Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Rack and pinion, motor selection method

  1. Apr 14, 2014 #1
    Hi!
    I'm stuck on a very basic problem while trying to design a product. Basically, I need to know how to calculate the gear ratios, pitch, motor specs etc for two rack and pinion mechanisms- one horizontal and one vertical. The pinion moves along the rack in both situations.
    Mass to be moved- 140 kg
    speed- 0.3m/s
    Distance: Horizontal- 0.65 m, and vertical- 0.45 m
    I realise this is quite a simple problem, but I'm have trouble figuring out the best way to go about calculating the specs. Could anyone tell me a basic route to go down- what steps to take in order to give me an approximate specification for my product? Any links would be helpful.
    Many thanks
     
  2. jcsd
  3. Apr 15, 2014 #2

    Baluncore

    User Avatar
    Science Advisor

    Do you include the acceleration and deceleration phase of movement in the average speed?
    Maybe you could specify the required acceleration or the minimum time to achieve the required maximum velocity.

    How will you counter the vertical axis force on the 140 kg due to gravity. Can you counterbalance that mass, apply a brake, or must you always provide a torque using a continuous current through an electric servo motor?
     
  4. Apr 15, 2014 #3
  5. Apr 15, 2014 #4
    I have a mechanism to counter the vertical axis force, the job of the rack and pinion will solely be to drive the load. The acceleration time is not an issue in this situation, but can be taken as a minimum of 1s for convenience.
     
  6. Apr 15, 2014 #5

    Baluncore

    User Avatar
    Science Advisor

    First you must accelerate 140 kg to 0.3 m/s. KE = ½ * 140 * 0.3 * 0.3 = 6.3 joule.
    That needs to be provided in 1 second, so power input is 6.3 / 1.0 = 6.3 watt.
    The mass on the vertical slide is balanced so no work needs to be done to lift that.

    Bigger gear teeth and wider racks are stronger, so;
    Select a commercially available material and rack pitch, p, that will handle the forces involved.
    Select the smallest available pinion for the rack, it will have probably about n = 20 teeth.
    Each turn of the pinion will advance ( n * p ) along the rack.
    To travel at a speed of 0.3 m/sec pinion will require 0.3 / (n * p) RPSec = 18 / (n * p) RPM.

    Assume that a DC electric motor is used that has “no load” speed of say 3000 RPM.
    For maximum power output, operate it at half speed = 1500 RPM.
    Gear ratio required is then (n * p) * 1500 / 18

    If p is 5mm and n is 20, then required gear ratio = .005 * 20 * 1500 / 18 = 18.0 : 1 ratio.
    That is probably best done with a commercially available enclosed motor with worm drive like used for car windscreen wiper or window winder units.
     
  7. Apr 22, 2014 #6
    I sorry I am doing a very similar thing, could I ask where 18 came from ?

    "To travel at a speed of 0.3 m/sec pinion will require 0.3 / (n * p) RPSec = 18 / (n * p) RPM."
     
  8. Apr 22, 2014 #7
    It just comes from converting RPSec to RPM. The 18 is 0.3 x 60
     
  9. Sep 30, 2016 #8
    Hello,
    I am making a project, in which i have used a rack which is in 'J' shape having teets outside and have a pinion of its respective pitch. The whole rack is attached with a 200kg load. Now i'm confused in selecting correct motor for the application.
    Help
     
  10. Sep 30, 2016 #9

    Baluncore

    User Avatar
    Science Advisor

    Welcome to PF.
    1. What length is the rack?
    2. Does the 200 kg load rise, fall or move horizontally?
    3. What is the maximum speed that the pinion must move on the rack?

    4. What is the tooth pitch of the rack?
    5, How many teeth on the pinion?
     
  11. Feb 7, 2017 #10
    i am working on rack and pinion system where rack is fixed and pinion has to move with 0.5 m/s and should carry 600 kgs load. i needed to know how to calculate the power required by a pinion to move from one rack teeth to other and the torque if possible please show how to evaluate
     
  12. Feb 7, 2017 #11
    i am working on rack and pinion system where rack is fixed and pinion has to move with 0.5 m/s and should carry 600 kgs load. i needed to know how to calculate the power required by a pinion to move from one rack teeth to other and the torque if possible please show how to evaluate
     
  13. Feb 7, 2017 #12

    Baluncore

    User Avatar
    Science Advisor

    Welcome to PF.
    Is the 600 kg load on the vertical axis pressing the gear onto the rack ?
    Or is it horizontal axial force needed to move a carriage by the spur gear meshed with rack ?
     
  14. Feb 7, 2017 #13
    rack is horizontal and load has to be carried along it in horizontal direction and the type of gear is spur gear only
     
  15. Feb 7, 2017 #14

    Baluncore

    User Avatar
    Science Advisor

    If I understand correctly, you have a carriage that weighs 600 kg.

    Energy will only be needed while the carriage is accelerating to the speed of 0.5 m/s.
    Once rolling, KE = ½·m·v2 = 0.5 * 600* 0.5* 0.5 = 75 joule.

    Power is rate of flow of energy. Watts = joules per second.
    A 75 watt motor will accelerate the carriage to 0.5 m/s in 1 second.

    Power will be torque times RPM.
    Torque depends on the radius of the drive pinion, or the number of teeth on the pinion and the pitch of the rack.
     
  16. Feb 7, 2017 #15
    you are correct
    a carriage weighing 600 kgs has to be moved by the help of rack and pinion mechanism
    i have a doubt that when a pinion try to move then there is a opposing force acting on the pinion by the rack teeth so how much force is the opposing and it acts tangentially or normal to the rack teeth.
     
  17. Feb 7, 2017 #16

    Baluncore

    User Avatar
    Science Advisor

    If the rack has a 20° contact angle then we can resolve the contact force into two different components.
    One component is horizontal, let's call it 100%, the other vertical = 100% * Tan( 20° ) = 36.4%.

    The vertical component is resisted by the mass of the carriage and there is no vertical movement, so there is no energy involved. The vertical force component is also acting radially on the spur gear so it generates no torque component on the spur gear drive shaft.

    The horizontal force component is in line with the carriage movement, so energy will be involved. The horizontal force is parallel to the rack, (tangent to the spur gear circumference), so the entire torque will be due to the horizontal force on the rack.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Rack and pinion, motor selection method
  1. Motor Selection (Replies: 4)

  2. Rack Capacity (Replies: 2)

Loading...