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I have built a little gearbox to reduce the RPM and increase the torque, that is composed of 8 gears that are all equal: they have 12 teeth "in the small side" that transfers the motion to the "big part" that has 36 teeth. So at the end I have (1st gear with only 12t, 2nd-7th 12(small) 36(big), 8th just 36t to receive the motion from the 7th gear.

Can anybody explain how can I calculate the gear ratio in this case?

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Image created in Inkscape by yours truly. reduction gearing

If I understand your description correctly, the diagram above should represent your question. Each "donut" is a pair of 12/36 gears joined to move simultaneously, while each donut drives the succeeding one with the ratios presented.

The text has been placed at the interface of each gear pair. The first circle drives the second one with a 1 to 3 reduction, while the center of the second gear drives the third. This is mechanically unsound, but suitable as a discussion diagram.

There are seven interfaces/reductions, which means your gear ratio is 1 to 3 raised to the 7th power. The final reduction is therefore 1 : 2187 If there is one more reduction, the ratio is 1 : 6561

If you did not have equal gears between each set, you would have to multiply the denominators of all the ratios together to calculate the final ratio.

From a mechanical standpoint, consider to create a planetary gear set. You can accomplish substantial reduction in a more compact design.

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