christhorn82
Mechanical
- Oct 24, 2005
- 12
Hi - thanks for taking the time to help.
I'm having trouble understanding how a speed reducer (Boston Gear HF738-30-B7-H-P39) and a helical multiplier (Boston Gear HMF2-5-B7-B7) can produce the required torque for an application.
Speed reducer
Single stage reduction, output rpm @ 1750rpm = 58.3rpm, ratio = 30:1, torque capacity = 362Nm, motor horsepower =2HP
Helical multiplier
Single stage reduction, output rpm @ 1750rpm = 350rpm, ratio = 5:1, torque capacity = 100Nm, motor horsepower =3HP
I have been tasked with selecting gearing to allow the manual opening of the lid of a pulverising mill. The lid sits horizontal and is opened to just past the vertical.
An identical machine works perfectly well with this equipment, and the lid is cranked open with a handwheel.
The lid's mass is 250kg and its centre of gravity is 0.53m from its pivot. Thus to hold it horizontal a torque of 1300Nm is needed.
The main thing that's confusing me is the torque capacity of the gear units. I have found the torque needed to hold the lid level is 1300Nm. The torque capacity (or "torque cap" as it's called on the Boston Gear site) of the speed reducer is only 362Nm. To look at, this seems very low and I don't understand how it can work.
I'm guessing this is related in some way to the speed of 1750rpm that is mentioned, and the 2HP motor. I will not be using a motor in this design and ultimately a person will be winding the lid open by hand at maybe 100-200rpm input speed.
I'm assuming the operator can apply a force of 60N on a 0.3m diameter handwheel, thus generating a torque of 60N x 0.15m = 9Nm. Then 9Nm x 150 ratio = 1350Nm. If it's that simple, then yes the operator can open the lid with relative ease. How does the torque capacity of the gear unit come into all this?
Could someone go through the calculations with me so that I can understand how the two gear components were selected?
Thanks very much for your help,
Chris
I'm having trouble understanding how a speed reducer (Boston Gear HF738-30-B7-H-P39) and a helical multiplier (Boston Gear HMF2-5-B7-B7) can produce the required torque for an application.
Speed reducer
Single stage reduction, output rpm @ 1750rpm = 58.3rpm, ratio = 30:1, torque capacity = 362Nm, motor horsepower =2HP
Helical multiplier
Single stage reduction, output rpm @ 1750rpm = 350rpm, ratio = 5:1, torque capacity = 100Nm, motor horsepower =3HP
I have been tasked with selecting gearing to allow the manual opening of the lid of a pulverising mill. The lid sits horizontal and is opened to just past the vertical.
An identical machine works perfectly well with this equipment, and the lid is cranked open with a handwheel.
The lid's mass is 250kg and its centre of gravity is 0.53m from its pivot. Thus to hold it horizontal a torque of 1300Nm is needed.
The main thing that's confusing me is the torque capacity of the gear units. I have found the torque needed to hold the lid level is 1300Nm. The torque capacity (or "torque cap" as it's called on the Boston Gear site) of the speed reducer is only 362Nm. To look at, this seems very low and I don't understand how it can work.
I'm guessing this is related in some way to the speed of 1750rpm that is mentioned, and the 2HP motor. I will not be using a motor in this design and ultimately a person will be winding the lid open by hand at maybe 100-200rpm input speed.
I'm assuming the operator can apply a force of 60N on a 0.3m diameter handwheel, thus generating a torque of 60N x 0.15m = 9Nm. Then 9Nm x 150 ratio = 1350Nm. If it's that simple, then yes the operator can open the lid with relative ease. How does the torque capacity of the gear unit come into all this?
Could someone go through the calculations with me so that I can understand how the two gear components were selected?
Thanks very much for your help,
Chris