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• # Compound planetary gear problem

Discussion in 'Calculations' started by Powell3943, May 14, 2021.

1. ### Powell3943Member

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Hi, I need a little help with this problem with compound planetary gears. I don't know how to apply the formulas for this specific type of placement. I need to know the final speed of the engine knowing the number of teeth z1 = -104 ring gear which is internal, z2,3,4 = 22 for small planets, z5,6,7 = 65 for large planets, z8 = 17 for the small sun, z9 = 95 for the big sun and z10 = 13 for spool gear. A weight of 12kg will be attached to the ring gear on one side and 2kg of counterweight on the other from a height of 2 m. I want to calculate the speed and torque that the engine will have if I want the weight to fall within 20 minutes from the 2m. And i also need the ratio that the system have. The modulus of the wheels is 0.8 and the pressure angle is 20. I also know that the spool from the z10 have diameter equal to 30 mm and the one from the motor have 15 mm. These are the data I know and don't really know how to apply it. Can someone help me?

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3. ### s.weinbergWell-Known MemberEngineeringClicks Expert

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You're missing a gear - after doing it a couple ways, I realized that it matters which.

There is likely a fixed gear. Probably a fixed Ring around your gears 4,5 and 6, but could be a fixed sun inside gears 2,3 and 4 (or maybe another option - who knows?)

Please clarify what is not shown. Thanks

4. ### Powell3943Member

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why is that? planets are fixed in a carrier but there are no more gears around them

5. ### s.weinbergWell-Known MemberEngineeringClicks Expert

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Imagine you held the output gear. You spin the input, the carrier will go around and around, and your output won't move.
Or hold your carrier. Now your output will be forced to move.
Or hold something else.

Your system is not kinematically complete. Without fixing something (or a more complex solution that fixes some part at a set speed, or links it to the input - but that's too complicated for this discussion) that isn't fixed, there is no set gear ratio

6. ### Powell3943Member

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O.K. I understand. In this problem, the carrier is fixed. The planets being forced to move after the inner wheel, the input in this case, and the output moves after them

7. ### Powell3943Member

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the planets stand in one place also thanks to the carrier. remaining stationary

8. ### s.weinbergWell-Known MemberEngineeringClicks Expert

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In that case, I wasted way too much time on this It becomes a simple gear train. It's not really a planetary.

Stage 1: Pulley, ratio is 30:15, same direction
Stage 2: Gears 10 and 9, ratio is 95:13, change direction
Stage 3: Gear 8 to 5/6/7, ratio is 65:17, change direction
Stage 4: Gear 2/3/4 to 1, ratio is 104:22, same direction

Total gear ratio is 264.17:1, same direction as the motor. Some torque loss will occur, but getting the exact value is way more complex than what can be done with what is presented here.

9. ### Powell3943Member

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sorry I wasted so much time on you because I didn't explain very well. I'm truly sorry. thank you very much for the answer and the time given. but can you tell me quickly how you came to these values? what formulas did you use? and if you have any idea how I can find out the speeds and torque for the motor I would be very grateful. what are the formulas for this system. if the string is attached to a 20-diameter spool, it is also attached to the 104-tooth inner ring. following what I said above with those weights

Last edited: May 14, 2021
10. ### s.weinbergWell-Known MemberEngineeringClicks Expert

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You didn't waste my time. It's fine.

For each stage, there are really only two gears. Multiple planets around a sun operate as a single gear, as far as determining the gear ratio. The ratio of each stage is the gear you start from divided by the gear you transfer to. Regular gears switch directions, and the pulley arrangement you have, or regular gears to ring gears and vice versa do not change direction.

Pulleys, the size ratio is 30:15, so that's the gear ratio. Your speed decreases by that much, and your torque increases by approximately the same.
Same applies for each set. To get the total, you multiply all stages by each other.

I'm not quite sure where your weights are in this case.

To calculate how far a weight will fall over a given time, you get one circumference of string every rotation of whatever it's attached to. Together with your speed and torque calculation, I think that gets you what you need.

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