a starting point when selecting drive motors.
Hi !
Thanks for the simple yet methodical calculation for the the torque required at the Wheels .
I wanted to confirm the effect of Gear reduction ratio to multiply the motor output torque to the wheel torque (since I could not see any reference to it in these tutorials ) - lets say the torque required at the wheels is calculated as 150N-m, and if we are using a gear set to reduce the motor output speed to axle speed by ratio of 10 ( driven gear teeth/ drive gear teeth )- then the torque needed at the motor output shaft can be calculated as 150 / 10 = 15 N-m, using the principles of torque ratio for gears , right ? So that we can select a Motor of 15N-m rating … Am I on the right track - Can you please confirm ? Thanks again.
@Anand correct; the torque output via the tool and described in the tutorial factor in the gear down. Since most people choose an assembled motor + gear down, you normally don’t need to care about the DC motor’s specs. However if you plan to purchase a normal DC motor and add gearing separately, you’ll need to do some additional calculations. 150Nm output via a 10:1 geardown means the motor itself is providing 15Nm.
sir i m going to make a robot for robowar. its weight will be around 50kg. i m going to use a iron disc, crusher and a spinner as weapons. please suggest me motor for my project.How many motors and which type. sir this is my first project.please guide me…
@Gursiddak Singh This tutorial is for the drive motors only; when it comes to weapons, we do not have any specific suggestions or guides.
Hi Sir,
Is the t for the c=It equal 3600?what value do i have to put for the t?
@Hafizul lower case ‘t’ is time, which is what gives you the capacity of the battery pack needed to run the robot for a certain time.
sir, if i want to make a robot which climb on pipe having its own weight of 20kg,
is there any further calculation required for it to determine the specification of motor ??
@shubham This tutorial is largely for robots on flat surfaces, so you’ll need to refer to the tutorial to know if there are any additional calculations which you need to consider for your specific project.
Hi Coleman. Nice post indeed. I like it. I believe there are some adjustments that could be made to refine it. Can you edit https://www.robotshop.com/blog/en/files/sunforcebal.jpg to change de direction of Mgx? As it goes opposite to acceleration.
On the other hand, the assumption on T=fR is simplistic. In a better analysis a sum of Moments must be done, so we get: Sum M= T-fR= I alpha, where I is the moment of inertia of the wheel (second momento of mass).In the end, instead of https://www.robotshop.com/blog/en/files/torqa.jpg we have: T = ax * (I/R + MtR) + MtRgsin theta, where Mt is the total mass, the wheel mass and the vehicle mass. If the Wheel mass is negligile then, for that specific case T= MR (ax + gsin theta).
Have a nice day.
Hi Clemson.
I have a Problem statement, and i am not yet able to find out the solution. But after reading your articles i am sure that you can answer it properly.
If a Locomative train is travelling at 160 KMPH of Speed. I want to put some accelerometers on its wheels, they are of 3g and 200g specifications (ADXL 337, ADXL377). The wheel dimension is 38 cm. How can i measure what is the net force acting on each point of its radius? Which forces should i consider in this scenarios?
Thank you
indeed a great forum to discuss things. i find it a great help in choosing a suitable dc geared motor for my two wheeled self balancing robot.
Thanks
@BSN You’ll need the forward (linear) acceleration of the center of the wheel, as well as centrifugal and centripetal forces.
I appreciate the post, it saved my day. I substituted a=1m/s^2, M=15Kg, R=3Cm, N=2, Theta=10 deg. Unfortunately i am getting negative torque value. I think i made mistake in acceleration assumption. Please comment on my assumptions also let me know how to decide the acceleration.
@Kishorekumar Can you enter the values here: https://www.robotshop.com/blog/en/drive-motor-sizing-tool-9698