Drive Motor Requirements for Gearing/Torque Transfer

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In summary, the conversation discusses the use of gearing and torque transfer in a real world application where a receiving stator requires 3600rpm/26 lb ft torque. The question is raised about what type of drive motor would be needed to turn the stator at those parameters using an 18:1 ratio. The participants consider the performance values and limitations of various motors and gearings, including the need to account for losses and inertia. Ultimately, the conversation concludes that a larger motor may be needed due to the inverse relationship between torque and speed in a gear-train. The idea of using a lower HP gear drive motor or a regular AC motor is also mentioned as possible solutions.
  • #1
strobe
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I have a question regarding gearing and torque transfer. If I had a receiving stator that requires 3600rpm/26 lb ft torque, and needed to get a drive motor to turn it at those parameters using an 18:1 ratio, what would the drive motor need to be capable of, performance-wise? This will be a real world application, under continuous-use, so longevity will be a large factor.

I have formulated 1HP @ 200RPM (1HP X 5252/200=26.26 lb ft), but this is at the drive motor. At 18:1 gearing, the drive motor will be capable of initial torque value required, but as it goes through the gearing, will the torque value increase at 18:1, meaning i can now drop to a different drive motor of approximately 1.45 lb ft/torque (26.26/18=1.4588). I know that the RPM transfers at 18:1, but does the torque do the same?

I am trying to determine the drive motor's performance values to apply it into this application. As long as the final drive at the stator is 3600RPM/ 26 lb ft, and gear ratio is a maximum of 18:1 (preferably lower), any advice on what 12V-BLDC motor would fit will be greatly appreciated.
 
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  • #2
In theory torque and speed are (inversely) proportional through a gear-train. Taking your 18::1 as output::input, you need to run 18x _slower_ at the motor to get 3600rpm at the drive shaft -- so 200rpm -- and your torque at the drive shaft would be 18x _higher_ than the motor -- so a motor of about 1.5 ft-lb could do. However there are losses in the gear-train due to friction and inertia. You might be able to measure these, or just do the usual seat/pants-engineering and double everything... You also might want to consider acceleration, which means having to figure out the inertia of the whole system and compare to the amount of power you need to get it going.

I think most BLDC motors like to run at higher than 200 rpm, but you would need to look at torque/rpm curves to see if you can find something that works. Or change your gearing to match the motor you find. Or...use a regular garden variety 1Hp AC motor?
 
  • #3
Thanks for the input, as that was what I needed to know. I am alternately thinking of using a lower HP gear drive motor to alternate as the BLDC motor, so cost is a factor in the whole thing :)
 
  • #4
Ooooopsy... I got the torque backwards... sorry. I usually gear things down, so I wasn't thinking clearly. The _product_ of torque and rpm is constant. So if you are gearing _up_ from 200rpm you need to scale the torque _down_ by the same ratio. Which means you need an even bigger motor. duh.

Brain death is a horrible thing to waste.
 
  • #5


Thank you for your question. To determine the performance requirements for the drive motor in this scenario, we need to consider both the RPM and torque specifications provided for the receiving stator and the gear ratio of 18:1.

First, let's calculate the torque required at the drive motor to achieve the specified 26 lb ft at the receiving stator. To do this, we need to divide the torque by the gear ratio of 18:1. This gives us a torque requirement of 1.44 lb ft at the drive motor (26 lb ft / 18 = 1.44 lb ft).

Next, we need to determine the RPM requirement at the drive motor. Since the gear ratio of 18:1 means that the drive motor will be rotating at 1/18th the speed of the receiving stator, the RPM requirement at the drive motor would be 3600 RPM divided by 18, which is 200 RPM.

Based on these calculations, your initial estimate of 1 HP at 200 RPM for the drive motor seems to be appropriate. However, it's important to keep in mind that this is the minimum performance requirement for the drive motor. In real-world applications, it is always recommended to have a buffer or margin of safety to account for any potential variations or unforeseen factors.

Furthermore, as you mentioned, longevity is an important factor to consider in selecting the appropriate drive motor. In addition to the initial torque and RPM requirements, it's also important to consider the motor's continuous duty rating and its maximum operating temperature. These factors will affect the motor's performance and longevity in your application.

In terms of specific motor recommendations, it would be best to consult with a motor supplier or engineer who can take into account all the necessary specifications and provide a suitable motor option for your application. They may also be able to provide additional guidance on optimizing the gear ratio and motor selection for your specific needs.

I hope this information helps. Please let me know if you have any further questions or if I can provide any additional clarification. Best of luck with your application.
 

Related to Drive Motor Requirements for Gearing/Torque Transfer

1. What is the purpose of gearing in a drive motor system?

The purpose of gearing in a drive motor system is to transfer the torque generated by the motor to the intended application or load. Gearing allows for the motor to operate at a lower speed while still providing the necessary torque to move the load efficiently.

2. How do I determine the appropriate gear ratio for my drive motor system?

The appropriate gear ratio for a drive motor system depends on the desired speed and torque of the system. To determine the gear ratio, you will need to know the RPM of the motor, the desired RPM of the output shaft, and the torque required for your application. Using these values, you can calculate the gear ratio needed using the formula: Gear ratio = (Output speed/RPM) x (Motor RPM/Input speed).

3. What factors should I consider when selecting a drive motor for my system?

There are several factors to consider when selecting a drive motor for your system, including the required torque and speed, the type of load, the operating environment, and the power source. It is also important to consider the motor's efficiency, cost, and reliability.

4. What is the difference between a geared and gearless drive motor?

A geared drive motor uses a system of gears to transfer torque and adjust speed, while a gearless drive motor uses a direct drive mechanism without any gears. Geared drive motors are typically more efficient and provide a wider range of speed and torque options, while gearless drive motors are simpler and require less maintenance.

5. How do I ensure proper maintenance and longevity of my drive motor system?

To ensure proper maintenance and longevity of your drive motor system, it is important to regularly inspect and clean the motor and gears, as well as regularly lubricate any moving parts. It is also important to monitor the motor's temperature and ensure it is not overheating. If any issues or abnormalities are noticed, it is important to address them promptly to prevent further damage.

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