Okay long break between now and my last post. I'm trying to set up a jackshaft completely made from parts I need to buy. Problem is, I don't know what I'm doing. Nor do I know what to buy. I have a 10 tooth clutch and a 44 tooth on the bike as of now. What do you all recommend I should get and where should I get them
I proceeded to remove the governor from my 212cc predator engine. Just need lots of help now.
Ok, to understand gear ratios and jackshafts, let go back to high school science. We learned about mechanical advantage (MA), using levers, screws, pulleys and gears. Focus on pulleys first. A 1" (drive) pulley belted to a 2"(driven) pulley gains an MA of 2:1. Conversely, with a 2" (drive)pulley belted to a 1"(driven)pulley, MA is lost, down to 1:2 0r .5:1.Like the bicycle's chainring drive sprocket and the driven wheel sprocket. (That's why the gears on the engine, jackshafts and wheel sprockets have to be sized correctly.)Gears(sprockets) and pulleys are leveraged the same.....by their CIRCUMFERENCE or number of gears along its circumference, not their diameters.
To determine the drive and the driven gear or pulley, the drive gear is where the power originates. The engine sprocket is ALWAYS the drive gear. It connects to the driven gear on the jackshaft. On the other end of that same jackshaft is another DRIVE pulley. It is chained to the DRIVEN pulley. The wheel sprockets are ALWAYS driven sprockets.
A single set of drive and driven sprockets is easy to calculate. a 6-tooth drive sprocket chained to a 72-tooth driven gear = 72/6 = 12:1. Remember, the drive gear is the DENOMINATOR of the equation.
The next example is with engine/jackshaft/wheel sprocket.
engine drive gear = 12t(tooth);
chained to a 36t on the jackshaft.
36/12 = 3:1.....
jackshaft drive gear = 9t;
wheel driven sprocket = 45t;
45/9 = 5:1.....
Now MULTIPLY both answers to get 3:1 x 5:1 = 15:1!
To complicate matters for a third scenario, run the engine thru a SECOND jackshaft, aka chainring to wheel sprocket.
Mechanical advantage is LOST when the drive gear(chainring) is larger than the wheel sprocket(driven).
chainring drive sprocket = 36t;
wheel driven sprocket = 18t;
18/36 = .5:1.....
Simply MULTIPLY all MA's to get final mechanical advantage.
3:1 x 5:1 x .5:1 = 7.5:1.....
There are oxymorons and misnomers in the issues of gear reduction. The higher the MA(number), the LOWER the gear reduction. The HIGHER the engine will rev at a LOWER mph, the SLOWER(but with stump-pulling, hill-climbing low-end torque) you'll travel. So, a 15:1 is considered LOW gearing, and an 11:1 is considered HIGH gear.
Ok, you have a 10t drive gear and a 44t driven sprocket.
That's 44/10 = 4.4:1
Say you need 13:1 MA. You already have 4.4:1
13/1 divided by 4.4:1 = 2.955:1, rounded off to 3:1
If you used your 10t engine gear and a 30t jackshaft sprocket, you get
30/10 = 3:1;
Place a 10t as your jackshaft drive sprocket.
You already have a 44t wheel sprocket.....
44/10 = 4.4:1
3:1 x 4.4:1 = 13.2:1
I hope this helps.
