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[on:]

What's a good way to determine how much weight a motor with a given torque can carry?

Given a basic (7''sq) setup:

_____
|       |
m     m
|       |

(where 'm' is a motor, with casters at the corners to support weight)

With, say, 5 lbs of weight evenly distributed across the top of the bot, how much torque would I need to move it?

[Reply #1 on:]

Sorry there is not the equivalent of ohms law for this as it depends on a lot of factors. The main one being the coefficient of friction between the coaster and the surface it is sliding against coupled with the friction between the wheel and the surface.

[Reply #2 on:]

If we were to assume the world is perfect, and the only friction my bot experiences is from the wheel's contact with the ground, where would be a good place to start?

[Reply #3 on:]

Can you rephrase your question.  "Torque" is what comes out of a motor shaft, not what force is required to move a weight across a flat surface.

Do you want to know what FORCE is required to shove a weight.  If so are the casters simple chair casters without bearings or are they decent pices of engineering with ball or roller races fitted.

Either way, as Grumpy says, there are many factors to consider, not the least of which is the quality of the surface - is it smooth flooring, rough concrete, carpet etc etc

With too many unknowns your best bet is to buy or borrow a spring balance as used by anglers, attach it to your weight and see what force is required to pull it across the surface.  

jack

[Reply #4 on:]

Infinite weight? :p

Sorry (but since it was a prefect world and all). Maybe not the best starting point. I would think it depends on how fast you want it to get to a certain speed. Not really my forté(?), I'm not sure how to calculate this. I think it would be dependent on things like wheel radius too.

T = F * L (in my simple mind at least, also I found it is equal to angular acceleration * inertia).
also
F = m * a

where:
T = Torque
L = arm length where force is applied
F = Force
a = acceleration
m = mass

so

a = F/m, up to a certain speed for your motor. Except minus some friction resistance, which is the big question as to how to get. Once you get that somehow, its probably just a matter to juggle the equations a bit to get a ballpark torque for your motor?

Or something like this. I'm sure a google could provide better answer. Or maybe a Wolfram Alpha


.... and since this was interesting, I googled up Rolling resistance

According to this, it is:

F = C_rr * N_f

And

F = N_f*b / r

and also

b = C_rr * r

where
C_rr is the dimensionless rolling resistance coefficient or coefficient of rolling friction (CRF)
N_f is the force normal to the surface.
F is the rolling resistance force
r is the wheel radius
b is the rolling resistance coefficient or coefficient of rolling friction with dimension of length


(The above basically taken from wikipedia.)



Now I don't know anything about what kind of wheels you use, the radius or substance they roll against, so this is a guesstimate. According to the rolling friction table at the above link, I guess on a C_rr of about 0.005 for your wheel thing.

It would then requre a force of:

5 lbs = 2.268 kg

F = 0.005 * 2.268 kg * 9.8 m/s/s = 0.111 N

This is just to be moving (rolling) horizontally. If you got 4 wheels I suppose they would add up to the same. Plus a bit for friction on the bearings.


That is part 1... Then part 2; how fast do you want to get up to a certain speed? Which is back to the F = m * a formula

Say 5 seconds to get to a speed of 1 m/s...just to pick some numbers. This is probably too slow an acceleration though.

Which indicates an acceleration of:

v = a * t
so
a = v/t = 1/5 m/s/s

Where
v = speed
a = acceleration
t = time

requiring a force of

F = m * a = 2.268kg * 1/5 m/s/s = 0.454 N


Add this to the rolling resistance force required:

F_tot = 0.457N + 0.111N = 0.565 N


Now for a wheel radius of say 5 cm (just to pick another number..), you would need a torque of:

T = F_tot * L = 0.565N * 0.05m = 0.0283Nm

Seems like a pretty low number, but I also picked a pretty low acceleration. At least I think so.

[glow]Be warned:[/glow] I got virtually no experience with motors and the like.. I don't even know if this is a practical number for torque, or indeed if this is correct-ish. But if it is, maybe it could be a "starting point" to fill in your actual numbers.

[Reply #5 on:]

I'm sorry if I was unclear.  What I am looking for at this juncture are motors, but I don't know what kind of torque to look for (as torque seems to be a defining factor, and at the moment I am not overly concerned with power consumption).

The casters I mentioned could be something like this: http://www.sparkfun.com/commerce/product_info.php?products_id=320

Just to throw out some numbers, I suppose it could be assumed that I want to use 2'' wheels and that a movement rate of 60 RPM is ideal.

If there isn't a simple equation somewhere that will give me the nice number I want, I may have to go with your idea jack (thanks!).

EDIT:

Wow.

You must have finished crafting that just as I made my post, which seems to answer some of the assumptions you made. Thank you very much for those equations! This will be very helpful!

[Reply #6 on:]

Just to throw out some numbers, I suppose it could be assumed that I want to use 2'' wheels and that a movement rate of 60 RPM is ideal.

Typically, when you buy a gear motor, it is going to be rated in something like inch-pounds, or foot-pounds (or cm-kg, etc) of torque, generally stall-torque.

What this means is that if you attached a lever to the shaft, extending out so many inches or cm, that's how many pounds/ounces/kilograms, etc it will take to stall (stop) the motor.

So you have a 2 inch wheel - that would be a lever of 1 inch. If you gear motor output 5 inch/pounds, you could move 5 pounds with that motor and your wheel (or 10 pounds if you had a wheel an inch wide - or 2.5 pounds if the wheel was 4 inches wide, etc).

See the relationship?

With that said, then, your best and simplest manner of determining things would be to build your platform (sans motors - but with the wheels set up like you want them - use temporary bolts or something), then load it up with the amount of weight you want to move, and drag it around with the fish scale as suggested. Note the number of ounces/pounds needed, then convert based on the size of the wheels and the distance to the shaft. Once you have that number, tack on a bit more (if using ounce/inches - tack on a pound, if you are over 16 ounce/inches, then that would be pound/foot or pound/inches - tack on a few pounds).

It will still be a rough estimate, but such eyeballing will work fine for most purposes, and is simpler than solving a bunch of equations and such (which you will still need to round up and/or 'add a bit' because they are for "ideal world" situations). I am not saying that those equations aren't useful; certainly, if you really need some accuracy, or you want to double-check your eyeballin', run the numbers.

 

[Reply #7 on:]

Sounds like a good suggestion, thanks for the help!