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  1. #1
    Join Date
    May 2010
    Posts
    0

    Question the rack and pinion machine motor torque

    I am a crazy hobbyist attempting to build a cheap and precise machine in my house.

    precise enough to cut parts for plexiglas enclosures at least

    I want to use rack and pinion drive, where the motor would drive the pinion directly over the rack.

    Can you check my math please:

    I am really not sure how I know the carriage acceleration. My machine will move the dremel tool over the material most probably.

    Thank you.
    Attached Thumbnails Attached Thumbnails NEWTON meters.GIF  

  2. #2
    Join Date
    Feb 2004
    Posts
    1086
    For any mechanical system, you will need to know the mass you want to accelerate, and the force you can impart to that mass. F=ma. For a rack and pinion system, this is not too hard, but it will require that you look at your stepper motor torque curve, and that you know the diameter of your pinion gear.

    Let's take an example with our components -- we'll do it in metric so we don't have to deal with slugs as a unit of mass . Our rack and pinion system offers a 25.4mm (1") pitch diameter pinion and a 3:1 reduction. At low rpm's, our stepper motors have around 2 Nm of torque (note, this is important to check on any motors you are considering -- pretty much everyone advertises stall torque, which is not a very useful number). With the 3:1 reduction, this is augmented to 6 Nm. At a radius of 12.7mm on the pinion, this means the system can exert a force of 6000Nmm/12.7mm = 472 N.

    So now you just solve F=ma for acceleration. If you have a gantry that weighs 47 kg, you should be able to get 10 m/s^2 of accleration, or about 1g (a N = a kg*m/s^2). If your gantry only weighs 23.5 kg, you should be able to get 2 g's. Of course, this isn't exactly right, as your motor will loose some torque as it accelerates, and there are some small frictional losses in the system, but this is a good approximation for machine design. Also, keep in mind that most people run two R&P systems (one on either side) on the gantry, so they will have double the force to play with.

    With a direct drive like you are proposing, your accel will be slower -- you will not have the 3:1 torque advantage in the example above, so your force will also be lower. This still may be workable for your application, but it will not have the optimal balance of speed and cutting force.

    Hope that helps!

    Ahren
    www.cncrouterparts.com

  3. #3
    Join Date
    Feb 2008
    Posts
    553
    Hi Ahren,

    Thanks for the helpful info you provide in your post, if you don't mind, I have some questions:

    Say you want to gear down the motor using a belt/chain pulley/sprocket, the motor's pulley diameter is "d" the bigger pulley's diameter is "2d", the bigger pulley is connected to the pinion by an axe(or axis i'm not sure about the language...I mean they turn at the same speed) the pinion's diameter is "d/2".

    1) In this case do I get four times the motor's torque at the pinion ?
    2) If the pinion's diameter is "d" do I get double the motor's torque at the pinion ?

    Sorry if my questions sound stupid, I don't do physics so frequently & I'm unsure of my ressults.

    Thanks !
    cnc2.

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