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Thread: Quadruped torque calculations - Confirmation Please!!!

  1. Quadruped torque calculations - Confirmation Please!!!

    Hi Everyone!

    A long time lurker here with a question.

    Firstly I'd like to say a big thanks to all those who contribute to this community and the great staff at Trossen. The information, opinions and assistance I've soaked up from my time lurking around this forum has been invaluable.Normally I like to nut out problems myself to keep my grey matter as active as possible, but I now have come to a point where I'd like the community to check a few things for me. Hopefully it may be of use to others too!

    OK, for a little/lot of background....I've dabbled with arduinos, an ARC-32 brained lynxmotion CH3 and a spattering of other little and large projects along the way.... custom building an over engineered RX-28 based quad is the next endeavor. To say that this is a long term project would be an understatement - My background has not lent itself to this hobby (think knuckle dragging Australian miner ) and as such I've had to learn what I don't know and then fill in the blanks as I go.

    Intermittently I've designed, repeatedly redesigned and started from scratch this project. I've now reached a point where I believe myself to have functional designs that I'm now ready to machine. Being as OCD as I am, I'd like to again dot my 'i's' and cross my 't's'. Which leads me back to the purpose of this post....

    Quadruped torque calculations - From the discussions and tutorials I've found online, I've done myself up an excel spreadsheet that, if I've gotten correct, may be of use to others deciding to embark on such undertakings. It's purpose is to report the required torques of a quad's leg joints for quads with up to 4dof per leg. (For 3dof legs or legs with no coxa length the Tarsus and Coxa length values can be set to Zero).

    The Word document should help clarify the formula's in the Excel cells calculating the reaction forces and the servo torque requirements for those interested in looking into this.

    Thoughts, suggestions and all comments are welcome!


    Attached Files Attached Files

  2. Re: Quadruped torque calculations - Confirmation Please!!!

    And..... bump.

    No takers, comments or suggestions?

    Further explanation of my intentions required to pip enough interest to garner a response?..... ok , here goes....

    I'm intending to build a quadruped platform capable of carrying a serious payload based around the RX-28.

    Each servo will be geared down 3:1 by means of a planetary gearbox. The gearbox will also be torsional load bearing and an integral part of each leg assembly. All parts in each gearbox, with the exception of pins, cap head screws and washers, I will be NC machining myself from either and or a mix of aluminium and POM and raw PTFE.

    I'm now basically happy with the balances of weights vs. strength of the gearboxes in the material combinations I could use but these choices will have major bearing on the design of the overall chassis, battery selection, etc.

    I've found and corrected too many errors in my formulas to really be confident I've them correct now, hence my asking the community for comment.
    Attached Thumbnails Attached Thumbnails Click image for larger version. 

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  3. #3

    Re: Quadruped torque calculations - Confirmation Please!!!

    That's a very ambitious project! Trying to dive into all the details would take a lot of time, so I think that's one reason you're not getting any immediate takers.

    If you measure the distance from each joint to the center of gravity, and multiply by payload, you'll get a rough estimation of torque needs (you want to divide by number of legs that contact the ground and multiply by a safety factor.) Do your calculations come up with numbers that are similar to the number you'd get through that quick-check approach?

    Also, in the exploded assembly picture, I don't see any provisions for an idler on the back of the servo; it kind of ends in the black bracket that reaches behind. Is that in another detail?

    Finally: milling PTFE. You're a braver (and more patient) man than me :-)
    Last edited by jwatte; 10-25-2012 at 01:18 PM.

  4. Re: Quadruped torque calculations - Confirmation Please!!!

    Hi Jwatte,

    Thanks for the reply/questions!

    I'll knock over the easiest to address first...

    No heart of hearts or patience here! PTFE is a menace to machine, no two ways about it - so I won't be machining it! I will however be making the punch/cutters to cut bushes from various thickness sheet, then adding a cutting board and a hammer.... I'm good with hammers... presto! almost instant low profile bearing bushes.

    There is an idler assembly there somewhere, not sure where it ended up with that exploded model though - It was probably exploded into the servo and then no longer visible. You'll be able to see two of the idlers in the leg assembly easily enough.

    The idler assembly could be done away with completely but that would severely increase the profile of the planetary gearbox to the point of creating conflict for leg positions... not to mention the extra weight and complexity. I am attempting to keep the part count to a functional (cough) minimum. The current design does require the support of an idler assembly for sure.

    Onto the harder Q and the crux of the matter..

    Quick approach results do appear more realistic (what I'd expect the joint torques to be) compared to the formulas that I've derived. Hence my initial post...... and now I remind myself that my assumptions that the community would see that as obvious, has me reminding myself that assumptions are bad, very bad. The results in the current version of the excel file appear to be out by an order of 10, which has me stumped.

    One of the earliest things that I picked up was that if the outermost section of leg is 90deg from horizontal, that joint's torque requirement should be zero when static, which holds true for all of the equations in the excel spreadsheet, (systematically reduce lengths to zero and set the next segment joint angles to 90 results in torque for that joint equaling zero) This holds true through out the spreadsheet but the results when leg segments are not vertical appear to be way too low for the weights being used.

    I've adapted the formulas from the tutorial on deriving a hexapod's torque requirement from the Robotshop website for use with quads. Actually I can do up a more visual version of the torque and resultant force equations which may help other visualise how I've come to those equations and where my possible problems lie. Initially that tutorial baffled me, so it is possible that my understanding derived from that tutorial could be flawed...

    I'll be back with more visually orientated/easier to understand/ more self explanatory formula workings.


  5. #5
    Join Date
    Feb 2012
    Sydney, Australia
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    Re: Quadruped torque calculations - Confirmation Please!!!

    I've done a similar thing for my hexapod, but before I say anything dumb, where do N1 and N2 act? I initially thought that N1 is on the left foot and N2 is on the right foot on your (very nice looking) graph, but then the fact that N1 and N2 are different makes me think that's not the case.

    Aside from that, all I can say at this point is that when I did my calcs I split the robot down the middle to take advantage of symmetry. If you're up to it you could do that and see if the answers tally up with the ones you've uploaded.

    Edit: I just ran your numbers through my method, and I'm getting:

    T1 = 11.94
    T2 = 52.41
    T3 = 98.95

    But these are still dependent on N1 being the reaction force on the left foot of your robot's leg.

    If N1 and N2 are the reaction forces on the left and right feet of your robot, then I see your first mistake, you're calculating forces like how you'd calculate torques. This section should just be N1 = N2 = mass/2.

    Incidentally, you're now the third Ferret I know online, and the other two have also been Aussies as well.
    Last edited by Th232; 10-26-2012 at 07:18 AM.

  6. Re: Quadruped torque calculations - Confirmation Please!!!

    Hi Th323,

    Nothing dumb said! "Always better to ask silly questions than make silly mistakes" (how I'd been brought up). All input is more than welcome!

    I also have used this in the past with a hex, plugged the numbers in and out pops the answers. Adapting and applying it all to a quad with another degree of freedom and coxas that have a measure has been the difficult bit for me.

    Calculations are 'ideal' with symmetry a focus of the entire design, as such the calcs take advantage of this. For those who may wish to use this for a bot that isn't square/round, the longest diagonal should be used in the calculations.

    Your on the money with N1 being the single reaction force at the single left foot.

    To explain why N1 doesn't equal N2, simply the calculations need to be done for three equal legs in contact with the ground. N1 basically equalling an entire half of the quads weight (only leg in contact on the left), where as each N2 will be a quarter each (two legs in contact on the right). Hence N1+2N2=Wbody+4(Legs).

    The third Ferret? Crap, then there's more poor feral bastards out there!!

    Visual representation of the equations is still on it's way.... Not all that used to a Mac yet...

    Care to share your method?



  7. #7
    Join Date
    Feb 2012
    Sydney, Australia
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    Re: Quadruped torque calculations - Confirmation Please!!!

    Hmm... several queries/observations then.

    1) How are you calculating W_body and W_(complete chassis)? Your equations are:

    Wbody = Wcomplete chassis +W1+W2+W3+W4
    N1+2 . N2=Wbody+4(W1+W2+W3+W4)

    This ends you up with:
    N1 + 2*N2 = W_(complete chassis) + 5*(W1 + W2 + W3 + W4)

    Where did the weight of an additional leg come in when you were calculating Wbody?

    2) You've taken N1 and N2 as different, yet you've put down T1 = T6, T2 = T5 and T3 = T4. T4, T5, T6 need to be different since the reaction force on that leg is different.

    3) In W1,2,3,4 you've said "weights of actuators at each joint". I don't see how you can say this, since the weight of an object acts through the centroid, or its centre of mass. Unless the centre of mass for your link is actually at the joint, the implications of this would manifest themselves in your calcs for T1,2,3, e.g. in T1 where you've calculated the torque exerted by W2 as W2*B instead of, say, W2*(0.5*B).

    As for my method as applied to your model:
    * Take robot as stationary, create a plane through one pair of opposite legs
    * Frame of reference is the robot's body
    * Calculate the reaction forces on the feet, placing the whole weight of the robot onto one pair of legs (absolute worst case scenario).
    * Method of sections at each joint to calculate the torque at each joint. So for your T2, I take the section containing L1 and L2:

    T2 =
    N1*(A + B) <<torque from reaction force>>
    + W1*(0.5*A + B) <<torque from weight of tarsus>>
    + W2*(0.5*B) <<torque from weight of tibia>>

    If you follow through with this you'll see that the answer is different to the one I posted above, I made an error when I calculated them before.

    This is actually a derivative of my full method, for the modelling of my robot I split the leg link masses into separate links and joints, with the links being symmetrical, or close enough to it.
    Last edited by Th232; 10-26-2012 at 09:33 AM.

  8. Re: Quadruped torque calculations - Confirmation Please!!!

    Hi again Th323,

    1) Good points regarding the body weight including a extra leg. Had carried that through from the robotshop tutorial. There is logic to including it's moment in the torque calcs.

    2) T1, 2 & 3 are the max required torques for their respective joints. T4 to 6 require only half the torque (2 vs. 1 leg). Have updated to reflect this. T4 to 6 would have been ambiguous with the heading I had above them, changed also. You'll note further fiddling also.

    3) Another valid point and yet another roll-over from the robotshop tut, I understand in most cases considering pivot points as segment cg's is too ideal. For my purposes for the moment it's not been worth clouding the waters with. Was something to be worked into the sheet at a later date as my designs have each segment's cg only 4mm from the pivot due to the nature of the of the current design. An added complexity to be worked in later for others.

    My designs currently are about as compact as I can make them. Their compactness compromises overall leg travel per step and I intend to balance this out with their carrying capacity with the aid of the spreadsheet - hopefully to end up graphing for a visual representation of the balance which is easy to rapidly update as the overall robot's design changes.....

    I like the way you think! re torque calcs for only two legs. For the moment I'd like to see if this excel sheet can in fact be worked into a useable tool for three legs. Two legs would require the absolute max torque at each joint and would be the make or break for max payload capacity.... Thus have begun playing with adding this to the spreadsheet.

    I'm still using pivots as Cgs and vis versa, you'll note in the attached version of the spread sheet I've fiddled with adding one and two leg weights (rough i know) and this change has been used in the two leg calc for N2 and hasn't been included into word doc either.... a work in progress that will have to wait until tomorrow.... bedtime for me.

    I think that I have found another issue with the logic inherited from the robotshop tut which i've also changed in the T2 and T3 calcs - W1(B) and W1(B+C) needed changing from plus to minus in the excel calcs... please correct me if i'm wrong.


    Attached Files Attached Files

  9. #9

    Re: Quadruped torque calculations - Confirmation Please!!!

    Another few points:
    - The worst case for each joint is likely that the weight it counteracts is horizontally located. Even with 45 degrees off, you only get about a 28% lessening of the acting torque.
    - The thing needs to be able to nimbly move even when loaded at the full torque, so multiply by a hefty movement margin.
    - The longest span that will generate torque is actually the distance between two opposed feet, not distance to center, but I think this only matters when distributing force among feet, because actual torque calculations use center-of-mass. But, consider a servo that's at the center -- it still sees the distance-to-feet times full-weight of torque.

  10. #10
    Join Date
    Sep 2010
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    Re: Quadruped torque calculations - Confirmation Please!!!

    I didn't realize it until I did a basic/rough test just now, but the MX-28 and AX-12A appear to be almost perfectly balanced - with the CG being very nearly dead center in all three axes, so one could probably model them as boxes with CG in the center and have a reasonably small error. I also did not realize that the main body of the AX-12 and MX-28 are nearly identical, with the biggest difference being the significantly deeper mounting ribs of the MX-28 (guess that just fooled my eyes a bit). Using this, the CG of the AX-12A will be ~13.5mm (along the long axis) from the center of the horn and the CG of the MX-28 will be ~13.3mm from the center of the horn. Just thought I'd share - it seemed interesting that I never thought to check that before.
    Please pardon the pedantry... and the profanity... and the convoluted speech pattern...
    "You have failed me, Brain!"
    gives free advice only on public threads

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