Figuring out gear reduction.
 

If I wanted to run a 26" or 27" tall tire on a gear drive how would I go about figuring out what gear reduction I would need to keep the mph the same or just slightly slow than stock?

Convert the # of revolutions for the drive gear and tire to distances by measuring the circumference of each at one rev of the tire, take the differance in circumference of the two tires and reduce the drive gear size by that differance divided by the # of revs the gear has to turn in order to turn the tire once.

tough to explain without confusing myself. LOL

What sort of cool and interesting project are you debating??

I'm still in the "that would be cool" stage of planning, if it ever turns into a reality I will start a thread.

Lol... ya we all have those eh.

If it matters at all... changing from a 23" osd tire to a 26" osd tire doesnt make a whole lot of differance as far as ground speed goes, its sort of noticable in the higher gears but not so much in the lower gears. If you were building a puller it is probably worth worrying about... but not for a basic yard machine. Ive put 26" ATV tires to replace the 23" turfs on all my personal toys/tractors.... top speed is maybe 1mph faster.

On my pullers, the one with 16 tooth second with 26" tires and the 19 tooth with 23" tires were very close to the same. It would be the cheapest way to go with taller tires for sure.

I would like to do a 16,17,19 tooth plowing tranny using all stock gears.


Here is a calculator from Midwest super cub.
https://gears.mwsc.co/

Randy

I guess my idea might turn into reality, I just scored a brand new pair of 26x12-12 Tru Powers off eBay for $153 and free shipping.:bigeyes: