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

[S pump]

So, great site.
I have a quick question that I think has a simple answer. On a snowmobile forum I visit there has an ongoing debate. The question is, all other factors being the same, does changing the lug height on a track, change the final drive ratio of the snowmobile? I think I know the answer but articulaying it correctly is difficult.

 

[Tom.G]

If you're driving on pavement, sure. The tips of the lugs, in contact with the pavement, are at a larger radius from the driving axle.

If you are on snow, my opinion would be "probably a little bit." It would depend on where the effective center of force occurs on the snow-lug interface. If you are on hardpack riding on the lug tips, it would be the same as driving on pavement.

On light, fluffy stuff I would expect the center of drive force to be anywhere from half way along the lug depth up to the body of the track. With the result that you would get between one half the expected drive ratio increase to no change at all.

In the real world you probably wouldn't notice the difference. When you take into account the radius of the drive pulley plus the track thickness and lug depth, adding a little to the lug depth is going to be a small percentage change.

 

[sophiecentaur]

You could get an idea of the different effective radius of the 'output' wheel by listening to the change in pitch in the engine over different surfaces or by measuring the speed over ground at a given number of revs. But there will be an added complication of slippage in very light snow because the snow will actually be moved backwards over the ground. The wheel revs will be higher than you 'd expect when there's slippage and you will be in the realms of Paddle Wheel motorboats.
I would guess that finding an optimum would best be done by trial and error - if you have access to different drive wheels.
Would it be possible for you to use a high speed film of the wheel's motion as it goes past a point? You could actually see (or imply) the part of the wheel that's stationary relative to the ground.

 

[S pump]

Ok, so I should add another detail and see if your assertions stick. Say that for the sake of argument, there are two tracks, both solid. One is 1" thick and one is 1' thick. The drivers of the vehicle are the same diameter throughout the drive trains. Does the thickness of the track change the final drive ratio? if the argument that the radius of the track being larger makes the ratio change, wouldn't a track that is 100" long be a lower ratio than a track that is 150" long?

 

[sophiecentaur]

I'm now visualising it right, I think. I was imagining the drive wheel actually digging into the snow to provide the drive. We never use those things in my part of the UK!
We're talking about a tracked vehicle (durr).
The speed of the drive wheel over the track will be the same as the speed of the vehicle over the ground (with no slip, the parts of the track in contact with the ground will be stationary relative to the ground). So I reckon that the only thing that governs the ratio is the radius of the drive. The thickness of the track may help or hinder the grip but, if there's no slip, it can't affect the ratio because there is no 'rotation' of the bottom section of the track (I assume).
The actual length of the track is irrelevant to the ratio because the only bit of track that's transferring drive to the ground is the horizontal bit. It's the equivalent of driving over a strip of track that's been laid over the snow (basically). The details of which bits of the track are actually pushed against by the drive wheel in only relevant in as far as where the teeth of the drive wheel actually bear on the links of the track (i.e. deep into the teeth or near the tips) but I guess the wheel and track are always matched. The ratio depends on the number of teeth on the drive. Each rev will push you forward by the number of teeth times the pitch of the track.
I've said the same thing is several ways. Dunno which one makes more sense to you.
If I'm still talking BS then give me a photo of the vehicle.

 

[S pump]

That is my assertion as well. There has been a long argued discussion on a site I participate in and now it has come down to placing wagers on who is correct. I'll wait for some more replies before I declare victory though.

 

[A.T.]

We're talking about a tracked vehicle (durr).

Yes:

The speed of the drive wheel over the track will be the same as the speed of the vehicle over the ground (with no slip, the parts of the track in contact with the ground will be stationary relative to the ground). So I reckon that the only thing that governs the ratio is the radius of the drive. The thickness of the track may help or hinder the grip but, if there's no slip, it can't affect the ratio because there is no 'rotation' of the bottom section of the track (I assume).

That is also my impression, unless the curved / rotating parts of the track transmit a substantial part of the driving force.

 

[sophiecentaur]

That is my assertion as well. There has been a long argued discussion on a site I participate in and now it has come down to placing wagers on who is correct. I'll wait for some more replies before I declare victory though.

I think you are right but it's not absolutely straightforward because of the bit around the wheel.
Looking at images of tracks, it seems that the track is fairly flexible rubber and that the teeth will not, individually, produce much force on the snow. That means that the relevant radius in the system is where the rubber belt lies and not the length of the teeth. The majority of the traction force is from the teeth on the horizontal section and the force there is definitely the same as the force on the belt (same argument as previous).

 

[sophiecentaur]

Yes:
That is also my impression, unless the curved / rotating parts of the track transmit a substantial part of the driving force.

I have thought about that over my dinner. The fact is that the tangential forces from the rotating teeth are not acting horizontally and that's what counts in calculating the effective gearing. The horizontal component is, I think, the same as for all the other teeth so the 'velocity ratio' (i.e. the gearing) is the same all the way along.
So the OP has to be correct about the teeth depth having no effect on the gearing.
It would be easy to find out by turning the engine over manually and measuring the distance traveled with two sizes of track.

 

[Tom.G]

The speed of the drive wheel over the track will be the same as the speed of the vehicle over the ground (with no slip, the parts of the track in contact with the ground will be stationary relative to the ground).

Ooops! You are right.

I hereby retract my post #2. Sorry folks.

 

[S pump]

Can some articulate the differences between a drive wheel diameter determining gear ratio versus a drive system of a tracked vehicle with more than one rotating axis?

 

[sophiecentaur]

Can some articulate the differences between a drive wheel diameter determining gear ratio versus a drive system of a tracked vehicle with more than one rotating axis?

If you mean more than one drive wheel, the drives must have the same peripheral speed (they are locked together by the track) and I would think there would have to be a differential between the two drives, to share the drive without 'contention'. If you had one large drive wheel and one small drive wheel, both driving the track then you would need to have the right gears to make those wheel speeds right. The overall 'gear ratio' - in terms of linear metres per engine rev would be the same for both wheels.
Does that answer your question? I wasn't sure what you meant.

 

[S pump]

If you mean more than one drive wheel, the drives must have the same peripheral speed (they are locked together by the track) and I would think there would have to be a differential between the two drives, to share the drive without 'contention'. If you had one large drive wheel and one small drive wheel, both driving the track then you would need to have the right gears to make those wheel speeds right. The overall 'gear ratio' - in terms of linear metres per engine rev would be the same for both wheels.
Does that answer your question? I wasn't sure what you meant.

No, not really but it is a good point that further reinforces my assertion. My question is how do you articulate the relationship between a drive system with one axis (ie Wheeled vehicle) versus a drive system consisting of 2 or more rotating axis (tracked vehicle)?

 

[sophiecentaur]

No, not really but it is a good point that further reinforces my assertion. My question is how do you articulate the relationship between a drive system with one axis (ie Wheeled vehicle) versus a drive system consisting of 2 or more rotating axis (tracked vehicle)?

HAHA. I thought that's what I just did.If you have two drive wheels of different diameters then the gears before each one must ensure that the peripheral speed of both wheels is the same.
Note: If you want to specify the 'gear ratio' for a system with a linear drive (i.e. a track), you can't express it as n:m where n and m are turns of a shaft. You have to specify it in terms of turns of the input shaft for a given linear distance.

versus a drive system consisting of 2 or more rotating axis

Whether some, one or all axles are driven, the revs of all wheels have to (will) follow the rule. and the fact that they are bearing on a different bit of the track makes no difference. They could be side by side and it still would make not difference because all that counts is distance per rev has to be the same for all.
OR . . . .are you wanting a statement about how the Power to each drive wheel is shared? That depends entirely on how they are driven and it's also very relevant to 4WD cars, where the front and back wheels may not have equal torque but their speeds have to be the same (natch). I can't see any point in things being that clever on a tracked vehicle because you cannot get one slipping but the other not. I made the point earlier that the two drives would probably need a differential drive of some sort to avoid stress on the drive and track is, for instance, there is some stretch in the track..
PS could you restate the question with a different word from "articulate" because I am not sure what you mean precisely.

 

[OldYat47]

If you mean the ratio of engine RPM to road speed then yes, changing the thickness of the track will change the ratio.

Look at where the center of the drive sprocket axle is in relation to the surface of the track. You can calculate the tangential velocity (essentially vehicle speed if you ignore track-to-road slip) of any point from the axle out. The tangential velocity is the rate of rotation times the radius from center of rotation. So for a given RPM the tangential velocity increases as the radius increases. If the surface of the track gets farther from the axle (thicker track) then the tangential velocity increases. Increasing tangential velocity relative to engine speed says the drive ratio has changed. It's just like putting larger diameter wheels and tires on a car, the vehicle will travel farther per axle revolution.

However, when talking about vehicles (e.g., cars, trucks, motorcycles) tire diameter is not included in quoted final drive ratios, just the ratio of engine RPM to output RPM. If you change wheel and tire diameters significantly you need to have your speedometer recalibrated or risk getting speeding tickets.

 

[A.T.]

Look at where the center of the drive sprocket axle is in relation to the surface of the track.

Ok, let's have a look:

Looks like the drive sprocket is nowhere near the ground.

The tangential velocity is the rate of rotation times the radius from center of rotation.

What is this radius, for the straight part of the track that contacts with the road? Where is the center of rotation for that linearly moving part of the track?

 

[sophiecentaur]

I'm glad some more people have joined this thread.

If you mean the ratio of engine RPM to road speed then yes, changing the thickness of the track will change the ratio.

You would need to justify that.
As far as I can see, where it is in contact with the road (where it counts) the top and bottom surfaces of the track are moving at the same speed. The periphery of the drive wheel is the thing that sets the speed of the inner part of the track that it's in contact with it. If the inner part of the track does not compress or stretch then that is what determines the speed of inner and outer surfaces on the straight section, in contact with the road. The pictures of tracks (Google has dozens and dozens of them) seem to have a continuous inner belt, in contact with the drive wheel and any added thickness of track consists of articulated transverse treads. These treads radiate outwards when going around a curve or drive wheel but are parallel along the straight section. So the distance per drive wheel rev is independent of the thickness of the track.
I can see that an alternative design of track could change things - for instance if the inner surface of the track had teeth that compress longitudinally as they go over a wheel - but I don't think that is the case on a snowmobile.

 

[OldYat47]

Justify? Now you've got me smiling. Let's get that out of the way. How are you defining final drive ratio? How are the people in your discussion defining this?

Where it's in contact with the ground it isn't moving at all.

Now, do you agree that in circular motion tangential velocity equals angular velocity times radius? If not, justify.

 

[OldYat47]

In response to A.T., above, I'm talking about the distance from the axle center to the surface of the track, not the road or ground. The straight part of the track is completely irrelevant. (Got me smiling with that one). Note that where the track is on the ground it isn't moving, whether or not the vehicle is moving.

 

[jbriggs444]

In response to A.T., above, I'm talking about the distance from the axle center to the surface of the track, not the road or ground.

I expect that a real track will have a fabric belt embedded within to sustain the tension load. This belt will likely be near the inner surface of the track. It is the speed of this fabric belt relative to the drive wheel's rotational speed that is crucial. Accordingly, it is the distance from the axle center to the fabric belt that matters.

 

[sophiecentaur]

Justify? Now you've got me smiling. Let's get that out of the way. How are you defining final drive ratio? How are the people in your discussion defining this?

Where it's in contact with the ground it isn't moving at all.

Now, do you agree that in circular motion tangential velocity equals angular velocity times radius? If not, justify.

Of course I agree with that but the speed over the ground is not the tangential velocity of the end of the lugs.
As I have already said, the "final drive ratio", stated in terms of wheel diameter (or gear teeth) is not so relevant as the linear distance per rev. Think about what happens to the lugs as they go round the rear (idler) wheel. They are stationary on the ground and so is the inner 'belt'. As they lift off the ground they start to rotate and end up going vertically, then forward and over the top of the wheel. On the way round the wheel, because they lie radially, they go faster than the belt but then they slow down again on the upper straight section and have the same speed as the belt. This applies whenever the belt goes over a wheel. So the effective drive 'ratio' depends on the diameter of the drive wheel and its peripheral speed.
This is not intuitive, I can see, but it's the right argument and drawing it out for yourself could help you 'justify' it.
BTW, although the track is stationary when it's on the ground, the velocity relative to the vehicle is what counts and that is equal and opposite to the velocity of the vehicle over the ground.

 

[A.T.]

The straight part of the track is completely irrelevant.

What is completely irrelevant is the speed of the lug tips around the drive socket, where they have no contact to the road.

What is relevant are the lug tips that contact the road. Their speed relative to the vehicle is the speed of the vehicle relative to the road. And their speed relative to the vehicle is not affected by the lug length, because these lugs are moving linearly relative to the vehicle, and not rotating.

 

[OldYat47]

It seems we've wandered far into the woods with this one, children. Let's ignore the lugs and let's ignore snow quality. Imagine a smooth track on the vehicle. What is the distance per revolution of the drive gear the vehicle will travel? Naming the radius from the center of the drive axle to the surface of the track R, it is 2*pi*R. If you install a thicker track the vehicle will travel farther. If you install a thinner track the vehicle will travel less far. The vehicles forward speed will be 2*pi*RPM*R.

Have we found a promising path?

 

[jbriggs444]

It seems we've wandered far into the woods with this one, children. Let's ignore the lugs and let's ignore snow quality. Imagine a smooth track on the vehicle. What is the distance per revolution of the drive gear the vehicle will travel? Naming the radius from the center of the drive axle to the surface of the track R, it is 2*pi*R. If you install a thicker track the vehicle will travel farther. If you install a thinner track the vehicle will travel less far. The vehicles forward speed will be 2*pi*RPM*R.

Have we found a promising path?

Which surface are you measuring to and why?

 

[A.T.]

Which surface are you measuring to and why?

Actually, the OP should clear up what is meant by "changing the lug height" with a diagram, that shows:
- Where the drive sprocket teeth attack
- Where the track links are connected
- What part is being changed / extended
We here and the people in the original thread might have different ideas on this, and thus might be talking past each other.

 

[sophiecentaur]

We here and the people in the original thread might have different ideas on this, and thus might be talking past each other.

That is quite possible if it were not for all the pictures of snowcat tracks and the OP is about snowcats. The tracks in all the images all appear to be driven by a wheel that bears on the inner surface of the track / belt. So, everything I have been writing must follow.
If anyone has drawn the situation out for themselves, instead of trying to think it out in their head, they will have found it fairly straightforward.
For another design of track. consisting of a proper drive sprocket and chain links within the track, the relevant 'drive radius' would need to include the depth of the track where the drive effort is applied. The lug length would still not be relevant.

 

[OldYat47]

jbriggs444, the surface of the track that comes in contact with the road. I fear that there's a lot of undefined uncertainty in this discussion so I'm trying to establish a footing we can all agree on. So look at a track with no lugs. What is the speed of a vehicle with that track? I propose that it's 2*pi*R*rpm, where R is the radius from the center of the drive gear to the "working surface" of the track.

 

[jbriggs444]

jbriggs444, the surface of the track that comes in contact with the road.

That answers half of the question. The other half is "why?"

As the track goes around the drive wheel, the inner and outer surfaces do not move at the same speed (obviously). After the track leaves the drive wheel the two surfaces do move at the same speed (obviously). At least one surface must change speed. Which is it?

Hint: My answer is in #20 above.

 

[S pump]

Think of a conveyor at your grocery store. Does the can of Tuna fish travel slower than the bottle of ketchup?
The tuna can is a small lug, the ketchup bottle is a tall lug. Does it change the gear ratio of the conveyor?

 

[OldYat47]

S pump, irrelevant.

A hand's on experiment? Take some car and truck toys with wheels of different diameters (toys are a great source). Choose some that you can "track" with a large rubber band (no interference from bodywork and stuff). Put the "tracked" toy on the ground. Where one wheel touches the ground mark the ground and the wheel (both marks line up). Now roll the toy forward one wheel revolution. Record the distance. Now put the vehicle in its starting position and mark the rubber band at the top of the "back" wheel. Now roll the vehicle forward until the rubber band mark is at the top of the "front" wheel (the axle to axle distance or wheelbase). Record the distance the vehicle travels.

Here's what you will find: The forward distance rolled for any vehicle will be pi*(diameter of the wheel plus the rubber band). This is sensitive enough that you can see the change if you stack two rubber bands together (adding a small amount of diameter). When the top section moves one wheelbase distance the vehicle will move 1/2 the wheelbase distance. The forward motion is defined by the wheel diameter. The top of the belt moves forward twice as fast as the vehicle moves forward. That's how tracked vehicles work, you can look it up.

Once we get that settled we can talk about those lugs.

 

[S pump]

I'll go on record saying I disagree Oldy. I know it's a tough concept but the gear ratio is not dependent on conveyor belt thickness. It ends at the drive sprocket. The part of the conveyor that conveys the vehicle is flat on the ground with the drive sprocket traveling the same distance with each rotation regardless of how thick the belt is, in this case the belt is merely a running surface for the drive sprocket to roll on.

 

[A.T.]

I know it's a tough concept but the gear ratio is not dependent on conveyor belt thickness.

I would say it depends on the belt, and how it deforms. The crucial parameter is the effective radius (around the drive sprocket) at which the belt keeps a constant length and thus constant speed throughout the cycle. That is the radius that determines the speed of belt/track on the straight parts and thus the gear ratio.

Adding lugs outside of this radius does not change the belt speed on straight parts. The outer lug tips just move faster around the sprockets than on the straight parts. But adding parts on the inside, which then contact the rollers, would increase that effective radius.

 

[sophiecentaur]

Here's what you will find: The forward distance rolled for any vehicle will be pi*(diameter of the wheel plus the rubber band).

You are telling us that is what we will find but did you find that in an accurate experiment?

 

[OldYat47]

S pump, I agree. Final drive ratio is generally the reduction between the engine and the final drive element. The original poster, however, has implied including the track. Her question is how far the vehicle will travel per revolution of the engine.

sophiecentaur, of course (high school mechanics demonstration project). I always was interested in tracked vehicles. Since no one believes me you should all go out and prove it for yourselves. You can get big rubber bands (the kind used to bundle 8 1/2" X 11" envelopes) at office supply stores. Those are good because you can stack them on top of one another easily. If you don't know any kids you can buy cheap toy trucks. All you need is one.

 

[S pump]

No!

 

[sophiecentaur]

No!

"no", what?

 

[S pump]

I don't agree with Oldy's assertion.

 

[sophiecentaur]

A hand's on experiment? Take some car and truck toys with wheels of different diameters (toys are a great source). Choose some that you can "track" with a large rubber band (no interference from bodywork and stuff). Put the "tracked" toy on the ground. Where one wheel touches the ground mark the ground and the wheel (both marks line up). Now roll the toy forward one wheel revolution. Record the distance. Now put the vehicle in its starting position and mark the rubber band at the top of the "back" wheel. Now roll the vehicle forward until the rubber band mark is at the top of the "front" wheel (the axle to axle distance or wheelbase). Record the distance the vehicle travels.

Here's what you will find: The forward distance rolled for any vehicle will be pi*(diameter of the wheel plus the rubber band). This is sensitive enough that you can see the change if you stack two rubber bands together (adding a small amount of diameter). When the top section moves one wheelbase distance the vehicle will move 1/2 the wheelbase distance. The forward motion is defined by the wheel diameter. The top of the belt moves forward twice as fast as the vehicle moves forward. That's how tracked vehicles work, you can look it up.

This description is hard for me to understand. Measuring what the track does round the top is not very relevant. What counts is the distances traveled by the vehicle for one rev of a wheel, with and without the track (or with different thicknesses of track. From your description, you have not done this. You seem to have been measuring distances on the track (or something) which is just an added confusion. The result you get will depend, as has been pointed out, by the detailed way in which the drive distorts the belt as it goes round the drive wheel. We have assumed that the drive is transferred to the contact surface of the belt and that the layers of the belt will stretch differentially as it goes round the wheel. (The belt has to distort in some way as it goes from straight to curved and back again.) But you can rely on the fact that there is no permanent 'creep' of one surface of the belt over the other surface. A given difference in the circumferences of the inner and outer surfaces on the curved section must disappear when that section becomes straight along the ground.
You could give us the reference where your idea is supported; that's the way things work on forums. And the reference needs to be of some reasonable quality to be believed. So far, you have done little more than make an assertion.
Try thinking in this way. Imagine you stood your vehicle on the end of its track so it was resting on the drive wheel plus belt (like in a wheelie). The peripheral speed (relative to the car) of the edge of the belt in contact with the ground would, indeed, be faster by a factor of (d+D)/D where D is the radius of the drive wheel and d is the belt thickness. This is because the outer face of the belt would have stretched. Once you lay the vehicle down flat, the upper and lower faces would be traveling at the same speed (relative to the car) - corresponding to D (the speed of the inner face and drive wheel edge). As with all forms of drive (gears, belts, chains etc) there is a certain amount of scuffing between the two contacting surfaces and it will happen here, too, where the belt shape changes from straight to curved at the transition and the outer surface stretches.
Edit: I could add that your rubber bands could compress on their inner surface, to some extent, as they go round the wheels and that would have the effect of transferring the effective drive to a point within the band. That could move the results in 'your' direction. There is another factor when using multiple bands and that is the possible relative slippage of the bands as they go round the curve. I suggest one thick band would produce different results from two or more layers of different bands.

 

[S pump]

This, I agree with.

 

[sophiecentaur]

I keep thinking of other angles on this one. Imagine that you gradually unwrap the track from the drive wheel so that it is bearing directly on the belt in one place. (Squeeze the belt between a drive and an idler) The track speed will be the same as the peripheral speed of the drive wheel. How much would you need to wrap the track around the drive wheel before the change in speed would kick in?

 

[OldYat47]

You could give us the reference where your idea is supported; that's the way things work on forums.

http://physics.info/rolling/

There you go. To convert all this to tracked vehicles just put two of the wheel drawings together, one behind the other, and imagine them tracked such that the outer surface of the track is at radius "r". Anything that changes that radius will change tangential velocity (pure translation) or forward velocity (pure rolling).

Any counter statements should be made with appropriate equations and calculations.

 

[jbriggs444]

http://physics.info/rolling/

That reference contemplates rigid rotation. A track does not rigidly rotate. It flexes.

 

[OldYat47]

OK, it flexes. So what? Relate that to changes in velocity and/or distance traveled per revolution.

 

[jbriggs444]

OK, it flexes. So what? Relate that to changes in velocity and/or distance traveled per revolution.

That has already been done repeatedly by various posters.

The most obvious model assumes that the track retains a constant thickness through out its cycle around the drive wheels, idlers and flat sections and that a "vertical" line drawn from inner surface to outer surface remains straight throughout the cycle. That is, such a line is radial when the track flexes around a wheel and is vertical when the track is flat on the ground.

Given these assumptions, it follows that when the track is moving around a a wheel, the inner and outer and outer surfaces are moving at different speeds. Given these assumptions, it follows that when the track is flat on the ground, the inner and outer surfaces are moving at the same speed.

It follows that the velocity of at most one surface of the track can remain constant throughout a complete cycle. Nothing in the model assures us that the outer surface of the track retains its velocity when it transitions from moving around the wheel to moving on the flat. It could just as easily be the inner surface. Or neither.

 

[OldYat47]

So according you your post, the inner and outer surfaces of the track are moving at the same speed where the track is on the ground and stationary. There are two points on the drive gear where the inner and outer surfaces of the track are moving at the same speed, though not the same direction. Those points are where the track is tangent to the drive wheel. No bend, no curvature, no distortion. What is the speed of the surface of the track at those points? (angular velocity) X (2*pi) X (radius from center of axle rotation to outer surface of the track).

Agree? If not, why? And what equation would you prefer for that tangential velocity?

"Nothing in the model assures us that the outer surface of the track retains its velocity when it transitions from moving around the wheel to moving on the flat".
That is true, but the fact that the track continues to operate implies that it isn't stretching infinitely in any of its segments, and isn't piling up anywhere. Over any period of time all points travel the same number of times around the drive gear.

 

[jbriggs444]

So according you your post, the inner and outer surfaces of the track are moving at the same speed where the track is on the ground and stationary.

It is unhelpful to think of the track as being stationary on the ground. Please shift to the reference frame of the snowmobile.

There are two points on the drive gear where the inner and outer surfaces of the track are moving at the same speed, though not the same direction. Those points are where the track is tangent to the drive wheel. No bend, no curvature, no distortion. What is the speed of the surface of the track at those points? (angular velocity) X (2*pi) X (radius from center of axle rotation to outer surface of the track).

The velocity of the track is UNDEFINED at those points.

Edit: I assume that you are referring to the points where the track transitions from flat to curved. There is a jump discontinuity in speed at those points.

 

[OldYat47]

Sheesh. No, there won't be a "jump discontinuity" at that point or at any other point on the track. I'm just trying to establish some common ground, which is beginning to look like an impossible task.

OK, looking at the frame of reference of the snowmobile, all points of the track make the same number of circuits around the track for a given number of rotations of the drive gear. How would you calculate the forward speed of the vehicle at any given RPM of the drive gear? Equations, please.

 

[jbriggs444]

Sheesh. No, there won't be a "jump discontinuity" at that point or at any other point on the track.

The speed of the track is not a single number. Since it has a non-negligible thickness, its speed varies from inner surface to outer surface as it passes over the drive wheel.

For a track of thickness R on a drive wheel of radius r, the tangential velocity of the outer surface is given by $\omega (r+R)$. At the inner surface the tangential velocity is given by $\omega r$. You have given no reason to equate the former with the speed of the track on the flat rather than the latter.

And yes, there is a jump discontinuity in speed at the transition from flat to curved for all depths within the track except, possibly, for one particular depth.

 

[Tom.G]

Hey guys. Can you all settle on a simplification of considering the 'Track' to be a chain and the 'Lugs' retaining the same attachment as in the OP?
That might bypass getting tangled up in side issues.

 

[sophiecentaur]

Sheesh. No, there won't be a "jump discontinuity" at that point or at any other point on the track. I'm just trying to establish some common ground, which is beginning to look like an impossible task.

OK, looking at the frame of reference of the snowmobile, all points of the track make the same number of circuits around the track for a given number of rotations of the drive gear. How would you calculate the forward speed of the vehicle at any given RPM of the drive gear? Equations, please.

In the simplest model, there will, indeed, be a jump discontinuity. In reality, there will be a change in speed from the section going round the wheel and the straight section (this applies to all the wheels and all the straight sections around the track). If you do not believe that the track flexes then take a rubber belt and stick some non-stretch tape along inner and outer faces and then bend it around a cylinder. You will, of course, notice that the outside paper will split and in inner paper may actually bunch up. You have two different distances around the wheels, traveled by inner and outer faces of the track / belt. That means two different speeds.
What equations did you want? The only relevant equations are
1. Speed of inner surface of track is constant all the way round
2. speed of inner surface = speed of outer surface
along the flat, and:
3. speed of inner surface = R/(R+r) X speed of outer surface
around the curve (R is wheel radius and r is track thickness.)
If both those equations are true then the outer surface must change speed at the transition between curve and flat section but the
4. speed of the straight track section = 2πRf (where f is the revs per second)
BTW, I notice that the reference you quoted was for a circular wheel. We are still waiting for a reference that agrees with your ideas about tracked vehicles.