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

[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.

 

[jbriggs444]

Let me try to flesh out some equations for this.

Position one wheel of radius R with its center at the origin, (0,0). Position a second wheel of radius R at a distance d to the right so that its center is at (d,0). Paint a dot on the inner surface of the track. At time t=0 this dot will be on the top of the first wheel at coordinate (0,R). Refer to the position of the inner dot as $(x_i(t), y_i(t))$. Paint a dot on the outer surface. At time t=0 this dot will be on a distance R above the top of the first wheel at coordinate (0,R+r). Refer to the position of the outer dot as $(x_o(t), y_o(t))$

As per the simple model, the dots move in lock step so that a line perpendicular to the track through the inner dot runs through the outer dot as well. As per the simple model, the distance between the dots is always equal to R.

Assume that the track has a velocity v in the clockwise direction. The dots begin by moving to the right on a stright line path from the top of the first wheel to the top of the second wheel. The position functions for the two dots are given by:

$(x_i(t), y_i(t)) = (vt, r)$ for 0 ≤ t ≤ d/v
$(x_o(t), y_o(t)) = (vt, r+R)$ for 0 ≤ t ≤ d/v

It is clear that differentiating either position function with respect to time will yield a constant speed equal to v for 0 ≤ t ≤ d/v.

At time t=d/v, the dots will have reached a point directly above the axle for the second wheel and will transition from the flat portion of their trajectory to a curved portion. Match the angular rotation rate to preserve the speed of the outer dot by setting $\omega = \frac{v}{R+r}$

NOTE: This is an arbitrary choice. Nothing in the simple model requires this choice. One could equally well choose to set $\omega = \frac{v}{R}$ or use any other arbitrary depth within the track by setting the divisor equal to something in the range [R, R+r]. One could even go a little crazy and imagine scenarios where depths outside that range would be appropriate, but let's keep things simple.

Substituting for omega and putting in an appropriate shift to match positions at t=d/v, the position functions for the two dots are then given by:

$(x_i(t), y_i(t)) = (d+R\sin(\frac{v(t-\frac{d}{v})}{R+r}), R\cos(\frac{v(t-\frac{d}{v})}{R+r}))$
$(x_o(t), y_o(t)) = (d+(R+r)\sin(\frac{v(t-\frac{d}{v})}{R+r}), (R+r)\cos(\frac{v(t-\frac{d}{v})}{R+r}))$

Both valid for $\frac{d}{v} \leqslant t \leqslant \frac{d}{v} + \pi\frac{R+r}{v}$

[I think I have those equations right]

Feel free to differentiate the above with respect to time and take the magnitude of the result to obtain speeds $v_i(t)$ and $v_o(t)$. Everything simplifies dramatically. As expected from the physical situation, the result should be a pair of constant functions:

$v_i(t)=v\frac{R}{R+r}$
$v_o(t)=v$

NOTE: the speed of the inner dot $v_i(t)$ has a jump discontinuity at the time of the transition from flat to curved from a speed equal to $v$ before the transition to a speed of $v\frac{R}{R+r}$ afterward.

Edit: Note that the relevant observation has been made previously in this thread, notably at post #32 by A.T.

 

[OldYat47]

There is no jump discontinuity because the belt doesn't suddenly change thickness, it changes thickness at some finite rate.

I give up. All I'm trying to do is establish the concept of effective radius. Let's try a reductio ad absurdum or two. These are just thought experiments for those unfamiliar with reductio ad absurdum.

Suppose belt A is 1/2" thick and belt B is 4' thick. Their elastic design is such that they actually work. For one rotation of the drive gear which belt will travel farther?

So here's another. Suppose the drive track is a total of 10' long. What if you install 20 flexible lugs, 1/4" thick, each lug 10' "tall" (standing up from the belt). Will that setup travel further per drive gear rotation than the original smooth belt?

 

[jbriggs444]

There is no jump discontinuity because the belt doesn't suddenly change thickness, it changes thickness at some finite rate.

There is, in the simple model, a sharp boundary between the stretched and unstretched state on the outside of the belt or, conversely, between the compressed and uncompressed state on the inside of the belt.

Suppose belt A is 1/2" thick and belt B is 4' thick. Their elastic design is such that they actually work. For one rotation of the drive gear which belt will travel farther?

How do you define distance traveled for a belt that has finite thickness?

So here's another. Suppose the drive track is a total of 10' long. What if you install 20 flexible lugs, 1/4" thick, each lug 10' "tall" (standing up from the belt). Will that setup travel further per drive gear rotation than the original smooth belt?

Will what component of that setup travel further per rotation? Will any component of that setup travel more rapidly than any other component over the flats?

 

[A.T.]

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.

Here is my simplified understanding of the scenario:

The speed of the vehicle on hard surface would be ωr and is not affected by the lug length.

 

[OCR]

The speed of the vehicle on hard surface would be ωr and is not affected by the lug length.

The tip speed... is not affected by the lug blade length.*

Would you say the same thing about an airplane propeller, or a helicopter rotor blade?* Note: My modifications to quote...

 

[A.T.]

Note: My modifications to quote...

If that's supposed to be my quote, please quote in full context. I was talking about the lugs on the straight part, which are not rotating.

 

[OCR]

If that's supposed to be my quote, please quote in full context.

The first quote IS in full context...

 

[OCR]

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.

OldYat47 is correct.....

 

[jbriggs444]

OldYat47 is correct.....

OldYat does not contemplate tracks of non-zero thickness.

 

[sophiecentaur]

Would you say the same thing about an airplane propeller, or a helicopter rotor blade?

Of course not. They do not have tracks.