Question about gear reduction ratio

amir81

Member
Join Date
Aug 2005
Location
Location
Posts
3
Hi all,



I have some questions about gear-motors and I hope you can help me to find their answers. We use some of Bauer gear-motor products in our factory for many years. Recently we need to change the application of a BG80/D13MA4 to hoist application, driven by an inverter.

On the nameplate of the motor, the following data are written:
P=7.5 kW
n2=65 rpm
n1=1420 rpm
I=15.8 A
PF=0.85
f=50 Hz

According to above data the gear ratio should be approximately 21.85. But in Bauer-CAT software, when I chose the above gear-motor (attached file "1.jpg") and click on the next screen button, it showed me a table with 13 rows of configuration (attached file "2.jpg"). Strangely, the gear ratio of SAME gear is changing, from 26 to 63. As far as I know, the gear ratio of the gear is always constant!!

Next, I changed the electrical supply combo-box to "mains duty" (attached file "3.jpg") and click on next screen button. The new table of configuration has 10 rows and in the P2 column, the value of 9.5 kW is written (attached file "4.jpg"). The hoist application is not S1 (continuous) type and in fact it is S3/S6 one.

The abovementioned scenario raised following questions in my mind that I want you to help me to find their answers, please:
1. Why there is numerous "gear ratio" for a single gear?
2. Why there is 5.5 kW configuration, even when I chose 7.5 kW motor type?
3. Which of reported torques is correct to assume for the hoist application?
4. We want to use a 7.5 kW inverter to drive the gear-motor of the hoist (the stated power rating on the motor nameplate). Should we use a 9.5 kW instead, if the data of the software was correct?

Thank you.

1.jpg 2.jpg 3.jpg 4.jpg
 
It looks like your selection of 'BG80' for the gear type is a generalized helical gear housing which can have many different ratios not specified by just 'BG80' - thus various torque outputs and speeds are listed depending on the exact gearing.
 
Hi
I think you are right. This can only describe why there is different reduction ratio.
But what about required power? Which of 7.5 or 9.5 is correct for hoist? and do I need to change the inverter to achieve more output power?
thanks
 
But what about required power? Which of 7.5 or 9.5 is correct for hoist? and do I need to change the inverter to achieve more output power?
thanks

Power required is determined by the load. You indicated this is a hoist:

hp = (F x S) / (33000 x eff)
F = force exerted by the load, pounds
S = speed of lift, feet per minute
33,000 ft lb per minute = 1 hp by definition
eff = decimal efficiency of the gear reducer and motor combined

You may need to oversize the drive because this is a constant torque load, potentially requiring full current output from the drive across the speed range. Check with your drive supplier.
 

Similar Topics

I have a virtual master that drives all servos. How do I make Virtual Master or any Servo to follow regular encoder which is driven by Regular...
Replies
5
Views
3,150
I have an HMI 2711R - T4T Series B, and I want to know which PLCs, besides Micro 820, can communicate with it.
Replies
2
Views
55
HI i would like to know how to get a variable that will store the amount of times a program has been executed. The issue is I have 3 DBs for 1 FB...
Replies
2
Views
61
I'm working with a project that contains some routines in ST but mostly in ladder. Am I correct in assuming this 'rung': DB1001DO._01AV55_OPEN :=...
Replies
4
Views
97
Is there a way to reset the count on the RS Logix BackUp?? XXXXX PROGRAM IN PROGRESS_BAK445.RSS XXXXX PROGRAM IN PROGRESS_BAK446.RSS XXXXX...
Replies
8
Views
249
Back
Top Bottom