Gear Train Ratio Calculation
How do you calculate Gear Train Radios?
Thanks 
The gear ratio is calculated simply by dividing the number of teeth on one gear into the number of teeth on the other. You can continue this process along a train of many gears to find a ratio for the whole train.
Prudens, I don't mean to brush you off, but have you looked at howstuffworks.com? A lot of the questions you ask here are explained extremely well on this site. I'm not asking you not to post here, but the website really is very useful and can often explain things a lot better than we can (I know one or two professional engineers who use the site as a regular source of reference!). Here, they explain gear ratios from the most basic principles through to some fairly detailed examples. Good luck! 
ive seen that website, didn't help me a lot.

What didn't you understand? Let us know and we'll try and help you out!

I understand that between two gears, the ratio is simply gearA:gearB,
but in the case of say 3 gears, or 4, or 5... how is it? 
PrudensOptimus its the multiplication of the ratios. To make math easy the ratios are typically expressed as a relationship to 1 because otherwise it is like fraction math and the back of the classroom is moaning already.
For example. My car has a first gear of 2.66:1 and the rear end differential has a ratio of 3.42:1. Since they are both referenced to 1 all I need to do is multiply the first numbers and get an overall ratio of 9.09:1 The key is that each gear is driven only by the previous one in the chain so the effect changes with each one in succession. Think of you and a row of friends in a movie theater. You take 1/4 the popcorn in the bag and pass it left. The next friend takes 1/4 of the popcorn left in the bag and passes it to his left, and so on. As you move left the people are getting less and less popcorn because the effect is cascading. 
so a 8t driving 40t then followed by 8t then 40t, 8t, 40t
produces 1:5 * 1:5 * 1:5, which produces 1:125, which produces a lot of torque at the cost of speed? on the other hand, a 40t driving, 8t, 40t, 8t, 40t produces 5:1 * 5:1 * 5:1, which is 125:1, produces a lot of speed at the cost of torque? 
btw is there worm gear trains?

Yes, your figures and assumptions about torque/speed are correct.
If you took apart a servo from an RC car you would find something similar to your first example. Most gear ratios are typically written x:1 with the 1 at the end. In laymen's terminology this would be gearup or geardown and is referring to RPM. So 125:1 gear down means very slow but high torque, 125:1 gearup means very fast but very little torque. Since most times its gear down since rotations are easier to produce than torque, it is assumed if not mentioned (like my transmission example). Gearing up for speed is typically called overdrive, like cars that have a 5speed overdrive transmission where 5th gear is going to have an output that turns faster than the input. Here is where it gets backwards with lay terminology, low gear (like first gear) has a high ratio and high gear (overdrive) has a low ratio. You could make any gear into a drivetrain or multiple gear reductions. The issue with worm gears is they require a 90 degree shift in direction (less efficient) so its hard to chain them together whereas regular spur gears all operate in one plane so its easy to drill a few holes in two plates and run the gears on axles inserted into the holes in the plates. Here, look at this PDF and you see the worm and the worm gear. They are seperated a little bit but you can see how the direction of rotation shifts. You can also see how the small worm can spin the large worm gear, and also how the large worm would not be able to spin the small worm. Its more like a thread of a bolt or more appropriately a wedge. This could be an advantage in a robot arm because the motor with the worm could be mounted where your bicep is and then the worm gear attached to the forearm section to move it up/down. https://sdpsi.com/ss/PDF/79001211.PDF 
Is it better to use 125:1 gear up for each of the two back motors driving a small lightweight lego car, or is it better to use 1:125 for better torque, or neither?
Is it relative easier to move a car in motion than to lift something up from the ground? 
Neither, one, or the other. Using a huge ratio like that could be like asking if you should use a 400lb hammer to pound in a simple nail or instead use a 40ft long hammer to pound in the same nail. No definitive answer.
With enough gearing you can move anything (the popular quote about a lever long enough/moving mountains) and then power is the question about how fast you can move it. Like I said, typically things are geared down to increase torque because most motors are more efficient at making power with RPMs. But like anything, there are exceptions but its an educated choice to do so. Its just going to be far more rare to speed things up, and especially by a large ratio. Rolling resistance would generally be an easier obstacle to overcome but there are ingenious solutions like discussed in your other threads of using counterweights to make lifting very easy too. Elevators are designed with counterweights so if its got a 2000lb limit and the car weighs 5000lb then its probably got a 6000lb counterweight. This way if the elevator is unloaded the most the motor has to lift is 1000lb (counterweight is heavier) or fully loaded the motor needs to lift 1000lb (the elevator is heavier) but it works on the difference. If the elevator is loaded to half capacity, the motor lifts no weight but then just accelerates the mass. 
Quote:
USE THE FOLLOWING EQUATION ND=nd where N= no of teeths on gear 1 D=dia of gear 1 n=no of teeths on gear 2 d=diameter of gear 2 hope you have got the consept or else you can mail me for furthur detalis 
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