Author Topic: Question on servo torque and lifting power  (Read 4897 times)

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Offline Robotboy86Topic starter

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Question on servo torque and lifting power
« on: July 26, 2007, 04:05:57 PM »
I am designing a 4 legged robot right now. Each leg has 3 segments, currently the total length 1 foot per leg total.  As such, I need to find something that can lift it from 90 degrees(Perpendicular with earth) to 0 and 180(parallel).  The issue is naturally weight though, I am not sure of the building material right now, but I am looking at Acrylic square tubing right now..  if that is the case, then it will be approximately 2/3 lbs a leg without the gears of servos.  Total leg weight will be nearly a pound each when it is done.  So in order for a servo to lift the leg, it must be able to lift at least 1ft/lbs. 


That would require an expensive servo.  So I was looking into just gearing them down..


then I came into another problem.  You can' necessarily gear down a servo, because of its range.  It can only move 180 degrees, If I geared down to a 2:1 ratio its movement would be 90degrees.  No good. 

So does anyone have any good suggestions for how to create 1ft/lbs of torque without breaking the bank on servos??

Thanks

Offline S. Karim

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Re: Question on servo torque and lifting power
« Reply #1 on: July 26, 2007, 04:41:32 PM »
1 lb/ft = 192 oz/in

if you want powerful and good servos, you have to break the bank, theres no other option besides getting cheap chinese servos that break within 5 minutes or arrive dead at your door.

im afraid you wont get 180 degrees of range at the size of a standard servo without offering at least 115 dollars at the table for each one of these little bombs.

you can get away with tons of torque for cheaper from Hitec, however, they'll be bigger than the standard size, they're called "large scale" servos.

Offline Robotboy86Topic starter

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Re: Question on servo torque and lifting power
« Reply #2 on: July 26, 2007, 05:18:33 PM »
Thanks :)

I was looking around and found this one:

http://servocity.com/html/hs-755hb_1_4_scale.html

Its size is much larger then typical, at 2.3" x 1.1"x 2.0" but I think  it would do the job very well :)

Now I just gotta finish modeling the robot.. and do the math on the torque for the 2 other joints and I can start building!

Offline hazzer123

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Re: Question on servo torque and lifting power
« Reply #3 on: July 27, 2007, 02:02:07 AM »
Each limb on the robot may be 1 foot long and weigh 1lb, but that doesn't mean that the torque created will be 1 foot/lb. Your biggest servo will no doubt be at the shoulder. This wont need to be lifted, so that means less torque.
There won't be a motor near to the tip of the limb, so this will shift the centre of mass down towards the shoulder.
Since torque is centre of mass * length, I think that the torque generated by the mass of the limb is less than 1/2 foot/lb.

But another thing you should take into account is the weight of the body of the robot. Which i dont have enough info to do any calculations.

Harry
Imperial College Robotics Society
www.icrobotics.co.uk

Offline Robotboy86Topic starter

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Re: Question on servo torque and lifting power
« Reply #4 on: July 27, 2007, 04:22:15 AM »
Right I came to the same conclusion as you as far as the mass of s that the legs.   I relize I could surviv on a much weaker servo..  but I figure with extra strength on the servo's I should be able to lift things more then the legs.. if you understand :P


Also the body weight is another reason for the servo's strength.  With something like 4 lbs of weight on the base, each leg would have to lift a pound each to "support it".  I'm going to post some rough images in a bit here..   feel free to look at em

Offline Admin

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Re: Question on servo torque and lifting power
« Reply #5 on: July 30, 2007, 07:12:31 AM »
Remember that your calculations are for just static lifting force (something that doesnt move).

To move the mass requires yet additional dynamic force:

F=m*a
force_dynamic = mass * rotational_acceleration

(rotational acceleration because the leg revolves about a point)

So your calculation should be something like:

torque = force_dynamic*(1/2 leg length) + force_static*(1/2 leg length)

If you arent careful, you could end up having a robot that can hold its weight, but not lift it.

You might want to consider reducing leg weight using some ultra light materials/design.

 


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