Any math guys here?

I have a cement mixer and pumper combo basically its a pool plaster machine for installing quartz aggregate. They want $4000 for a new transmission for the little 22HP drive system (not the vehicle that transport whats on back which is mixer pumper which has it's own transmission and engine) I can buy a fucking $4000 transmission for a car. But these specialized parts on this equipment are outrageous.

Anyway the mixer engine 22HP has a 1 1/4" shaft going into transmission - transmission only consists of a 6 inch gear turning on the 1 1/4" shaft coming out of engine which then goes to a 3" 3 belt pully.

Can I bypass tranmission and just insall a 1" belt pully?



Does math work out to same ratio bypaassing?

I could mill that shaft down and put a 1" pully with a woodrift key in 15 min if close. Actually i'd just mill the 3 1" V's right into shaft


I';ll probably have to bite bullet and just buy this overpriced part but still curious about proper ratio conversions.

More of an engineering question than math.

Why? You have 3 circles turing one aanother

a 1.25" to 6" to 3"

If you eliminate the middle man (transission 6")

What ratio does 1.25 to X have to be to maintain same rotation speeds at end.

Seems like one of those math question I got in HS. Which I got a D in geometry. Not proud of it just sayin not good at it.

if it was in metric i would help you. Joke aside(or not) with turns and length this is kinda easy. Your new belt(as the last thing interacting with the shaft) must be at the same speed as the old belt. Widdling down the shaft accelerates the system, the exact opposite you probably need... Without the transmission you need to loose speed at the belt.

Is the 3 in pulley welded / bolted to the 6in gear, so that 1 turn of the 6in gear is 1 turn of the 3in pulley?

If so every 4.8 turns of the 1 1/4 in drive will turn your 3in pully drive once.

Youd need to figure out a 4.8:1 pully reduction system.

Cheers,
Matt.