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Can someone please define gear ratio before I buy

I am looking at the gearboxes i.e Tamiya 70103 universal gearbox
When I see this number >> 719:1 I assume that’s 719 turns of the motor vs 1 turn of the drive-shaft.

What I really need is the opposite … which is that the motor gear turns a very small number of times per rev of
the drive shaft. I want to be able to monitor the number of revs of the drive shaft (the things with the wheels on them)
and translate a hundred turns into a much small number of turns on the drive shaft (motor) … which will not have a motor on it.

This is for a science fair and the goal here is to count off revs and take that information (mechanically only)
and trigger the brakes for the vehicle when it has traveled a specific distance … like say 10 meters.
I was hoping to have the monitor gear encounter a mechanical stop at a specific distance in its turn … thus braking the vehicle.

The contest requires no electronics (such as an arduino) so the control system must be purely mechanical.
Also no motor … its a gravity vehicle …
All ideas from Mech-Heads greatly appreciated.

Unless a worm gear is involved, it doesn’t matter which end of a gearbox is considered the input, and which end is the output. Just pick a gearbox with the right ratio, and use it.

One thing to watch is how much force and friction comes into play.

A ratio of 1:719 sounds like you only have to turn the input shaft once to get 719 rotations of the output shaft and that is true. BUT, I think you will find it takes considerable INPUT force at that ratio to move a vehicle.

Remember EVERYTHING is amplified/modified by the ratio of the gears. Force IN/OUT in particular becomes possibly a problem.

In general you can “sort of” ignore friction but as the gear ratio gets higher friction is also multiplied by the ratio.

If you are using the 1:719 ratio gearbox then one pound of force IN = 1 / 719 pounds of force OUT. One pound is roughly 453 grams. At that ratio one pound IN = 453 grams / 719 = 0.63 grams of output force. That may not be enough to move your vehicle.

Of course the leverage also comes into play. If the input shaft has a pulley where the one pound weight is connected to a string and wrapped around a 6 inch pulley then the force IN is MULTIPLIED by the radius of the pulley BUT you have to travel farther so your weight will have to move a much longer distance. The circumference of a circle = 3.14 X Radius Squared. So 3.14 * (3 inches squared) = 3.14 X 9 = 28 inches of linear travel for the one pound weight to achieve one rotation of the input shaft if the “string” attached to the weight is wrapped around a 6 inch pulley. This might allow you to use an input weight of 4-5 pounds if you can construct a vehicle that will allow the weight to drop that far. You also have to remember you could bend the shaft of the gearbox if you use too much weight and it “hangs” directly on the shaft.

You have several approaches. One is find the right ratio and “dangle” your motive power weight at the end of “string” wrapped around the input axle/pulley. If your car needs to move 10 meters calculate the number of wheel rotations and translate that into how far the weight must drop to get that many wheel rotations.

One other options is use a spring in place of a weight. Long (12-14 inch) old style screen door springs come to mind. Not huge force (probably several pounds) but a 6-12 inch (or maybe more) stretch distance to provide that force.

The other option is create a vehicle with a drive mechanism that is “free wheeling” when the input force limit is reached. When the weight drops to it’s fullest travel or the spring recoils to it’s shortest length then the vehicle would then “coast”. This design would have to launch your vehicle to a fast speed quickly (without spinning wheels), dis-engage the drive mechanism to the wheels, and momentum would allow it to coast until the friction in the wheels brought it to a stop.

My “gut” tells me that “coasting” vehicle might work the best but your drive mechanism would be more complex. If you use this approach then reducing friction of the wheels and axles of the freewheeling vehicle is VERY important.

The other approach with a “coaster” would be a spring that wrapped around the axle. You would wind the wheels backwards and then place the vehicle at the starting point. When released the spring would return the energy to the wheels and, if designed as freewheeling, it could coast after the all of the spring rotations were reached. Lots of little toy cars, sometimes called pullback cars (search on the Internet), work with this type of mechanism. Pull it backward about 6 inches and it will travel forward many dozen feet.

This should be enough to get your “brain spinning”.

I would suggest using a mechanical counter instead.
Or maybe even a spring-based timer, rather than relying on very large gear ratios.