A= bXz /12 ???? Any math students out there?

[D4570]

I have an old snowmobile track and bogeys and the motor.
I want to make a "Snow Dog"

What I'm wondering is.
Here we go.
I want it to go about 20 mph full out.
The track is a little over 8 feet long/ around.
What is the RPM to do 20 miles in an hour?
Then I need to do two reductions
One from a belt than one to the track with a chain.
5000 rpm on the motor.
My head hurts. :doh:

 

[madtom]

20 miles is 105,600 ft in 1 hr

Track is 8 ft long

Track needs to turn 105,600/8 = 13,200 times in 1 hr to travel 20 miles

13,200/60 = track spinning 220 rpm to go 20 mph

I think..?

If that’s correct, the track will spin 3.6 rev per second. That seems fast

 

[Carnage2011]

That is hauling ass for one of those...

 

[Fullquiver]

What kind of motor are you using to get 5000. Certainly not like the engine in your picture as typically that type engine has a max governor speed of 3600 rpms.

 

[D4570]

There is a Co that makes them and they claim twenty MPH plus.
The Pic is just one off the computer, I have not made anything yet.
I Have the motor that came off the snowmobile, a 71 Johnson skee horse 30 HP, and red lines at 5500 rpm.

 

[BuckRut]

The length of the track makes no difference in the speed. What you need to calculate is the motor RPM multiplied by the maximum ratio of the drive pulley diameter to the clutch pulley and then multiply that by the number of teeth on the drive gear to the number of teeth on the driven gear then multiply that by the diameter of the track hub that is being driven.

 

[D4570]

Ultimately the length of the track makes the speed...
Each revolution the unit moves 8 feet there are 5280 feet in a mile.
so the track has to turn 660 times to cover a mile. If I want it to go 20 miles per hour
it had to turn 13,200 times in an hour, which is 220 times in a minute?
Is this right? That is a little over 3 1/2 turns a second, that sounds very fast.

 

[BuckRut]

Ultimately the length of the track makes the speed...
Each revolution the unit moves 8 feet there are 5280 feet in a mile.
so the track has to turn 660 times to cover a mile. If I want it to go 20 miles per hour
it had to turn 13,200 times in an hour, which is 220 times in a minute?
Is this right? That is a little over 3 1/2 turns a second, that sounds very fast.

The track is not a wheel and there for it doesn't have revolutions to make per second. The track is driven by a wheel and the RPM of that wheel times the diameter equals the speed of the track.

 

[D4570]

The track is a wheel just not round

 

[BuckRut]

The track is a wheel just not round

A track is not a wheel! It is a belt with traction!

 

[Dave N]

The drive sprocket turns at your ground speed. It uses the track to roll over instead of contacting the ground directly.

 

[vanish]

Ultimately the length of the track makes the speed...
Each revolution the unit moves 8 feet there are 5280 feet in a mile.
so the track has to turn 660 times to cover a mile. If I want it to go 20 miles per hour
it had to turn 13,200 times in an hour, which is 220 times in a minute?
Is this right? That is a little over 3 1/2 turns a second, that sounds very fast.

The track is driven by a wheel and the RPM of that wheel times the diameter equals the speed of the track.

You're BOTH right.

The incoming RPM and the diameter of the drive wheel on the track is going to determine how many RPMs the track makes. How many RPMs the track makes isn't necessary to calculate the answer, since we could just use the diameter of the drive wheel and come to the same result.

Edit: The resulting thing we've learned is we don't have all the information needed to answer the OP's question. We need to know the diameter of the drive wheel.

 

[grizzly_]

Drive wheel diameter is the way to go

 

[TexanSam]

a squared plus b squared equals c squared

 

[Schaaf]

NERDS

 

[Fullquiver]

There is a Co that makes them and they claim twenty MPH plus.
The Pic is just one off the computer, I have not made anything yet.
I Have the motor that came off the snowmobile, a 71 Johnson skee horse 30 HP, and red lines at 5500 rpm.

A snowmobile engine would work but so would a snowmobile. Why reinvent the wheel? How do you plan on turning your machine. A single tracked machine seems as though it would be a bear to turn without ski's. Especially at any speed.

These seemingly simple machines aren't nearly as simple as one would believe. Just a couple of things that you might want to take into consideration as you build..

 

[VOLhunter]

You actually need to know the circumference of the drive wheel (2 x pie x radius of the drive wheel). Each revolution of the drive wheel moves the track by the distance of the circumference. (20 miles / 1 hour) * (1 hour / 60 minutes) * (5280ft / 1 mile) * (12inches / 1 ft) = 21120 inches per minute. Divide this number by the circumference (ie. multiple 21120 by 1 revolution/x inches). This leaves you with revolutions per minute (RPM). Example of a 4 inch radius drive wheel (8 inch diameter) would equal 840.34 RPM for 20mph.

 

[Bob-WY]

Is the track 8 feet long or 8 feet around? Makes a difference

 

[D4570]

OK, I'm starting to get a picture in my mind.
If I took off the track, and if you could drive on just the couged sprocket. the diameter of the sprocket is how many inches/feet you move in one revolution. The track would have to turn 220 times but the drive wheel will have to turn many more times to cover the distance in the allotted time.
OK.
The drive wheel is 23" around.
There is 63360" in a mile.
It would take the drive wheel 2755 revelations to cover a mile.
That makes it turn 55,096 times for 20 miles then divide that by 60 minutes.
That is 920 rpm, to get it to 20 mph.
Same thing but Now I know the final RPM needed I can work backward to get the ratios.

 

[madtom]

I say just build it and see how fast it goes.