Marcus,
The gearbox isn't using any of the power or horsepower so the speed it runs at doesn't matter. A gearbox does have an efficiency rating but that rating is constant though the speed range. You do have to be careful not to run the gearbox too fast or too slow. Running it too fast could explode the bearings. Running it to slow could require too much torque.
Calculating speed an torque through a drive train is pretty easy. Lets say you have a 14 HP motor. That HP rating should give you about 20.5 ft/lbs of torque and 3600 rpm at the drive shaft. More than likely you won't have the engine running full throttle, but we'll use those numbers anyways.
Speed and torque are inversely proportional. If the speed goes up the available torque goes down and visa-versa.
Now that we know that lets look at the snow thrower as an example. Starting with the PTO and the gear box pulleys. You should take the drive pulley inches and divide it by the driven pulley inches. I think the stock pulleys are the same size so there isn't any reduction. So lets use the 3 inch pulley mod as an example. Lets say the PTO pulley is 4.5 inches and the gearbox pulley is 3 inches. Then that is 4.5/3 or 1.5. You multiply your speed by that number and you divide your torque by that number. The input shaft is spinning at 3600*1.5 or 5400 rpm and has 20.5/1.5 or 13.6 ft/lbs of torque.
The same holds true with the sprocket teeth. If the drive is 18 teeth and the driven is 36 teeth the speed is multiplied by 18/36 or .5 and the torque is divided by .5 (I never counted the teeth on my thrower I am guessing at the number of teeth). Using the calculations above your auger would spin 5400*.5 or 2700 rpm and the torque would be 27.2 ft/lbs.
To answer your question. If you go from, lets say a 18 tooth on the drive sprocket, to a 20 tooth you would be decreasing your speed by 10% and increasing available torque by 10%.
If you give me your exact teeth and pulley sizes I can calculate it for you.
The gearbox isn't using any of the power or horsepower so the speed it runs at doesn't matter. A gearbox does have an efficiency rating but that rating is constant though the speed range. You do have to be careful not to run the gearbox too fast or too slow. Running it too fast could explode the bearings. Running it to slow could require too much torque.
Calculating speed an torque through a drive train is pretty easy. Lets say you have a 14 HP motor. That HP rating should give you about 20.5 ft/lbs of torque and 3600 rpm at the drive shaft. More than likely you won't have the engine running full throttle, but we'll use those numbers anyways.
Speed and torque are inversely proportional. If the speed goes up the available torque goes down and visa-versa.
Now that we know that lets look at the snow thrower as an example. Starting with the PTO and the gear box pulleys. You should take the drive pulley inches and divide it by the driven pulley inches. I think the stock pulleys are the same size so there isn't any reduction. So lets use the 3 inch pulley mod as an example. Lets say the PTO pulley is 4.5 inches and the gearbox pulley is 3 inches. Then that is 4.5/3 or 1.5. You multiply your speed by that number and you divide your torque by that number. The input shaft is spinning at 3600*1.5 or 5400 rpm and has 20.5/1.5 or 13.6 ft/lbs of torque.
The same holds true with the sprocket teeth. If the drive is 18 teeth and the driven is 36 teeth the speed is multiplied by 18/36 or .5 and the torque is divided by .5 (I never counted the teeth on my thrower I am guessing at the number of teeth). Using the calculations above your auger would spin 5400*.5 or 2700 rpm and the torque would be 27.2 ft/lbs.
To answer your question. If you go from, lets say a 18 tooth on the drive sprocket, to a 20 tooth you would be decreasing your speed by 10% and increasing available torque by 10%.
If you give me your exact teeth and pulley sizes I can calculate it for you.
