cyccommute 

The front wheel isn't under power but it still has torque applied to it through the drivetrain. The spokes transfer the power from the drivetrain to the contact patch. Therefore when you reverse the process and apply force to the rim, that force transfers through the spokes to the hub to stop the wheel from spinning. You'd have trouble finding a better definition of torque. Here's one from Hyperphysics out of Georgia State U



The force of the brake in both a hub mounted rotor and a rim brake acts through the lever arm of both rotors (the rim brake is just a larger rotor) to change the rotational motion of the wheel. The tire patch acts as a friction pad on the ground so that the wheel doesn't slide. But the forces on the wheel are acting from the brake pads to through the spokes to the hub to change the rotational motion of the wheel. In other words, a torque is being applied to the hub by the rim through the spokes.

cyccommute 

Yes, he says it very clearly and he is just as clearly wrong. Further nothing I've said shows that I agree with him. I don't.

dabac

For there to be torque load on the spokes in a rim brake front wheel, the hub has to resist turning with the rim.
And in a rim brake front wheel the only thing resisting the hub from turning is the minute resistance created by the bearing drag.

Is that what you are referring to?

The only place where a rim brake front wheel sees any important amount of torque is between rim and tire.
There are no important amounts of torque load on the spokes in a rim brake front wheel as the force occurring at the contact patch is perfectly balanced by the force occurring at the brake caliper.

FBinNY 

I guess I was giving you far more credit than I should have. Either this is a case of extreme nitpicking, or you have failed to consider Newton's 3rd law of motion.

Simply put -- there cannot be a change in spoke tensions on a rim brake front wheel because that would require resistance at the hub. Since that's a freespinning unit, any change in torsional force applied by the rim couldn't be resisted. It's comparable to trying to tension a rope that's not tied to anything at the other end.

I was careful to keep referring to changes in spoke tensions from torsion because, of course, there are torsional loads in a tangent spoked wheel, with clockwise spokes torque one way, and counterclockwise spokes the other, but the hub moves to a neutral position and the system is in equilibrium.

As for the nitpicking --- There has to be a tiny amount of torque applied to the hub to stop it from continuing to spin on it's own momentum, but I think that any reader here would agree that raising that point would be nitpicking.

So to use the same bold terms you did, it's plain to anyone with an understanding of mechanical principles that you're wrong here. You might draw a freebody diagram if you want to see where you made the left turn, but I'm out.

mconlonx

Wheels: Your bike does not work real well without them.

cyccommute 

Let's break this down as simply as possible. The wheel is a rigid unit. It is spinning. Applying a force to the rim to cause the wheel to stop spinning is a negative torque by the definition of torque: "A torque is an influence which tends to change the rotational motion of an object." What else would you call it? If you don't think that there is any torque applied through the rim, then how can there be torque applied through a hub mounted rotor. It's the same principle...i.e. applying a force to stop rotational motion of the wheel...and the same mechanism.

Nope. Didn't fail to consider Newton's 3rd law.

The spokes connect the rim to the hub which makes the whole unit a freespinning unit. The rim isn't moving independently of the rest of the wheel.

Spoke tension may change but that doesn't matter. The unit is rigid. The rim is attached to the hub. Force applied to the rim changes the rotational motion of the whole wheel which is a torque.

FBinNY 

This is classic. I won't waste another moment arguing. Feel free to believe what you will.

Readers are invited to consider the difference between external forces (torques) acting on the wheel as a whole, and internal forces within the wheel which transfer torque from rim to hub and vice versa.

As per usual, I now cede the last word to you, because that's how it'll turn out anyway.

davidad

There you go. Violating rule #2 again.

hueyhoolihan

bet there's more than one intelligent well meaning forum member wondering "how the hell did i get into this... again".

davidad

After these drag on for a while they become "how can I pee on his feet and keep mine dry?"

achoo

If there's a net torque on the wheel, it has to be accelerating/decelerating relatively rapidly.

A front rim-brake wheel while braking has effectively three forces applied to it:

1. Backwards force from the ground against the contact patch.
2. Backwards force from the brake pads rubbing against the rim.
3. Forward force at the hub from the deceleration of the rest of the bike/rider mass (assuming zero rear braking, and ignoring drive train losses and drag)

If the wheel isn't significantly accelerating (and given the magnitude of the forces involved at the brake pad and contact patch compared to the mass of the wheel, it ISN'T accelerating significantly), net torque on a rolling wheel is about zero.

The backwards force from the brake pads is just about the same as the backwards force from the contact patch.

Assuming the rim is completely rigid (probably a real bad assumption...), the spokes at the front of the wheel see a drop in tension and the spokes at the rear of the wheel see an increase in tension.

Given the small moment of inertia of the hub, if there's a net torque on that free-spinning hub it'd spin like crazy.

dabac

I've been trying to talk about the effects on the parts of the wheel, primarily the spokes.

You, apparently, is taking another tack.

Maybe I should let it go at that.

I can see why you've gotten hung up on this. Braking does change the rotational motion. But I think the definition lacks something, or it's taken for granted.
There has to be a reference. You don't know if anything is rotating or not until you have something to compare it with.

And it doesn't say that torque is the only thing that can influence rotational motion.

So let's look up rotation:

Wikipedia has rotation defined as "A rotation is a circular movement of an object around a center (or point) of rotation".

Thing is, rim brakes don't work through rotation.
Rim brakes work by applying friction to a motion that is darn close to linear and aligned tangentially.
As far as the brake pad/rim interface is concerned, the rim might as well have been a perfectly straight rail being endlessly pulled through the pinch of the calipers.

And the contact patch don't work by rotation either. It's just providing a tangential push at the tire, the force of which is equal and opposite the tangentially aligned force created by the rim brake.

No rotation required, no torque invoked.

It doesn't become torque until something is trying to rotate, within or around something else.

Now, in a rim brake wheel, the hub is free to rotate around the axle.
If the hub is free to rotate, there can't be any torque loads on the spokes. Can't just tighten a string at one end.

The only place in a rim brake wheel where something is trying to rotate around something else and meeting resistance is where the tire is trying to rotate around the rim.

There is torque, between rim and tire. But not between hub and rim.

It isn't the same mechanism. A hub motor (with drive engaged) isn't free to move around the axle, it is forced to move around the axle, dragging the rim with it. It fulfills both definitions of torque and rotation.
Rim brakes might fight rotation, but they don't don't work through rotation.

A better comparison of how motors and rim brakes influence the wheel would be to compare a rim brake with a friction-drive motor RoadBug Bicycle Motor Kits - Roadbug Bicycle Engine Kits rather than a hub motor.
(Although an even better comparison would be spoon brake vs friction drive.)
The friction drive doesn't rely on rotation of the wheel either, and would just as happily spend its days feeding a sufficiently stiff garden hose past its drive roller. Again, the force is tangential, linear. No rotation, no torque on the spokes.

cyccommute 

With you so far.


The wheel is accelerating significantly...it's just that the wheel is acceleration is negative and we call it deceleration. It's still an acceleration and the magnitude depends on the amount of braking being applied.

So? Given the structure of the wheel, the drop in tension front to rear is probably on a par with the drop in tension from top to bottom when loading the wheel. In other words, too small to be measurable. The rim itself isn't rigid but the wheel is rigid.

The hub isn't an independent member of the wheel. The wheel is a whole system that is rigid and connected from the rim (tire, really) to the center of the axle. Apply force to one part of the wheel and the rest of the wheel has to react. You can't break a wheel down into its parts and consider how each one reacts without dealing with the whole system.

dabac

But it isn't a rigid connection to the center of the axle.
Around the axle there sits a bearing, allowing the rest of the wheel to move freely around the axle.
If you had a rigid connection to the axle, the wheel wouldn't be able to turn at all. (unless you're thinking about unicycle-style fork ends.)

achoo

No, it's not accelerating significantly. Take a bike with a wheel-mounted speed sensor and the computer running, lift that wheel, and smack it so it rotates.

How much force did you apply? Yeah, not much, especially compared to the forces applied when braking.

How fast is that wheel rolling? Oh, it could easily be rotating at the equivalent of 15-20 mph just from one half-decent whack.

It's not that hard to get a bike wheel rotating at 20+ mph just by whacking a tire a few times with your hand.

That acceleration from that one whack from your hand is multiple orders of magnitude greater than the acceleration of the wheel under braking, yet the force is also multiple orders of magnitude less.

When a whack from your hand can accelerate a bike wheel to the point it goes from zero to rotating at 15-20 mph in 1/5th of a second, the force required to decelerate that same wheel while braking from 15 mph to 0 in four seconds isn't significant.

True, but the force has to be transferred somehow, and that's where that happens.

The rim is part of the wheel. If the rim isn't rigid, the wheel isn't rigid.



Doesn't matter how you define wheel rigidity - if there's a net torque on the hub, it has to accelerate. There can be no significant torque applied to a freewheeling front hub through the bearings, and it's not accelerating. Given a front hub is 200g at most that's close to the rotational axis, it has a very low moment of inertia - if there is any net torque on the hub it'd be accelerating rapidly.

"For every action, there is an equal and opposite reaction."

With a freewheeling front hub, were spokes to apply a torque to the hub there's nothing to provide that opposite reaction.

cyccommute 

But I thought you (and dabac) are the ones who say there is no torque on the rim from braking. Now you seem to be saying that there is torque acting on the wheel as a whole...which is what I have been saying. Get your story straight!

How the torque from the brakes acting on the rim get from the wheel to the hub is details but the wheel as a whole unit experiences torque because, well, it's a whole unit.

That doesn't seem to me what you said. And I quote "...I think it's generally agreed that rim brake wheels doesn't carry ANY brake torque..." and then you went on to describe the tire moving on the rim as the primary way that deceleration of the wheels happens.


I'm pretty sure that nearly everyone can tell if something is rotating or not. It's not all that hard to observe.

Yes, braking changes rotational motion through the application of force on a rotating object. That application of force, whether to cause a positive change or negative change in the rotational motion, is called torque just to differentiate it from linear motion and linear forces.

Yup. See it was easy to understand rotation. Now, go a step further. If you want to change the circular movement of an object around a rotating center, you apply a force to that object to either accelerate it or decelerate it. Name a force that can influence a rotational motion that isn't called a torque.

The friction is applied tangentially but it is acting on a rotating object. Don't get hung up on the "mechanism" but concentrate on the result. It's a tangential force applied to a rotating object. A hub mounted disc rotor is exactly the same force applied tangentially to a rotating object. If you don't have torque at the rim, you can't have torque at the hub. Both "rotor" are solidly attached to the hub and, therefore, each acts in the same way.

That is a result of the torque applied by the brakes to the wheel which is composed of the rim, spokes, hub and tire as a single unit. Act on any part of that structure and you act on it all.

The entire wheel from the hub shell to the tire is trying to rotate. It is rotating around the axle. With the exception of a small amount of bearing drag, the axle isn't experiencing any torque because it isn't connected to the rest of the wheel. It is locked into position and the hubshell, spokes, rim and tire are rotating. A force applied to the rim or to a rotor attached to the hub to stop that rotation is putting torque on the whole system.

I'll agree that there might be a minuscule amount of torque between the tire and the rim during braking (or acceleration) but, generally, movement of the tire is minimal to nonexistent. Tire creep usually results in damage to the valve stem and the valve stem doesn't move that often unless the tire is under low inflation.

And I'll agree that there isn't any difference in torque between the hubshell, spokes and rim. There can't be because they are a solid unit but to say that the rim brake wheels don't carry any torque is wrong.


You aren't grasping that braking and driving the wheel are two side of the same coin. Under drive, the rotation of the wheel is being changed around a fixed axle. Under braking the rotation of the wheel is being changed around a fixed axle.

You aren't grasping that these friction drive and a spoon brake are exactly the same just with opposite signs. It doesn't matter that the friction drive would " happily spend its days feeding a sufficiently stiff garden hose past its drive roller". It if it did feed hose all day long, it would be a drive unit for a bicycle wheel. It would be a hose feeder. It does matter that it is a tangential force applied to a rotating wheel. The wheel is the object experiencing torque, i.e. a change in rotational movement through the application of force.

Consider further how the drive unit is working. I doubt that the drive uses a paddle of some sort to drive the wheel. It uses a wheel of its own against the tire of the bicycle. It uses torque to induce a torque on the wheel of the bicycle. If you reversed the torque of the drive unit to make it a braking force, it wouldn't be a different force, it would be the same force with a negative sign.

cyccommute 

Yes, you are right that it isn't rigidly connected to the axle and I miss spoke. However, that just means that the axle isn't experiencing torque. The rest of the wheel still is.

cyccommute 

So is anyone forcing either of you to click on this thread? Or are you just trying to inflate you post numbers?

cyccommute 

If you spent all your time just whacking the wheel with your hand, you would have a valid point. But the bike as a whole system...i.e. bike, rider and any extra stuff... is experiencing significant deceleration and significant force on the front wheel under braking. All the weight of the system is transferred to the front wheel which results in significant force on that wheel.

In fact you can calculate the maximum possible deceleration before being pitched over the bars, force put on the tire patch and the torque the wheel experiences during maximum deceleration. The maximum possible deceleration is 0.56g or 5.5 m/s^2 (See Bicycle Science by Gordon for details). For a 90kg rider and bike, the braking force on the front wheel is 494N. With a 622mm wheel that's 148 N*m of torque at the contact patch.


The force is applied to the wheel as a whole structure. The individual components may react in their own way but the wheel should be considered as a whole unit.

You are the one who brought up the lack of rigidity of the rim, not me. The rim itself isn't rigid but once part of the whole wheel structure it is rigid. As such, it has to be considered as a whole unit.

Under braking, the wheel is accelerating. The acceleration just has a negative sign. The axle isn't accelerating nor experiencing any significant torque but the wheel, i.e. tire, rim, spokes and hub shell is experiencing torque from the brakes.

You are also misapplying Newton's Third Law. The action is the application of the brakes causing a torque on the rolling wheel which results in the rolling wheel stopping...hopefully in a controlled manner.

dabac

I'm grasping it just fine, that's why I brought it up.

I have far greater doubt about your grasp of the difference between how a hub motor and a rim brake affect the wheel.

Since this all started out with parts failure, how would you plan to analyse that if you insist on seeing the whole wheel as one single structure?
You have to look at the subsets to find out why they would fail in a certain manner.

I've never been that interested in the wheel as a whole, as it all started with parts failing.
You're talking about the wheel as a whole, I've been going on about torque influence on spoke tension. And you still fail to explain how you'd introduce any torque load to the spokes.

That'a very clever use of selective quoting there.
But I understand why you edited it like that.
The full bit, perhaps from a later post:"There is no torque transfer from rim to hub via the spokes in a RIM BRAKE front wheel." wouldn't have been as useful to you.

Again, I'm talking spokes and you're talking whole wheel.

Tires do move on rims occasionally, but movement is not required for there to be torque.
I never said the tire had to move on the rim, only that it would like to.
The brake is acting on the rim in one direction, and the contact patch is acting on the tire in the opposite direction.
The force between rim and tire to seems to match your definition of torque rather well.
And I wouldn't think of it as minuscule either, as this interface has to carry all the braking force. But sure, its got a decent amount of circumference to soak it up.

Deceleration happens because the rim is experiencing a friction force from the brake pads equal to the friction force appearing at the contact patch, which doesn't require any torque loading of the spokes to occur.

You can have your whole wheel torque concept if you wish.

As long as it doesn't involve torque loading the spokes, I don't care.

cyccommute 

I look at the whole wheel as a single structure only insofar as the wheel reacts to forces on the wheel. If some part of the wheel fails, I would address how it works with the other parts of the wheel to ascertain the mode of failure.

No, you weren't talking about spoke initially. You may have changed you tune later but you were talking about braking torque influence on a rim brake wheel or lack there of. That's a different issue.

I engaged in no selective quoting. I copied that quote directly from post 37. I changed nothing and simply cut and pasted it into place.

As for your above quote, that's wrong too. The hub shell (since the axle is fixed and doesn't experience significant torque) and the rim aren't moving separately. The only means of transferring the force of braking to the rest of the wheel is through the spokes. The spokes themselves don't move relative to the rim or hub shell but, then, the hub shell and rim don't move relative to one another either...unless there is something wrong like a broken or loose spoke. But that is a separate issue.

You have to look at the wheel as a whole because it moves and reacts as a whole.

Yes, the tire experiences torque during braking. It is rotating after all. But that torque is translated from the brake pads to the rim to the spokes and to the hubshell as well. You really don't want the tire to have any significant movement independently of the wheel or the bike wouldn't stop. The tire would just slip around the rim.

Airburst

I am not an easy person to render speechless, but the sheer level of misapplied physics involved here managed to leave me temporarily lost for words...

FBinNY 

Way back in post No.57 I suggested that people consider the difference in forces acting on the wheel from the outside, and forces acting on the various parts within the wheel. By extension, I suggest that people consider the very material differences in scale between forces necessary to overcome momentum of the wheel (in isolation) and forces necessary to overcome momentum of the bike and rider.

Unless there's agreement about these key questions, and what they're discussing, this "discussion" can drag on forever.

If anyone is actually interested, I suggest they draw a freebody diagram for themselves and consider what's happening. Otherwise, as far as this thread is going, they might be better entertained by this

hueyhoolihan

be careful what you say! at least one contestant is willing to take time out to swat at anything resembling a ringside heckler.

BTW, although i find you well meaning and intelligent, and i hope you take that as a compliment, your post count is in desperate need of inflating... my accountant says i'm good for 2015 already.

SquidPuppet

You are grabbing the RIM in order to stop the RIM. The hub isn't even involved.