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

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!