Helping out with pulley diameter calculation

AI Thread Summary
To calculate the required pulley diameter for reducing motor speed from 1800 RPM to 80 RPM, the relationship n1 d1 = n2 d2 can be used, where n1 is the motor RPM and n2 is the desired RPM. The ratio of n1 to n2 is 22.5, indicating that the driven pulley diameter should be 22.5 times larger than the motor pulley diameter. However, this large ratio raises concerns about potential belt slippage, suggesting the need for a toothed belt or multiple pulleys to achieve the desired reduction more effectively. Alternatives like a gear reduction system or chain drive can also provide more flexibility in adjusting the speed ratio. Proper selection of components is crucial to ensure efficient power transmission without slippage.
Giannakoulis
Messages
8
Reaction score
0
Hey guys.. Sorry for my english, i'am only 16 and it's not my parent language. I need help about a problem on mechanical engineering, and i really appreciated if you can help me out. I assume that I've got a moter currents which gives me out 220-230 Volt trend, and has 1800 bends per minute. Now i want to know how much diameter should have a pulley to put it on another machine. I want this machine to give 80 bends per minute. And what pulley should i put in the original current moter to suit them. The starting transmission from the moter will be done with a belt.. For more details or clarifications, just ask me..
 
Engineering news on Phys.org
Hello Giannakoulis, :welcome:

Basically you have ##n_1 d_1 = n_2 d_2## (because the belt speed is the same on both pulleys). In this expression
##n_1 = ## motor rpm (revolutions per minute)
##d_1 = ## pulley diameter on motor axis
##d_2 = ## pulley diameter on driven axis
##n_2 = ## driven axis rpm​

In your case you have ##n_1/n_2 = 22.5## so you want ##\ \ \ \ \ ## pulley diameter on driven axis = 22.5 times pulley diameter on motor axis
 
  • Like
Likes Giannakoulis
thanks a lot!
 
BvU said:
In your case you have n1/n2=22.5n1/n2=22.5n_1/n_2 = 22.5 so you want pulley diameter on driven axis = 22.5 times pulley diameter on motor axis

That is a large ratio. It is possible that the belt will slip on the small pulley. The maximum power you can transmit is limited by friction and the "contact angle" (how far around the pulley the belt wraps). You might need to use a toothed belt?
 
CWatters said:
That is a large ratio. It is possible that the belt will slip on the small pulley. The maximum power you can transmit is limited by friction and the "contact angle" (how far around the pulley the belt wraps). You might need to use a toothed belt?
Or a series of pulleys with a smaller reduction in each. If my maths is right three pairs with a reduction of 3, 3 and 2.5 gets you 22.5:1.
 
I would probably look at a 10:1 or 25:1 gear reduction.. with the 10:1 reduction you'd need a 2.25:1 belt drive (2" to 4.5" or so), and with the 25:1 the driven pulley would have to be just a bit smaller than the motor pulley.. If you use a chain drive from the gear to the machine, you can adjust the ratio there as well (may be easier and more accurate than changing pulley sizes)
 
Thread 'Turbocharging carbureted petrol 2 stroke engines'
Hi everyone, online I ve seen some images about 2 stroke carbureted turbo (motorcycle derivation engine). Now.. In the past in this forum some members spoke about turbocharging 2 stroke but not in sufficient detail. The intake and the exhaust are open at the same time and there are no valves like a 4 stroke. But if you search online you can find carbureted 2stroke turbo sled or the Am6 turbo. The question is: Is really possible turbocharge a 2 stroke carburated(NOT EFI)petrol engine and...
I need some assistance with calculating hp requirements for moving a load. - The 4000lb load is resting on ball bearing rails so friction is effectively zero and will be covered by my added power contingencies. Load: 4000lbs Distance to travel: 10 meters. Time to Travel: 7.5 seconds Need to accelerate the load from a stop to a nominal speed then decelerate coming to a stop. My power delivery method will be a gearmotor driving a gear rack. - I suspect the pinion gear to be about 3-4in in...
Back
Top