RocketPenguin
New Member
Now, i have already asked both of these questions before, and never got any real answers/gave incorrect information.
Lets start with the jackshaft.
I am working on converting a bicycle into a chainsaw powered moped. For this, i will have the chainsaw mounted somewhere between the seat and the pedals, probably in the triangular opening. I will then have the clutch of the chainsaw lathed so that i am able to fit a #35 chain onto the already existing teeth (The clutch has 6 teeth). I will remove the pedals This is where i need suggestions/reassurance) and place two bearings into the pedal housing. I will then get a jackshaft of some sort (I was thinking, would a really big bolt work?) and mount two sprockets on it. I would then have the original bicycle chain go to the jackshaft (Onto the smallest sized sprocket possible, 17t. I am thinking of using 5/8 or 3/4 diameter jackshaft). The chain from the chainsaw engine would then also go to its own sprocket on the jackshaft.
To simplify:
6t sprocket on clutch (Drive) goes to
X sprocket on Jackshaft (Driven)
17t sprocket on Jackshaft (Drive) would then go to
28t sprocket on 24-26" bicycle tire.
The chain between the engine and jackshaft would be #35, and from jackshaft to bicycle would go back to #25.
How many teeth would be needed on sprocket X in order to get the ratio down to 25:1?
And for the jackshaft, i was thinking this:
In the pedal housing, which can fit a bearing with an OD of 1.3-1.4" i would have two bearings, one on each end. They would either be pressure fitted, or i could have a a tension bolt or two holding them in. I would then find a bolt with either a 3/4 or 5/8 OD and around 6-10" long, and use that as the jackshaft. Possibly key it myself if needed, or lathe the OD down a little if needed (Would this weaken it? Reason i would like to use a bolt is because it is A, Cheaper, B, solves the problem of how do i prevent it from sliding out (Both sprockets will be on the same side), and C, one is already threaded, so i can simply put a nut on the other end to lock it from sliding out.) Both sprockets would be mounted onto the bolt/jackshaft via a key, or by tension screws. Probably keys. Or both.
So, to simplify:
Two bearings placed into pedal housing
Bolt placed into bearings
Both sprockets placed onto bolt, possibly with some spacers/nut before, locking the bolt and bearings into place
Sprockets would be anchored to the bolt via key/tension screws
a final nut would be placed onto the bolt to hold everything on and together.
Would this method work?
Any good place to purchase bearings that can handle ~5-10K rpm?
What sized sprocket would i need?
If any more information is needed, or further explanation, let me know. I can even attempt to draw a diagram.
Thanks!
PS: Didn't know where to squeeze this in, but is probably important, the engine runs at about 10-14K rpm, with around 2-4HP, and is 42cc
Lets start with the jackshaft.
I am working on converting a bicycle into a chainsaw powered moped. For this, i will have the chainsaw mounted somewhere between the seat and the pedals, probably in the triangular opening. I will then have the clutch of the chainsaw lathed so that i am able to fit a #35 chain onto the already existing teeth (The clutch has 6 teeth). I will remove the pedals This is where i need suggestions/reassurance) and place two bearings into the pedal housing. I will then get a jackshaft of some sort (I was thinking, would a really big bolt work?) and mount two sprockets on it. I would then have the original bicycle chain go to the jackshaft (Onto the smallest sized sprocket possible, 17t. I am thinking of using 5/8 or 3/4 diameter jackshaft). The chain from the chainsaw engine would then also go to its own sprocket on the jackshaft.
To simplify:
6t sprocket on clutch (Drive) goes to
X sprocket on Jackshaft (Driven)
17t sprocket on Jackshaft (Drive) would then go to
28t sprocket on 24-26" bicycle tire.
The chain between the engine and jackshaft would be #35, and from jackshaft to bicycle would go back to #25.
How many teeth would be needed on sprocket X in order to get the ratio down to 25:1?
And for the jackshaft, i was thinking this:
In the pedal housing, which can fit a bearing with an OD of 1.3-1.4" i would have two bearings, one on each end. They would either be pressure fitted, or i could have a a tension bolt or two holding them in. I would then find a bolt with either a 3/4 or 5/8 OD and around 6-10" long, and use that as the jackshaft. Possibly key it myself if needed, or lathe the OD down a little if needed (Would this weaken it? Reason i would like to use a bolt is because it is A, Cheaper, B, solves the problem of how do i prevent it from sliding out (Both sprockets will be on the same side), and C, one is already threaded, so i can simply put a nut on the other end to lock it from sliding out.) Both sprockets would be mounted onto the bolt/jackshaft via a key, or by tension screws. Probably keys. Or both.
So, to simplify:
Two bearings placed into pedal housing
Bolt placed into bearings
Both sprockets placed onto bolt, possibly with some spacers/nut before, locking the bolt and bearings into place
Sprockets would be anchored to the bolt via key/tension screws
a final nut would be placed onto the bolt to hold everything on and together.
Would this method work?
Any good place to purchase bearings that can handle ~5-10K rpm?
What sized sprocket would i need?
If any more information is needed, or further explanation, let me know. I can even attempt to draw a diagram.
Thanks!
PS: Didn't know where to squeeze this in, but is probably important, the engine runs at about 10-14K rpm, with around 2-4HP, and is 42cc