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Calculating tq for incline / best practices for gear reduction

  1. Nov 14, 2016 #1
    I am calculating the amount of torque required to give mobility to a 7000 lbs oilfield service vehicle. The vehicle has a rectangular frame with fairly even weight distribution. The radius, from wheel center to edge of tire is 11.5 inches. These tires are treaded with various half-inch tall "cleats" along the tire. for lack of a better word. I'm mentioning this because I do not know how to accurately characterize the coefficient of rolling friction for the Force calculation. For my example, I have used 0.04, which seemed like a fairly coarse road tire:

    F = (c)ma = (0.04) (7000 lb) (32ft/s^2) = 280 lbf.

    The above would presumably be the force required to roll the unit along the grade of surface described by the coefficient; while it will be on generally flat surfaces, it must be loaded into a dovetail trailer, so an incline, 45 degrees assuming the worst, will be present for a short period of time. Is this the correct equation? This is where I begin having trouble profiling the additional torque required.

    F = (280 lbf) (sin 45) = 198 lbf = the gravitational force acting upon the object at 45 degrees?

    This value is lower than the original value, so that's where I get a little lost. Must I add the original (280) + 198 + a cushion amount to guarantee good acceleration up the ramps?

    Once I figure that out, I am trying to determine how a gear reduction to the (electric) drivetrain would ease up my power requirements. One luxury is that the unit only needs to move 5mph, max. I would prefer it be mechanically limited from a safety perspective (wireless electronics will be controlling the throttle, through a receiver, to the motor controller)

    With this speed in mind, is there an expression I can reference to determine the torque required to move this object, of this weight, with this friction loss and pitch requirement, with respect to X gear ratio?

    Regards,
    Mel
     
  2. jcsd
  3. Nov 14, 2016 #2

    Mech_Engineer

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    Regarding your incline calculation, you might take a look at this equation set: http://www.engineeringtoolbox.com/inclined-planes-forces-d_1305.html. Keep in mind that to pull the vehicle up the ramp, you need to take into account the rolling resistance (probably low on a metal ramp) and the weight of the vehicle multiplied by sin(angle).

    Regarding rolling resistance, I would recommend overestimating your rolling resistance or including a safety factor in your analysis. An oil field vehicle will probably have to be able to navigate through loose soil and mud, which according to this source may result in rolling resistance as high as 0.2.

     
  4. Nov 14, 2016 #3
    I think I understand this thus far. However I am having a lot of trouble getting a relative feel for the power requirements. According to the incline calculation using some correct values (6000lbs object), and using as relatively low rolling friction coefficient (0.04), it says I'd need about 14,000N of force to go up a 30 degree incline:

    Fp = W (sin α + μ cos α)
    Fp = m ag (sin α + μ cos α)
    Fp = (2721 kg) (9.81 m/s2) (sin(30)+0.04cos(30)) =
    14271.18 N
    or 3208 lb-f

    This seems like a lot especially knowing that in previous builds, a gasoline engine with a 25HP rating linked to a hydraulic drivetrain got them up these inclines fine.
     
  5. Nov 15, 2016 #4

    Mech_Engineer

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    Your numbers seem ballpark correct to me; I would say you're on the right track. It's easy to imagine a hydraulic drive train could put out a lot of torque, so try starting there?
     
  6. Nov 16, 2016 #5
    Calculation of fluid motor torque for the 25HP legacy installation:

    Tq = (Pressure * Motor Displacement) / (2 * pi)
    Tq = (3000 psi * 1.14 cu. in) / (2*pi)
    Tq = 544 in.-lbs
    Tq = 45 ft.-lbs

    Why are the numbers so different?
     
  7. Nov 16, 2016 #6

    jack action

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    Wasn't there a gear reducer between the motor and the wheels?

    The power requirement is the force times the speed. For 5 MPH (2.2 m/s):

    ##14271.18\ N \times 2.2\ m/s = 31397\ W \equiv 42 hp##

    You can't get around that, no matter the type of drivetrain you are using. It might help you reverse-engineered the previous builds.

    For the rotating components - not counting inefficiencies - that same amount of power (##P##, W) will be torque (##T##, N.m) times angular velocity (##\omega##, rad/s) or:[tex]P_{[hp]}=\frac{T_{[lb.ft]}\omega_{[rpm]}}{5252}[/tex]
     
  8. Nov 17, 2016 #7
    Jack is right, there MUST have been additional gear reduction.. Excavators, etc have hydraulic drives, and some of them are over 100:1 planetary reductions.. It sounds like a similar case in your application.

    If you can reuse the old reduction drives, you have a lot of your work already done for you, you just need a motor capable of the same torque and speed as the old hydraulic unit, and you already figured that out at 45 ft lb
     
  9. Nov 17, 2016 #8
    it's also possible the old unit had a 2 stage pump, delivering additional flow at lower pressures where the engine power couldn't handle full flow at high pressue
     
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