how do u figure out the torque req'd to move a robot?

[Salik Syed]

Hi guys i've got a maybe simple question...
how does one go about calculating the torque req'd to move
a 130 lb robot with 6" wheels.... i know the whole force=mass* velocity
or whatever the heck they taught freshmen science doesn't seem right
... i mean what about friction etc.. i would think all of this goes into
the calculations....
btw i'm talking about torque @ the shaft not @ the wheel.although
i guess this can be calculated from the whole 3" lever arm distance or
whatever...

i don't know if its so simple actually.... now that i think of it.....

[patrickrd]

This is a great question but the answer is complicated. There are a couple things I think you might be wondering about.

1) Stiction. This is a made-up term for how much force is required to get an object to transition from static (not moving) to dynamic (moving). All wheels tend to dig into the ground and at a microscopic level, interweave with the abnormal properties of the ground material. This creates a bond between the ground and wheel. A certain amount of force is needed to break this bond, or overcome the "stiction." Stiction comes from many places in a mechanical system other than the wheels, too: ball bearings, grease in gearboxes, and many others. The more steps you have in your transmission, the more stiction you probably will have. Stiction -- to my knowledge -- is not a real scientific term, but it is commonly used. Generally, if you want to know how much force is required to overcome stiction, you've got to build a prototype and measure it. It is good practice to assume very large stiction when designing a robot or any mechanical device, until you can measure what it actually is.

2) Dynamic friction. When a robot starts to move (after overcoming friction), the robot will accelerate depending upon how much torque it transmits to the ground. The more torque, the faster it accelerates. Remember, torque equals force times distance, so force equals torque divided by distance. Since F = m*a, a = F / m. Substituting in what force equals, a = T / (d * m). With a wheel, you use the radius as d. Also, when there is more than one force, the SUM of the forces equals mass * acceleration. Thus, if there are losses in the system due to friction, then you have to subtract all the friction forces to get your net force. Again, you normally have to measure friction.

3) Gear ratios. Remember your gear ratios. The torque on the wheel is equal to the torque of the motor times the gear ratio. The speed of the wheel is equal to the torque divided by the gear ratio. The force that causes acceleration of the robot is the torque of the wheel, divided by the radius of the wheel.

Hope this helps.

Patrick

[dlavery]

Originally Posted by patrickrd

1) Stiction. This is a made-up term for how much force is required to get an object to transition from static (not moving) to dynamic (moving). All wheels tend to dig into the ground and at a microscopic level, interweave with the abnormal properties of the ground material. This creates a bond between the ground and wheel. A certain amount of force is needed to break this bond, or overcome the "stiction." Stiction comes from many places in a mechanical system other than the wheels, too: ball bearings, grease in gearboxes, and many others. The more steps you have in your transmission, the more stiction you probably will have. Stiction -- to my knowledge -- is not a real scientific term, but it is commonly used.

"Stiction" is actually a legitimate term. It is a contraction of "stick-slip friction" and is characterized as the force required to overcome static adhesion/bonding forces that may occur when two objects are at rest with respect to one another. It is also sometimes called "static friction" although this is a slightly less accurate definition of the term.

Salik - to get to the heart of your question, there is not a simple, practical way for the vast majority of FIRST teams to accurately calculate the answer you seek. While classic physics equations will give a first order approximation (see "f = m * a"), drag on the system must also be factored in. Mechanical drag can come from many sources such stiction, dynamic friction, gear train efficiency losses (all outlined by Patrick), but also from binding due to bearing misalignments, bearing housings clamped a smidge too tight, rough finishes on bearing surfaces, burrs on gear teeth, metal chips falling into gears, misalignment of gear teeth, binding in chain linkages, wheels being slightly under inflated, wheel camber slightly off spec, adhesion characteristics of the wheel material, and about a zillion other causes.

What all this means is that rather than spend way too much time trying to characterize and something that cannot be completely characterized, just cheat and go straight to the answer. Do this by either measuring or estimating. If you are doing the analysis on a pre-existing robot or mechanism, the easiest solution is to just measure the mechanical drag. Connect a spring scale to the front of the robot, pull it over a known distance in a measured time increment, and calculate the force needed to do the task (disconnect and remove the motors so you don't have any back EMF concerns). The difference between the measured force and the theoretical force will be the measured drag.

If you are trying to do this for a future build, and are not characterizing an existing machine, then the problem is slightly harder. The best suggestion is to still measure one of your existing robots, as that will give you a ballpark estimate of the drag on a robot built with your production capabilities. Then add 10-15% to give a conservative estimate.

Note that a drag number that seems very high is not all that unusual. Don't be surprised if your drag coefficient is in the range of 40% for a typical FIRST robot. Once you have determined the drag on the robot, the fun part is trying to figure out where it is coming from, and how to reduce it. Here is one hint to get started: a friend who is a professional bicycle builder once told me that over 50% of the mechanical drag on a typical high-end racing bicycle comes from the interaction between the chain and the sprockets, and the chain itself. Get rid of the chain, and your mechanical drag coefficient goes WAY down.

-dave

p.s. I really, desperately want Dr. Joe or Andy Baker to come along and tell me that I have forgotten about some really simple, accurate way to predict mechanical drag - I'll even buy a case of Dew if they do! But I don't think it exists.

[Al Skierkiewicz]

Salik,
Dave presented a great explanation but I would like to weigh in on the motors. The bearing and brush friction can be considerable when the motor is new or if the output shaft should get bent during handling or running. If you don't measure force with the motors in place but disconnected from anything, add in a little fudge factor just to be on the safe side.
In my business, a real world example of stiction (we have always thought of it as "STicky frICTION") occurs between the rotating video head and the polished surface of the tape. Although every attempt is made to force a cushion of air between the two surfaces to overcome the stiction, other factors and Murphy will intervene. A fluid, for instance, in the form of high humidity or a liquid will actually turn into an adhesive under such tight tolerances. The result is tape damage as the tape sticks to the rotating head and is sucked out of the cassette at light speed. That is why all consumer video tape recording devices have humidity sensors on the surface of the video head. When the humidity gets too high, the transport is disabled.

[Salik Syed]

thanks.... i figured it would probably be quite complex...
I wanted to know because i designed a transmission and "it looks about right" however i did not know if the torque it supplies would be sufficient to move a robot... i guess the torque req'd to actually move it changes based on the speed the bot is currently moving at.... etc etc... but the torque figures i calculated for the output of the tranny seem irrelevent with out a "required torque " or something like that.... i'm guessing it will work since our older trannys worked w/ appx the same amount of torque etc... (those were designed guess & Check )

[Tom Bottiglieri]

My $.02...

In this game, wouldn't you want a design that produces far more torque than that required just to move the robot?

[Max Lobovsky]

Originally Posted by Tom Bottiglieri

My $.02...

In this game, wouldn't you want a design that produces far more torque than that required just to move the robot?

Perhaps he is trying to determine the torque neccesary for his highest gear which he intends to use only when driving free of obstruction?

I think the best suggestion anyone can give you on calculating torque is too look at the numbers on other robots. Making any accurate calculation would probably be way to difficult.

[Andy A.]

Originally Posted by Tom Bottiglieri

My $.02...

In this game, wouldn't you want a design that produces far more torque than that required just to move the robot?

Not Always.

The torque/speed trade off we all face when building a drive train pushes us in different directions. While the majority of winning bots have had a great deal of torque (many have multiple gears, but not all), some favored top speed. I suppose that, given the speeds at which we are limited just by drivabilty, even the fastest bots still had enough torque to accelrate up to speed rather quickly.

I surprised that the issue of turning hasn't come up yet. Assuming for a moment that you are building a 4 wheel bot, getting it to move forward requires a certain amount of torque, but turning it will likely require more. This topic has been covered very throughly, but it's still common for teams to find that they over geared their bot and it can't overcome the stiction of it's wheels being pulled across the carpet during a turn. This usually results in tripping breakers and short battery life along with no/poor turning.

I highly recommend you take a look at Chris Hibner's White Paper on drive train basics. I don't know that it will shed much light on your original question, but will probably give you an idea of the forces involved in turning a 4 wheel bot. This might be a good starting point.

-Andy A.

[CrazyCarl461]

In some cases, where pushing power is vastly more important to you than speed, you can gear for as much torque as it would take to cause the wheels to barely slip in the most extreme condition of pushing (as if the robot were driving up against a wall). You can even gear for more than that if your robot does something clever like lifts up another robot or a goal (when allowed) for more down force. If you are going to gear your robot using this approach, be prepared to use lots of powerful motors if you want any appreciable speed.

I agree with Andy, Chris Hibner and others have written some extremely useful whitepapers about this.

[sanddrag]

Torque is directly proportional to current. For a certain motor, their will be a certain stall torque for which their is a stall current. The stall current for the FIRST drive motors is usually well above what the breakers can supply before tripping. So, you want to design your drive system around not tripping your breakers. When your motors are drawing about 40 amps, whatever torque they are making (determined by the curve graph), should translate through your drivetrain into (at least) just starting to make your wheels slip. Now, how much force does it take to make the wheels slip? Your guess is as good as mine. The main case in point is that you want the wheels to slip before the breakers trip.

[Al Skierkiewicz]

Originally Posted by sanddrag

Torque is directly proportional to current.

Sanddrag et al,
This would be the case if there weren't so many other variables in the system. Please remember, that unless the speed controllers are at full throttle (I don't recommend this except for short bursts) any current measuring you would use is likely averaging the current supplied to the motor. Remember that the controller outputs a PWM signal except when at full throttle. Each device will use a different method of trying to average the current so the results are really inconclusive under normal operating conditions. Averaging varying pulse width, fast rise time signals, with a complex component as a load is a daunting task for designers. Add to that the different brush/commutator designs, production variations, temperature induced resistance changes and transmissions losses and one can see that relying on current to torque conversions will become misleading. Dave's method is a simple, ball park approach to aid in design.

[Doug G]

Originally Posted by dlavery

"..... the easiest solution is to just measure the mechanical drag. Connect a spring scale to the front of the robot, pull it over a known distance in a measured time increment, and calculate the force needed to do the task (disconnect and remove the motors so you don't have any back EMF concerns). The difference between the measured force and the theoretical force will be the measured drag.

If you are trying to do this for a future build, and are not characterizing an existing machine, then the problem is slightly harder. The best suggestion is to still measure one of your existing robots, as that will give you a ballpark estimate of the drag on a robot built with your production capabilities. Then add 10-15% to give a conservative estimate.

Salik,

I suggest you do just as Dave suggests. We have an accelerometer and a force probe that can easily measure the Force required to travel at various speeds and accelerations (hmmm.. 2nd Law) using the two robots from past years. Then compare the results to the theoretical torque output by those transmissions. This should get you close to a mechanical drag coefficient.

Regarding the efficiency of the chain drive. Can anyone comment on the differences they've experienced between a direct gear driven setup vs. a Chain driven setup? I remember the Bionic bulldogs (60) using a direct gear drive in 2001(?) and 2002 and then went to chain in 2004. I suspect the efficiency of a chain drive depends on the size of chain and how well it's tensioned.