2
$\begingroup$

I saw this picture today:

enter image description here

They give an ratio of 5/352. I think this is not possible with a simple gear. Maybe the upper sides ratio is the cause for it.

Which formula do I need to calculate the ratio for first half and which one for the upper half?

I want to use it for a stepper motor (nema 17) as a gear box to increase the torque.

$\endgroup$
10
  • 3
    $\begingroup$ I'm voting to close this question as off-topic because this has nothing to do with Electrical Engineering. $\endgroup$
    – brhans
    Jan 4 '18 at 12:01
  • $\begingroup$ Hi @user8886193. This is unfortunately not an EE.SE related question, suggesting that is is migrated to engineering.SE $\endgroup$
    – MrGerber
    Jan 4 '18 at 12:01
  • $\begingroup$ Why off-topic? Stepper motors and gears are part of electrical engineering. Do you never calculate gears for steppers? $\endgroup$ Jan 4 '18 at 12:04
  • 1
    $\begingroup$ Once the motor starts turning it's mechanical engineering not electrical engineering! $\endgroup$
    – JeffUK
    Jan 4 '18 at 12:09
  • 1
    $\begingroup$ Have a watch of this video (sometimes the old ones are still the best!), and let us know if you are still having trouble understanding the system. I can't provide you with "the formula you need to calculate the ratio", since the ratio depends on which of the three components you hold still... youtube.com/watch?v=SrkzaQRDtuM $\endgroup$ Jan 4 '18 at 15:31
1
$\begingroup$

Just count the cogs teeth on both sides that is your ratio. Keep dividing by 2. You can cross check it by counting the teeth on the housing and dividing it by the number of teeth on the cog or spindle.

$\endgroup$
5
  • $\begingroup$ So which gear are you holding stationary? $\endgroup$
    – Solar Mike
    Jul 3 '18 at 20:42
  • $\begingroup$ @SolarMike the white housing stays stationary the rest moves $\endgroup$
    – user4139
    Jul 3 '18 at 21:03
  • $\begingroup$ @SolarMike can I get a up vote??? $\endgroup$
    – user4139
    Jul 3 '18 at 21:04
  • $\begingroup$ do you think you have the correct result? $\endgroup$
    – Solar Mike
    Jul 3 '18 at 21:07
  • $\begingroup$ @SolarMike the more I look at the picture this is the reverse drive and there is no gear ratio its 1.1 . $\endgroup$
    – user4139
    Jul 3 '18 at 21:10
0
$\begingroup$

There are simple steps to solve this.

  1. Count the number of sun gear teeth and the number of ring gear teeth if the planetary gear is set in underdrive with the carrier as the output. Add the two numbers together.
  2. Divide the resulting number by the number of teeth on the driving member for the gear in underdrive when the carrier is the output. The total is the gear ratio for that gear setting.
  3. Divide the number of teeth on the driving member by the sum of the sun gear and ring gear teeth on the planetary gear for a system set in overdrive with the carrier as the input. The resulting number is the gear ratio for that transmission setting.
  4. Count the number of teeth on the driven gear and the number of teeth on the driving gear if the planetary gear is set in underdrive with the carrier held.
  5. Divide the number of driven gear teeth by the number of driving gear teeth to obtain the gear ratio for this system setting.

    Read More : https://woodgears.ca/gear/planetary.html

$\endgroup$
3
  • $\begingroup$ I tried your steps. The gear in the picture consists of two ratios. The first one is between sun (input) and carrier (output) (step 1 and 2). The second one is between carrier (input) and ring (output). I can not find a suitable step for the second one. Which one? $\endgroup$ Jan 4 '18 at 17:47
  • $\begingroup$ Did you read the info via the link? $\endgroup$
    – Solar Mike
    Jan 4 '18 at 19:19
  • $\begingroup$ @Solar Mike I read the site. If I use the formula i do not get the number of 5/352. I do not know much about the planetary gears but this ratio from the picture looks for me not correct. $\endgroup$ Jan 4 '18 at 20:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.