Newbie needed torque/moment arm design assistance

  1. Folks, hopefully someone can assist me. I am a EE by education and am in the process of designing a mechanism that I need some assistance on.

    The basic issue I face is determining the torque needed for a miniature drive motor/gearbox for this mechanism. The problem, I think, might be fairly complex and to be honest I am not sure where to start. The small motors I need to use (22 to 26mm diameter) are a constraint of the design and mechanism size.

    The basic problem involves a motor driven "shuttle" that traverses a curved profile guide rail system over a length of about 12.5". Attached to the shuttle is an arm of about the same length with about a 1lb mass at the opposite end to where this drive motor will be mounted - the shuttle (motor/shuttle is mounted at the base of the arm).

    When the shuttle is at one end of the guide rail profile, the arm rests in a vertical position with respect to the shuttle (directly above it) while at the other end of the guide rail profile the arm rests in a near horizontal position and the shuttle is located 12.5" behind the "top" of the arm. The shuttle needs to traverse the curved rail profile over a period of 5 to 6 seconds and have sufficient torque to handle the transition of the arm from vertical to horizantal and vise versa.

    I have attached a simple drawing of the mechanism and was wondering if anyone could assist me in determining the motor and gearbox (planetary type) torque needed to authoritatively run this mechanism.

    Thanks in advance
     

    Attached Files:

    Last edited: Aug 22, 2007
  2. jcsd
  3. Danger

    Danger 9,878
    Gold Member

    Welcome to PF, Skyman. I don't actually know how to figure this out. Going by your diagram, though, the maximum torque requirement appears to be about 1/2 shuttle-length short of the vertical resting position. It's hard to be sure; it looks as if at that point the shuttle/arm assembly will be angled backwards trying to tip over.
     
  4. Danger - yes the arm does tip backwards (to the right in the picture) for a short period of the traverse and then back to vertical and the tips the other direction (to the left in the picture) all the way down to the more hoizontal position.

    In my mind however, the peak torque to move the shuttle/arm would be required when the arm is in its horizontal position and the shuttle/motor drive mechanism is to the far right in the picture.

    This is the arm position where the full weight of the 1lb mass is acting along the 12.5" arm and therefore on the drive motor gear in the shuttle. I believe the effective moment arm is 12.5 inch lbs at this point with respect to the drive gear noted in the picture (right side). When the arm tips backwards from it's vertical position as you note, it does so by about 2". If my memory serves me well from my collage days (28 years ago) the resultant vector of force acting at the shuttle is considerably less at this point than the 12.5inch lbs when the shuttle is all the way over to the right - if my memory serves me correctly!!

    Of course, my memory may not be serving me well and I could have it all wrong!!

    If not, it seems to me that it is a question of deriving the rotational torque of the motor drive gear to overcome the load presented at the end of the arm to traverse the shuttle from the far right in the picture to the far left. As the shuttle traverses from the right to the left, my mind also tells me that the required torque to continue the traverse reduces to a minimum up to the point where the arm first becomes vertical (due to the arms rotation), then increases slightly as the arm passes the vertical and starts to tip backwards and then reduces once more and the arm finally rests vertical at the left most position.

    Here lies the main question - how do I calculate this. Right now, the motor gear is a 0.7" pitch diameter sprocket (15 tooth) that runs on a chain link rail that parallels the profile curve noted in the picture. The motor and gearbox delivers 2 inch lbs at the output shaft and uses a 1:1 gear ratio to the main drive shaft where the 15 tooth sprockets are.

    I am wondering if the torque required to move this arm is simply a case of calculating the equivalent torque at a radius of 0.35" (half the PD of 0.7 of the sprocket) by dividing the 12.5 inch lbs down - but I'm not really sure.

    Tony
     
  5. Danger

    Danger 9,878
    Gold Member

    You might well be correct about the maximum torque; I'm really no good at figuring things out from real engineering principles because I've never studied anything like that. My reasoning regarding my idea about the maximum point is that gravity, inertia, and friction are pretty much the only forces that you have to overcome. Since the inertia and friction shouldn't change, the vertical position is the one in which gravity would have the most effect. While there's a horizontal component of movement, won't the rails, rather than just the drive train, be supporting part of the weight?
    Regardless, I really don't know what formulae would be involved.
     
  6. FredGarvin

    FredGarvin 5,087
    Science Advisor

    It's tough to see how the arm would actually pivot around a motor to allow that kind of motion. It does indeed appear to be a dirty little dynamics problem (albeit 2D).
     
  7. Danger

    Danger 9,878
    Gold Member

    Hmmm... I'm not having a problem with that. Maybe it's because I'm used to sketching up stuff like that, but I can pretty clearly see the setup as a 3D construction. (At least, I think that I do.)
    If I'm reading it correctly, the arm itself doesn't pivot at all. It's solidly attached to the shuttle. So are the motor and drive gear system. It appears to work similarly to a 'donkey engine', but with a chain to follow rather than a rope.
    Unfortunately, that doesn't do a damned bit of good toward me figuring out the problem. :grumpy:
     
  8. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    One simple place to start (to get some insight into what's going on, not an "accurate" answer) would be to assume the guide rail pins are frictionless, and calculate the static torque needed to hold the device steady at different points on the track.

    Obviously that's leaving out a lot of details, but at would give you an order of magnitude idea how powerful the motor needs to be.

    If all else fails, step 2 would be build a prototype with a motor 5 times as powerful as that estimate, and start testing :smile:
     
  9. Fred, Danger is correct. The arm is fixed to the shuttle as is the driver motor and the two sprockets that drive the shuttle over the curved profile. The shuttle is about 70mm wide and there is a sprocket on either side of it in the location shown in the drawing.

    At the two locations showing the guide pins are two pins that go all the way through the shuttle and stick out about 15mm either side where shown. The profile itself in the picture is a representation only. On the final assembly, the profile represents two guide "rails" as such, either side of the shuttle - essentially "sandwiching" the shuttle with applicable clearances for the shuttle to move between them.

    The guide pins at the lower left front and lower right rear slide in slots in the guide rails and are directed/positioned by these slots in the rails as the shuttle traverses the distance from end to end. The Sprockets engage a length of chain embedded into the guide rails just above the slots for the pins and provide the translational motion of the shuttle from end to end with the guide pins controlling the trajaectoy.

    I have added a second picture with some additional positions of the arm/shuttle assembly. You will notice that the two guide pin locations follow the profile curve and that the black circle above each position represented shows the drive sprockets that are attached to and on either side of the shuttle.

    I just also saw a reply in my mail as I was typing this from AlephZero regarding static torque. I can calculate the theoretical static torque at many positions along the profile with simple T=FxL and T=Fx(L x cosa). What I really need to understand is the next step. The moment arm and static torque is for the arm. The motor however, translates torque to the rail through a 0.7" PD sprocket. Do I assume that if the transmission of a force through the spockets is made at a radius of 0.35" that the torque required REDUCES a proprotional amount - is this correct theory? If it is I know how to proceed.
     

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  10. AlephZero - ooops!! I just reread your e-mail. "Hold it steady" is the key I think and I believe you mean put the shuttle/arm on the rails and attached a spring scale to see what force is needed to hold the unit in certain positions along the curve.

    This, to me, implies a prototype of the shuttle with arm and rails. Since the materials for each are exotic (carbons predominantly), I will need to construct a prototype from alternate materials and estimate but this is a good idea - thanks. Just wish I knew how to crunch the numbers to find the starting range I should consider.

    Thanks again.
     
  11. FredGarvin

    FredGarvin 5,087
    Science Advisor

    That's what wasn't clicking for me. Now it makes sense. I'm slow but I'm....slow. As the sprocket grows in size the force transmitted through it decreases but since the output sprocket is directly on the motor shaft, it will have the same torque as the motor (if I am understanding your question correctly. It's been a long day).

    Ideally you would need to calculate the moment of inertia of your rotating hardware and also the angular acceleration you would wish to impart on the arm. This is a pretty tough dynamics problem. You have differing angular accelerations of the shuttle arm and mass plus a varying acceleration in the translation of the whole thing. As a first run I would pick the worst case section on the track and hold the entire assembly rigid at that point in time. At that point calculate the required force to produce a certain linear acceleration and back that to the motor via the sprocket size.

    I would really need to look back at my old dynamics notes on relative reference frames, etc...I am seriously rusty in that department.

    It's a cool problem to work on though.
     
  12. FYI, I have added a couple 3D renderings of a very early version of the concept here for those that may need a visualization. The two pictures cover the "up" position and the "down" position and shows the rails either side of the shuttle/arm assembly. Note that the sprocket positions have since changed relative location to more reflect the positions in the 2D drawing.

    The 2D drawing shows the motor position with respect to the drive sprocket and the rotational motion of the motor is translated to a bevel gear on the drive shaft for the two sprockets. The new profile curve (orange) is the profile the sprocket follows and is offet from the guide pin slot profile.

    Any other help anyone can give - even detail mathematical concepts and ideas would be gladly welcomed. I would really like a base torqu to work with as there are many motors and gearboxes available to choose from and they can be fairly expensive.

    Again, thanks in advance
     

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  13. Apologies for over doing this but one more picture that shows the trajectory curve of point A noted at the top of the picture as the shuttle moves from the left most position to the right most position. This is the motion of this point over the whole shuttle movement from arm vertical to arm horizontal.

    Again, thanks
     

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  14. Q_Goest

    Q_Goest 2,976
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    I’d agree with Fred, that doing a really accurate job on this would be very difficult. Doing something simple such as Aleph and you have suggested is probably sufficient anyway.

    Just ‘thinking out loud’ here… I was going to suggest creating your CAD model and show about 10 points between the two extremes of motion – kinda like what you’ve shown in your “Torque Diagram-2.jpg”. You want the motion of the shuttle or arm (not sure what you’re calling it) to traverse the entire length in 5 or 6 seconds. If you assume the motion along the slider is constant velocity, then put those 10 points half a second apart for example.

    Take a look at the 10 points now and consider what power is needed to transverse the various points.
    - The weight of the shuttle/arm is one issue. It rises some distance per unit time which directly equates to power. (ie: Power = weight * height / delta-time)
    - Also, the entire arm has to go through some rotationial acceleration. If you can estimate the moment of inertia of the unit around that point, you can then calculate power needed to for this rotational acceleration.

    Not sure how much each might contribute to the overall power needed, but I’d bet the lifting of the weight is the primary contributor. Do that first and maybe add 50% to 100% and you’ll probably be in the ball park. . . . <still thinking>
     
  15. AlephZero

    AlephZero 7,298
    Science Advisor
    Homework Helper

    The power to lift 1lb through 1ft in 5 seconds isn't much. I make it about 4N x 0.06m/sec = 0.25W. A 2 in-lb torque motor could do it as a straight lift with a 4in diameter drive, so that doesn't seem like the problem.

    Looking at the picture in post #8, one thing that could be worrying is the reactions at the pins when the arm is horizontal. In position 6 the 1lb weight is going to give forces of about +5 and -4 pounds on the pins because of the lever effect. Compare with position 2 where you will get about +0.5 pounds on each pin. So the friction force at position 6 is going to be about 9 times what it was at position 2 - i.e. (5+4)/(0.5+0.5).

    One way to reduce those forces is a have a longer "wheelbase" on the track, but that might lead to other complications.

    Actually I was thinking of calculating the forces to hold the structure steady, but there's nothing wrong with measuring them instead!
     
  16. AlephZero - thanks for the additional help - I am reading intently!!!!!

    Your thoughts are somewhat simialr to what I thought regarding position 6 being the problem area. I have spent quite some time trying to figure out a method of reducing the friction forces between the pins and the slots they ride in. The two concepts were lining the slots with a dry lubricant laminate strip which I think would be a construction issue and secondly some form of bearing on the pin. My current concept is a 2 stage bearing on the pin, the first being a needle roller bearing on the pin that would be retained on one end by a secondary small thrust bearing. The slot would then be machined about 0.005" larger than the outer diameter of the needle roller bearing and the needles would roll on the pin and the outer casing would in turn "roll" on one or the other of the slot upper and lower surfaces. The thrust bearing would then be placed between the roller bearing and the shuttle framework and be used as a bearing surface for one end of the roller bearing.

    The problem area is the other end of the roller bearing/pin inside the slot which I cannot attach any bearing surface to - it would essentially "rub" against the inner face of the slot potentially causing friction and partical generation leading to increased wear. I guess a dry lub laminate of some slippery material like PTFE could be used here but I'm not sure. This was the best I could come up with to date.

    If you have any alternate methods I could consider to reduce the frictional interface. I added a drawing.

    Also, if there is any way you could explain your formulas I would really appreciate it. Power for a EE is volts x amps which I think does not work here!!

    I really appreciate you taking the time to describe this for me.

    Thanks
     

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  17. Danger

    Danger 9,878
    Gold Member

    Now that I've seen the 3-D version, the purpose is obvious. Sheer genius! I never would have thought to make an electric ice-cream scoop.

    Anyhow, as for your 'problem area'... could you use tapered roller bearings to take care of thrust and rolling friction at the same time? That would minimize the end-play.
     
  18. Hmm, tapered roller bearings - didn't think of these.

    The purpose of the thrust bearing on the pin is two fold

    1). To provide a true thrust bearing function to control rotation in the XY plane and
    2). as a rotating surface for the outer casing of the roller bearing to bear on as it rotates, if it rotates, during the shuttle transit from one end to the other.

    From the research I did on the web, I believe that tapered roller bearings would only provide the function of the thrust and/or roller function but the outer casing of the roller bearing would not have a rotating surface to bear on as the shuttle traverses from one end to the other.

    Now, if the tapered roller bearing had a tapered outer casing, that would be a different story. In this case, the inner race would lie on the shaft and if the slots in the rails could also be machined with the same taper as the outer casing it would work as I require it to.

    In my limited experience with bearings, a tapered bearing could possibly work in combination with a standard thrust bearing as a rotating surface for the roller bearing outer casing and a tapered sleeve over the outer casing to engage in a tapered slot. A picture paints a thousand words!!

    Now, I see another problem with this design. If there is play in the XY plane, I can see a condition where the top and bottom suface of the tapered sleeve coming into contact with the top and bottom surface of the guide slots therefore stopping rotation of the roller bearing. This would be bad from a frictional perspective but I only thought of this after I sketched it - erhaps there is further development on this idea!!
     

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  19. Anyway, going back to the question of friction forces noted by AlephZero, can someone please help me understand the magnitudes of these forces in relation to the torque selection for the motor?

    I have multiple motor/gearboxes I can select from as well as output torque capabilities however, the more torque I have inside the mechanism, the higher the cost of the motor/gearbox combination. 2 inch lbs is about the minimum I was considering and I can go up to about 4 in lbs directly from the gearbox with minimal cost increase. Combine this with the drive sprocket only being 0.36" radius provides me with about 11 lbs of force on the drive chain.

    Thanks again for your help.
     
  20. Danger

    Danger 9,878
    Gold Member

    You're right. I didn't think it out completely. The only solution that I see right now (after 10 beers) is to build a 'double trolley' for your shuttle. You could have a subframe with idler wheels that go between your tapered bearings and the top of the track.
     
  21. AlephZero

    AlephZero 7,298
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    Homework Helper

    In position 6, imagine "flattening out" your device so you have a see-saw pivoting about the left hand pin. You have a weight of 1 lb a long distance away balanced by the force on the right hand pin a short distance away. I estimated the distance ratio as about 4:1 from your picture, so the force applied to the right hand pin by the track will be 4lb downwards to make the see-saw balance.

    So, you have a total force downwards of 4 + 1 = 5 lb. To stop the whole structure falling downwards, there must be a force of 5 lb (upwards) applied to the left hand pin by the track.

    On the other hand at position 1, the 1 lb mass is in between the two pins, so the forces applied to the pins by the track are both upwards and add up to 1 lb. If the mass was central, there would be 0.5 lb on each pin.

    Assuming a simple model of friction (Coulomb's friction law) the (horizontal) friction forces will be proportional to the (vertical) reactions, and all in the same direction (opposing the motion) So the friction at position 1 is proportional to (0.5+0.5) and at position 6 is proportional to (4 + 5).

    Re the power, for a ME work = force * distance, so power = work/time = force * (distance/time) = force * velocity. You need to use consistent units - i.e. a mass of 1 Kg under earth gravity gives a force of 9.8 Newtons. (By strange conincidence, a force of one Newton is approximately equal to the weight of one apple)

    Best to use SI units, since in SI there is a clear distinction between mass units (Kg) and force units (N). "Pounds" is used ambiguously for either "pounds mass" with "pounds force".

    Hope this helps.
     
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