Author Topic: Hexapod design and torque  (Read 2760 times)

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Offline mark11originalTopic starter

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Hexapod design and torque
« on: January 11, 2011, 07:39:48 AM »
Hi All,

So I have my hexapod designed (I won't take all the credit as I have look at many others and used ideas and designs from those) I drew it up and assembled it using SolidWorks.
Now I really need to get odrering parts, I know all the electronics I need but i'm having trouble working out the torque of the servos required. The mass of the Hexapod at the moment (structure and servos) is 1.3kg the batteries and pcb's have not been added yet as I don't know the weight but I'm guessing when finished it should be 1.5kg approx.

I have attached some pictures so you can seen the design and some measurements of the legs. I know torque is T= Force x distance but how do I work it out for the hexapod legs?

Thanks in advance for all the help :)

Offline hardmouse

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Re: Hexapod design and torque
« Reply #1 on: January 11, 2011, 08:28:49 AM »
Hi,
Don't know much about the math :P. But I notice the body part you have top and bottom in 2 separate portion. I assume you gonna connect them later right? Cuz it looks not stable to have them separated. Sorry about my English.

Offline mark11originalTopic starter

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Re: Hexapod design and torque
« Reply #2 on: January 11, 2011, 10:56:50 AM »
Ya they will be connected, I want to start ordering the electronics for it and once I have them I will be able to position some struts to hold the top and bottom together more firmly. The electronic components will determine the positions of these struts.

Offline photomark

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Re: Hexapod design and torque
« Reply #3 on: January 11, 2011, 04:19:42 PM »
nice looking design Mark .

You are correct in the way you are working out the torque , M=Fd . M=moment (torque) to work out for each of your legs you need to know how much force there is on each leg at any one time , you would probably need to assume that only 3 of your legs would be in contact with the ground at any one time so what ever the total weight is of your bot divide by 3 to give you the force on each leg at that time .

The torque need for the leg is calculated by measuring the distance from the pivot point on the body out to where the leg touches the ground (not the length of the leg itself but how far out from the pivot it is strait out then down 70mm as in your drawing    ) that will give you the distance d .

Thew force on the leg is worked out by the weight on the leg x 9.81 for example say your bot is 3kg / 3 legs give 1kg per leg x 9.81 ( this is the gravitational constant) = 101.9 N (N=newtons)

70mm or 0.07m so with the formula M=Fd we get 101.9x0.07= 7.133Nm .

Truth is it is a lot more complicated that that to work out precisely but that is it in a nutshell and you would really need to study the operation of your bot closer   .

BTW as your leg straitens out the distance (d) increases and so does the moment around the pivot  
Have fun and I hope this helps a bit
« Last Edit: January 11, 2011, 04:29:15 PM by photomark »

Offline mark11originalTopic starter

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Re: Hexapod design and torque
« Reply #4 on: January 11, 2011, 05:34:07 PM »

The torque need for the leg is calculated by measuring the distance from the pivot point on the body out to where the leg touches the ground (not the length of the leg itself but how far out from the pivot it is strait out then down 70mm as in your drawing    ) that will give you the distance d .


Thanks Photomark thats a great help to me. Ya i know what you me about the distance d, I have attached another picture which may be correct or should the distance be from the bottom of the leg to the coxa servo?

Thanks again for your help, much appreciated.

Offline waltr

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Re: Hexapod design and torque
« Reply #5 on: January 11, 2011, 06:59:30 PM »
The Force (F) at the end of a leg is straight up and the distance (d) is perpendicular to this force.

 


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