Rear wheel with rotor brake substitute

Well, I finally took the Kawaverdasaki out this arvo after the conversion and I couldn't be more pleased with the result. It's like a new bike. Fair enough,it does have a new tyre on but the real improvement was getting rid of that heavy wheel, all 5kg+ of it. Turning control is very responsive, acceleration has picked up, and the rear brake feels great in trail braking.
The old alloy wheel used to feel so sluggish like you had to drag it around corners. I think it had to do with the rotational inertia of the heavy wheel acting like a stabilising gyroscopic force resisting change of turning direction. Anyway I couldn't help thinking while riding that Laverda took a perfectly good wire-wheeled bike in 1975 and degraded it with the heavy alloy wheels. Anyway, I think we've worked over this thread enough to help anyone interested in the KLR650 mod.
Thanks for all your input guys, made the job much more enjoyable (and entertaining!)
View attachment 59155
Good result, Chuck. Still blown away by that weight reduction! Where is all thew weight in the cast wheels?

I'll weigh the Triumph and Z1000 wheels and compare - definitely heavier, as they are a lot beefier. I initially chose the KLR hub because it's not a super-light dirt bike but it is dual purpose so should have a reasonable degree of inherent strength and reliability. I might have to build a 3rd 17 inch wheel using the KLR hub now!
 
Could be Marty. My Tiger XCX runs wire 17 rear 21 front!

And my friends Triumph Scrambler runs wire 17 rear, 19 front (and chain on the right like the Lav.)
Interesting thing on that factory built model, Triumph used a right hand brake caliper assembly and flipped it over onto the left side upside-down!
Triumph are odd - the Legend hub I'm using ran LH chain std, so flipped for the Lav have more cush rubber on trailing than on accelerating, but many other models I've checked out use RH chain. Maybe those XCX hubs would be a nice fit, but doubt VERY much you'd find (m)any for sale, let alone at anything like the price of the KLR.

Is the Scrambler a modern bike? What width rear rim? And on the XCX? I do like looking at spoke wheel options - some lightweight dirt bikes have very light rear hubs but don't have enough cush for my liking on a heavy road-going Laverda.
 
Triumph are odd - the Legend hub I'm using ran LH chain std, so flipped for the Lav have more cush rubber on trailing than on accelerating, but many other models I've checked out use RH chain. Maybe those XCX hubs would be a nice fit, but doubt VERY much you'd find (m)any for sale, let alone at anything like the price of the KLR.

Is the Scrambler a modern bike? What width rear rim? And on the XCX? I do like looking at spoke wheel options - some lightweight dirt bikes have very light rear hubs but don't have enough cush for my liking on a heavy road-going Laverda.
I just mentioned the Triumph scrambler and the Tiger because I was looking at the Scrambler (2015??) yesterday and talking about front sizes, but in both cases the Triumph wheels are too wide for the Laverda.
I'm convinced now the heavy Lav alloy was acting like an un-sprung weight gyroscope,
degrading turning response. I immediately felt the difference. Anyway I recommend the KLR wheel in the configuration shown in the pic here to anyone interested.
Wheel diameter at the rubber PXL_20211002_073757307.jpgPXL_20211002_073837734.jpgPXL_20211002_074248674.jpgis almost the same as the old Lav alloy.
The brake caliper position is fine. No issues whatsoever. And the bracing rod was changed to 16mm dia steel. The only thing left to do is put a 40T sprocket on if Red can make one.(he's looking into it for me).
 
Thanks Marty, I'll do a bit of research at Titmans.
(I haven't heard back from Red, I think he might be busy with other things right now.)
 
Don't bother trying to get any Borani or Morad rims from them, they are always out of stock. Be interesting to see how they are with sprockets.
 
Hi Chuck. That's a neat wheel conversion. Well done to get it all together.

Steve has given you a bit of sensible engineering input regarding the aspect ratio of a strut loaded in compression. He also said that "an engineer would analyse the loads and calculate the 'critical stress' and yada yada yada.... "
Naturally, as an engineer, I took that as a challenge. :)

I had to make a few assumptions: I allowed 150kg (half of the bike /rider combo) for the weight on the rear tyre.
Coefficient of friction between tyre and road of 1.0 max, so the breaking force at the tyre is also 150kg force (1500N).
I guestimated the lever arm radius from wheel axle to the torque arm mounting on the caliper to be 70mm, and the rear tyre radius to be 600mm.
A simple torque calculation reveals that the compressive load on the torque arm is around 12900N (roughly 1.3 tons force).

Now to the complicated bit. Using Steve's estimate of 200mm M8 rod for the torque arm, and applying Euler buckling formula for a pin-jointed bar, the buckling load for that rod comes out at around 5.8 tons, so you have a safety factor of more than 4. That would be considered ample for most engineering applications.

Of course all of the above is contingent on the assumptions being somewhere near reality. I think the torque arm looks longer than 200mm. I'd guess 400mm between the heim joints at each end. In that case the safety factor reduces to about 1, which means it's in imminent danger of buckling (a saferty factor of less than 1 means it'll fail). However, it has a threaded sleeve in the middle that would stiffen it up a bit so maybe you'll be OK.

BTW, the bolts through the ball joints are marginal. The shear load capacity of an M8 x grade 8.8 bolt is 14000N, so you only have a 1.1 safety factor there.

The torque arm attachment at the caliper end looks too close to the wheel axle. So an obvious improvement that you could make would be to attach the torque arm to the caliper mount at a greater radial distance from the axle. That would massively resuce the torque multiplication between tyre and torque arm, which would also reduce the load on the M8 bolts in shear. But that looks easier said than done because I can't see a handy attachment point in your photos.

There is a nice attachment point further from the axle on the rear side of the caliper. I presume the wheel was the other way around in the Kawasaki (brake on right side) so maybe that's the bolt that Kawasaki used as the anchor point.
Is it possible to un-bolt the caliper from its mounting plate, flip the plate over and reassemble? That way, you'll have that bolt available in front of the caliper to fix your torque arm to.
If flipping the caliper plate isn't an option, then you could swing it around so the caliper is under-slung as someone else suggested. That'll allow you to use that attachment point further from the axle, with the rod in tension.

If neither of the above suggestions are acceptable, how about re-using the original Laverda/brembo mounting plate and caliper? You might have to modify it a bit to suit the new disk diameter and axial spacing, but that shouldn't be too difficult for a man of your ability.

Alternatively, you could take on Vince's timber engineering rule of thumb and use a lump of 4"x 2" hardwood. :)

Good luck with the project.
Thanks for going to all that effort Cam. I haven't done this analysis yet, but I think if we do we should approach it by estimating the force on the rotor from momentum, m*v, not weight. That is, calculate worst case scenario like: starting to brake at, say 160km/hr, rear brake only.
Then we have momentun, mass*velocity. Then estimate stopping time with this force, so equivalent Force*time, road is equal and opposite force on tyre, then work back to rotor torque. The distance from axle centre to rotor at pad centre is 100mm, and to tyre surface (wheel radius) is 325mm.

I'm starting here, so if you agree:
RwWheel radius =0.325m
RrRotor radius =0.1m
MMass =290kg
SSpeed =44m/s
DsStop Distance =200m
TsStop Time =20s
drate of deceleraton2.2m/s/s(S-0)/(Ts-0)
FfForce =638NM*d
TwWheel Torque =207NmFf*Rw
TbBraking Torque =2,074NmFf*Rr/Rw

If I dust off my slide rule I'll have a go. For now, I've put a rod on that will stop a train!😉. The rod is 210mm and 16mm dia.
Chuck
 
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Thought I'd wade in and add my 2 cents worth...I agree with earlier opinions regarding the tie rod in compression, you need to be careful there.

I wouldn't bother changing the caliper/carrier combination as it works so why make extra problems for yourself. There is an advantage to using the single sided sliding calipers in that they don't extend very far on the inside so don't have clearance issues with the spokes.

However you may have notice that the caliper has two pistons of different sizes. This is done to help equalise pad wear. The smaller piston presses on the leading edge of the pad - which is most prone to wear - with less pressure than the larger piston, which presses on the trailing edge of the pad. However it was designed to work when installed on the right side of the wheel, standard modern practice, but the Laverda has it on the left side. So the small brake piston is now bearing on the trailing edge of the pad and the large piston on the leading edge of the pad. This will work, but you may expect the tendency of the pads to wear faster on the leading edges than the trailing edges will be increased. I don't know the long term implications of this but I'm sure that you'll keep us posted!

cheers,

bazzee
That shouldn't be any problem in my case bazzee, both pistons in this KLR650 brake caliper are exactly the same size, 1 in. diameter.

I did change the rotor direction today (opposite the picture) so the drilled holes throw dirt/water inside-to-outside correctly. Forgot to do that when I first put them on.

Chuck
 
That shouldn't be any problem in my case bazzee, both pistons in this KLR650 brake caliper are exactly the same size, 1 in. diameter.

I did change the rotor direction today (opposite the picture) so the drilled holes throw dirt/water inside-to-outside correctly. Forgot to do that when I first put them on.

Chuck
Ah, ok, I couldn't tell from the photos.

cheers,

bazzee
 
I think if we do we should approach it by estimating the force on the rotor from momentum, m*v, not weight ... so if you agree
Interesting approach, but you shouldn't have asked if I agree. I don't. :)

I think treating the rear wheel as an isolated spinning flywheel without considering road forces is not representative of the true picture. The rotational inertia will certainly add to the braking forces applied by road/tyre friction. So perhaps adding both components (rotational inertia + tyre forces) will give a more complete answer.

Anyway, your maths is flawed. Brake rotor radius is not relevant to the force applied at the torque arm.
Also, you can't apply Newton's laws of motion to rotational systems.
For example, calculating the force by applying F=m*a is fine for linear motion, but it doesn't work for a rotational system. You need to work out the rotational moment of inertia of the wheel in kg.m2 (not just its mass in kg), angular speed and acceleration, etc.

Sorry for giving you a fail mark on your homework ;)

Cheers,
Cam
 
Jeez, this is getting technical….. if it works, and looks in keeping, works for me. I doubt that the brakes on a chuck wagon (pardon pun) had that much technologically theory behind it.

Then again, they fucked up…. Big time.

I bet they didn’t take into consideration how fast the wagon wheels go backwards, when going forward … I’ve seen it for myself on TV. :whistle:
 
Interesting approach, but you shouldn't have asked if I agree. I don't. :)

I think treating the rear wheel as an isolated spinning flywheel without considering road forces is not representative of the true picture. The rotational inertia will certainly add to the braking forces applied by road/tyre friction. So perhaps adding both components (rotational inertia + tyre forces) will give a more complete answer.

Anyway, your maths is flawed. Brake rotor radius is not relevant to the force applied at the torque arm.
Also, you can't apply Newton's laws of motion to rotational systems.
For example, calculating the force by applying F=m*a is fine for linear motion, but it doesn't work for a rotational system. You need to work out the rotational moment of inertia of the wheel in kg.m2 (not just its mass in kg), angular speed and acceleration, etc.

Sorry for giving you a fail mark on your homework ;)

Cheers,
Cam
Ah, hold on Cam, I'm not sure you understand this approach. Let me explain a bit further. 😉 you have to take into consideration the speed (and thus momentum) of the bike when you apply the brakes. I don't see anywhere in you approach where you do that.
The bike/rider are in linear motion and are decelerated by the counter-force of the brake pad on the rotor as it is transferred as an opposing torque on the wheel, and consequently to the road via the friction of the tyre. That friction couples the bike to the road to cause the bike/rider to decelerate and (hopefully!) stop. So I make some assumptions (wild-ass guess) about speed, coefficient of friction, etc and simplify the problem by translating force of forward motion into opposing force by the torque of the brake rotor. I didn't go further yet to transfer that force at the rotor/pads to the bracing arm.
Anyway, food for thought at least until this lockdown is over and it stops raining!
 
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My head hurts, bloody engineers. I like practical testing. As long as it's not me when it brakes. Did I spell that right?
 
Still like my cardboard template, and suck-it-and see approach - but that's mainly because i have no idea about rotational gravitational inertia waves!! I understand the principles of force, stress zones ... and design accordingly
 
you have to take into consideration the speed (and thus momentum) of the bike when you apply the brakes.

I understand what you're trying to work out, but none of that stuff is necessary. For the purpose of your brake rod design, you only need to know the braking forces involved, not the amount of energy dissipated or how fast it happens.

To figure out the load on the brake rod, you just need to know the torque applied to the wheel. The maximum torque that can be applied to the wheel under brakes is at the point when the tyre breaks traction, regardless of how fast it happens to be rolling along the road. Torque is just force x distance. It has no time component to its units, so it's independent of speed. The only relevant variables are the coefficient of static friction (because the tyre contact patch is not moving in relation to the road) and the downward force on the tyre.

Obviously speed has an effect on stopping distance. But all that means is that for higher speeds, the braking force has to be applied for a longer period.

Another way of looking at it is to assume the bike decelerates at a constant rate, then force applied to the bike by the road (and consequently the torque applied to the brake) is constant through the whole process. We're back to F=m.a. If mass and acceleration (or deceleration in this case) are constant, then the braking force must also be constant. Speed and momentum are not part of the calculation.

May the braking force be with you - as distinct from the breaking force. :)
 
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