How can I calculate final gear ratio when the drive train contains multiple gears with some reductions on the same shaft?

Question

I am trying to calculate some rather long and complex drive trains, this is a project that I have worked up to, but its the first problem of its kind I have come across, to try and explain the position I am in please follow my example of how I am trying to calculate the final ratios and if I am going wrong, please can you point out where I have gone wrong and if you have the time, why.

Example: In the image below is a simple drive train, it consists of three shafts with four gears turning.( I know it doesn't look it, but please assume all teeth are of the correct size and mesh with no problems)

• Shaft 1
• • Input Gear 1 (Purple) has 24 teeth
• Shaft 2
• • Driven 2 (Red) has 12 teeth
• • Driven 3 (Blue) has 15 teeth
• Shaft 3
• • Output Gear(Green) has 10 teeth.

So the power travels through Purple,Red, Blue Green.

Now for my Math.

though a 2:1 is connected to a 3:2 I have been calculating the ratio like this, but I am sometimes getting different answers with long (up 12 gears) trains in my project.

I would calculate the final ratio like this:

(12/24)x(15/12)x(10/15) = 0.416(RECUR.)

This is where I am going wrong Is the Final ratio 100:416 (1:0.416) or is it 1/0.416 = 1.666 = ~16:10 or is it something else?

Any help is greatly appreciated, im stuck at the moment and need to get on!

Answer 1

According to the Woodgears web site, one takes the gear ratios independently and "adds them together" as shown in the video within the linked site.

In your example, the progression is as follows:

24:12, 15:10 which reduces to 2:1 and 3:2

Two (first ratio) times one and one-half (second ratio) is 3, which provides for a 3:1 increase or a 1:3 reduction if the reverse is created. There is no math between the red and blue gears as one rotation of the red is one rotation of the blue.