hi i was wondering where i can get a Arduino capable motor that can pull 40 pounds??
First find a motor that will pull 40 pounds (whatever that means) then look for a controller that will drive the motor current and interface with an Arduino. Which Arduino is important cause some are 5V and some 3.3V.
thank you but all i need to know it where i can get a motor that has enough torch to lift 40 pounds
dtotino:
thank you but all i need to know it where i can get a motor that has enough torch to lift 40 pounds
Problem solved!
These will easily lift 30 pounds not sure if they will lift 40.
https://www.google.ca/images?sourceid=navclient&ie=UTF-8&rlz=1T4GZAZ_enCA411CA411&q=window+motor
Do you want to pull the 40 pounds upwards or sideways ?
A motor that pulls could be a linear actuator, winch, servo lever arm - you must clarify the
geometry for your question to have any meaning.
How fast do you want the load to move - the force and the velocity are both needed to
estimate the size of motor (which depends on the power).
The worlds puniest motor can pull 40 pounds if you have a enough reduction gearing,
it just won't pull it fast.
sorry for the delay i am making a robot that need to be able to drive (slowly is fine) that will be able to pull 10 pounds
Motors have speed and torque ratings, do you know what speed and torque you need?
Gearboxes allow trading one for the other, at the expense of reduced efficiency (friction
losses).
If you don't know how to do mechanics calculations you could try describing the
physical arrangement/dimensions of any existing wheels, axles, gears, etc.
at the moment i do not have any existing anything
i do not now what torque i need
it would be great if i did know though
how do i find out??
dtotino:
how do i find out??
MarkT will hate me for this - he's a stickler for certain details, and I have a feeling that he does something with FIRST robotics, because he always mentions the way of calculation of torque yielding a particular result in units which I have only ever seen specified in the datasheets of motors meant for FIRST competitions. Actually, such units are also specified on manufacturer's motors - if you can get the real datasheet for them - but generally, such datasheets aren't as easily available as one would like, unless you purchase such motors brand new. Note that I am not saying that MarkT is wrong in his assertions...
...it's just that in the world of low-cost gearmotors (ie - surplus) - ratings for torque are given in something like inch-pounds, or cm-grams, or similar "non-SI" or whatever you want to call it "units". Call it laymen's units, call it bubba units (I can say that, being a redneck myself) - call it whatever you want; I just know that when you start looking for lower cost motors (such as surplus gear motors), that is how you will find the torque specified - if it is specified at all!
So - what are these "mystery units". Let's take a common bubba-unit: "inch-ounces". Let's say you find a gearmotor that is rated at 24 in-oz torque; what does that mean?
Well - what it means is that if you attached a 1 inch lever to the center of the output shaft of the gearmotor - so that the length of the lever from the center of the shaft to the end is one inch in length - it could lift 24 ounces of weight.
Or - if the length of the lever was 12 inches, it would only be able to lift 2 ounces of weight. If it were 6 inches - it could lift 4 ounces. See how that relationship works?
W = T / L
Where W is the amount of weight, T is the torque in bubba-units, and L is the length of the lever.
Now - what is a wheel? It's nothing more than a "lever" that forms a circle. So the radius of the wheel becomes "L"...
Of course, that would only apply to a winch lifting a load straight up - it doesn't apply to a wheel moving horizontally.
So - you can re-arrange that equation so that:
T = W x L
Take the weight of your robot, multiply by the radius of your wheels, and that's your torque value (in bubba units) - find a motor of that value.
So - if you had 6 inch wheels (3 inch radius), and your robot weighed 10 lbs (ie 160 ounces) - you would need a motor that had a torque rating in bubba-units of 30 in-lbs (or 48 in-ounces) - I hope you are following the math there...
But how do you find that weight? Well - the above is actually overkill - for a number of reasons. The first being that the value given is more the "lifting weight" - not the static "move horizontally" weight. For that, the easiest way to find it would be to again use a bubba method: Build a simple frame with the wheels you are using, add the weight needed for the robot (whatever you think the robot will ultimately weigh), then attach and pull the weight horizontally using a fish or luggage scale on the surfaces you think you will traverse - while noting the "pull weight" you read on the scale. Then use that value in your calculations.
If you think you will be going up inclines or such, do the same test on those inclines. Maybe you might want to add 5 or 10 percent more to your final calculation - but you probably won't need to; why?
Secondly, it is overkill because your robot isn't likely going to use a single motor and wheel to move itself; most homebrew robots typically will use two (or more) motors driving the wheels (or tracks) in a "differential steering" manner (like a tank or bulldozer), in which case both motors are typically moving the robot around. So - halve (or more) the number you get (or keep it as-is if it won't break your budget - it just means you can move more weight - but it also might mean lower efficiency in power usage, as larger motors can using greater current) - if you use 4 motors, use only a quarter of the figure - and so forth.
So - there's the bubba method of figuring out your torque needs. It's low tech, it will probably be slight overkill (heck, it may be completely wrong if you do the -real- math) - but it will get your platform needs in the ballpark, and allow you to pick out a motor from a surplus supplier without you wondering what "those other units" are that MarkT mentions...
Again, though - check out that book I mentioned in the other thread - it will essentially cover the calculation version for the above bubba method (including inclines). As I said, MarkT can guide you on a different method (again, though, you will need to likely purchase your motors new in order to get that information in some manner, as surplus motors tend to be rated in the "bubba-units" I mentioned - heck, I've even seen mixing of the units - like inch-grams and the like - real bubba there! You'll also find the opposite - ounce-inches!).
I don't claim my explanation is exact or correct - just that it will be "good enough" to get you close, if you want to take a more "hands-on" approach without needing more complex math and understanding...
I did physics, where SI units are the only units... So torque is N-m and the
equations have few if any arbitrary constants (so you can easily remember
them all!)
For mechanics the only constant you need to remember is pi and the gravitational acceleration at the earth's surface, namely 9.8 m/s/s (though that is planet-dependent
of course).
The benefit of using N-m (newton-metres) for torque is the simple relation
power = torque x angular velocity
Which is needed all the time when choosing motors and gearing.
thanks so much guys i will do the work and see what i get
MarkT:
I did physics, where SI units are the only units... So torque is N-m and the
equations have few if any arbitrary constants (so you can easily remember
them all!)
Ok - I'll remember that in the future.
As I stated though, I have only rarely seen N-m used for the torque measurement in listings for motors being sold; typically, I've only seen them at sites selling new motors for FIRST robotics (AndyMark, for instance), or on datasheets for motors (again, typically new motors).
I've never seen them on RC servos (not even on the spec sheets). Then again, I haven't seen every servo or spec sheet out there; perhaps on higher-end RC servos they give such information?
If you purchase surplus motors, though - invariably, if the torque is listed at all - it will be in those so-called "bubba-units". Perhaps SI units for the torque could be found in the spec sheet for the motor - if such a sheet is still existing (your chances of finding a spec sheet for a surplus motor - even direct from the manufacturer - tend to be fairly low, at least in my experience - it will depend on the manufacturer, and whether the motor is a "standard" motor and not something that was custom spec'd and designed for a third-party).
That said - perhaps the calcs could be done using SI units and formulae - then the result converted to those bubba-units at the end?
Perhaps that would be the best of both worlds: Simple calculations, "real world"-applicable results...
