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Question about velocity and motors

  1. Aug 18, 2010 #1
    I'm working on choosing an appropriate motor. I have the calculations I need for the required torque, power, max velocity, force, and acceleration. Now I need to know the maximum velocity that a specific motor can move 159 kg 0.45 m across a horizontal plane.

    In all of my calculations for which motor was needed, I neglected to take friction and inertia into consideration because I did not need to be so exact. The friction is quite minimal anyway. It is a seat moving along a metal rail.

    The motor specs are as follows:

    Torque - 0.304 Nm
    Speed - 4000 RPM
    Power - 0.13 kW
    Gearing - 60:1 (15*4)
    Inertia - .031 kg-cm^2

    I don't yet know what the radius of the gear attached to the gearbox will be. I imagine something like 5 cm. I was told that the equation would involve something like (2pi*r(RPM))/15, but I don't know how that will work in with the power, mass, and distance when calculating the max velocity.

    I was thinking if using 1/2*130*0.45 would be possible to use for the kinetic energy and then solve for v with 29.25=1/2*159*v^2. But again, I'm not sure how to work in the Speed and gearing of the motor.

    I appreciate the help. If I'm missing any info, let me know.
    Last edited: Aug 18, 2010
  2. jcsd
  3. Aug 20, 2010 #2
    Does anyone at least have some tips or ideas?

  4. Aug 20, 2010 #3

    Gearbox is 15:1. Pulleys are 1:1.
    Radius of pulley is 1"

    The pulley's RPM speed is (circumference*RPM)/gearing, right? In other words, (2pi*4000)/15=1676 RPM. The torque would then be 0.304*15=4.56 Nm.

    So then how do I get the maximum velocity of the 159 kg along 0.45 m that this setup can provide?
    Last edited: Aug 20, 2010
  5. Aug 20, 2010 #4
    Hi FES, if this is your homework, it should go to the homework section.

    Add a sketch or drawing if possible.

    Think different, if you had to lift the same mass 0.45m in the vertical plane, frictionless and disregarding inertia, how would you solve the problem ? What would be the acceleration and kinetic energy ? And the maximum vertical speed ?

    Now turn back to the horizontal plane and consider friction and inertia, what top speed would be attainable considering also the transmission efficiency?
    Last edited: Aug 20, 2010
  6. Aug 21, 2010 #5
    Hi John,

    Thank you for your response. No, this is not homework, this is real-world for my job. I simply am unfamiliar with working with motors and have not done physics in several years, so I am having a mental block. Unfortunately, I cannot disclose this device, but think of those Concept 2 rowers you would find in a gym.

    I am running out of time. The motor guy will be in on Monday and I must have a spreadsheet complete with this same calculation for every motor in a certain class by then for my CEO.

    Ergo, if you wouldn't mind feeding me a bit more than a crumb, I'd greatly appreciate it. I am thinking what I need to do is use Force=torque/radius and find the acceleration from there with F/m.

    4.56 Nm/.0254 m=179.53 N
    179.53 N/159 kg=1.13 m/s

    1.13*0.45 m=0.51 s

    Do the RPMs still need to factor in somewhere?

    Thank you.
    Last edited: Aug 21, 2010
  7. Aug 21, 2010 #6
    It looks like this is a gear\rack mechanism, a very simplified approach will be:

    Considering the mass path is perfectly horizontal, negligible rotational inertia, no friction, starting at max power and max rpm...

    Fmax = Tin x i x n / r

    F -> Max Traction Force [N]
    Tin -> Motor Input Torque [N.m]
    i -> Gear ratio
    n -> Efficiency
    r -> Gear Radius [m]

    amax = Fmax / m

    amax -> Max Acceleration [m/s²]
    m -> Mass

    converting rpm -> rad/s => 1 RPM = 2*PI / 60s

    w2 = w1 / i

    w2 -> Gear speed [rad/s]
    w1 -> Motor speed [rad/s]

    vmax -> w2 x r

    vmax -> Maximum Travelling Speed [m/s]

    tmin = (2 x d / amax)^0.5

    tmin -> Minimum Acceleration Time
    d -> Travelling Distance [m]

    P = Fmax x vmax / n

    P -> Power [W]
  8. Aug 23, 2010 #7
    I typed up a lengthy reply that failed to post and is now lost, so all I will say is thank you for all of your assistance, John. I'm grateful!
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