Effects of lug height on final gear ratio for a snowmobile....

[Tom.G]

An image and two videos will help clarify. They show that the driving force to the track is a positive drive, as in a chain & sprocket, NOT a friction drive as has been assumed in this discussion.

This from post #7. Note there are lugs on both the interior and exterior of the track.
http://image.fourwheeler.com/f/9369...apex_rtx_snowmobile+rear_suspension_track.jpg

See time 12:22 - 12:38


See times 5:31 and 6:34

 

[OldYat47]

Track runners? Not relevant. Go back to the site http://physics.info/rolling/, the third drawing of the wheel rolling motion. How do you calculate the forward speed of the axle (and therefore the vehicle)? How would you calculate the distance traveled in one rotation of the wheel? It's all calculated by using the effective radius as "r" in those equations. What is the effective radius? It's the distance from the center of rotation to the outside surface of the wheel (or track). You could include some small change in radius if you're running on a soft tire, for example. We have no data for any change in thickness on the track so that's moot. On a tracked vehicle, assuming there is some stretch, the stretch would be at the top of the drive gear and top of the track all along to where the track is stationary on the ground. Once the track reaches the point where it leaves the drive gear it is either slack or under idler tension. So the forward speed of the tracked vehicle can be calculated using the drive gear as the rolling element and the outer surface of the track as one end of the effective radius.

If the lugs originally proposed add to that effective radius (radius of belt without lugs + added thickness caused by lugs on the ground) then the forward speed and forward distance traveled per RPM will change. If not (sunk in the snow) then you've got no change. If you start discussing lack of traction then the whole discussion is out the window.

 

[jbriggs444]

What is the effective radius? It's the distance from the center of rotation to the outside surface of the wheel (or track).

Wheel, yes. Track, no.

 

[sophiecentaur]

It's the distance from the center of rotation to the outside surface of the wheel (or track).

Why do you keep asserting this without any explanation or proof? (And don't say "it's obvious" because it appears that it's only obvious to you.)
How would you square your above statement if the drive wheel were just bearing down onto the track, half way between front and back rollers. How fast would the vehicle be travelling? Now lift the rear section of the track by a degree or so, so it does not contact the road. Would the vehicle speed suddenly increase because the back section has just lost contact? How far would you need to lift the back section to make it the vehicle go as fas as you claim?
If you tried to understand this instead of being scared to be wrong, you might get somewhere and we could all go home.
Also, you have not yet given a reference that proves your hypothesis, nor given any appropriate Maths.

 

[OldYat47]

Let's forget about effective radius and just use radius, although I fear someone will immediately throw in thinning of the track.

Let's look at that situation where the drive wheel is actually on the bottom section of track. Now go back and look at that third picture again in the referenced physics site. Imagine that wheel is the drive gear and the outer surface of the tire is the outer surface of the track. Now look at the formulas. The velocity of the axle center is r*w, where w is in radians per time period. One revolution would be 2*pi*radians, so the distance traveled in one revolution of the drive gear would be r*2*pi. Anything that changes r would change that distance. The velocity of the center of the axle would be r*2*pi*(revolution/time period). Any physics or mechanical textbook will give the same equations for rolling motion.

 

[sophiecentaur]

Any physics or mechanical textbook will give the same equations for rolling motion.

That is blindingly obvious but the motion of the straight bit of the track is not "rolling", it's linear and the clue is in the word "straight". When it leaves the circumference of the wheel (or joins it) the motion kind of motion changes instantaneously (if the track is totally rigid and the contact with the wheel is truly circular or non-slipping.
Read the following carefully. You have two drive wheels, driven in perfect sync, with a single chain or gearbox. One has the track going round it and the other is on a flat section over the ground. Both wheels are going forward at the same speed (bolted to the frame) and both wheels are rotating at the same rate. Is one of them slipping on the track? How does your conjecture fit those conditions?

 

[A.T.]

That is blindingly obvious but the motion of the straight bit of the track is not "rolling", it's linear and the clue is in the word "straight".

Maybe OldYat47 thinks about a snowmobile doing a wheelie, and thus driving on the sprocket!

 

[S pump]

So, this seems like such an easy thing to explain. A few of you did explain it very well, better than I ever could as I lack the terminology needed to articulate the theory. In my experience with tracked vehicles, I can say with confidence that changing the thickness of the track does not change the final drive ratio of the vehicle. It may change many things but not the gear ratio. On a tracked vehicle, the track acts as the surface that the vehicle is propelling itself on. Just like laying down a road ahead of itself. A question for Oldy, assume you have two wheels with one driving the track. That wheel is 6 inches in diameter and the rear wheel is 24 inches in diameter. Only the drive wheel is driving the track. Now reverse the diameters, 24 inch drive wheel and 6 inch idler. Does that change the gear ratio? Why? or Why not?

 

[OldYat47]

No, A.T. Refer to sophiecentaur's comments about imagining the drive gear being on the ground. And doing a wheelie would not put the drive gear on the ground. Look at where they actually sit.

sophiecentaur, why are you referring to the portion of the track that isn't moving? That is irrelevant. Focus on your proposed drive gear. Let's add some details. All the idler wheels are the same size as the drive gear so the top and the bottom of the track are perfectly parallel. Imagine that the vehicle is moving left to right. Imagine the drive gear is at the front of the vehicle. Now imagine the only the front half of the drive gear with the track wrapped around that front half. Now look at the third sketch in that physics site I referred to. The front half of the drive gear (on the right in this thought experiment) with the track wrapped around it is identical to the right half of that pictured wheel.

Note that as a wheel rolls there is always one point that is stationary (on the ground) and one point 180 degrees opposite that is traveling "forward" at twice the vehicle speed.

And I would like to see your equation(s) for the distance a tracked vehicle travels per revolution of the drive gear, and the equation(s) for forward speed per RPM of the drive gear.

 

[jbriggs444]

Note that as a wheel rolls there is always one point that is stationary (on the ground) and one point 180 degrees opposite that is traveling "forward" at twice the vehicle speed.

This obscures the important point that the speed of the outer surface of the track as it lays flat on the ground will not, in general, match the speed of the outer surface of the track as it curves around the drive wheel.

The point on the track that is "stationary" as the track leaves the wheel will not, in general, be the point touching the ground. It will, in the usual case, be a point at some depth within the track. The point on the track that is traveling forward at twice the vehicle speed will not, in general, be the point exposed to the air on the track's highest point. It will, in the usual case, be a point at some depth within the track.

 

[sophiecentaur]

Now look at the third sketch in that physics site I referred to.

Correct me if I am wrong. I think you have only posted a ref to one site and the whole of that web page was dealing with wheels and wheeled vehicles - which do not have tracks. You are asking us to extend what's written about wheels and to "imagine" what happens to a track, wrapped around the wheel. That is precisely what you are not doing. You are making assumptions and making assertions on a false basis, rather than following all the arguments that have been put to you.

The 'forward traveling part of the wheel' that you have mentioned is totally irrelevant to the argument. It happens that all wheels have a 'top half' which is always traveling forwards but a vehicle with 90° worth (or even less) of lower wheel sector would still have the same velocity ratio and the effect of the track, for 90° of rotation would be the same as in my argument.
You have still not replied to this, which I wrote earlier:

Read the following carefully. You have two drive wheels, driven in perfect sync, with a single chain or gearbox. One has the track going round it and the other is on a flat section [of track] over the ground. Both wheels are going forward at the same speed (bolted to the frame) and both wheels are rotating at the same rate. Is one of them slipping on the track? How does your conjecture fit those conditions?

Perhaps you would give us your thoughts on that.

 

[OldYat47]

Of course it won't. The entire track, every little bit of it that lays on the ground is not moving at all. This is not relevant. On a rolling wheel the point in contact with the ground is stationary. This also is irrelevant. And yes, the point on the track that is stationary as the track leaves the wheel will be identically the point touching the ground. More on this below. And the entire track, every little bit of it at the top in this example will be traveling forward at twice vehicle speed.

Where is the distance between the drive gear and the ground the least? At the very bottom. Any space is filled by the track. If you look just a degree or two farther back on the drive gear the distance between the gear and the ground is greater, so the track won't be thick enough to fill the gap, so the track is not yet in contact with the ground. The track starts at the top traveling twice vehicle speed and when it reaches the bottom it's speed is zero. You can resolve the tangential velocity into two components, one relative to the ground and one relative to the forward speed of the vehicle.

What's your alternative?

 

[jbriggs444]

And yes, the point on the track that is stationary as the track leaves the wheel will be identically the point touching the ground.

No. In general, it will not be.

 

[OldYat47]

I should add that the equations of motion for tracked vehicles and wheels are the same. If you think not then post links to a site that explains the difference in the math between the two.

 

[A.T.]

And doing a wheelie would not put the drive gear on the ground.

It doesn't matter which sprocket you do the wheelie on. If the contact to the ground was on a curved part of the track, where the outer track surface moves faster than the inner, then the thickness would affect the gear ratio. But in reality the contact to the ground is on the straight part, where inner and outer track move at the same speed, which doesn't depend on the track thickness.

What about the below diagram don't you understand / disagree with?

The entire track, every little bit of it that lays on the ground is not moving at all.

We are talking about velocities relative to the vehicle here, where the axis of the sprocket is at rest. The speed of the lower track relative to the vehicle is equal to the speed of the vehicle relative to the ground.

 

[OCR]

The entire track, every little bit of it that lays on the ground is not moving at all.

The track starts at the top traveling twice vehicle speed and when it reaches the bottom it's speed is zero. You can resolve the tangential velocity into two components, one relative to the ground and one relative to the forward speed of the vehicle.

Again, OldYat is correct... watch this video ... all of it!

Watch this one, too ...

 

[A.T.]

Again, OldYat is correct...

About the track speeds relative to the ground, sure. About the rest, nope.

 

[sophiecentaur]

. watch this video ... all of it!

Watch this one, too ...

Two good videos. Thanks for finding and posting them. I watched all of both of them and it's actually quite apparent that the visible, outer parts of the track are going faster around the wheels than on the flat sections (more to the point, faster than the vehicle is moving forward). The track treads appear to 'whip round' the end wheels visibly faster than anywhere else. Compared with the wheel radius, the depth of the track appears to be at least 10%. That would imply the rotational speed would be at least 1.1 X the linear speed.

What's your alternative?

My alternative is not to look at the irrelevant parts of the track (over the top) but to look at what happens where the curved section mets the flat section. You clearly have not read my arguments (nor any of the others, I suspect) because you are so convinced of your own. None of us has a problem between talking about velocities relative to the vehicle so why do you keep introducing the fact that the track is actually stationary on the ground. Of course it is.
I would really appreciate a response to my previous point:

Read the following carefully. You have two drive wheels, driven in perfect sync, with a single chain or gearbox. One has the track going round it and the other is on a flat section over the ground. Both wheels are going forward at the same speed (bolted to the frame) and both wheels are rotating at the same rate. Is one of them slipping on the track? How does your conjecture fit those conditions?

AT's diagram, above really demands a comment from you, too. Why do you never respond to such 'details'?
If you really are right about this thing then you should have an easy, valid response. If not then you need to consider the possibility that you are wrong.

 

[OCR]

You are asking us to extend what's written about wheels and to "imagine" what happens to a track, wrapped around the wheel.

I'll help you with the "imagine" part...[COLOR=#black]..[/COLOR] lol

Help at all... even a little?[COLOR=#black] ...[/COLOR]Again, OldYat is still correct...[COLOR=#black]..[/COLOR]

 

[jbriggs444]

Help at all... even a little?

There is no track flat on the ground between those two wheels and no indication of the thickness that such a track might have. The crux of the... vigorous debate here is on the behavior of the inner and outer surfaces of such a track at or near the boundaries between its curved and its flat segments.

Can you please stick to mathematical and physical arguments. Statements of opinion about who is right or who is wrong are unhelpful.

 

[OCR]

There is no track flat on the ground between those two wheels and no indication of the thickness that such a track might have.

That's where the 'imagine' part of imagine comes into play...[COLOR=#black]..[/COLOR]

Statements of opinion about who is right or who is wrong are unhelpful.

Lol... well, "are unhelpful" is an opinion isn't it ?

Stay on track, now... don't derail ...[COLOR=#black]..[/COLOR]

 

[berkeman]

This thread has run its course, and is now closed.