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DC Electric Motor Tourque/Power Usage

  1. Jul 14, 2011 #1
    DC Electric Motor Torque/Power Usage

    Hello Everybody,

    I am mid way through doing an electronic engineering degree, I fancied as a summer project to build 6 wheel search and rescue robot. Idea being to use wheel chair motors to get the thing going. The problem is I am not an mechanical engineer, so am rusty with power and torque, coef of static friction etc. From what I remember from a-level physics I made the quick calculations shown in the image below:


    Originally I tried do to this in metric...however after looking at a couple of sites it seems to only work in imperial and I was getting over 2000HP!

    ....now I think a 50HP car can manage a motorway, so I was expecting a low figure. This seems quite high? Can someone give me their take on this cheers!


    [PLAIN]http://img841.imageshack.us/img841/321/robotpng.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 14, 2011 #2
    All the power is going to do is determine the acceleration. You need to look at the zero speed torque of the motor to determine if it will overcome the static friction, and then provided there is enough power, the motor and/or any gearing you use will determine the top speed.
     
  4. Jul 14, 2011 #3
    Did you mean static coefficient or kinetic coefficient? The static coefficient only matters when the robot is stopped. Once the robot is moving the rolling resistance (kinetic coefficient) is what impedes motion.
     
  5. Jul 14, 2011 #4
    This is where I show how little I know about such things, I thought the static coefficient was the force the robot had to overcome to change from being stationary to in motion. I presumed there would various other variables that would effect speed such as air-resistance etc. Honestly I know nothing about the kinetic co-efficient. All I am aiming for at the moment is to make a robot that can go 20kmph and pull 40kg. Once I have the thing built and working I will worry about packing it full of extra's.

    I'm a fan of paintballing, I've played with programming wireless cameras before, my plan was to use a several PIC's. One to control the robots main locomotion, and one to possibly control an arm with a paintball gun attached and a camera ;) and a few different ideas in mind
     
  6. Jul 14, 2011 #5
    The static coefficient applies UNTIL the object starts moving. Once the object is moving the amount of force required to maintain a constant speed is referred to as kinetic friction, or in the case of a wheeled vehicle, the rolling resistance.

    Measuring the rolling resistance is simply using a force gauge to measure the force needed to maintain a known speed. In most cases of slow speed the rolling resistance remains constant, so simply find the force required to maintain ANY arbitrary constant speed. You could also calculate it based on the weight of the vehicle and the kinetic friction coefficient of the tires (if you have that information). By the way, air speed only applies at higher speeds as well.

    Here's what you can do:
    For the simple case you're working with you can easily find the force needed to maintain a constant speed, let's call it "f"

    (Work) w=f*d (units in Joules (J)), d=20km=20000m for this example
    (Power) P=w/t, (units in Watts (W)), t=1h=3600s for this example

    so if f=1N, the power required by your robot to maintain 20kmph ON LEVEL GROUND is:
    P(-)=(1N)*(20000m)/(3600s)=5.6W

    That's on flat ground, and not including acceleration up to full speed. The negative sign is there because this power is a loss in the system (once you find the actual rolling resistance you can find the actual power loss).

    Calculating the power required to overcome hills is more complicated, but not by much:

    (Weight)=40kg*g=390N, the component the motors are required to overcome becomes:
    F=390sin(angle of hill)
    P=(F+f)*d/t - the power required to climb a certain distance vertically in a given amount of time

    Say we want this to climb up a vertical wall at 20kmph (not necessarily practical but let's see what it looks like):
    F=390N
    f=1N
    P(total)=391N*5.56J=2.2kW

    The conversion from mechanical power to electrical power should be nearly 1:1, but I might be wrong.

    Hope this helps!
     
  7. Jul 15, 2011 #6
    Thats fantastic! really helped a lot, thank you very much! This is the first time I have used this forum, very impressed. I was wondering how to HorsePower compares to Power in watts, and you have answered that before I even asked. Cheers mate
     
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