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sprocket vs gear box 3

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harrinshotmail

Automotive
Jul 7, 2008
26
The reduction ratio needed im our application is 1800.
Following are the ways to do it:

1. Use an AC invertor and bring its speed down to 140rpm. Then use worm reducing gear 60:1 ratio. Then use a sprocket to reduce the speed further by 4.5 ratio to get a final rpm of 0.5 rpm. The motor chosen is 865rpm
2. Use a 1740 rpm motor and use 2 sets of sprocket reduction to get 0.5 rpm and an ac invertor.
3. Use a double reduction worm gear and 865 rpm motor to get the final rpm to 0.5 rpm

Which of the above 3 is the best choice. The output torque required is 16000in-lb at 0.5rpm. Please suggest a choice from the above 3 or any other choice. The motor only runs few seconds only to rotate a dumper manually. So there is no heat problem. Please help.
 
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Sounds like 0.5 rpm is your goal. Can't you just use an ordinary 1725 rpm induction motor with a worm gear box? Hub City has standard helical worm gear boxes with a ratio of 3460:1 and horsepowers from 1/4 Hp up to 23 Hp.

Don
Kansas City
 
I like option 1, with a 4.5 to 1 spur mesh, followed by a 60:1 worm.
 
Matt51- It is actually 60:1 worm gear and then sprocket. The main question was whether the sprocket transmits effeciently or worm gear. The dumper has to rotate 360 degree.

Second question is what is the moment of inertia of a 4000lb cuboid with the axis of rotation offest by x inches.
Will the torque required is same if the axis of rotation is x, y or z axis. The cuboid is the cage which tilts/rotates 360 degree. I need the dumper to accelerate from 0 to 0.5 rpm in 3 seconds.
 
Hi,
I like the idea of using as little mechanical reduction as possible, and I like to avoid chains.
 
I was proposing modifying option 1. It looks like you have a number of choices.
 
The main question was whether the sprocket transmits efficiently or worm gear.
In my view a worm gear is the most inefficient, and it gets more inefficient as the input speed decreases.
 
I agree with Peter. Thanks.

What about AC Invertor? Is it a good idea to use it to reduce the speed by 1/6?
 
I beleive the main reason for the expense of an AC invertor is for it's variable speed capabilities. I have often seen variable speed drives used in place of engineering, as the user could not easily calculate the speed needed. The drives are started up, the correct speed is found, and then the speed is never changed for the rest of it's service life. What a waste of the extra money for the vari-speed drive. If you want to vary the speed, get an invertor, and a gearmotor, and choose your gear ratio to attain your absoulte maximum speed. If you do not need var-speed, just get the gearmotor. You did not mention how this needs to be connected to the machine. You may need a chain drive or dirct coupling to the driven shaft. Does it require overload protection? Many more things to be considered.

Russell Giuliano
 
You're overcomplicating this. You're worried about the efficiency of something turning at 0.5 rpm, which should take no more than a little 1/4 hp motor. The acceleration torque would be minuscule...and there's nothing efficient about a VFD, believe me.

Unless you need variable speed, you should just run the motor at its full rated rpm, gear it down and be done with it.

Don
 
I agree with Don above. This is an intermittent operation. Look at minimizing your first costs because it will take a long time to re-coup energy savings with occasional use.
 
Russell: How did you calculate the horsepower please?

Don: I calculated the torque to be 16000in-lb by using the formula:

Torque = mass moment of Inertia * ang. acceleration

Inertia =1/12* mass* (sq. of a + sq. of b) where a and b are the sides of the basket.

The I calculated Horsepower using the formula:
Torque = 5250 * hp/rpm

After I know the torque I can select a gear motor, sprocket or double reducing worm gear. This is why i was asking what is the best choice.
If I just choose a worm gear or sprocket, then I have to use a coupling or chain respectively. What is the easier option in a coupling or chain and sprocket?
 
I was assuming that your torque figure was correct. I got the approximate hp (neglecting efficiency) from:

P = Tw (Power is torque x angular velocity)

P = (16000 #-in)*(0.5 rev/min)*(1 min/60 sec)*(2*pi rad/rev)*(1 ft/12 in)*(1 hp/550 ft-lb) = 0.127 hp

Don
Kansas City

 
Don: Is my torque calculations right?
Assuming the basket is 34" high and 40" wide. Assume the mass of the full basket is 4000lb?
 
A fully-loaded basket may not always be the worst-case torque. I have seen instances when an empty, or partially loaded, container required more torque to rotate.
 
What is the location of the trunnion? Is it at the center of the rectangular ends? Also, what is the substance being dumped?

In addition to the torque caused by the inertial reaction (angular acceleration) you need to consider what torque the contents might cause at different levels of fullness as you tip. Depending on the substance you are dumping, there could be a considerable torque at different fill levels and angles of tip.

A particulate solid, like sand, would only spill at an "angle of repose" so that more sand would end up on one side of your pivot point and its weight would cause a torque. A non-viscous liquid, like water, would be self-levelling, but depending on the shape of the tub and the location of the trunnion, it could still cause a torque. If your pivot was down at the bottom of the bucket, for example, you could have a hefty torque as you tip the entire 4000-lb load!

Don
Kansas City
 
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