MERM Practice Problems 52-5 need help!

[MechGuy]

I don't have any experience in machine design so I'm having a lot of difficulty with this problem.

The question asks "A gear train is to have a speed reduction of 600:1. The gears have no fewer than 12 teeth and no more than 96 teeth. a) how many stages are needed? B) how many teeth should be in each gear train?"

My main problem is that I can't seem to find any equation for the number of stages needed. The solution suggests that the speed reduction (600) = gear ratio^number of stages.

Does anyone know where this equation comes from? I've looked in MERM, Marks, Shigley, Machinary Handbook... I just can't seem to find it.

for B), the solution calculates the gear ratio for each stage = speed reduction^ recipricol of # of stages. Again, no idea where this comes from.

Then the solution seems to randomly choose the number of teeth per gear with no reasoning.

If any of you mechanicals with gearing experience can help me, I'd most appreciate it!

Dave

 

[jmbeck]

Do you have Shigley's "Theory of Machines and Mechanisms"?

 

[ODB_PE]

While not a ME I often look at these questions to see if I would have a clue. I like gear ratios so this one caught my interest. Knowing this, please oblige the potential for nonsensical reasoning on my part.

Since the limits of the gearing permit a maximum reduction of 8, it seems like you could get the required reduction in 4 stages.

Then, I took 600^(.25) to find the ballpark ratio needed (4.95:1) but I have no idea how to proceed from there - but it seems like it would be easy to check if given 4 or 5 choices. Or do they just give you total number of teeth? In my head it seems like you could do 8:1 * 5:1 * 5:1 * 3:1 using 12 teeth on the small gear would give 96+12 + 60+12 + 60+12 + 36+12 or 300 teeth (half of 600 - coincidence?)

Also, I don't know if there is a certain efficiency you're looking for (e.g. if you need a 5:1 ratio is 60:12 better than 75:15, etc - or is that just a question of space/material constraints?

I'll crawl back under my rock now......

 

[MechGuy]

Do you have Shigley's "Theory of Machines and Mechanisms"?

No, I was referring to Shigley's Standard Handbook of Machine Design. I don't have the Theory of Machines and Mechanism's book.

 

[MEPE2B]

Machine design is not my strongest area either, but I was looking just out of curiosity.

This stuff is covered in the MERM under "Dynamics and Vibrations", not "Machine Design". Look on pages 57-4, 57-5. What is written there is fairly sparse and does not seem to directly answer your question. However, reading through it quickly, I think you could find enough there to solve this problem with some thought.

It is mentioned in MERM that "Finding the number of teeth that each gear should have in order to achieve a particular train ratio is time consuming, as a trial and error solution may be needed." I don't have the whole problem or solution to look at, but it seems like the solution you mentioned may have been assuming equal stages as a beginning point, factoring the velocity ratio, Nout/Nin into its stages. That would be 600=Nout/Nin=N4/N1=(N4/N3)(N3/N2)(N2/N1), if there are 3 stages. If each stage has an equal ratio, then it becomes (N2/N1)^3 for 3 stages, or (N2/N1)^number of stages. I am guessing that the final selection for number of teeth seems to pop out of nowhere in the solution because it is found trial and error.

I'm sure a dynamics of machinery text book, such as the Shigley one mentioned, would give some detailed sample problems similar to this.

 

[buick455]

The answer was in front of you all along and you just did not see it but this happens to me as well. Every book you will pick up will have some version of the same formula. Some use different symbols and some use different substitutions which can get confusing. In some cases they just make up their own formula. In any event to try and answer your question:

The equation (gear ratio to the 3rd power) is nothing more than taking the standard ratio equation, i.e. (N-input/N-output) x (N-input/N-output) ...) in this case 96/12 = 8 for one gear set or a reduction of 1 to 8. Now to get a reduction of 1 to 600 simply multiply 1/8 x 1/8 x 1/8 and so on until you get 1/600. So knowing 8 to the power of 3 is less than 600 and 8 to the power of 4 is more than 600 you know the number of gear sets using the ratio above is >3 but <4 with the maximum ratio given but 4 gear sets will work but the ratios will have to be less than 1-8.

Now you can simply take 600 to the 1/4 power to determine what the ratio of each set must be. The answer is 4.95. You now plug this into the gear ratio formula for one set using the minimum number of teeth given which is 12. So (12/N) = (1/4.95), or N=59.4 or 60 teeth. Now multiplying (12/60)x(12/60)x(12/60)x(VR)=(1/600) and solving for the last ratio (VR) = .208. From there you can simple plug in a few gear combinations like 15/72 which gives you a ratio of .208. You now have (12/60)x(12/60)x(12/60)x(15/72)= (1/600) or a ratio of 1 to 600.

To be honest I believe the answer in the book is incorrect. If you use a planetary gear set with the number gear teeth given you can get VR=9 for one set therefore (1/9)x(1/9)x(1/9) = (1/729) so a 1 to 600 ratio can be achieved with three gear "sets" and not four. 10940623:

 

[MechGuy]

The answer was in front of you all along and you just did not see it but this happens to me as well. Every book you will pick up will have some version of the same formula. Some use different symbols and some use different substitutions which can get confusing. In some cases they just make up their own formula. In any event to try and answer your question:
The equation (gear ratio to the 3rd power) is nothing more than taking the standard ratio equation, i.e. (N-input/N-output) x (N-input/N-output) ...) in this case 96/12 = 8 for one gear set or a reduction of 1 to 8. Now to get a reduction of 1 to 600 simply multiply 1/8 x 1/8 x 1/8 and so on until you get 1/600. So knowing 8 to the power of 3 is less than 600 and 8 to the power of 4 is more than 600 you know the number of gear sets using the ratio above is >3 but <4 with the maximum ratio given but 4 gear sets will work but the ratios will have to be less than 1-8.

Now you can simply take 600 to the 1/4 power to determine what the ratio of each set must be. The answer is 4.95. You now plug this into the gear ratio formula for one set using the minimum number of teeth given which is 12. So (12/N) = (1/4.95), or N=59.4 or 60 teeth. Now multiplying (12/60)x(12/60)x(12/60)x(VR)=(1/600) and solving for the last ratio (VR) = .208. From there you can simple plug in a few gear combinations like 15/72 which gives you a ratio of .208. You now have (12/60)x(12/60)x(12/60)x(15/72)= (1/600) or a ratio of 1 to 600.

To be honest I believe the answer in the book is incorrect. If you use a planetary gear set with the number gear teeth given you can get VR=9 for one set therefore (1/9)x(1/9)x(1/9) = (1/729) so a 1 to 600 ratio can be achieved with three gear "sets" and not four. 10940623:

Thanks for all your help! You're right, it was in front of me the whole time... I guess I was looking for an equation to be laid out in front of me and I just didnt think about it hard enough. I think I'm happy I'm taking the HVAC depth and not the MD depth