Mecanumbot Help : Need Motors with more Hauling Power

VEX V5 General Discussion Technical Discussion
#1

Hey all,

  I am a student, currently pursuing Mechanical Engg in India. I am building an Omnidirectional Vehicle based on Mecanum Wheels as my Final Year Project. 

  The main constraint I have, as of right now is to figure out the Motors that will Haul the whole vehicle whose weight will be in excess of 5 kg :( as I am mounting a Robotic Arm/ Forklift on the Vehicle to focus on presenting the vehicle as suitable for Industrial Application. The Motors that VEX provides, I estimate, will not be suitable for the Vehicle. Also I can order the VEX package only once, as it will be really expensive and time consuming to order a single item later on. :(

  So basically, the idea boils down to this. Can I possibly find a Substitute for a Vex Motor with more torque and still capable to work with the VEX Micro Controller ? :confused: Will an ordinary Servo Motor that I can buy from India work efficiently with the Vex Micro Controller ? (By Breaking the 'Lock' to convert the Servo into a Continuous Rotation Motor ?)

Please Help Me. As I have to order the package in a week’s time.

Also, as I am a total amateur in the field of Robotics and has been refrained to experiment with those awesomus :smiley: Vex Robotics Kits, I might need a suggestion about what exactly I should order for making the Vehicle.

Thank You. :slight_smile:

#2

5kg is really lightweight for a VEX robot. I don’t know anything about the surfaces you intend to run on or how fast you want to go, but you can easily carry that weight with four VEX motors driving four wheels. I know of a few robots that weigh that much with two motors driving, but the designers were careful with how the robot turned. Many competition VEX robots weigh 10kg.

Good luck!

#3

I think part of the issue is with getting full omni-directional drive from the mecanum wheels. Depending on the friction of the rollers, it may be difficult to get the little VEX motors to turn them at the desired speed.

I’d be careful about plugging a more powerful servo directly into the VEX robot controller as the reason the servo is more powerful is that it draws more current. You don’t want to be blasting too much current through the RC. Perhaps the power expander would be useful.

Alternatively, there are a number of DC speed controllers and small gear motors that could be adapted to work nicely with VEX… they aren’t part of the VEX kit, and aren’t legal for VEX competitions, but that isn’t what you are doing. Keep in mind that their output shafts will not be VEX compatible and will need to be adapted, as will their mount points, but you’ll have that problem with non-VEX servos, anyway. Or, you could just use two VEX motors to drive each wheel…

You might get some ideas from www.banebots.com

Good luck.

#4

Thank You for the help ‘DT Engineering’, ‘Chuck Glick’ and ‘Rick Tyler’.

Sorry was busy and will be busy :frowning: with my term examinations for a month or so, so expect a slow response from me. But in the month of Jan, my package will arrive and with all your help I’ll complete my Project.

Thank You all for responding, and asking you all a lot more questions. :smiley:

Bye.

#5

Change gear ratio for more torque, if you need more power…but remember it WILL BE SLOWER the higher the torque.

#6

Abed,

Only if you have magical gears. :rolleyes: Power is the rate of energy, that is, energy per unit time. If you want to satisfy Tim Allen’s fantasy, you need to add (or increase the power of) an active element, in this case probably a motor.

As you state, if you change gear ratio, you can get higher torque at lower speed or lower torque at higher speed. Not coincidentally, power is the product of torque and rotational speed.

Torque is force exerted tangentially at the end of a radius. Torque, therefore, has the dimension of forceXdistance (newton X meters [Nm] for example).As the element turns, that radius endpoint moves. The product of the force and the tangential distance is the work, or energy, provided. In one radian of rotation, the radius endpoint moves a circumferential distance equal to the radius, so the energy provided is tangential force X circumferential distance. As the latter is the same as the radius, the energy per radian of rotation is equal to the torque.

Power is energy per unit time. So the rotating element provides the amount of energy equal to the torque in the time it takes to turn through one radian. Rotational speed is expressed in units of angle / time, such as radians/second or revolutions/minute (RPM). (A revolution equal 2pi radians.) Therefore, the power of the rotating object is the torque multipied by the speed (in radians/unit time), such as Nm X (radians/sec).

A perfect gear train multiplies the torque by some value and divides the speed by the same value. (If you prefer, you can consider it to divide the torque and multiply the speed by a value that is the reciprocal of the value if you look at it the other way.) The consequence of this is that the power at the output end of the gear train is equal to the power at the input end, multiplied by the gear ratio, then divided by the gear ratio. That is, the output power is the same as the input power.

As a practical note, gears are not perfectly efficient, so every stage of gearing costs you a few percent of the power.

Eric

#7