Disk brakes cause spoke breakage!

[carlucci1106]

Originally Posted by reuben

No, no, no, a thousand times no. Why is it these alternative facts never die? I blame the internet.

It's totally the opposite. Broken spokes cause the wheel to go out of true, the disc rotor rubs against the brake pads, and they heat up like crazy, almost glowing hot.

Speaking of internet-disseminated mechanical fallacies... the rotor wouldn't rub, as a result of spokes breaking. The cocentric mounting point(s) of the rotor is in a fixed plane. Loss of rim true will not affect this. The hub shell would have to flex as a whole. In theory at least, the whole wheel tension could be diminished such that the rim is grazing the fork legs, yet the rotor would still be straight.

As for this scenario, clearly not a heat issue. Likely the tensions were whack-a-doo, and the braking forces caused the spoke failure. Roval wheels are not the best built out there. They are decent; probably good enough for most amatuers.

I am firmly part of the save the rim brake group of luddites, but I think that engineering will eventually prevail, to where disc is the obvious choice. However, to me (the mechanic as well as the rider) the hassle, money, increased weight, and increasing complexity with mounting, adaptors, and all related ilk keep me away. It's still not there yet.

In fact, it probably will never be there for me. I do not like the way it looks or feels. Simplicity is what makes road bikes beautiful in my eyes.

[oldpotatoe]

Originally Posted by KonaSS

The guy lost 10 minutes on a stage, was understandably frustrated and looking for a reason, blames equipment as an excuse.

His explanation doesn't pass logic either in mode of failure, or the amount of time lost (unless it was a 10 minute wheel change).

Nothing to see here unless you want to believe in bogeymen and that all disc brakes are bad.

Ain't it the truth. I've known a few pro team bike wrenches..When the rider has a great stage, it's all about him, when he has a crappy stage it's the bike or...the mechanic...

It was a bike change and with the ton of 'standards' out there, hubs, rotors, thru axles, speeds, etc, reliance on team cars/bikes/wheels is an even more essential part of this, rather than just any old 'neutral' support wheel...

As for this

Broken spokes cause the wheel to go out of true, the disc rotor rubs against the brake pads, and they heat up like crazy, almost glowing hot.

The rim and spokes could be gone and the hub/rotor is in the same place. A MAJOR reason why MTBs event to disc..wack a wheel and still have brakes.
If the rim is SO wacked that it rubs on the frame or fork..that may be an issue but the rotor/caliper relationship doesn't change with any amount of broken spokes.

[Mark McM]

It looks like Kelderman has been given a talking to about who pays his salary (and what they expect from him), and has now recanted his accusation about disc brakes causing his spokes to break:

https://www.cyclingnews.com/news/kel...-giro-ditalia/

[tomato coupe]

Originally Posted by Mark McM

It looks like Kelderman has been given a talking to about who pays his salary (and what they expect from him), and has now recanted his accusation about disc brakes causing his spokes to break:

Or ... one of the team mechanics told him he was full of 5&#@.

[PacNW2Ford]

*On the rear wheel, spoke stresses from drive torque are greater than brake induced spoke stresses, so the number of spokes on rear wheels is often the same between equivalent rim and disc brake rear wheels.[/QUOTE]

I am wondering about this. A disc brake can easily produce enough torque resistance to exceed the tire’s traction e.g. lock up. But even the strongest riders cannot go around doing burnouts (exaggerating for humor) e.g. breaking traction. Can a rider pedaling on a trainer overcome a locked disc brake?

[dddd]

Originally Posted by PacNW2Ford

*On the rear wheel, spoke stresses from drive torque are greater than brake induced spoke stresses, so the number of spokes on rear wheels is often the same between equivalent rim and disc brake rear wheels.

I am wondering about this. A disc brake can easily produce enough torque resistance to exceed the tire’s traction e.g. lock up. But even the strongest riders cannot go around doing burnouts (exaggerating for humor) e.g. breaking traction. Can a rider pedaling on a trainer overcome a locked disc brake?[/QUOTE]

The limit on rear wheel braking torque is defined as much by weight transfer acting on the bike's short wheelbase.

So the question might be more about whether the rider's torque (at the hub) going uphill (with weight transfer helping traction) is more than or less than the braking torque (at the hub) that can be applied (with weight transfer hindering traction).

There is also the matter of the number of fatigue cycles from pedaling, versus the number of fatigue cycles from braking (they are not likely the same).

It might also be be that the peak of torque loading at the hub from braking is greater than from pedaling, due to the greater (momentary) static friction of the tire on the road during sudden/abrupt, hard braking, and which thus might over-stress the spokes on a particularly light wheel having fewer and/or lighter-gauge spokes. The peak torque from braking is made yet higher by the weight of the rim and tire acting on a hub when the disc brake suddenly locks up during abrupt application of the rear brake.
Braking on very rough ground also tends to create higher peaks of braking torque between the hub and rim (i.e. through the spokes).

[Mark McM]

Originally Posted by PacNW2Ford

I am wondering about this. A disc brake can easily produce enough torque resistance to exceed the tire’s traction e.g. lock up. But even the strongest riders cannot go around doing burnouts (exaggerating for humor) e.g. breaking traction. Can a rider pedaling on a trainer overcome a locked disc brake?

As dddd says, it is about the forward weight transfer during braking. As we know, weight is shifted forward during braking, reducing the weight on the rear wheel and increasing the weight on the front wheel. As more brake torque applied to a wheel, more traction is required by that wheel - but at the same time, as brake force is increased, the traction available to the rear wheel decreases, due to forward weight shift. When rear wheel traction is reduced below what is required for the rear wheel brake force, the rear wheel locks up and skids, and rear wheel braking force can not be increased further. Depending on bicycle geometry (wheelbase, height of CG, etc.), the limit of rear wheel braking is generally between 1/4 - 1/3 g-force.

In contrast, as as weight is shifted forward under braking, the traction of the front wheel increases. On clean dry pavement, the limit of front wheel braking isn't the traction limit of the tire, but is generally the tip-over point - the point at which there is so much forward weight transfer, that the rear wheel leaves the ground and the bike starts to go end-over-end. This is also geometry dependent, but tends to be about 1/2 - 2/3 g-force. So, the maximum front wheel brake torque is roughly twice the maximum rear wheel brake torque, so there can be much larger spoke loading in the front wheel due to braking.

But the forward weight shift has another affect on spoke stresses. The stresses on a wheel's spokes are from all the wheel loads combined - this includes drive torque, brake torque, and weight bearing forces. When braking, the brake torque spoke stresses are added to the weight bearing torque stresses. As braking increases, the brake torque stresses increase, but due to the forward weight shift, the weight bearing stresses decreases. Depending on wheel design (rim stiffness, flange diameters, spoke angles, number and thickness of spokes, etc.) it is possible that the spoke stresses under maximum rear wheel braking may be less than from weight bearing loads alone when simply coasting.

Also consider the case of spoke stresses under drive torque. With very low gear ratios, it is possible to lift the front wheel (wheelie) under pedaling force alone. Depending on geometry, this tends to occur at around 1/4 - 1/3 g-force, or roughly the same as maximum rear wheel braking. But unlike when braking, lifting the front wheel under drive torque shifts 100% of the weight to the rear wheel, and the combined loading will result in more spoke stress than when braking.