Ok the gbm is building another LMG (Lawn mower generator) this time he has a brand new 5 hp Honda horizontal shaft engine and a 3g 130 amp internally regulated alternature.
I have one of these alts installed in my truck so this is where i am getting my figures from. i know this alternature puts out damn near full 130 amps at idel and is putting out full output at 1500 rpm.
I am going to play it safe and go with 1750 rpm
the pulley sizes on the truck are 8 inch crank pulley and 2.5 inch alt pulley.
so using the formula driven/drive
2.5 / 8 = .3125
so 1750rpm / .3125 = 5600 rpm(alt speed)
so with my honda engine spinning at 3600 rpm and me wanting my alt spinning at 5600 rpm i use the formula driven / drive
5600 rpm / 3600 rpm = 1.55 to 1 ratio
so if i use a pulley on the honda engine having 30 teeth and a pulley on the alt having 20 teeth (#35 series chain) that should spin my alt around where i want it right?
hows my math?
-gbm-
I have one of these alts installed in my truck so this is where i am getting my figures from. i know this alternature puts out damn near full 130 amps at idel and is putting out full output at 1500 rpm.
I am going to play it safe and go with 1750 rpm
the pulley sizes on the truck are 8 inch crank pulley and 2.5 inch alt pulley.
so using the formula driven/drive
2.5 / 8 = .3125
so 1750rpm / .3125 = 5600 rpm(alt speed)
so with my honda engine spinning at 3600 rpm and me wanting my alt spinning at 5600 rpm i use the formula driven / drive
5600 rpm / 3600 rpm = 1.55 to 1 ratio
so if i use a pulley on the honda engine having 30 teeth and a pulley on the alt having 20 teeth (#35 series chain) that should spin my alt around where i want it right?
hows my math?
-gbm-