Number of gear teeth in two-stage reduction gears

In summary, a two stage reduction gearbox is possible to have more teeth on the gear pair (helical-parallel axis) of the 2nd stage. If everything is matched to the load and speed requirements, this is permissible.
  • #1
cabellos6
31
0
In a two stage reduction gearbox is it possible to have more teeth on the gear pair (helical-parallel axis) of the 2nd stage? I ran through a specification for gear sizes and I am wondering if this is ok.
 
Engineering news on Phys.org
  • #2
I can't think of any reason not, as long as everything is matched to the load and speed requirements. Better wait for a more informed opinion, though.
 
  • #3
i might not be understanding your point correctly, but here goes. In 2 stage units, since the distance between the two shafts(assuming the output shaft is in line with the input shaft, & i mean its distance from the counter shaft) is to be minimized, optimum gear ratio for each stage comes out to be the geometric mean of the required gear ratio for the gear box, ie same gear pair ll be used(obviously the pair carrying the higher torque will be thicker). So the number of teeth on both should be equal.

Even if the output shaft isn't in line with input shaft, the same thing should hold.
 
  • #4
The assumption that the input and output shafts are aligned (a reverted train) is an unnecessary design hindrance that can drive the cost up and is not justified except in very special cases where the layout requires this. It is rarely done as a matter of routine.

It is certainly permissible to have a different ratio in one pair versus the other pair; this is the norm. It is important to assure that neither pair involves a pinion with less than the minimum number of teeth to avoid undercutting.
 
  • #5
Dr.D said:
The assumption that the input and output shafts are aligned (a reverted train) is an unnecessary design hindrance that can drive the cost up and is not justified except in very special cases where the layout requires this. It is rarely done as a matter of routine.

true. but then the whole designs based on constraints which can vary very much from application to application. since the op didnt specify any constraints, i only assumed a set of constraints.
probably a not so useful assumption, it only messed up in the end.
It is certainly permissible to have a different ratio in one pair versus the other pair; this is the norm. It is important to assure that neither pair involves a pinion with less than the minimum number of teeth to avoid undercutting.

true again, but the size should be minimized anyways(weight, inertia, space etc). surely undercutting is the deciding factor for minimum number of teeth.
 
  • #6
While it is never good practice to grossly oversize, it is also rarely worthwhile to truly minimize in the strictly mathematical sense. Even in aerospace work, where weight is at a premium, this is almost never done. Instead, people use good sense and use the smallest, lightest parts that are practical without seeking a true minimum.
 
  • #7
how should we go about finding the true minimum Dr.D?(i am not being sarcastic)
 
  • #8
Because of the great number of variables involved (strength, wear, dynamic response requirements, etc.) just about the only way that I know of to find the true minimum is to define all of these criteria, and then make an exhaustive computer search while maintaining each of these at the minimum acceptable level. This is a very costly - and rarely cost effective - thing to do. I have never done it, and I do not recommend it.
 
  • #9
ank_gl said:
how should we go about finding the true minimum Dr.D?(i am not being sarcastic)
In "the real world" you are given a basic set of criteria (variations are a certainty):

- weight
- $/part
- performance

You and your team/management decide what goes into the requirements of a successful design. If you can meet all of your requirements, you are done. You may go back and revist a design in the future for any more refinements that can be made (usually weight savings).
 
  • #10
Fred's answer is exactly why I said that I do not recommend this minimization at all.
 
  • #11
While it is never good practice to grossly oversize, it is also rarely worthwhile to truly minimize in the strictly mathematical sense. Even in aerospace work, where weight is at a premium, this is almost never done. Instead, people use good sense and use the smallest, lightest parts that are practical without seeking a true minimum.

:eek:duh Dr.D., I misread your post, i thought you were suggesting the true minimum based upon optimization of all the constraints.

:redface:need to read carefully
 
  • #12
the number of teeth should increase since the 2nd gear is bigger

the relation is d=[tex]\frac{N}{p}[/tex] where
d is the diameter of the gear
N is the number of teeth
P is the diametral pitch ( number of teeth per unit length )

the P for all your gears should be the same ( other wise it won't work) and you should have no problem finding the number of teeth

of course you should have already decided the gears properties i.e ( material, width, tooth type & number of teeth in your smaller gear )
all of thees things can be determined by revising the catalogs of the gear manufacturer, based on load, speed of rotation and temperatures
 

1. What is the purpose of having two stages in gear reduction?

The purpose of having two stages in gear reduction is to increase the overall gear ratio, which allows for a larger speed reduction and greater torque multiplication. This is especially useful in applications where a high amount of torque is needed, such as in heavy machinery or vehicles.

2. How is the number of gear teeth calculated in a two-stage reduction gear system?

The number of gear teeth in a two-stage reduction gear system is calculated by multiplying the number of teeth in each gear of the first stage by the number of teeth in each gear of the second stage. For example, if the first stage has a gear with 20 teeth and a gear with 30 teeth, and the second stage has a gear with 25 teeth and a gear with 40 teeth, the total number of gear teeth would be (20x30) + (25x40) = 900.

3. What factors determine the number of gear teeth in a two-stage reduction gear system?

The number of gear teeth in a two-stage reduction gear system is determined by the desired gear ratio, the size and strength of the gears, and the available space for the gears within the system. Other factors may include the type of application and the desired torque and speed output.

4. What is the relationship between gear teeth and gear ratio in a two-stage reduction gear?

The gear ratio in a two-stage reduction gear is directly related to the number of gear teeth. As the number of teeth in the driven gear increases, the gear ratio also increases. This means that a higher number of gear teeth in the two-stage reduction gear will result in a larger speed reduction and greater torque multiplication.

5. How does the number of gear teeth affect the overall efficiency of a two-stage reduction gear system?

The number of gear teeth in a two-stage reduction gear system can significantly impact the overall efficiency of the system. A higher number of gear teeth can lead to more friction and energy loss, resulting in a lower efficiency. Therefore, it is important to carefully consider the number of gear teeth and their size and placement to optimize the efficiency of the system.

Similar threads

Replies
7
Views
1K
Replies
2
Views
1K
Replies
12
Views
2K
  • Mechanical Engineering
Replies
2
Views
2K
Replies
1
Views
4K
Replies
3
Views
935
  • Mechanical Engineering
Replies
23
Views
2K
  • Mechanical Engineering
Replies
9
Views
2K
  • Mechanical Engineering
Replies
15
Views
4K
  • Mechanical Engineering
Replies
2
Views
2K
Back
Top