Longer lasting cush drive material available

[dlaing]

Is "noticeably smoother" a good thing.

If you put a softer spring on your shock absorber, you may get a smoother ride over the smooth bumps, but the bumps that will break your spine are the nasty unexpected potholes that bottom out the shock and move your tailbone closer to your skull.

If you simply soften up your cush by drilling, you may get a smoother feel over the smooth normal shifts, but the shifts that will break your spline are the nasty unexpected shifts and aggressive clutch actions that make the front wheel come up or the rear wheel lock.

Is this compressing the rubbers to a bottomed out state?

I stuck one of the wedges in my door jam and started closing the door and watched as the rubber bushing wedge that is not chrome-alloy hard effortlessly mushed into a flattened out pancake.

Without some sort of meter measuring movement in the cush housing, I ain't sold on the drilling.

Or maybe someone can do the math to convert the torque at the cush to the torque measured on the dyno (about 60 foot pounds, although I get with clutch action you get the torque to momentarily be a bit higher).

If the wheel base is about 58.66 inches and the front of the bike with rider weighs 250lbs and one can pop a wheelie, doesn't that mean that we can divide 58.66 by 12 and get roughly 5, and multiply 5 times the 250pounds we get about 1250 foot pounds of torque.

Please correct me if I am wrong, but I think that would be like putting a wedge in the crook of a one foot lever and squashing down with about 208lbs of force. (1250 foot pounds spread over 6 wedges.(there are 12 wedges altogether, right?))

Assuming I got the math and theory right, it should be easy to test a wedge and see if the holes help.

Any "holes" in this theory?

[Phil A]

Any "holes" in this theory?

Yep heres one.......

I stuck one of the wedges in my door jam and started closing the door and watched as the rubber bushing wedge that is not chrome-alloy hard effortlessly mushed into a flattened out pancake.

 

The rubbers are contained and cannot escape to be squished flat.

[Guest ratchethack]

Any "holes" in this theory?

Well, Dave, I reckon you put the question out here expecting a response.

If the wheel base is about 58.66 inches and the front of the bike with rider weighs 250lbs and one can pop a wheelie, doesn't that mean that we can divide 58.66 by 12 and get roughly 5, and multiply 5 times the 250pounds we get about 1250 foot pounds of torque.

Let's see if I've got this straight. Divide the wheelbase in inches by the number of cush drive rubber blocks. Multiply by the laden weight of the front end of the motorcycle in pounds. 1250 lbs./ft. of torque. Hm. Yes, yes. . . all this makes perfect sense.

 

1250 lbs./ft. at the cush drive from a motor that produces 65 lbs./ft.

 

Nope, nuthin' faulty with any o' the applied math or physics here.

 

I guess I never realized we've all got the pulling power of a D9 Cat!

 

 

But why stop there? Why not multiply somethin' by the bore, and throw in stroke while y'er at it??

 

Then we could have the torque of a tugboat. Ever check one o' those things out?? They've been described as a massive pair of deisel powered locomotives in the water wrapped in a tiny hull with a little pilot house on top. Simply amazin' things.

 

 

Think I'll fire up the ol' Guzzi and pull a few stumps this mornin'. While I'm at it, maybe I'll see if there are any freighters run aground outside the channel cut in the harbor that I can pull free.

[dlaing]

Yep heres one.......

 

The rubbers are contained and cannot escape to be squished flat.

Not true. I thought they might be contained and that I might use the containment as a mold, but there is indeed room for the rubber to expand.

But there may be a good point that the torque deflection would have to be measured while in the containment of the cush housing, as there is some containment.

Thanks!

[Guest ratchethack]

Well, the return of this subject finally got me off my lazy ass and in the garage to drill my pucks. I found the situation inside my bike was really strange. Things started to get scary when all the button head screws were properly tightened and came out with no heat or damage to the screw heads. Then I opened things up and found almost no rust and everything in pretty good condition for a bike that has been on the road for five years. Once I got over the shock (Guido actually greased everything and torque the screws correctly ) I cleaned things up, drilled the pucks, greased things and put it back together. I can't say it was night and day but the drive system is noticeably smoother. The bike is more pleasant to ride and it has to be easier on the U-joints and spines. Maybe another material would be even better, I can't say since I lack Ratchethack's physic engineering ability, but the drilled pucks are better then stock. I used a normal twist bits so my work doesn't look quite as spiffy as Greg's but it is pretty close. Greg didn't give a speed but I used 700 RPM, the slowest setting on my press, and it worked well. I found it easiest to just hold the puck in my hands while I drilled it. If something better comes along I'll look at it again but for now I'd call it and hour or so well spent. Make that two hours if you include the time to clean up the big pile of rubber drilled out of the pucks. I would recommend anybody who hasn't done so to read Greg's post and follow his directions.

 

Lex

Lex, may I offer contratulations. It seems that you've managed to extract a worthwhile and satisfying result from wot's become a mountain o' some o' the most incredible windage ever posted hereabouts -- and I reckon that's sayin' something! Yes, indeed. One more successful case, mercifully closed, to add to the long long historic list of triumphs of common sense over a mighty big fresh heap o' seemingly neverending folly.

 

From your account and others recently, It seems there's no predicting wot kinda sauces & condiments the Luigi's might've been inclined to use (or not use ) on cush drives on any particular day in the happy little workshop on the shore of Lake Como. I reckon it might've had somethin' to do with wot Mama Luigi happened to be cookin' up at the time? Any telltale evidence of canoli's in there?

 

But d'you feel guilty for not making an open-ended science project out of this? Did you find there was a Great Wall of fear and anxiety you had to scale before actually penetrating virgin cush drive blocks with a sharp drill bit, knowing that so many have so many times (and so earnestly) suggested that there's an "obviously better" alternative (which no one seems to've ever witnessed on a V11)? Any remorse a-tall over not spending any money, and only a few hours of your time? Incidentally, I think it was 20 minutes for Yours Truly for the drilling part, but I already had the wheel out for a tire change when I did mine 3 years ago. It was such a plain, simple, easy thing that I didn't even think about posting wot I'd done here at the time, any more than I'd have thought about posting about a tire change. Any controversy wotsoever over this wasn't even on my radar screen at the time. Shame on me for my prosaic lack of imagination. Why, just look wot's happened!

 

Enquiring minds. . . (well, you know)

 

Oh wait -- Have I demonstrated "physic engineering ability"? All this time I just thought wot I'd done was simply follow another boring, common sense "cookbook" procedure here, expecting the same kinda results as everyone else who'd already done it for decades.

 

Ain't human nature just a never-ending source of wonder and fascination? I often find myself whelmed. Not overwhelmed, mind you. But truly whelmed.

[dlaing]

Well, Dave, I reckon you put the question out here expecting a response.

 

Let's see. Divide the wheelbase in inches by the number of cush drive rubber blocks. Multiply by the unladen weight of the front end of the motorcycle. 1250 lbs./ft. of torque. Hm. Yes, yes. . . all this makes perfect sense. And 1250 lbs./ft. at the cush drive from a motor that produces 65 lbs./ft.

 

Nope, nuthin' faulty with any o' the applied math or physics here.

 

I guess I never realized we've all got the pulling power of a D9 Cat!

 

But why stop there? Why not multiply somethin' by the bore and/or stroke while y'er at it??

 

Then we could have the torque of a tugboat. Ever check one o' those things out?? They've been described as a massive pair of deisel engines in the water wrapped in a tiny hull with a little pilot house on top. Simply amazin' things.

 

Think I'll fire up the ol' Guzzi and pull a few stumps this mornin'. While I'm at it, maybe I'll see if there are any freighters run aground outside the channel cut in the harbor that I can pull free.

Is the dyno measuring the torque the 60 foot pounds at the axle?

I don't think so.

That would mean that if put a little more than 60 pound bag of cement on a foot long lever coming off the rear axle, that front end of the bike would lift the bike into the air!!!!

Something is wrong with your math.

But, assuming you are correct that there is only 60 foot pounds of force(plus a little from momentary bursts of a popped clutch), I don't see how that could effect the massive splines, and I don't think we need any cush unless it is not torque that is threatening our splines, but rather vibration eating at the splines and the whole purpose of the cush is to dampen the vibration, not absorb the force of the rather miniscule 60 foot pounds of axle torque.

If that is the case we should be looking an elastomer with excellent vibration suppression.

But I think your math is off.

I think that while the dyno is measuring RWHP, it is not measuring axle torque.

The measured torque is probably an estimate of the torque at the crank.

Geared down to first gear where we can pop a wheelie that torque is multiplied.

But how much?

Can we simply use the ratios from the owner's manual?

first gear is 1:11.7589 (engine-wheel)

So, 60 foot pounds would be multiplied by 11.7589 to give us 705.534.

I know my bike with roughly 60 foot pounds of torque can't lift the front wheel without a bit of coercion.

How much more torque can we get by popping the clutch?

We know it is enough to bring the front wheel up, and I know that my bike won't wheelie from simply giving it 705 foot pounds of torque at the rear axle.

It does seem unlikely that we would have to escalate the force from 705 to 1250 just to make the front wheel come off the ground.

Assuming my 1250 number is correct, that would correspond to 106 foot pound of engine torque following the owner's manual's 11.7589 ratio.

That does seem a bit high. The true force to lift the front wheel is probably somewhere between 1250 and 705 foot pounds.

[Guest ratchethack]

Something is wrong with your math.

I'm clearly on a different planet altogether. Help me.

. . . I think your math is off.

Set me straight, please. I'm begging you.

 

Hilfe, Mr. Wizard!

 

Aufenthaltsort Wernher? Aufenthaltsort Albert?

[dlaing]

Set me straight, please. I'm begging you.

I already did.

I have quite probably put a major hole in the drilled cush rubber bushing theory.

With one of my formulas for popping a wheelie, we get 1250 foot pounds of torque at the axle. With the dyno measured engine torque applied to the gear ratios I get 705 foot pounds of torque.

But you never really offered any math, just your usual hyperbole and sarcastic approval of my math.

Ratchet, please answer these questions:

Approximately how much torque is each acceleration bushing enduring when one pops a wheelie?

Approximately how much torque is at the rear axle for our bikes when in first gear if the engine puts out 60 foot pounds of torque?

And please show the math you used to arrive at the answers.

 

The splines are quite likely more prone to damage after drilling the cush rubber bushings.

But to prove it, we might need to set up a jig to apply 1250 foot pounds of force, divided by six bushings....assuming that the 1250 foot pounds of torque is about the maximum pressure that the six acceleration bushings would have to endure.

Then again, maybe it is not that big a deal for it to "bottom out" at wheelie popping torque. Perhaps if it bottoms out at only 600 foot pounds or so, it has already protected the splines from nearly half the 1250 foot pound force, and maximum torque can be delivered more efficiently?????

Is the lost protection of the splines worth trading for perceived smoothness (vagueness)?

What really concerns me is the people chucking out half the bushings and drilling the other half.

[Ryland3210]

On dyno charts which show graphs of torque and horsepower versus RPM. the torque is not the torque delivered by the rear wheel. For example, in the case of the typical inertial dyno, its computer calculates the force exerted by the tire by measuring the angular acceleration of the dyno's known rotational inertial mass. Power can be calculated by various means from that information. One way is to use the relationship: dyno torque = dyno rotary inertia X dyno angular acceleration and the relationship power=dyno torque X dyno RPM. Once it knows the power, it can use the relationship engine torque=power/engine RPM to determine the equivalent engine crankshaft torque. This is not the actual crankshaft torque, since it has been reduced by the friction losses in the drive train.

 

The cush drive material debate pops up from time to time simply because there has been insufficient quantifyable data to permit an objective comparison and lay the question to rest. Until that happens, we should not be surprised or perplexed to see it return someday, along with opinions in place of logic. Repetition of strongly held views is likely to occur when one's cherished views fail to persuade another. I occasionally remind myself of the phrase: "A man convinced against his will is of the same opinion still". That's life.

[dlaing]

The cush drive material debate pops up from time to time simply because there has been insufficient quantifyable data to permit an objective comparison and lay the question to rest. Until that happens, we should not be surprised or perplexed to see it return someday, along with opinions in place of logic. Repetition of strongly held views is likely to occur when one's cherished views fail to persuade another. I occasionally remind myself of the phrase: "A man convinced against his will is of the same opinion still". That's life.

So, is the logic beyond our capability?

I'll stick to my 705 and 1250 foot pounds of torque until proven otherwise.

This may be worth pursuing.

A few hundred dollars in R&D to sell a few sets of rubbers that only the truly blessed can understand the benefits of.

Maybe I could prove to myself exactly how many holes it takes to optimize the OEM rubbers for 1250 foot pounds of torque? Assuming they are not already optimized for less torque?

[Guest ratchethack]

But you never really offered any math, just your usual hyperbole and sarcastic approval of my math.

You already set me straight?? But. . . but Dave. I thought there was something wrong with my math?!?! You said so twice above.

Ratchet, please answer these questions:

Well, this is a bit of a silly academic exercise without much point, as far as I'm concerned, and it's strictly foolin' around for sh!ts & giggles, but I'll take a rough swag at it with the disclaimer that there's only so much o' the great massive goofy balloon barrage you've launched that I can shoot down now, since there's not enough ammo in this part o' the entire countryside for that, I'm down to only half an ammo can now, and I gotta go re-arrange the old spice rack real soon. I'm not about to take my cush drive apart for measurements, so calculations are out o' the question today, or any day until I'm convinced there'd be even the slightest justification to have any interest in the first place. But here goes an approach to the problem at least, straight off the top o' me head.

Approximately how much torque is each acceleration bushing enduring when one pops a wheelie?

I'm gonna focus in on the word, "each" here, because I b'lieve this is actually wot you're interested in. But I've gotta make some fundamental corrections before I get any further here.

 

Torque is a measure of rotational force. It's expressed in force over distance. More technically, it's the vector product of the radius vector from the axis of rotation to the point of application of the force and the force vector. Though it's perfectly legitimate to consider the torque force on ALL cush drive blocks at the same time (that's a given here, it's 65 lbs./ft. max), when considering the force at an individual block, there are no rotational forces applied on each of them, since there is no axis of rotation through any block. A measure of force on each block would involve a relatively complex calculation of vectored angular (again, not rotational) force properly expressed either in weight/area, or weight/volume at each block, measured in lbs./sq. inch, kg/cm^2, or in lbs./cu. inch, kg/cm^3, or on scales such as these. One approach would be to consider the blocks to be solid for purposes of force calculation. This would be the weight/area calculation. In this scenario, the angular force at the small end isn't the same as at the large end, or at any point between. An accurate calculation for the irregular area of the block, perpendicular to the force vector, would be far beyond my training and ability to perform.

Approximately how much torque is each acceleration bushing enduring when one pops a wheelie?

Popping a wheelie is entirely irrelevant to calculation of max force on cush drive rubbers, since you can achieve a wheelie with an infinite scale of motor torque at torque values below max torque and full load, all the way up to the max torque of the motor under full load.

 

There can be no force greater on the cush drive blocks than the engine can deliver at max torque capacity and full load (at approx. 5K RPM, as on a brake dyno, or on the road, whether a wheelie is in process or not). Since there are 6 drive blocks, and the max torque at the rear wheel is 65 lbs./ft., you'd have to take radius measurements at each end of the block in place in the hub to determine min and max radius, and determine the pressure on each block accordingly.

 

An alternate (but I b'lieve equally valid) way to calculate the pressure on each block would be the weight/volume calculation. In this scenario, one would assume each block to be contained in a walled chamber equal in volume to the volume of the block, since the compound will behave somewhat like a liquid under compression, and then calculate the overall pressure of each one this way. This method is an approximation that requires an assumption that the blocks cannot be deformed beyond their relaxed shape, yet be considered a fluid, with equal pressure exerted on all points of its surface (container). This is not quite corrrect, but it would get you "close" for comparison purposes, and it seems lots simpler for purposes of calculation than using all the advanced calculus of the method above. But that's just me. Oh yeah. You'd have to forgo the ability to use this method with drilled blocks for comparison purposes, since holes considerably alter the compressibility of the material, as well as the volume.

 

Though I've had the course, so to speak, on Physics at undergrad level and minored in it, I've never taken engineering, and I'm no mechanical engineer (that'd be me ol' man, who taught me a great deal of this stuff). But I b'lieve there are a few ME's hereabouts. I ain't smart enough, nor would I be qualified to attempt wot to me would be a horrifically complex calculus to properly compute this -- even if it were important to me. I happily defer to those so trained -- and so motivated.

Approximately how much torque is at the rear axle for our bikes when in first gear if the engine puts out 60 foot pounds of torque?

Max torque at the rear axle may be achieved in any gear, as long as full load can be achieved at max torque output.

 

There is no doubt in my mind that somebody's gonna want to take a few repeated, long bursts with the ol' 88's at wot I've just run up the flagpole above the ol' aerial bombardment field. But I'll stand by the basic logic of it until corrected by superior physics and logic, which most certainly exists out in all directions from here.

 

That's all I got. Gotta go now.

[dlaing]

Thanks for making the effort!

I'm gonna focus in on the word, "each" here, because I b'lieve this is actually wot you're interested in. But I've gotta make some fundamental corrections before I get any further here.

 

Torque is a measure of rotational force. It's expressed in force over distance. More technically, it's the vector product of the radius vector from the axis of rotation to the point of application of the force and the force vector. Though it's perfectly legitimate to consider the torque force on ALL cush drive blocks at the same time (that's a given here, it's 65 lbs./ft. max), when considering the force at an individual block, there are no rotational forces applied on each of them, since there is no axis of rotation through any block. A measure of force on each block would involve a relatively complex calculation of vectored angular (again, not rotational) force properly expressed either in weight/area, or weight/volume at each block, measured in lbs./sq. inch, kg/cm^2, or in lbs./cu. inch, kg/cm^3, or on scales such as these. One approach would be to consider the blocks to be solid for purposes of force calculation. This would be the weight/area calculation. In this scenario, the angular force at the small end isn't the same as at the large end, or at any point between. This calculation would be far beyond my training and ability to perform.

All I am interested is the torque at the rear axle divided by the number of bushings.

So, your answer of 65foot pound maximum is about what I expected from you based on your earlier reply. So, I shall stick to the idea that your math is wrong.

Technically you make a good point about the actual force on the bushing. It is not torque.

But I don't need to know the exact force at each bushing. All I need to know is how much torque is generated at the axle. The torque can then be reapplied in a test environment with the bushing the same distance from the center of the axle.

Once I know what the torque is, I can take a torque wrench and fit it to the splines (not trivial), insert one bushing into the cush housing(or maybe two for balance) and turn the torque wrench measuring the compression of the bushing at various points of applied torque.

Given your 65 foot pounds of maximum force, the deflection of a full set of bushings would be pretty minimal.

This is where we don't see eye to eye, and I believe the force can get up to about 1250 foot pounds of force at the rear axle when popping a wheelie.

 

Popping a wheelie is entirely irrelevant to calculation of max force on cush drive rubbers, since you can achieve a wheelie with an infinite scale of motor torque at torque values below max torque and full load, all the way up to the max torque of the motor under full load. Once the bike is balanced on the rear wheel, no load need exist, and no torque would be required whatsoever.

 

There can be no force greater on the cush drive blocks than the engine can deliver at max torque capacity and full load (at approx. 5K RPM, as on a brake dyno, or on the road, whether a wheelie is in process or not).

It is not irrelevant.

If I apply WOT in first gear at 3000 RPM and hold it up to about redline, my front wheel does not lift unless I hit a bump in the road.

However if I pop the throttle at about the point of maximum torque, I can exceed maximum torque as I get added force from the momentum of the driveline as it un-slackens. Likewise popping the clutch can help exceed the rated engine torque applied to the rear wheel.

Yes, once the front wheel is up in the air, I imagine there is no significant torque.

I am concerned with the torque that got it up, not what keeps it up.

 

Since there are 6 drive blocks, and the max torque at the rear wheel is 65 lbs./ft., you'd have to take radius measurements at each end of the block in place in the hub to determine min and max radius, and determine the pressure on each block accordingly.

An interesting academic exercise, but I am just looking torque at the rear axle.

Sorry I did not make that clear earlier.

 

An alternate (but I b'lieve equally valid) way to calculate the pressure on each block would be the weight/volume calculation. In this scenario, one would assume each block to be contained in a walled chamber equal in volume to the volume of the block, since the compound will behave somewhat like a liquid under compression, and then calculate the overall pressure of each one this way. This method is an approximation that requires an assumption that the blocks cannot be deformed beyond their relaxed shape, yet be considered a fluid with equal pressure exerted on all points of its surface (container). This is not quite corrrect, but it would get you "close" for comparison purposes, and it seems lots simpler for purposes of calculation than using all the advanced calculus of the method above. But that's just me. Oh yeah. You'd have to forego the ability to use this method with drilled blocks for comparison purposes, since holes considerably alter the compressibility of the material.

Interesting, but way over my head.

 

Though I've had the course, so to speak, on Physics at undergrad level and minored in it, I've never taken engineering, and I'm no mechanical engineer (that'd be me ol' man, who taught me a great deal of this). But I b'lieve there are a few ME's hereabouts. I ain't smart enough, nor would I be qualified to attempt wot to me would be a horrifically complex calculus to properly compute this. I happily defer to those so trained -- and so motivated.

I would have pursued physics or engineering but I was too undisciplined.

I wish the Mechanical Engineers would speak up, but we may have bored them away by not fighting over unmentionables.

But no calculus is necessary.

We just need to know how much torque at the rear axle is required to lift the front wheel.

 

Max torque at the rear axle may be achieved in any gear, as long as full load can be achieved at max torque output.

I believe the torque at the rear axle is greater in lower gears.

This is why it is easiest to wheelie in first gear.

There is no doubt in my mind that somebody's gonna want to take a few repeated, long bursts with the ol' 88's at wot I've just flown up above the ol' aerial bombardment field. But I'll stand by the basic logic of it until corrected by superior physics and logic, which most certainly exists out in all directions from here.

Sincere thanks for giving it a shot.

That is certainly something I like about you.

You don't back down from something just because it requires you to read, research, or use the ol' grey matter!

[Phil A]

Well, before the testing begins for squishibility, practicability, power transmission ability and durability I would like to register my design. :D

 

 

[dlaing]

Well, before the testing begins for squishibility, practicability, power transmission ability and durability I would like to register my design. :D

You stole my idea!!!!! <_>

Just kidding, Your idea certainly is not swiss cheese overkill.

My idea was very different and even more minimal in effect, so much so that it did not make a noticeable improvement.

I did one single drilling from the outside of the pie towards but not all the way to the tip of the triangle.

Something like this with drill hole in red

| ..\\

| ........\\

| ..................\\

| ............................\\

|...........................................\\

 

===================:::::>

|...........................................//

| .............................//

|...................//

|..........//

| ..//

 

Is that abstract enough?

The genius theory behind it is that as the wedge is pushed toward the the outer wall, it seals the hole, and then as the wedge is squashed, the cavity retains air and acts as an air cushion.

FWIW I only did it to the decelerating wedges.

Patent Pending....

[Troy]

So, 60 foot pounds would be multiplied by 11.7589 to give us 705.534.

I know my bike with roughly 60 foot pounds of torque can't lift the front wheel without a bit of coercion.

How much more torque can we get by popping the clutch?

We know it is enough to bring the front wheel up, and I know that my bike won't wheelie from simply giving it 705 foot pounds of torque at the rear axle.

It does seem unlikely that we would have to escalate the force from 705 to 1250 just to make the front wheel come off the ground.

Assuming my 1250 number is correct, that would correspond to 106 foot pound of engine torque following the owner's manual's 11.7589 ratio.

That does seem a bit high. The true force to lift the front wheel is probably somewhere between 1250 and 705 foot pounds.

 

Dave, I think you should consider that a. The force at the rear wheel contact patch is generating a moment around the CG of bike and rider and b. Rear wheel torque then needs to equal this force at the contact patch times wheel radius.

 

Assume the height of the CG is 2ft, (wild guess). The force at the patch would be (1250ft-lb)/(2ft) for 625lb. The rear tire has a radius about 13", so torque at the wheel would need to be 625lbX((13/12)ft or 677ft-lb. Your 705ft-lb then puts you in wheelie land. (But this is too easy. Maybe the front wheel gets more than 250lb and/or maybe the CG is lower than 2ft.)