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Thread: need dc motor

  1. #1
    Senior Member Android
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    Default need dc motor

    i need 2 dc motors with ratings nearly as given
    voltage: 6v(preferable) but 12v also acceptable
    current: about 750mA to 1A when loaded
    speed: 300rpm

    if anyone want to sell his motors please reply...... also if anyone know where i can find them (online shop, or in mumbai).....also include the approx cost.....

    thanks.......

  2. #2
    Member Android
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    Default

    Hi,

    How much torque you are looking for

  3. #3
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    Default hi

    i am not good in those mechanical calculations...... some friends on this site gave me some formulas.... just cross check if i am doing right

    torque=(voltage*current*60)/(2*pi*N)
    i think N here is linear speed so
    torque=(4.5W*60)/(2*pi*2m/s)
    =21.49 ( i dont know unit..... i think its kgm)

    basically i want that both motor can together carry 1kg (excluding motors) with a maximum linear speed of 2m/s..... and sufficient acceleration (about 2m/s sq)

  4. #4

    Default

    Hey friend you can get DC geared motor of 12v 300rpm with roland electronics hyderabad he send it by courier to any destinationin in india check it out with him..09849070411 is his contact no.

  5. #5

    Default u need a moter

    torque of 3.8 kg m or 380 kg cm
    N rpm

  6. #6
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    Default hi

    @abdulmateen
    thanx for correcting me........ i think 380kgcm here is the torque required to carry my 1 kg requirement......

    now i have 2 dc motors of 6v, 2200rpm, no load current 30mA, full load current 300mA.....
    so torque of motor=7.81kgcm
    so if i use a pulley+rubber belt to reduce the speed to 440rpm the torque with pulley= 39.05kgcm...i think its too low....

    the only shop here have only one more type of motor whose rating is 12v, 2200rpm..... i dont know current rating....... but even if i assume it to have same resistance then also power will be 4 times and so torque with pulley= 156.2kgcm..... still too low.... he say he can provide me geared motors using same base motors..... will gear have any advantage over pulley?????

  7. #7
    Senior Member Cyborg
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    http://www.onlinetps.com/
    Online Component Shoppe
    Microcontrollers and Solar Robotics Stuffs at cheap rates

  8. #8
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    Default hi

    @tpsbpl
    i have seen this page many times but how can one order a part online without knowing its specification?????

  9. #9
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    Default

    Hi,
    I don’t know about the formula for calculating torque of motor but as per my personal experience if you are using Differential Drive system with Robot weight 1.5 Kg including both motors then minimum torque required is 3 Kgcm for each motors.

    http://www.warbotsxtreme.com/wheels.htm

  10. #10
    Senior Member Cyborg
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    @ Ashish
    hi ashish this may help you and all


    Some people might be curious about how a low speed, low power gear motor is able to generate a high torque. For example, a 60W output, 50 RPM gear motor is able to generate a 11.5 N-m (8.5 ft-lb) of torque, but yet, a 60W output, 2500 RPM motor is only able to generate a 0.23 N-m (0.17 ft-lb) of torque. The trick is speed reduction. Torque is very important to many applications because by knowing torque, one can determine how much of load the motor is able to lift an item, to drive a wheel, to turn a conveyor and so on. For example, if a 50 rpm motor has a torque output of 8.5 ft-lb (102 in-lb), a 4 inch diameter pulley should be able to lift 51 lb (102 in-lb / 2 in of pulley radius) of weight at 50 rpm; a 2 inch diameter pulley should be able to lift 102 lb (102 in-lb / 1 in of pulley radius) of weight at 50 rpm. Having that being mentioned, it is NOT recommended to put weight directly on a gear motor's shaft! An excessive radial bearing load may damage a motor's gear box quickly! It is always safe to isolate the load by introducing a pulley or sprocket system.

    One vital formula that is needed to calculate torque of a motor or its reduced speed output is:

    Power (watts) = Torque (N-m) * RPM * 0.105

    and of course,

    1 N-m = 0.74 ft-lbf = 8.9 in-lbf
    1 hp = 0.00134 Watt

    Here's an example on how to use this formula. Joe purchased a 12v gear motor from eBay that runs 60W at 50 RPM. He wants to know if he can lift a 200 lb of weight at 20 RPM with this motor.

    Here is how he can find out:

    First, he needs a belt and pulley speed reduction to reduce the motor's 50 RPM to his desired 20 RPM. That is, the speed reduction ratio of 50RPM to 20 RPM or 5 to 2 ratio or 2.5 to 1 ratio. He decides to get a 2 inch diameter pulley mounted on the motor's shaft to belt drive the 5 inch diameter pulley for the final output (of 20 RPM). Now he has a system that outputs 60 watts at 20 RPM instead of the motor's original 50 RPM. Also, he isolates the radial bearing load from the gear box. With 60 watts at 20 RPM, Joe now wishes to find out the torque of his system at 20 RPM:

    60 Watts = Torque (N-m) * 20 RPM * 0.105
    Torque (N-m) = 60W / (20 RPM * 0.105)
    Torque = 28.6 N-m = 254 in-lb

    Joe knows that the smaller the pulley size he has, the more weight his system can lift. He decides to go with a 2 inch diameter pulley and mounts it to his 20 RPM system. That is, the output shaft that is connected to the 5 inch pulley is mounted with his 2 inch diameter pulley to lift his load. Torque / Radius of Pulley = 254 in-lb / 1 inch = 254 lb. Joe is pleased to know that his system is able to lift 254 lb of weight that is more than his targeted 200 lb of load at 20 RPM.

    I hope this guide helps many of you save your design time and money

    http://reviews.ebay.com/How-to-Buy-t...dZp3286.c0.m17
    http://www.onlinetps.com/
    Online Component Shoppe
    Microcontrollers and Solar Robotics Stuffs at cheap rates

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