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Rack and pinion, motor selection method 
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#1
Apr1414, 09:36 AM

P: 13

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
Apr1514, 09:57 AM

Sci Advisor
Thanks
P: 1,923

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? 


#3
Apr1514, 10:00 AM

P: 13

I have found a product with a similar rack and pinion application as mine.
http://www.drivemedical.co.uk/files/...aintenance.pdf How would I go about calculating the gear ratios for the reduction gears based on the speed I want the pinion to move at, while carrying a load of 140kg? thanks 


#4
Apr1514, 10:02 AM

P: 13

Rack and pinion, motor selection method



#5
Apr1514, 07:24 PM

Sci Advisor
Thanks
P: 1,923

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. 


#6
Apr2214, 08:41 AM

P: 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." 


#7
Apr2214, 01:09 PM

P: 13




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