Need some advice Please

[SacramentoNathan]

Hello Everyone, I am new here and hoping to get some advice on a couple of topics:

I bought a custom heavy duty go cart that had an arctic cat 440 motor in it that was not running. I am replacing the engine with a predator 420cc from Harbor Freight. I cut out the old engine mounts and am installing new cross bars with an engine mounting plate (see pic below(



I am using a Series 40 torque convertor and plan on connecting the #40 chain 10T sprocket from the torque convertor to the jack shaft that the original snow mobile torque convertor was connected to. The jackshaft has an end with multiple small groves (see pic below and ignore the sprocket currently sitting there) It looks like a motorcycle front sprocket may fit onto it and my first question is are motorcycle front sprockets universal in terms of the groove pattern that slides over the axle shaft? Also do they sell front motorcycle sprockets that are a #40 chain size that will match the 10T sprocket on the series 40 torque convertor?

My next questions is about gearing as this cart has the following sprockets/gears:

10T on the series 40 Torque convertor that goes to the Jack shaft
I am not sure how many teeth to put on the Jackshaft axle, maybe 13T (this is a variable as I have not bought one yet)
The jackshaft has a sprocket on the other end with 11 teeth that feeds into the Gear Box (not in the picture as I removed the jackshaft)
The Gear box has a receiving sprocket with 21 teeth on it and the output sprocket from the gear box to the drive axle as 11 teeth on it (does this mean that I can ignore the 21 tooth sprocket in my calculations as the Jackshaft has an 11 tooth sprocket going to the gear box and an 11 tooth sprocket coming out of the gear box?See pic below)


Finally the Rear Axle Sprocket has 41 teeth:


So my big question is how do I calculate my gear ratio? The Arctic cat engine had about 50 HP and revved around 6,000 RPM. The predator motor has 13 hp, maybe a little more if I bypass the governor, but will only rev around 3,600 - 4,000 so I am pretty sure I am going to need to change my gear ratio. As mentioned above, can I ignore the 21 tooth sprocket at the gear box in my calculations?

If so I found this calculator that I can use: https://www.gokartguide.com/gear-ratio-chart-speed-calculator/
I assume that the input jackshaft input sprocket is the one coming from the Torque convertor and the jackshaft output sprocket is going to the wheels, is this correct?

Also any advice on where to buy sprockets and chains would be appreciated,

Thank you and Have A Great Day!

Nathan



Any advice that anyone can give regarding what the formula is would be very helpful, or even as simple as what is the impact of changing the receiving jackshaft axle sprocket larger or smaller.

 

[karl]

I would start out by figuring out what the kart is, and what was the factory motor.

Looks like some prior welding / mods were done to the frame.

Just because it had the sled engine bolted to it does not mean it worked right.

Then if nothing else, break out the sharpie, spin the input and figure out what the gear reduction
if any, those two gearboxes have. Then we can work on calculating a final ratio

 

[ezcome-ezgo]

What Karl said. Reconnect everything you have and see how many rotations of the splined shaft you get from turning the wheel one rotation.

 

[SacramentoNathan]

Thank you everyone for your input. I am trying to find a sprocket for the jackshaft, so as soon as I do that I will reconnect everything and if I understand correctly, I should rotate the tires 1 full rotation and count how many rotations the jackshaft rotates. Is that correct. Once I have that number, what do I do with it? Is that the gear ratio? For example if I rotate the tire one full rotation and the jackshaft rotates 5 times, does that give me a 5:1 rotation?

Thank you again for all of your help. I will report out once I have made some progress.

 

[redflash]

Is that a gear reduction box the chain is hanging off of ? factor thT IN....

 

[madprofessor]

rotate the tire one full rotation and the jackshaft rotates 5 times, does that give me a 5:1 rotation?

Yes it does, but what you really need to know is how many engine revolutions it takes to turn the rear axle one time, ignoring any gearing in between. If your engine's output shaft is directly coupled to its crankshaft inside (is literally the end of the crankshaft, one piece), then every rotation of the output sprocket equals a rotation of the engine itself. If there's an internal gearbox, let's just say 2:1 ratio, you have to figure that into determining your engine rpm's. The direct engine rpm's is what you need for determining your engine/axle rotation gear ratio, not what it is after being changed by an internal gearbox.
With a jackshaft being fed from the engine output sprocket, as opposed to a direct chain to the rear axle, you then have to figure the ratio of engine revolutions/jackshaft revolutions. With that, you then have to use that ratio to figure from there to the axle to get your final gear ratio.
Just wild numbers here: Engine revolutions = 10. Internal gear reduction of 2:1 = 5 revolutions of the output sprocket. 10-tooth sprocket on that final output shaft feeds that 2:1 ratio as if it was only what a 5-tooth would be without that internal reduction. That 10-tooth engine output sprocket feeds to a 30-tooth input sprocket on the jackshaft as if it was only a 5-tooth on the engine output due to internal reduction, so you really have an engine/jackshaft ratio of 6:1. If that 6:1 ratio jackshaft input sprocket turns a matching 30-tooth jackshaft output sprocket so there's no change in ratio across the jackshaft, then you're already feeding that 6:1 ratio to the rear axle. If the axle sprocket was also a 30-tooth like the jackshaft output sprocket, then your final ratio would still be 6:1. If you want a 5:1 final ratio for more power and less top speed, then you could make the axle sprocket just a 25-tooth to get that 5:1 engine/axle ratio.

 

[mmhmmtellyuwhut]

Yes it does, but what you really need to know is how many engine revolutions it takes to turn the rear axle one time, ignoring any gearing in between. If your engine's output shaft is directly coupled to its crankshaft inside (is literally the end of the crankshaft, one piece), then every rotation of the output sprocket equals a rotation of the engine itself. If there's an internal gearbox, let's just say 2:1 ratio, you have to figure that into determining your engine rpm's. The direct engine rpm's is what you need for determining your engine/axle rotation gear ratio, not what it is after being changed by an internal gearbox.
With a jackshaft being fed from the engine output sprocket, as opposed to a direct chain to the rear axle, you then have to figure the ratio of engine revolutions/jackshaft revolutions. With that, you then have to use that ratio to figure from there to the axle to get your final gear ratio.
Just wild numbers here: Engine revolutions = 10. Internal gear reduction of 2:1 = 5 revolutions of the output sprocket. 10-tooth sprocket on that final output shaft feeds that 2:1 ratio as if it was only what a 5-tooth would be without that internal reduction. That 10-tooth engine output sprocket feeds to a 30-tooth input sprocket on the jackshaft as if it was only a 5-tooth on the engine output due to internal reduction, so you really have an engine/jackshaft ratio of 6:1. If that 6:1 ratio jackshaft input sprocket turns a matching 30-tooth jackshaft output sprocket so there's no change in ratio across the jackshaft, then you're already feeding that 6:1 ratio to the rear axle. If the axle sprocket was also a 30-tooth like the jackshaft output sprocket, then your final ratio would still be 6:1. If you want a 5:1 final ratio for more power and less top speed, then you could make the axle sprocket just a 25-tooth to get that 5:1 engine/axle ratio.

lost me after wild numbers but i returned at 6:1 but if you did 5:1 for more power and less top speed get a smaller sprocket ...uhhh dont teach people stuff with big elaborate logic when you dont know what your saying please . smaller sprocket in rear gets less torque and more top end . but wild numbers is if engine is 3600rpm and your tire is 10" tall 10"x3.14 is circumfrence so every rev of that tire covers 31.4 inches. x 3600rev per minute time 60 so revs per hour and eventually youll get a number that says your lil engine wants to go a million mph and you say well half of 3600 rpm then and it says half a million and you do that until you get to a number tbat says a reasonable speed and figure how much did you do the "well try half of that "thing if number is3600 ÷50 because that gave you a reasonable speed then you want about a 50:1 meaning spin motor 50 times to get one rev from the tire. because of weight and variables if it says a "6hp"rated 3600 is gonna go 10 mph you better believe that its got enough torque for that to not be much of a variable and its probably gonna go 10mph but if it says 100 you gotta think of the variablesand use a little intuitionon the effect of variables.. "give me a long enough lever and i will move the planet" and thats all gears are is levers on a fulcrum and only the tips of the levers contactthe other "tips" aka "teeth". a very good way to learn is realise they are just levers getting leverage. gaining distance sacrificing power . gaining power sacrificing distance to move a house with a lever youd probably have to walk 500000 mm to move 5 mm lol, but thats a ratio of 100000:1