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* = *mr*^{2}

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·m^{2}

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.