About Torque in DC gear motor.

[sysysy]

Hi all,

i am doing a mobile robot project. I am using 2 dc gear motor to control my car movement. May i know what will be the maximum load of my robot to make sure my robot can run smoothly.

For example,
The at load information given by motor datasheet is,

current: 0.5A speed(RPM):112 then the Torque 2.16Kg.cm

for what i know is, if my radius of my wheel is 1cm, then the load muz be lower then 2.16kg. but now i using 2 dc motor, 1 as a right wheel and 1 as a left wheel.

so in this situation, may i know more accurate or precise way to calculate how weight my robot can carry. (* Basically my robot load just including wheel, robot chassis, breadboard circuit, variety of component)

Instead of implement experiment, i hope i can know some correct design method for my robot carry weight.

hope can get some knowledge from u guys.

Thanks.


Regards,

sysysy

 

[dknguyen]

Calculations for uneven ground or inclines or soft wheels is complicated.

But for flat, level/even, hard ground it's very easy. With hard wheels (that don't deform very much) on flat, level/even, hard ground (where all wheels are always touching the ground) , the rolling friction (force on the edge of the wheels, not the torque) is probably less than 10% (for flat ground with no incline, this is 10% of the total weight). On an incline, it's X% of the normal force (the normal force is the the component of the weight that is 90 degrees to the incline) plus the component of the weight that is parallel to the incline which wants to pull the robot down the incline. Since it's flat ground and all whels always touch the ground, two motors means you can move twice as much. For reference, outdoors on rought terrain with soft wheels might have a rolling coefficient of friction of 0.3 (30%).

Larger wheels means you a longer radius which is like a longer lever that the motor must turn. This means less force on the edge of the wheels so to move the same weight you need a motor with more torque (or a gear ratio that gives you more torque). But larger wheels also travel more in one rotation and can go over bumps more easily than small wheels. You can get the same speed by using a small wheel and high speed, low torque motor or a large wheel with a low speed, high torque motor.

Just don't use the stall torque rating from the datasheet. The continuous operating torque is what you need (or rule of thumb is 1/7th of stall torque for best efficiency). Using what I said above (flat hard ground, no incline, hard wheels) you could probably move a 40kg robot in your case.

 

[sysysy]

hi, thanks a lot for the sharing

With hard wheels on flat even hard ground, the force (not torque) to spin the wheel is probably 10% or less of the total weight.

i am not really understand the above statement. i am weak in language understanding...

Just don't use the stall torque rating from the datasheet. The continuous operating torque is what you need (or rule of thumb is 1/7th of stall torque for best efficiency).

for the above statement is it mean always use 1/7 of stall torque will give our robot better and smooth movement? if exceed that, then some movement difficulties will raise up or become not optimum?

thanks.

Regards,

sysysy

 

[dknguyen]

Force is force at edge of wheel (how hard you need to push it if you pushed it from behind). It is Newtons/kilograms. Torque is rotational force of wheel (Newtons-cm, kg-cm) and is how hard you need to turn wheel if you turned it from the center.

Math Example: If 10kg-cm of torque used to turn wheel, then 1cm radius wheel will have 10kg of force on wheel's edge, 5cm radius wheel will have 2kg of force on wheel's edge, 10cm wheel will have 1kg of force on wheels edge. But wheel travels more when spun once. Radius of wheel is like lever lets you use same amount of torque for either more force with less movement, or less movement with more force.

Real Example: Grab bag of potatoes and hold in front of you with arms close to your body. Now lift bag of potatoes with arms close to body. Very easy because lever length is short so shoulders don't need a lot of torque to lift potatoes but shoulders must swing a lot to lift potatoes up by a certain amount (like a small wheel). Now hold bag of potatoes out in front of you with your arms straight. Very difficult to hold up same bag of potatoes because now lever is longer so shoulders need to make much more torque to hold up same weight of potatoes. But swinging shoulders a little bit will move bag of potatoes much farther, but takes more torque in shoulders (like a large wheel). Bag of potatoes is like rolling friction of robot (proportional to weight), arm length is like radius of wheel, and shoulders like motor at center of wheel. Understand?

1/7 of stall torque is a "rule of thumb". General crude rule that seems mostly true for most motors to be *most efficient*. Seems to hover around 1/6 to 1/8. It will work just fine at 1/4 stall torque or maybe even more. It will just not be as efficient. But don't go too high or motor will start to have a hard time and be overloaded. I would not try and go above 1/2...that's a lot. 1/2 stall torque is maximum power anyways so you don't want to go above that.

 

[sysysy]

Ok. i am more understand for the concept now.

but 1 very simple question...juz for example,
if let say now the rate torque is 10kg-cm. then my wheel radius is 1cm. then what is my maximum weight that robot that can be carry in theoretically? is it 10kg?

in other words, to ensure my robot perform well, then i dun exceed 5kg load that u mention juz now the 1/2 ratio.

am i correct? i hope i can catch the concept more clearly

so if the robot use 2 dc motor in control each of the wheel...then the theory oso can apply in this?

thanks.

regards,

sysysy

 

[dknguyen]

Theroetically, maximum weight on flat ground is infinite because zero friction. Takes zero force to get any amount of weight moving like pushing something in space. In theory with zero rolling friction on incline you need zero force/torque to spin wheels, but you need enough force to counteract force of gravity pulling weight of robot down inline:

Force = Weight*cos(degrees)
Weight = mass x (acceleration of gravity)
Torque = Force x radius

Yes, do not exceed 1/2 stall torque. Yes simple theory applies to this. On incline or uneven ground it is more complicated. But flat ground 10% is usually conservative (need 10% of robot's weight at edge of the wheel).

 

[sysysy]

hi friend, i hope this will be my last question

have to let u know i am a Chinese, then i am not very good when understanding sentence,s maybe u already explain very clear, but just my problem.

But flat ground 10% is usually conservative (need 10% of robot's weight at edge of the wheel).

what does above statement mean by? how to make 10% of robot's weight at edge of the wheel? this is what i am

thanks.

regards,

sysysy

 

[dknguyen]

Conservative means safe design not trying to push limits.

Assume coefficient of rolling friction = 0.1.

Force at wheel edge = [Weight of robot] x [coefficient of rolling friction]
Force per wheel = [Force at wheel edge] / [number of wheels]
Torque per wheel = [Force per wheel] x [radius of wheel]

kg is mass, Newtons is force/weight. Weight = mass * acceleration of gravity = mass*9.81m/s^2
On Earth 1kg = 9.81N. Might be easier to visualize if you use kg instead of Newtons even though kg is technically wrong to use...just make sure to stay on the same planet because force/weight changes on other planets with different acceleration of gravity, even though mass remains the same.

 

[blueroomelectronics]

If it's one of those Tamiya gearbox and motor combos it's not going to support 5Kg. The sleeve bushings aren't designed for it. You need a ball-bearing gearbox.

 

[sysysy]

Thanks a lot for the patient explanation.

I'm really appreciate it...

 

[sysysy]

Hi, today i have another question.

May i know any hint to calculate the maximum weight of robot for uneven ground or inclines plane?

Thanks.

Regards,

sysysy

 

[vinith paul]

i want to make a mauilty robo.............

 

[RMMM]

i want to make a mauilty robo.............

Really? I want to make a keyboard that shocks people that use too many periods in a row!

 

[Gary B]

I think you might have overlooked one aspect. The torque will control maximum acceleration. The force is not used to support the weight, as was pointed out, but is used to overcome friction and momentum. As speed increases, the amount of force required to overcome friction and drag increases leaving less for acceleration.

 

[sysysy]

Hi Gary B,

Thanks for the reply..
may i know what is the significant influence in supporting the robot weight?

Can u give further explanation or example?

Thanks.

Regards,

sysysy

 

[dknguyen]

On an incline, the force is used to supprt some of the weight because some of the weight is trying to pull the robot down the slope.

This is the minimum force to keep the robot moving or hold it on a slope (you need more to actually accelerate)
https://www.electro-tech-online.com/threads/motor-sizing-for-moving-robots.23264/#post155131

 

[sysysy]

Hi Dknguyen,

After read your explanation of Motor Sizing For Moving Robots, i am 90% understand about the design already.

I am very excited and happy now. Thanks a lot and i do appreciate it.



Regards,

sysysy