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How much weight can following Brushed Motors carry together?

  • 4X 12 Volt 100rpm supplied voltage => (5/17)*12

  • 2X 12 Volt 100rpm supplied voltage => (7/17)*12

These motors are basically the wheels for my battle bot.

Here is the link to the Motor.

I am new to this so please tell me if I'm wrong somewhere.

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    $\begingroup$ Running torque is the parameter you should understanding and looking at! $\endgroup$ – MaNyYaCk Jan 9 '18 at 18:30
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Well, to answer this question for your special case, one needs more information about the motors. Maybe, you can supply a product code. And one needs more details of the involved mechanics... is it a car-like application? Is it going up-hill or down-hill? How long needs the motion to last? How well is the cooling? And so on...

I'll try to give you an idea on what all these data means which you can find out about a motor...

You can approximately calculate some kind of upper limit for the "power" a motor can deliver for an infinite duration, due to thermal design limits at lets say 'standard conditions' (normal room temperature, normal ventilation, ...):

A Motor has one or multiple phases (coils that will induce a magnetic field), which have:

  • I_max - a maximum design current (in the unit: Ampere or milli-Ampere),
  • R- an electrical resistance (given in the unit Ohms)
  • U_max - a maximum design voltage (given in Volts)

At least two of the above must be known. The relation between them is: R = U_max / I_max, so you can calculate the third one. You can measure the current for a given voltage or the restance of the coil, if you need to. But, make sure the motor doesn't move while measuring (e.g. by fixing the axle).

If you have I_max and U_max you can calculate P_max = I_max * U_max [Unit: Watts]. This will be the maximum power the Motor will -consume- without destruction under normal conditions! It can stand more power, if you improve cooling. And you can overload a motor for a limited time, until it gets too hot and finally breaks...

Each motor has an efficiency which depends on the motor itself and how fast it is spinning (see datasheet). This means the ratio of energy in form of motion or acceleration that is produced relative to its consumed energy. At optimum speed this can be as low as 50% or even higher than 90%, this depends on the motor.

Each good motor datasheet has some kind of curve that gives you an value for efficiency (%), torque (typically in: Ncm, Nm, mN*m) and motor current. At speed=0 the motor current is the highest (I_max) and the torque (=holding torque) is also at its highest value. The torque is the value for the rotational force. The holding torque is achieved when applying the nominal/design voltage of your motor at stand-still (=very short period until it moves faster) and/or blocked motion (=until not blocked anymore, good for measurements). If you use a higher voltage (which is valid but might damage the motor), the holding torque will also be higher. You can measure the holding torque, by mounting a little lever to the motor (e.g. r=10 cm=0.1m long) which the motor pushes onto e.g. a kittchen scale. The mass m (in Kilograms, e.g. m=34g=0.034kg) measured by the scale can then be used to calculate the force, which is F = m * g, where gis the earth gravity constant (approx. 9.81 m/s²). Then you can multiply the length of your lever by the measured force, you'll get the torque: M = F * r = m * g * r = 0.034kg * 9.81 m/s² * 0.1m = 0,033 Nm ... where "Nm" means "Newton-Meters" or if you use the lever length in cm it is: M = 0.034kg * 9.81 m/s² * 10cm = 3,33 Ncm (Newton-Centi-Meters).

Mechanical systems usually need to overcome the stiction first, which usually depends on the mass of it (which pushes it onto a surface), the materials of the material that has contact to the surface and the material of the surface itself.

What is Stiction? Here is a little experiment: You might have noticed that your smartphone will not slide downward an slightly inclined surface... the gravitational force is not strong enough, yet. But, if you push it a bit, it will overcome stiction ("sticking to the surface") and will slide on its own. As said, how strong the stiction is depends on the material of the thing... e.g. rubber on wood has an higher stiction than flat, polished plastic on polished glass. So also the roughness of the materials have an influence. The force to overcome Stictions can be calculated by the formula: Fs = mu * Fg, were mu is the constant for the materials and Fg is the force that pushes the thing to the surface. In a simple case (e.g. a smartphone laying on a flat and horizontal surface, no inclination) this can be Fg = m * g, where m is the mass (e.g. of the phone) to be moved and g = 9.81 m/s² (earth gravity constant) again.

The needed torque depends on the mechanical system. In a simple case you just mount some wheels on the motor. Then the radius (half of the diameter) of your wheels are r. And your needed torque to overcome stiction will then be Ms = Fs * r. If you use larger wheels you need a higher torque. If you use gears you can also influence the needed torque (in both directions), but you'll decrease the efficiency of the system and - at the same time - introduce a now stiction torque of the gears (which might or might be significant).

As soon we've overcome stiction, you will have still friction that might stop things from moving... that means deaccelerating it until it finally stops again. But, theses forces are usually much smaller (for slow speeds only!). And remember you can - for a short period of time - drive your motor with a higher current/voltage than specified!

All those friction, which depends on a lot of other factors, limits how fast it can move (top speed)... and finally how fast it can accelerate to that speed (maximum acceleration). This all falls under the topic "dynamics" and there is lots of it to learn... but, the basics can be learned quite fast.

So your implicit questions actually are:

  1. How much holding torque do my motors have? (needed: Motor Datasheet)
  2. How much power can my motor controller provide (for a short [or infinite] period of time)?
  3. How much torque do I need to overcome stiction? (needed: Info on mechanics, as kind of wheels, surface, style of movement)

I hope you got some information to start your further investigation on that topic. If you have more questions, just ask.

Useful Links:

Edit

You provided a link to some technical data, which are:

   Weight: 100 gm(Approx.)
   RPM: 100
   Operating Voltage: 12V DC
   Gearbox: Attached Plastic (spur)Gearbox
   Shaft diameter: 6mm
   Shaft Length: 21 mm
   Torque: 1.2 Kg-cm.
   No-load current = 60 mA(Max)
   Load current = 300 mA(Max).
   Dimensions: Body Diameter: 38mm; Motor Length with shaft: 77 mm;
   Same size motor available in various RPMs.

As we can see, the motor already has some gearbox attached to it, which reduces the maximum rotational to 100 rpm (=rotations per minute) and increases the available torque at the same time. The maximum torque is given as 1.2 kg*cm (which includes the gearbox-transition), which means having wheels with an radius of 1 cm (or 2cm in diameter) will give you a force that is equivalent to 1.2 kg in weight at nominal voltage, as long as your electronics can provide a current of 300 mA (load current).

However, 1.2 kg "weight" force doesn't mean, that you are limited to a weight of 1.2 kg for e.g. a robot, as 1.2 kg do not mean the weight of the "thing" which is moved but an equivalent force. It means a force that is equivalent to the force that gravity causes at 1.2 liters of water (in container which we asume to have no weight) to be pulled downwards. If you wanted to pull 1.2 kg in upwards direction, you would need a bit more than this force...

If you just need to drive a robot on a 100% horizontal surface you need to overcome Stiction (see formulae above), which for a robot with a weight of 1.2kg on wheels is much less than a force of 1.2kg "weight" (=1.2kg * 9.81m/s² = 11.8N). Anyhow, your robot will also need to climb slightly upwards.

Lets assume 20% inclination, which means you need a force equivalent to sin(20°) times the weight of the robot. And let us assume this force needed is much more than Stiction. In that case you can pull 1/sin(20°)=2.92 times 1.2 kg = ~3.5 kg upwards on a surface with an inclination below 20° with this motor using wheels with an radius of r=1cm. If your wheels are bigger, the maximum robot weight will be less (e.g. r=2cm ==> 3.5kg/2 = 1.75kg).

However, your speed will get near to zero if you inclination is near to 20%. So every weight that is less than this limit ensures you some reserve for accelerating your robot, even at 20% inclination. If our maximum robot weight is 3.5 kg according to the above calculation and the real robot has a weight of 2kg, then your acceleration force will be like 1.5kg "weight" pulling it forward on a horizontal surface. As stated above, 1.5 kg "weight" means in fact a force of 1.5 kg * 9.81 m/s² = 14.715 kg*m/s². Our example robot having a weight m=2kg will be accelerated by a force of F=14.716 kg*m/s². The formula that relates mass, force and acceleration is: F = m*a, where F is the force, m is the mass and a is the acceleration. So a = F/m would be 14.715 kg*m/s² divided by 2 kg and therefore: ~7.358 m/s². This means within 1 second your robot would reach a speed of 7.36 m/s (or 26.5 km/h) if there was no friction. Also the maximum speed is 100 rpm; so having the wheel radius of r=1cm the robot will move forward U=2*r*3.14=6.28 cm. Having 100 revolutions per minute gives us 628 cm = 6.28 m per minute or 0.1046 m/s as our top speed, if there is just the friction of the motor itself within its gearbox and so on... so it moves 10,46 cm forward per second. If you need more speed you need bigger wheels or some gears... but as said, your maximum weight will drop then.

IMPORTANT: Calculation of all the friction correct is even more complicated, so dont expect the full 10,46 cm per second! This is just an upper limit that is known for sure...

And of course having 4 motors will give you 4 times the torque, but will also increase the weight of the whole robot and the overall current needed to drive the motors.

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