Wheelbuilding / Wheel Building / Wheel build help

Energy is not the question in this case but yeah, I think you're right. The bike slows down through backwards force from the tyre on the ground. That force must pass through to the axle to slow the frame/rider and therefore must always be transmitted through the spokes. Torque/force at the tyre at the point of skidding is always the same thus stress on spokes is always the same.

Damn, I’m just planning a dick-breaked bike, my first, just when the forum has consigned them to history once and for all. Then again I always was a luddite.

dick-breaked bike

Ouch. Sounds painful. Sooner you than me.

dick-breaked

Duck-beaked?

The kinetic energy removed maybe the same but due to the rotor size difference between a rim and the disc brake rotor the forces on the spokes are quite different. The higher forces on the spokes applied at the hub flange (set by the hub flange diameter to rim diameter) do not mean higher retardation force on the bike. I have corrected my first post on this. There is a glaring error.

As I have shown torque = force applied * diameter of rotor.

Energy in one revolution is actually work done=torque*2*pi

Assuming tangential spokes
So W.D = (braking torque)2pi = force(rotor diameter). That could be rim or brake.

The work done here is fixed. The rotor diameter is not. It's 10 times larger for the rim than the hub flange pcd ( assuming 62mm Shimano 6 bolt flanges) so the force on the spokes has to becl different. Small disc brake hubs just make matters worse.

So bringing energy into this changes nothing.

So when you apply physics/naths the answer falls out. I used to teach this subject and maths too.

The problem is you are misunderstanding the relationship between torque and energy.

when you apply physics/naths the answer falls out

It does. Not necessarily the answer you've given, but the right answer.

Since neither of us has done the full calculation, the exact answer is not yet determined, but that's not really the point. The point is that people who have done no thinking about the question imagine that disc brakes give the spokes an incredibly hard time, but that the torque from rim brakes magically finds its way all around the rim from brake caliper to contact patch without ever bothering the spokes.

Would rim brakes acting on the rim, and so both sides of the wheel relatively evenly and disk brakes acting on one side of the hub, so acting slightly differently on each side of the wheel change the forces involved to affect the spokes differently?

So when you apply physics/naths the answer falls out. I used to teach this subject and maths too.

I know. I'm feeling increasingly sorry for your former pupils. None of the equations you've posted show the relationship between retardation rate at the tyre and the load on the spokes. If you can't see that I really don't understand how you could teach physics.

Yes, because the hub shell does not have infinite torsional stiffness, the rotor side spokes do the heavy lifting. This is exactly the same as the situation with the drive side of the rear hub.

Yup, so there's different forces on the spokes with disc brakes, but that's more to do with them acting more on one side of the wheel than over the whole wheel differently. I can't remember what the first point was but maybe that's relevant to someone's thinking, I skipped over all the mathsy stuff.

Yeah this situation is a bit more complicated than just a single equation. If you actually want to understand all the forces in play, you need to draw a free body diagram of the whole situation - and the specific force acting on one spoke is not that easy to determine even then. That's with making the assumption that a spoke can only take tension and no compression: otherwise it's more complicated still.

it's more complicated

And that's before you even throw in the preloads and components which are not fully kinematically constrained.

Isn't it more the case that because there are a lot of parts that are kinematically constrained (though in complicated ways), the entire system (once you take into account the spokes) becomes massively overdetermined, which is what makes the exact load calculations so difficult?

But then also when you want to get into the details, you need to take into account second order effects because it's not actually a rigid body system, rims and spokes etc. can flex as well.

So, tl:dr ...?

The tester thinks people are mad to think that dick-breaked bikes have significantly different/worse spoke-based worries than other kinds of bike-stopping methods, correct?

The tester is always right, ergo I can carry on building my dick-breaked dreambike with nary a care for such spoke-worrying frippery.

I just got...if there is a difference it didn't matter in the real world.

Dick breaks for all

Eggmazactly

Break a dick.

Dicks are death.

Isn't it more the case that because there are a lot of parts that are kinematically constrained (though in complicated ways), the entire system (once you take into account the spokes) becomes massively overdetermined, which is what makes the exact load calculations so difficult?

I'd say so, yes. So it's easier, rather than trying to model the individual components of the wheel, to work on the basis that you have a decelerative force at the contact patch and this is being resolved by the wheel as a whole to the axles, which are the only parts of the wheel system rigidly attached to the frame and forks. The only things connecting the axle and the contact patch is the spokes, so for the same decelerative effect at the contact patch and the same number of spokes the average spoke loading must be roughly the same.

The only things connecting the axle and the contact patch is the spokes

I like building my wheels with rims and hubs.

I follow the overall approach, but I'm not sure about the details. Essentially, braking deceleration (which is the only thing different when comparing rim brakes vs disc brakes) is introduced at the rim in one case, and on the hub (via the disc that is supposed to be rigidly connected to the hub) in the other case. Without looking at it in detail, I wouldn't say it's entirely ridiculous to assume that could result in a different amount of force being transmitted through the spokes, or a different distribution?

Isn't the real point here though that modern spokes are usually strong enough anyway? I'm usually above the weight wheels are really built for, wheels have fewer spokes than ever before, and spokes breaking on quality wheels is not something I hear about every day, or that has happened much to me... apart from the really cheap, really shit wheelsets I used to buy once upon a time.

Isn't it more the case that because there are a lot of parts that are kinematically constrained (though in complicated ways), the entire system (once you take into account the spokes) becomes massively overdetermined

Yes, if you simplify your model to fix all the joints and treat the materials as inelastic, and as you say if you model it realistically you have to consider that the shape changes under load, and that alters the way the load distributes among the different components. A radially-laced front wheel is neither circular nor radially laced once you load it, whether that's rim brake loads or just pushing down on the axle and up on the tyre contact patch.

Yeah, those were called 'second-order effects' when I studied this kind of thing (in a civil engineering context though, and that was in German, so might be called something else in English). If I remember correctly, unless the deformations go beyond a certain limit, it doesn't impact the overall load distribution that much though. The thing is that here, a full model is already difficult to 'solve' fully for a rigid wheel.

unless the deformations go beyond a certain limit, it doesn't impact the overall load distribution that much though

A wheel designed to be only just up to the job will have the spoke tension drop to zero for at least one spoke under maximal load condition, unless the maximal load causes the tension of at least one spoke to rise to the yield point. A perfectly designed wheel will have at least one spoke at zero tension and at least one spoke at the yield point under maximal load condition 🙂

Ok that sounds good in theory, but that's not really how engineering works though. In practice I would not trust anyone claiming to have designed a wheel like that, because you don't really know the 'maximal load condition', and you always want some kind of safety factor anyway. In actual fact, almost everything in life is overengineered to a varying degree. The stuff that isn't is the stuff that occasionally breaks way too early and people get annoyed about. So you want to make sure you're not going to the limit on anything. I'd also rather pay for stuff to have a safety factor, than brake my bike with my face.

In any case, the tension in spokes being different is not what second-order effects refer to, that's just what happens due to the preload you mentioned earlier, and the way any wheel is built. That's too complicated to work out quickly on a sheet of paper, but there are computer programs that would allow you to figure it out relatively 'easily'. The really complicated stuff happens when you start to consider how the non-rigidity of the material subtly changes the way the forces apply, and how that in turns changes the deformations themselves again. You can then move on to third-order effects, but honestly it gets a bit pointless at that point.

But the thing is, because you always put on a safety factor and because those second-order effects are comparatively small unless your rim was to flex significantly, it's not really worth putting too much time into working out their exact impact, you don't want to be that close to failure anyway.