R1100S - GS length torque arm (365 mm).

[Corvus]

I'm always in doubt, which why I haven't smugly hung up the slide rule and I'm still plugging away to try to get to the bottom of this. It's an interesting one.

My current line of thinking is that maybe the link is never under much in the way of compression and spends a reasonable amount of time pretty neutral. At least on hard Tarmac. My riding against a wall thought experiment lead me to think that maybe it is off road where the link can be subjected to the most compression. We need a scenario where there is no rearward weight transfer whilst full power is being called for. Apart from riding against a wall, off road conditions can give this circumstance.

Make sense or not?

Yes. Lots of sense. Off road or on bumpy tarmac will cause the greatest shock loads under braking or acceleration. Off road is the most likely place to get a sideways whack sufficient to make the 'column' collapse if it's under compression at that moment.

Your thought about the link never being under much pressure during steady riding etc is spot on. Is why I'm happy to tootle to the shops at 35 etc. , but would not consider a long trip etc.

But if I was stuck miles from anywhere with a broken stay arm, say on a an R11 GS, I'd have an idea of how to fix it 'enough' to get moving (slowly).

Throttling up against a wall would be a good one to see. How would Sir like his clutch? Medium rare or well done?

For me, the driving against a wall thing was just a thought experiment, to help me get my head around things. It helped, believe it or not! I was rather hoping you might do the actual field testing on our behalf?

Actually the test would work just fine by turning the crankshaft with the engine off, should there be a suitable nut available (don't look at me!).

The back driving torque thing is also a little tricky to get ones head around. We had a little discussion along those lines a while back, mainly with BMbler. My interpretation is that the directions of rotation don't change (the bit that we can all attest to by experience!), but the load switches to the opposite tooth flank. The reactions also swap direction. Agreed?

[Corvus]

Working loads: I weigh 120 kg. Bike 240 kg. Total: 360 kg. Actually kg wt.
Ratio of wheel radius to brake-stay-mounting-point to axle = 2.14

Just another observation. The link arm is not set at 90 degrees. This will increase the load through it.

[HerrFlick]

Working loads: I weigh 120 kg. Bike 240 kg. Total: 360 kg. Actually kg wt.
Ratio of wheel radius to brake-stay-mounting-point to axle = 2.14

Just another observation. The link arm is not set at 90 degrees. This will increase the load through it.

Hi Corvus,

Yes it will. (Bad miss on my part).




Will get back to you later re your other post.

Cheers

[Corvus]

I'd be worried about shock compressive loads on that arrangement. A small lateral deflection, and the integrity has gone ........

...........

I think this is it in a nutshell, which, I'm learning, is the blackal trademark style! Although he never seems to venture any deeper. What have you got to loose, apart from your reputation! I lost mine ages ago. In fact I don't think I came with one.

The slightest shimmy or wobble in that 3mm ally and its game over. As I mentioned earlier on, there's a metalastik bush in the system. It may help the cause when it comes to mitigating shocks, but it may be the undoing when it comes to a momentary one sided and offset load sent into one of the links.

Herrflick has been accused of not having a clue about dynamics. That doesn't seem to be the case.

Would I now, with further exploration into what may be happening, use herrflicks 3mm bars? Nope. Not a chance.

Looking forward to herrflick's load test once he gets the pukka item on board.

[Blackal]

You don't have to kick something to death - to express your thoughts.

Al

[Corvus]

Working loads: I weigh 120 kg. Bike 240 kg. Total: 360 kg. Actually kg wt.
Ratio of wheel radius to brake-stay-mounting-point to axle = 2.14

Just another observation. The link arm is not set at 90 degrees. This will increase the load through it.

But the angle relative to the paralever joint looks nearer 90degrees.......

[Corvus]

You don't have to kick something to death - to express your thoughts.

Al

No, not if you know. But if you don't know, or are exploring to want to know or for reassurance of your knowing, it ain't so simple. If you seek to help someone else know you'll also need words. I feel.

[HerrFlick]

Working loads: I weigh 120 kg. Bike 240 kg. Total: 360 kg. Actually kg wt.
Ratio of wheel radius to brake-stay-mounting-point to axle = 2.14

Just another observation. The link arm is not set at 90 degrees. This will increase the load through it.

But the angle relative to the paralever joint looks nearer 90 degrees.......

H Corvus.

'F' is at rt angles to the radius line from the centre of the hub (tangential), and always will be, because the backing plate or boss ... whatever ... is being forced to rotate around the rear axle (by the pinion climbing reaction).

If the paralever arm was long enough to have 'F' along the centre line of the arm, 'F' would be the compression force. The angle 'A' = 0 degrees, and Cos A = 1, so C = F/1 = F.

Next (inportant bit), because the arm has pins at each end, and is free to rotate, the reaction forces at the pins must always be along the centre line of the arm.

Now envisage a situation using an arm short enough that the arm and the radius line to the rear axle are almost in a straight line, but not quite. Say 175 degrees, making 'A' = 85 degrees. (5 degrees from the rt angle position, or 175 from the 180 deg straight line)

You would know from your own day-to-day experience that it doesn't take much leverage to get all three points to a straight line, but in doing so you would break either the arm or the hub, because the leverage aspect gets really HUGE.

Compression force 'C' = F/Cos 85 degrees = F/0.087 = 11.5 times 'F'.

The closer we get to 180 degrees, say 179 deg, the greater the effective leverage, and therefore compression force, because Cos 0 degrees = zero, and anything divided by zero = infinity. Yo baby.!

At 'A' = 179 degree, Cos 89 degrees = 0.017, 'C' = F/0.017 = 58.8 times 'F'.

At 'A' = 179.5 degrees, Cos 89.5 degrees = 0.0087, 'C' = F/0.0087 = 114.6 times 'F'.

And at 'A' = 179.9 degrees, Cos 89.9 degrees = 0.0017, 'C' = F/0.0017 = 574.1 times 'F'.

You can see why the compression forces become so large. If there was a big spring and slider in place of the strut it would be the basis of an over-centre mechanism.

Back to the existing situation where the angles are as per the pic, the compression force 'C' is 1.3 times the force 'F', 'F' having been calculated from the inertial reaction of the weight of the bike/rider when the clutch is dropped.

Does this make more sense?

Cheers

Herr Flick.

[Corvus]

Quote herrflick: "does this make more sense?"

Not a lot more, no!

I can visualise things though.

Help me out by talking me through what will happen in my little thought experiment.

We glue the rear tyre to the floor. I put a long spanner on the engine crankshaft and turn the usual direction. You watch what happens at the paralever/bevel box. If you think it helps me understand better, we can put a spring balance in place of the link arm. Now tell me what happens as I turn the crank.

It will be a series of levers, but you can explain to me what levers against what. Mathematics are out I'm afraid.

Can you help?

[HerrFlick]

Quote herrflick: "does this make more sense?"

Not a lot more, no!

I can visualise things though.

Help me out by talking me through what will happen in my little thought experiment.

We glue the rear tyre to the floor. I put a long spanner on the engine crankshaft and turn the usual direction. You watch what happens at the paralever/bevel box. If you think it helps me understand better, we can put a spring balance in place of the link arm. Now tell me what happens as I turn the crank.

It will be a series of levers, but you can explain to me what levers against what. Mathematics are out I'm afraid.

Can you help?

Yeah Corvus. Well I think so.

But I'll have to get back to you in a day or so. Up to my armpits in crocodile commitments at the moment.

[Corvus]

........

'F' is at rt angles to the radius line from the centre of the hub (tangential), and always will be, because the backing plate or boss ... whatever ... is being forced to rotate around the rear axle (by the pinion climbing reaction).


......

In the case of the old solid swingarm boxer I can see the relationship between bevel gear torque reaction and swingarm pivot, but the paralever system complicates things.

There's a lot of pivots here so it's easy to become distracted by them all! I'm speaking for myself here.

With respect to the relationship between the wheel centre point, the paralever bearing centre point and the link bar end centre point. The three pivot points in your photo, basically.

Isn't the torque reaction felt at the paralever bearing point and then magnified into the end of the link bar by virtue that the paralever bearing point lies pretty much at 90 degrees to the link bar end point, but the distance is shorter?

You can visualise either the paralever bearing or the link bar pin as the "primary" pivot point for torque reaction. It doesn't (seem to me to) matter? The resultant force value (but not direction) is the same either way.

If I'm on the right lines then im afraid it's looking a lot worse for your 3mm ally.

But I confess, this needs careful scrutiny because it's complicated, or at least there are distractions in all the various pivot points, and I can only devote small intervals of time to it.

I certainly look forward to your further thoughts and indeed the thoughts of anyone who can help me fully get to grips with this.