Inertia and getting the right stepper motor

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Hello everyone.

As this is my first post I would like to introduce myself. I am very new to robotics in the sense of the real world. I am very used to writing autoIT scripts for my work place. And I just absolutely love automating stuff. I have a masters in cell biology so that wont help me at all here heh.

Okay to the question:

I am looking to build a large semi-crane which takes water from a typ of waterfall and moves it over my crops. So basically what I'm thinking is a simple crane looking robot roughly 7 meters long driven by a stepper motor to turn it.

My problem is, since I havent studied much mechanical engineering since my first year at the uni I am unsure to which amount of torque the motor needs to turn a crane.

In short:

7 meters of crane weight ca 100kg. 

Ballbearings to reduce friction.

Stepper motor Nm=unsure.

Gear system if needed

So I started calculating the amount of inertia without factoring in friction. Using I = mr2

For simplicity I split the crane into 7 segments of 1 meter each. Then its Inertia = (100/7)*7^2+(100/7)*6^2+(100/7)*5^2+(100/7)*4^2+(100/7)*3^2+(100/7)*2^2+(100/7)*1^2. 

 

14,28571 7 700
14,28571 6 514,2857
14,28571 5 357,1429
14,28571 4 228,5714
14,28571 3 128,5714
14,28571 2 57,14286
14,28571 1 14,28571
Inertia summed up:   2000

Sorry if it's not pretty, I just did it now in excel. Basically it adds up to 2000 kg·m2

What I know is that the amount of inertia is reduced exponentially when gears ratio is applied. But since I have very limited experience with how strong a motor needs to be...I'm looking for help.

My thoughts: since 2000 kg*m2 is basically a 1 meter long arm loaded with 2000kg? to move such an arm I need 20000 Nm of torque (for simplicity assumed that 10Nm = 1 kg of force and not 9.80665).

The strongest stepper motors I can find are rated at 28 Nm (help me out here if there are stronger ones commercially available). So now im thinking, would it be possible to apply a fairly large gear ratio to reduce the torque needed. If I have 100:1 gear ratio, i.e 1 revolution of the motor reflects 1/100th revolution of the crane then the reflected load inertia = Load inertia / (gear ratio^2). So 20000/(100^2)=2 kg·m2

So with a 100:1 gear ratio I should easily be able to move the crane with a 28 Nm motor? As it is 10 times stronger than required (Not compensating for friction). And it should also be able to overcome the friction in the system as I don't think it's that large?

Please help me out if I did anything completely wrong or if the friction of the system is going to completely ruin my calculations :)

I would also like to excuse any incoherens in the text and/or if my english is terribad:)

In the future if this works well I might mount on another crane to build a type of SCARA-robot. To get more degree's of freedom.

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This is a cool project but you need to brush up on your physics

You don’t need a torque of 20000Nm to move an object with the moment of inertia of 20000Nm2. 

If you apply a torque of 20000Nm to an object with the moment of inertia of 20000Nm2 you will get a fairly decent acceleration of 1 radian/s2. That means in the first second of applying that torque, the tip of your 7 meter long arm will move about 3.5 meters, in two seconds it would have moved 10.5 meters - that fairly fast and probably much faster than you need it to be.

The problem is that you a approaching the problem from the wrong angle :).

The absolute minimal torque required to rotate the object is given by the torque generated by all the friction forces acting on the object. Since the friction forces depend on the actual construction of your arm, this will be pretty hard to calculate, and you will have to determine it experimentally or approximate it. With a little effort I think you could get a fairly low friction coefficient so maybe you can achieve less than 100Nm.

When you have determined or measured your friction torque you can start thinking about how fast you need your arm to move. Then you can think about how fast the arm has to accelerate (and then decelerate) to achieve that average speed, and you can determine your “acceleration torque”. Since you probably only need a few centimeters per second to water the plants you will need a lot less torque than you thought, on the order of 100Nm.

If you take a wiper motor of let’s say 24Nm and gear it down 10 times (with a bike chain and sprocket system) I think it will work in normal weather conditions. I also suggest that you build it as a servo system (with a dc motor and encoder) instead of open loop (servo). You may want to add a mechanical brake to the system or have some “parking” position where the arm can be locked so it doesn’t flop around in strong winds.


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This is what I am talking

This is what I am talking about! I knew that there was a major flaw somewhere in my calculations. Guess reading on wikipedia for 20 hours its still possible to miss the simple fact:

Moment of Inertia is just how much an object resists acceleration. And the motor just needs to overcome the friction of the system to get it moving.

 

Thanks a ton Antonio :slight_smile: