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Torque distribution to thrust?

  1. Apr 25, 2012 #1
    I'm working with a 3000 RPM @ 0.85 Nm motor in the transaxle from a powered scooter (like for disabled/elderly people). The differential has a 67:4 (16.75) reduction, which puts the output shafts around 179 RPM @ 14.23 Nm.

    I'm passing this to a 25:9 chain drive, which makes the final output 64.4 RPM @ 39.5 Nm, or 29.13 pound-force-feet.

    Each tire has an 11 inch radius, for a top speed of 6 feet per second, and a max thrust of 31.77 pound-force.

    My question is, is the thrust at this point distributed evenly among 6 tires, or is it additive per tire? If I hooked a load to this this robot/ATV I'm building, would it stall at a 31 pound load with each tire providing 5.3 pound-force, or a (6*31.77 = ) 190 pound load with each tire providing 31 pound-force?

    Thanks.

    This is the axle carriage as I have it right now in my CAD program. Slowly getting more drawn.

    https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-ash4/461089_10150962129059606_771454605_12935442_879036768_o.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
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  3. Apr 25, 2012 #2

    haruspex

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    It's shared by the six tires. Power = force x speed. You have the same power and the same speed whether it's across six tires or one, so the total force must be the same.
     
  4. Apr 25, 2012 #3
    Thanks. At this I hypothesize: in "stock" form, with 9 inch wheels,

    3000 @ 0.85nM -> 16.75:1 -> 179 @ 14.23 Nm = 179 RPM @ 10.5 pound force feet, 9" tire, equals 7 feet per second at 14 pound-force-feet of thrust. How would this be able to move, say, a 180 pound rider, and not stall?
     
    Last edited: Apr 25, 2012
  5. Apr 25, 2012 #4

    haruspex

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    I guess you mean 14 pound-force of thrust.
    That's ok, ignoring rolling resistance, as long as the upward incline is not too steep, i.e. less than 1 in 13 (sin(slope angle) = 14/180). A bit limiting by the sound of it.
    Might need smaller wheels or a bigger gear ratio.
     
  6. Apr 25, 2012 #5

    If my approximations are correct, 82kG person * g = 804.42 newtons, with an approx mu/rolling resistance of 0.30, yields 241 newtons. To do this on a 9 inch tire, would take 55.5 newton-meters of torque, or 3.3 newtons from the electric motor.

    Is that right?

    or working mu off the .85Nm rating on the sticker,

    14.23 Nm / .23m = 61.87 newtons of thrust, against a load of 804.42 newtons, would require a mu of 0.077


    -------

    The motor is made by Electro-Craft, a division of Reliance Electric.
    Model number E675
    Serial number G006497
    Date code week 20 of 1997
    Part number E675-07-042

    3.375" or so diameter casing with a 1/2" diameter shaft

    E_675_Pride_800_344.jpg
     
    Last edited: Apr 25, 2012
  7. Apr 25, 2012 #6

    haruspex

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    So.. you're never going uphill, so you only care about rolling resistance?
    In your last calculation you seem to have left out the chain drive.
     
  8. Apr 25, 2012 #7
    I'm trying to figure out how it functioned in it's designed life as a handicapped scooter motor. I am thinking the 0.85Nm on the label is an average/standard value, not a stall torque? Because I don't see how (per that calculation) it could have moved an 82Kg/180lb person generating only 62 newtons of thrust from a dead stop. I could understand it during constant motion, but not under acceleration.

    The robot I will be putting it in will only weigh 150 pounds max, but will be traversing property which is not safe to walk through due to snakes and other critters (A wetlands area) where the grass/weeds/vegetation is as high as 4 feet, as part of a research project. It will be running on 2 car batteries in series in a watertight aluminum hull with 24V/50A speed controller, controlled over R/C with remote camera broadcasting. Extreme case design is for moderate surface speed with significant amount of torque for pulling itself through the vegetation. The tires are 22"H x 11"W ATV tires. The 6 tires alone could support a { [[V(22,11)-V(8,8)]/1728]*7.48*8.34 }*6 = [[4181-402]/1728] = 2.18 CF * 7.48 gal/CF * 8.34 lb/gal * 6 = 818 pound vehicle if fully submerged. And make for good surface flotation over thick/muddy/cakey water.
     
    Last edited: Apr 25, 2012
  9. Apr 25, 2012 #8

    haruspex

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    As I read it, you are supplying the drive chain. Maybe in its designed role the gearing was much higher. In your first description, you reckoned 6f/s, or 7km/h. That's rather fast for a handicapped scooter. Probably would be half that, and limited to shallow inclines and decent surfaces.

    To go through vegetation you're going to need a lot more thrust. I'd be surprised if you could get 3km/h with 150 pound load (including the vehicle?).
     
  10. Apr 25, 2012 #9
    Found some new data. The batteries are 22V/55A or 24V/70A depending on options, and the chairs were rated for a 20-25 mile range per charge at up to 4.5 miles per hour.

    55A/20Mi = 2.75A/Mi * 4.5Mi/Hr = 12.375 A/Hr @ 22V = 272.25W
    55A/25Mi = 2.20A/Mi * 4.5Mi/Hr = 09.900 A/Hr @ 22V = 217.80W
    70A/20Mi = 3.50A/Mi * 4.5Mi/Hr = 15.750 A/Hr @ 24V = 378.00W
    70A/25Mi = 2.80A/Mi * 4.5Mi/Hr = 12.600 A/Hr @ 24V = 302.40W

    = Tao * Radians/sec * efficiency (arbitrary 0.8 sound good?)

    3000 RPM = 314 rad/sec

    0.8 * 314 = 251.33

    272.25 = 251.33 * tao = 1.083 Nm
    217.80 = 251.33 * tao = 0.866 Nm
    378.00 = 251.33 * tao = 1.504 Nm
    302.40 = 251.33 * tao = 1.203 Nm

    Of which the second value matches the 0.85 on the motor, for all effective reasons.
     
  11. Apr 25, 2012 #10
    Original design is 4/67 differential gearing and 9" tires. That's it. I posted a picture of the entire motor assy a couple posts up.

    I'm taking that assy, removing the wheels, and putting 9 tooth sprockets there instead. These then go to 25 tooth sprockets on the center axle. 12 tooth sprockets distribute to the outboard axles (1:1 of course).

    The scooter weighed 200 pounds on it's own, with spec for a max 250 pound rider. My entire robot will be under 150 pounds when done (80 of that will be the car batteries).

     
  12. Apr 25, 2012 #11
    Just to be on the safe side, I'll change the 9:25 chain to a 9:32. "Only" a 9 pound-force difference in thrust for a 1.4 foot per second decrease. But as you said, Torque is what I'll need given the conditions.

    40.64 pounds of thrust / 6 tires = 6.77 pounds of thrust per tire.

     
  13. Apr 25, 2012 #12
    Sorry if I write something redundant here, I'm in a small hurry.

    Rolling friction is not the same as sliding friction. You need a horizontal force of about F = k·mg where k ≈ 0.01 for car tires on asphalt, give or take ±0.005 or so.

    So a 100 kg vehicle needs about 10 N to move. Multiply with velocity (in m/s) to get the required motor power output (in watts).
     
  14. Apr 25, 2012 #13
    Thanks. So that would make the numbers indeed make sense, since I came up with a MuK of about .077
     
  15. Apr 25, 2012 #14

    haruspex

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    Two key questions:
    1. what slope the worst terrain you encounter will be equivalent to. You won't want to risk it's getting stuck often. I would guess something like 30 degrees.
    2. How the motor responds to increased load. There'll be a stall torque, but even before that there may be a falling off of power? If that's unknown, just take the stall torque.

    With the above assumptions, the fundamental need is to ensure you can get up a 30 degree slope, however slowly, at stall torque. What speed you manage on easier terrain is somewhat secondary.
     
  16. Apr 25, 2012 #15
    For safety/overhead tolerances, I say 45 degrees is a good design margin, especially for the size of tire.

    For MuK of 0.03 (arbitrary for knobby tire on dirt), on flat ground,

    0.03 * 9.8 m/s^2 * 68kg = 19.992 kilogram-meter per second squared = 19.992 Newtons of thrust.

    friction force going down = Mu * cos(theta) * m * g

    thrust going up = friction force + sin(theta) * m * g

    For MuK of 0.03 at 45 degree angle,

    friction force = 0.03 * 0.707 * 68 * 9.8 = 14.13N

    thrust force = 14.13N + (0.707 * 68 * 9.8) = 14.13 + 471.1448 = 485.27 Newtons

    For MuK of 0.03 on 20 degree plane:

    friction force = 0.03 * 0.93 * 68 * 9.8 = 18.59N

    thrust force = 18.59 + (0.34 * 68 * 9.8) = 18.59 + 226.57 = 245.166 Newtons

    For MuK of 0.03 on 5 degree plane (handicap ramp is 4.73 degrees or shallower per ADA)

    friction force = .03 * 0.996 * 68 * 9.8 = 19.92N

    thrust force = 19.92 + (0.087 * 68 * 9.8) = 19.92 + 58.080 = 78.00 Newtons

    ------------------------------------------

    Now that said, I o back to the designed spec for this motor. A 250 pound rider with a 200 pound dry weight is 450 pounds total- 204.5 Kg. I'm assuming they designed a minimum spec of going up a wheelchair ramp, ADA limited to 1 foot fo rise per 12 feet of run, or similar. Round off to 5 degrees.

    friction force = 0.013 (arbitrary for smooth tire to concrete) * 0.996 * 204.5 * 9.8 = 25.95N

    thrust force = 25.92 + (0.087 * 204.5 * 9.8) = 25.92 + 174.36 = 200.27 Newtons of thrust.

    To generate that, 200.27 newtons * (4.5" radius/39.25" per meter) = 22.96 Newton-meters of torque, through the 16.75 gearing, the electric motor had to generate 1.37 Newton Meters of torque to ascend that handicap ramp.

    --------------------

    1.37 Nm * 16.75 = 22.96 newton-meters at the output shaft * (32/9) = 81.635 newton-meters at the axle shaft. 81.635 newton-meters / (11" radius/39.25" per meter) = 81.635 / .280255 = 291.289 newtons of thrust

    If that new formula and calculations hold,

    friction force going down = Mu * cos(theta) * m * g

    thrust going up = friction force + sin(theta) * m * g

    thus, thrust force = (Mu * cos(theta) * m * g) + (sin(theta) * m * g)

    finding theta for a Mu of 0.03 = (0.03 * 68 * 9.8 * cos(theta)) + (68 * 9.8 * sin(theta)) = 19.992cos(theta) + 666.4 * sin(theta) = 291 newtons

    19.992cos(theta) + 666.4 * sin(theta) = 291 newtons

    plugging that in my calculator, shows a 289.31 N force at 24 degrees and a 299.75 N force at 25 degrees. So a 24 degree slope would be the upper limit.

    ---------------------

    using the 0.85Nm rating on the sticker,

    0.85 Nm * 16.75 = 14.24 newton-meters at the output shaft * (32/9) = 50.622 newton-meters at the axle shaft. 50.622 newton-meters / (11" radius/39.25" per meter) = 50.622 / .280255 = 180.63 newtons of thrust

    finding theta for a Mu of 0.03 = (0.03 * 68 * 9.8 * cos(theta)) + (68 * 9.8 * sin(theta)) = 19.992cos(theta) + 666.4 * sin(theta) = 180.63 newtons

    Calculator shows a result of 180.61N at 14 degrees. So with a Mu of 0.03 and a weight of 68Kg/150 pounds, It could climb up to 14 degrees, possibly up to 24.

    ---------------------

    finding theta for a Mu of 0.018 (arbitrary value for knobby tire on concrete) =
    (0.018 * 68 * 9.8 * cos(theta)) + (68 * 9.8 * sin(theta)) = 11.995cos(theta) + 666.4 * sin(theta) = 291 newtons = 24 degrees

    finding theta for a Mu of 0.018 =
    (0.018 * 68 * 9.8 * cos(theta)) + (68 * 9.8 * sin(theta)) = 11.995cos(theta) + 666.4 * sin(theta) = 180 newtons = 14 degrees


    ---------------------

    finding theta for a Mu of 0.013 (arbitrary value for smooth tire on concrete) =
    (0.013 * 68 * 9.8 * cos(theta)) + (68 * 9.8 * sin(theta)) = 8.6632cos(theta) + 666.4 * sin(theta) = 291 newtons = 25 degrees

    finding theta for a Mu of 0.013 =
    (0.013 * 68 * 9.8 * cos(theta)) + (68 * 9.8 * sin(theta)) = 8.6632cos(theta) + 666.4 * sin(theta) = 180 newtons = 15 degrees

    -------------------

    So the Mu isn't very effective of the total angle- So with this system of equations, 14 and 24 degrees are the lower and upper inclination limits.

    180 newtons = 40.5 pound-force

    290 newtons = 65.2 pound-force

    ------------------

    The batteries of choice would be Group 26 size for a 2000-2001 dodge neon. Each battery is 28 pounds, 675CA, 530CCA. Based on an 18AH load, would be about 30-37 hours of runtime between charges.
     
    Last edited: Apr 26, 2012
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