@Cameron It’s really up to you - it’s part of the equation so it needs to be considered. Looking into the moment of inertia may be a bit much (but certainly not out of the question if you want to be precise). Just think of the distance you want it to cover in order to reach a certain speed (from a dead stop).
Yeah that’s what I have done. Looked into inertia calculations but can get a bit nightmare trying to find CG of individual components and such like. Glad you’re still around though, noticed this was originally posted in 2012!
Why do we multiply by100/efficiency. What efficiency are we talking about. The torque produced by motor . I am having some problems understanding this part, can u please explain?
@crawler The efficiency is the overall efficiency of the system, most notably the batteries -> motor controller -> DC motors -> Gearbox. If you have data as to the efficiency of each section, multiply them all to get a total efficiency. If not, you will need to estimate the value. For example spur gears are around 65% efficient.
Should add dynamics so that Torque T = T = fR + Inertiaa/r^2 instead of T = f*R, which is only for statics ?
@kelvin If you know the inertia of the system (not easy to do especially if you’re still in the process of choosing the parts), then you can certainly refine the equations.
is that the inertia of wheel since it is the only rotational part?
If in case a robot needs to ‘pull’ a weight of 2000Kgs. Is that similar to ‘carrying’ a weight of 2000Kgs when it comes to calculating motor’s power. If that is correct, can I simply substitute 2000Kgs in place for mass in your calculations ??
@Harshesh Gokani Yes and no; if the load is being dragged as opposed to rolling (i.e minimal friction), there will be significant resistance which needs to be factored into the equation. You also need to take into account the fact that the robot itself will need to weight enough to be able to get traction when pulling a load. The tool / tutorial are really for smaller to medium sized robots - nothing close to 2,000Kg.
Thank you sir
The pulling will be done by placing the load on carts having 4 wheels and assuming the weight of the driving robot is enough to get the required traction to be able to pull, what more (if any) changes would I have to make in the calculations given in the tutorial ?
If there needs to be many more concepts to be taken care of, please can you also tell me what I can refer to ?
If the cart has wheels, then you may be able to treat it as a payload, but 2,000Kg may require some additional considerations - it’s essentially the weight of a full sized pickup truck, and significantly more than a full-sized pickup truck can carry. If you’re seriously considering creating a robot to pull 2,000Kg, you should do all of the math and calculations.
Thank you sir
We are trying to build an AGV that can pull 2000Kgs. Can you please tell us what are the calculations that need to be considered ?
Excellent article! I am trying to build a tennis bowling machine with 2 motors having dia of 4 inches. I want to use 12V DC motors and preferably with < 2.0A peak current for convenience sake. I want to find what is the required torque for (each of) the motors - to get the maximum throw distance of 25m and speed of 120 km/hr. Tennis ball is around 60gram in weight I am thinking of having the throw done from the height of 1.1m height. If you can recommend a motor or specification of the motor, it will be helpful - Thanks again!
@Ram Your application is quite different than the intended use of the Drive Motor Sizing Tool. You would primarily calculate the speed at which the motor + wheel need to rotate in order to launch a ball at a specific tangential speed. The torque calculations is a bit harder since the wheel needs to compress the ball.
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.