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Epicyclic Gear Verification/Measurement Failure, Needing Help Please?

cadkins
7-Bedrock

Epicyclic Gear Verification/Measurement Failure, Needing Help Please?

Hi,

I'm doing a standard planetary gear system analysis (pictured), however when verifying the results I find my planetary gear speed is coming up as 378 deg/sec when by hand calculations it should be 270 deg/sec.

However what I did notice while writing this is that the given answer of 378 deg/sec - 108 deg/sec (the mechanisms calculated arm rotation) = 270 deg/ sec. So I think my measurement is adding the rotation of the arm and the rotation of the planetary gear, otherwise that's a huge coincidence.

For the measurement I have chosen Type->Velocity and for the Motion Axis as one of the rotation axes that the pin joint connects the planetary gear.

So my question is; does anyone know how to measure the local rotational velocity of the planetary gear (assuming that it's not a coincidence)? The answer should be 270 deg/sec

Many Thanks,

Cameron

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I come up with 378 as well on the hand calculation.

Input to planet:ratio 3:2

Back rotation on the ring: 108 = 0.3*360

360 input to planet = 540 (360*1.5) - if stationary

cycles along the input is 0.7 (1 rotation - 0.3 back rotation)

540*.7=378

I'm no math wiz, but if I input this in my planetary system, I'm pretty sure that is what I get.

A quick test is to put a servo motor to each axis and confirm visually.

View solution in original post

4 REPLIES 4

I think you answered your own question.

The velocity of any of the 3 axes, (stationary, input and output) is measures at the sun gear's axis.

What you are seeing is the planetary axis which is over-driven in this arrangement.

Here is a good formula and calculator site for this.

http://www.torcbrain.de/uebersetzungsrechner-planetengetriebe/?lang=en

I come up with 378 as well on the hand calculation.

Input to planet:ratio 3:2

Back rotation on the ring: 108 = 0.3*360

360 input to planet = 540 (360*1.5) - if stationary

cycles along the input is 0.7 (1 rotation - 0.3 back rotation)

540*.7=378

I'm no math wiz, but if I input this in my planetary system, I'm pretty sure that is what I get.

A quick test is to put a servo motor to each axis and confirm visually.

Thank you.

See the 0.3 you have, is that from the website you linked in the previous message where the ratio= 1 + Nring/Nsun = 1 + 42/18 = 3.333

-> so 360/3.333 = 108

or is it to do with 12/42?

It was from the inverse of the 3.3333... 1/3.333333...Essentially the 108 was also given in the calculator. So the 108 (carrier) divided by 360 (sun) gave me the .3.

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