SClaraPokeman

For several decades I've believed that brakes work by converting a bike's forward momentum into heat through friction. I've never thought about torque going through spokes, hubs or forks. Is my thinking all wrong? Isn't a lot of energy being lost heating rims or disks?

Trevtassie

Think of it this way.... in a rim brake the rim is braked by the brake pads acting directly on the rim. In a disc braked bike the braking force on the disc can only get get to the rim via the spokes. Another way to think of it: Imagine the brakes are locked but you are trying to push the bike forward. The braking force from rim brakes is transferred mainly around the the rim to the contact patch. That's why you can have radially laced spokes because all the spokes are doing is holding the rim central to the pivot of the axle (ie resisting the brake force parallel to the ground) . In a disc braked bike the force has to be transmitted from the disc, down into the hub and then back along the spokes to the rim. If it were laced radially the forces in the spokes as the hub tries to twist in the rim would be extreme. The tangential angle of the spokes in two and, even better, three cross spokes reduces the forces on the spokes from the twisting of the hub because the trailing spokes are already angled to resist twisting... a long winded way of saying the same thing as cubewheels...
In answer to the OPs question though, some single wall alloy rim brake rims with low spoke count could be pretty sketchy with disc brakes I suspect, since the spoke bed wouldn't be designed to cope with braking forces.

tomtomtom123

ost people here have it wrong. Rim brakes do not transfer the (edit: torque load from) braking force to the spokes, while disc brakes do (through only half of the spokes, trailing, possibly more on the brake side). The Sheldon Brown article is correct. If you draw a force diagram, it may help you to understand. Deceleration when braking though can change the radial load on the spokes the same way on both rim and disc models, but that's not the same as the braking force.

Some rims are I identical between their rim vs disc models, with the only difference being the milled brake sidewall. For rims that only come with a rim brake model I could imagine that some have been engineered thinner and therefore cannot resist the extra pull on the spokes/nipple hole when braking with disc brakes. But I wouldn't know the thought process of the people who made it, so if you want to know then you would ask the manufacturer.

Similarly if you looked at the force diagram, because the disc brakes are much lower than rim brakes, and only on one side, there is greater torque on the fork.
Edit: I changed the diagram.

cyccommute 

Sheldon’s got this one wrong. The hub isn’t “free to turn” because the hub is laced to the rim through the spokes. Apply force to the rim and the force translates down to the fork and contact patch through the spokes. Apply force to a disc hub and the force translates to the fork and the contact patch through the spokes as well. Perhaps a slightly different path but the force is the same.

We don’t use radially spoked wheels on hub mounted disc because we are afraid that the hub will twist under braking forces but it really can’t. The spokes would have to be much more elastic then they are for that to happen. Any twisting of the hub allowed by the leading spokes under braking would be countered by trailing spokes. The flanges on the hub probably take a beating, however.

None of this has much to do with the rim, however. Nor would it negate using a rim brake rim with a disc hub.

cyccommute 

No it is not considerably greater. It is the same. There is no additional force. The wheel has to be considered as a whole. The deceleration is of the same magnitude and direction regardless of what type of brake is used.

As for thicker walls for rims using hub mounted disc vs rim brakes, compare the wall thickness of a Velocity A23 to a Velocity Blunt 35. The wall thickness of the A23 is greater in the sidewall (expected since it is a brake track) and in the spoke bed. The weight difference between the A23 and the Blunt is 140g. That 140g is accounted for by the greater cross sectional area of the Blunt. 140g of additional metal isn’t going to result in much more strength either.

But, trying to move this back to crankholio’s original question, there is still no extra forces that would negate using a rim brake rim with a hub mounted disc brake.

cyccommute 

The argument that rim brakes don’t transfer braking force seems to be based on the idea that there is some kind of air gap between the rim and the hub. The hub is attached to the frame and the rider is sitting on the frame. Any braking force applied is to slow or stop the entire system. The force of the brake pads on the rim goes through the spokes to the fork and frame regardless of what kind of braking system you use. The force is the same for both systems.

Rim brake rims aren’t engineered thinner. Look at the links I provided above. The rim brake is thicker at the side walls where you’d expect it but they aren’t thinner in the spoke bed.

Your force diagram is wrong because you are thinking that the spokes don’t do anything in the rim brake model. I’ve only put one arrow but each spoke would have the same arrow of force towards the hub. The force of braking isn’t just translated through the contact patch.

tomtomtom123

You are wrong. The hub runs on bearings, so it is effectively a roller system. It is not fixed. Therefore the axle fixed to the frame does not transmit torque through the bearings to the hub and to the spokes. The force of the brake pads on the rim sidewall does not go through the spokes to the fork to counteract the force of friction from the ground (other than deceleration of momentum of the bicycle as a whole while braking, through a radial load on the spokes ). The brake arms are already attached directly to the fork through the brake bosses. The brake arms transmit the force directly to the fork. The brake force (in this sense) is for resisting the rotation of the rim caused by friction of the ground on the surface of the tire near the rim.

What is transmitted through the axle/frame to the spokes (with the hub in between) is the (static) vertical load from the pull of gravity.

Basic physics example on fixed vs roller beam would help to explain this. A roller joint does not transfer torque/moment. A fixed joint does. The connection between the frame and the hub is a roller joint.

In my previous post, I crossed out the line about deceleration because i was going to rewrite it, but it actually makes sense at it is: "Deceleration when braking though can change the radial load on the spokes the same way on both rim and disc models, but that's not the same as the braking force." The deceleration would apply the same radial load for both rim and disc brakes, at the axle, like in the diagram below.
(edit: actually I was wrong, the radial force on the spokes of the disc brake wheels while braking would be greater than on rim brake wheels, because you would need to apply a greater force to the disc brake since it is closer to the center of rotation than the rim brake.)

Imagine if you turn the bike vertically on it's rear wheel, and then you hang 2 weights:
Rim brake - one weight on the tire at the ground, and a counterweight on the rim at rim brake. There is no torque through the spokes.
Disc brake - one weight on the tire at the ground, and a counterweight on the disc. There is torque going through the spokes.

(edit: the spokes on the disc brake wheel experiences greater radial load from deceleration while braking plus the torque from the brake force. The spokes on the rim brake wheel experiences lesser radial load from deceleration while braking, and no torque from the brake force)

A disc brake is the mirror opposite of a freehub. So if you imagine the torque that a freehub while pedaling applies to the rear wheel on the leading spokes, then the disc brake applies torque to the trailing spokes when braking. That's why you see in some manuals saying opposite things about whether to put the leading spokes on the inside or outside of the flange, depending on whether you use rim or disc brakes. And also why you cannot use radial lacing with disc brakes. Also, the leading spokes would decrease in tension on the wheel with disc brake while braking, so extreme dishing would be problematic since the non-drive side spokes are already at very low tension (assuming you decelerate faster when braking than accelerate when pedaling). That's why you need less dish for disc brake wheels than compared to a rim brake wheel.

(edit: after thinking more, disc brakes don't create radial loads on the axle, so no radial loads get transmitted to the spokes. Ignore the part of the diagram about radial load from disc brakes)

RocThrower

I always wondered why these Bontrager disc wheels, that came with my Domane Disc, have the 'rim brakes only' sticker...

70sSanO

Well that is hilarious.

Somewhere in this thread is an excellent re-write of Alice’s Restaurant.

John

fietsbob

Peel the sticker off and it will shut up.




(why would this question elicit so many people expounding on it, for 4 pages & counting.. boredom I suppose)







....

tomtomtom123

I made another diagram. Hopefully it's easier to understand. Imagine the cross section of the wheel is a see-saw (represented by the translucent black box), and the bicycle/world is rotated 90 degrees vertically.


The amount of torque transferred from Fb2 disc brake force to each trailing spoke will be Fb2 divided by the number of trailing spokes, times the proportion of the radius of disc divided by the radius of the spoke hole circle of the hub flange (or the effective diameter of the exit angle of the spokes from the flange hole, which is why greater number of spoke crosses is better for resisting torque). And also the proportion of transmission between the drive side and non drive side, since most of the disc brake force will be transmitted to the hub flange on the non drive side, and much less to the drive side.

The diagram is not to scale, but if you take it as scale, assuming the disc is 1/4 the diameter of the wheel, then Fb2 is 4 times greater than Fb1. So the disc brake spokes will have 2.5 times the radial load as the rim brake spokes (let's assume distributed to the 4 nearest spokes behind the axle). The additional torque on the disc brake spokes is maybe (on the non drive side, with 32 spokes in total) [Fb2]x[2 ratio of disc to hub flange spoke hole circle] / [8 trailing spokes] x [90% transmission] per spoke (assuming 10% transfers to the drive side spokes) = 0.225 x Fb2 = 0.225 x Fb1 x 4.

Therefore, the (trailing non-drive side) spokes on the disc brake wheel will experience approximately 4.3 times the load when decelerating while braking, when compared to spokes on the rim brake wheel.

At the same time, the spokes on the opposite side of the disc brake wheel will lose more tension, causing greater material fatigue, when compared to spokes on the rim brake wheel (assuming the same dish). But that's why disc brake wheels are recommended to have greater number of spokes, crosses, or spoke diameter than rim brake wheels (at least for the front wheel).

phughes

You do understand that the force stopping the bike is applied at the rim, and the tire, and the hub still turns don't you? You see, the hub contains a bearing, which allows it to turn with less friction than not having a bearing, that allows it to turn freely. There is no braking for being applied to the hub when using rim brakes. There is load applied to the hub due to the "massive" G-forces from the weight being shifted forward due to braking, but not the same amount of torque as with a disc brake.

Sheldon also simply used the article to show a rim used for a disc brake needs to be laced differently. There is a good possibility that some rims cannot handle the lacing method, doubtful, or extra torque applied to the holes for the spokes. Of course I am not sure I would buy that rim.

AlmostTrick

Interesting theories here. I believe in the Sheldon/phughes/tomtom theory: rim brakes do not transfer force to the hub, and discs do.

It seems clear to me:
Rim brake force is transferred to the fork, where it is attached. (or frame in the rear)

Disc brake force is transferred to the hub.

Set up two identical bikes and rims, with the type of brake being the only difference. Loosen all the spokes a little at a time and do some power braking. First wheel to fail will be one with the disc.

royphotog

Interesting experiment, and one that would prove the theory.

BoraxKid

Rim brake calipers are mounted to the fork for the front wheel, and to the frame for the rear wheel. Disc brake calipers are mounted to the fork for the front wheel, and to the frame for the rear wheel. Your argument is invalid.

Sheldon Brown's website, specifically the article written by John Allen, have this topic incorrect. The article in question neglects the connection between the rim and the hub (the spokes!) and the fact that both braking systems described are connected directly to the fork. Even my Sturmey Archer XL-FDD has a reaction arm clamped onto the fork, and after 3000 miles, the fork is not damaged, because the braking forces are being transferred from the hub body, THROUGH THE SPOKES, and then to the rim and so on.

FiftySix

Following the thread with interest.

---------------------------------

• A rim brake works directly on the rim and the attached tyre; a disc brake applies a potentially large torque moment at the hub. The latter has two main disadvantages:

1. The torque moment must be transmitted to the tyre through the wheel components: flanges, spokes, nipples, and rim spoke bed. An engineered disc brake would reduce weight by not having most of the metal rim components
2. A front disc brake places a bending moment on the fork between the caliper anchor points and the tip of the dropout. In order to counter this moment and to support the anchor points and weight of the caliper, the fork must be of a certain size (most likely heavier).

https://en.wikipedia.org/wiki/Bicycle_brake#Disc_brakes

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I think that the critical difference between rim brakes and disc brakes, that the addition was trying to convey, is that with disc brakes the torque of braking occurs between the hub and the rim, while with rim brakes the only time torque occurs between the hub and the rim is at the rear hub during pedaling. It is true that rim brakes can work with radial spoke lacing on the front wheel, while disc brakes require cross-lacing (usually 3x). So while the addition was somewhat inarticulately expressed, it is indeed based in fact. Matt Gies 17:25, 11 May 2006 (UTC)

https://en.wikipedia.org/wiki/Talk%3...kes_and_spokes
----------------------------------

With rim brakes, forks are loaded in bending at the fork crown that similarly supports road shock, while a disc brake places an equal bending torque at the tip of the fork and only on one blade. A fork can only be slender and light because it carries no bending loads at the dropout. With disc brakes, forks would require a substantial increase in cross section (and weight) and the brake would be heavier.

https://www.sheldonbrown.com/brandt/brakes.html

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holytrousers

roger musson on radial lacing :


jobst brandt on rim brakes then hub brakes

AlmostTrick

Thank you, 56 and HT. You would think that the simple, well known fact that rim brakes can run radial spokes and discs can't would quickly put this discussion to rest. I'm kinda glad it doesn't though, because reading these explanations is fun!

79pmooney

OK,some really crude math. Let's say we are stopping hard and almost going over the handlebars. Let us say also that the line from the tire patch on the road to our center of gravity is 63 degrees from vertical. Now, we are stopping our weight plus bike. Say the total is 200 pounds. The relationship between our weight, that 63 degrees and maximum possible stopping power is cos (63) / sin (63) = .51. In other words, we can develop a braking force of 1/2 our total weight, ie 100 pounds.

Now, the brake acts on the hub. The force that stops us is developed 13.5" away from the hub center. That's 100 X 13.5 = 1,350 ft-pounds of torque that must be transmitted from hub to rim. All of this is done by 1/2 of our 32 spokes. (The other half are laced in the wrong direction and do nothing here. Let's say each spoke comes into the rim tangentially 2" from the hub. Now we can convert the braking torque in to forces acting along the spokes.

So: 1,350 / (2" X 16 spokes) = 42 pounds additional pull per spoke. (Note,this is completely separate from the job the spokes are doing to transmit out stopping force into the bike and therefore, us. Likewise the job of supporting all of our weight plus bike weight. Remember, the rear tire is one inch off the ground. It's doing nothing. In fact its own weight is on the front wheel, loading those spokes.)

So we know the additional spoke loading. Easy, straightforward, exact calc. Tough are "normal" spoke loads, both JRA and stopping. To be crude, let's say 1/2 the spokes are used to either hold us up or slow us down. Using high school math (Pythagorean theorem), we get a total force that is 1.12 X the 200 pounds of weight = 224 pounds total force being supported by the wheel between gravity and deceleration. So each spoke l is resisting 224 / 16 = 13.9 pounds. Interesting! The disc brake loads every spoke to 42 /14 = 3 times the load a rim braked wheel sees doing the same braking. (Granted, the load / 1/2 the spokes is crude; the highest load of the wheel rotation will be higher. Still you can see that disc brakes make very real difference.

Now for a reality check. Everyone in the racing world knows the 200 pound muscle men who race the velodrome matched sprints ride track bikes built as strong as bridges and never race with light spokes. They'd simply tear those wheels apart So what do they do to the rear wheel spokes when they take off from a stand still with quads like my waist. Say they apply 1000 pounds to the cranks. Crank length 7".

1000 pounds X 7" = 7000 ft-lbs. Now this torque is modified by the gear ratio. Say he is riding a 52-15. 7000 ft-lbs X 15 / 52 = 2019 ft-lbs delivered by the chain. We calculated our somewhat lighter rider developed 1350 ft-lbs just stopping. So you and I can develop 2/3s the force on the spokes using just our fingers that a velodrome animal can apply when he tries to rip his bike apart (say when the starter's pistol goes off for the Keirin).

Not hard for me to believe there are light road rims where they never expected or designed for that. On top of that, the failure that would result could be happening at the worst possible moment. If I were selling tose rims, I"d sticker 'em.

Ben

tomtomtom123

(edit: after thinking more, disc brakes don't create radial loads on the axle, so no radial loads get transmitted to the spokes.)

My force diagram clearly shows how disc brakes apply greater forces to the spokes compared to rim brakes. Anyone who remembers college physics or structures 101 should be able to figure it out (most of the basics is in high school text books). How much extra compared to the pretension and static loads? I don't know. But from my estimate disc brake would apply at least 4 times as much extra (braking) force to the spokes (rotor) than what rim brakes would apply (to the trailing spokes behind the axle to the rim), on large diameter wheels.

What the naysayers are forgetting is that the braking force needs to be transmitted from the pads to the ground, and the difference is that the rim brakes act at the exterior of the wheel system at the rim so there is only radial load and no torque load, while disc brakes act at the interior of the wheel system at the disc/hub body which creates an additional rotational torque on the (hub and) spokes. and the fulcrum of the disc brakes is much shorter than with rim brakes so then the radial load (braking force) on disc brake spokes is greater.

Not only the spokes, but the fork will experience much greater bending with disc brakes, because the force will be at least 2.5 times greater at the drop out and 4 times greater at the brake pads and mostly on one leg of the fork.

Skulking

I'd like to circle back to the original question and see if I've understood everything correctly. It seems like the answer is actually that there is no reason that a rim would be for rim brakes only, but because of how the spokes are laced, a wheel may only be suitable for rim brakes. If a wheel is sold fully assembled with the spokes laced in such a way that is only appropriate for rim brakes, then it would make sense for the rim to be labeled as only being suitable for rim brakes, but if it was desired the spokes could be re-laced appropriately for disk brakes because there is nothing specific to the rim which makes it only suitable for rim brakes. Have I correctly understood the gist of this thread?

tomtomtom123

It's not only the spokes that experience differences in forces between disc and rim brake wheels. The connections between the ends of the spokes will also experience equivalent amounts of force. That includes the nipples and spoke holes in the rim and hub flange. How much difference compared to the total load of the entire system? You'd have to calculate that yourself. If you want to know if rims labeled for rim brakes only can be used for disc brakes (even after relacing), I think only the manufacturer can tell you.

Bike Gremlin

Interesting discussion.

I would worry about buying rims that can't stand "disc brake force", such rim may not be strong enough to last under pedalling torque and other stresses that pile up with mileage.
Though "rim brake only" does sound like some lawyer/marketing nonsense.

As far as mechanics and engineering go - some have already quoted Jobst Brandt:


Drive force disk loads (for disc brakes it is the same principle, only in reverse direction)


Wheel load when rim brakes are used

Will give myself the liberty to quote relevant paragraphs from his book (that English was way better than mine ):

Driving torque explained (I would say that disc brake loads are similar, only in reverse direction)
"In a wheel with cross-laced spokes, torque, unlike other loads, affects all spokes
equally but in opposite ways. Half the spokes become tighter and half become
looser. All spokes are involved, not just the pulling ones. Torque is equal to the
tension change, times the number of spokes, times the effective flange radius.
Spokes that become tighter pull, and the ones that become looser push the rim
around. The pulling spokes stretch and become longer, and the pushing spokes
compress and become shorter. The rim bulges inward at the pulling spokes and
outward at the pushing spokes while the average tension, and therefore average
rim compression, does not change.

Of course, the pushing spokes don't push in the usual sense because they are
wires and are not in compression. In the tensioned wheel, however, they have
exactly the same effect as pushing. In an unloaded wheel without torque, all
spokes are in equilibrium and at the same tension. When torque is applied,
spokes become tighter and looser in pairs and, except for their pretension, push
and pull.
In the Figures 11 and 12 the effects of torque are first shown alone, then
combined with a vertical load. Torque causes changes in tension that appear as
waves in the rim. The waves in the left side of the wheel are above the average,
and the ones on the right below the average. This difference arises because, in
the diagram, the road pushes to the left. The rim responds to the pulling and
pushing spokes as if they were rigid columns. The pulling spokes pull the rim
inward, and the pushing spokes push it outward. Because they are tangent to the
hub but point in opposite directions, they pull and push to produce torque in the
same direction at the rim.
Because only radial dimensions are exaggerated in the figures, rotation displace-
ments are not visible. Pulling spokes should appear longer than pushing spokes,
but, because hub rotation is not magnified, this is not visible."

Rim brake loads explained:
"Braking with a caliper brake causes a small but significant radial load that affects
spoke tension. Under hard braking, the brake shoes ****** the rim with a force
of up to 500 N by pushing rearward with 2 50 N force and pulling on the front
half of the rim equally. This increases compression in the rear half of the rim
and decreases compression in the front half about the same as the increase from
tire pressure.

Spokes in the forward half of the wheel become about 5 % looser and ones in the
rear, 5 % tighter. At the caliper and the ground contact point, where forces act
on the wheel, there is little effect so tension remains unchanged. The bending
stiffness of the rim and the direction of the braking force cause a smooth
transition in spoke tension as the rim passes through the brake caliper. Of all the
loads on a wheel, braking is the only one that causes an significant increase in
rim compression, and severe braking can cause an overtensioned wheel to
collapse into a saddle shape (pretzel)."


My thoughts (and less than perfect English):
Braking loads can be stronger than the pedalling power humans produce. And can be rather sudden/sharp.
It is clear that Jobst hadn't addressed disc brake loads in the book, considering the time when it was written. So (in my opinion), even though the loading principle of drive torque loads vs disc brake loads is the same, the magnitudes might not be.
However, I also think it is correct to assume that tension increase of all the pulling spokes (and decrease of all the pushing ones) is pretty much uniform, and evenly distributed around the whole wheel.

To put it differently:
With rim brakes - top and bottom part of the rim are practically being pushed hard towards the rear end of the bike (hence the front spokes are loosing tension, while rear facing ones are gaining it).
With disc brakes, rim is being forcefully turned, against the highly resisting hub. Even lightest of rims are rigid enough when facing force that goes in that direction (parallel with the ground, not pushing/pulling sideways, or towards the centre - like when hitting a bump). So any spoke tension is evenly distributed along the entire wheels circumference.

While I am certain that disc brakes put much more load on the fork, I'm not sure they put more stress on the rim, or the spokes for that matter, compared to rim brakes. I'd say it's the opposite.
Could be wrong - I'm not a mechanical engineer. Hence the long post - to explain my line of thought as good as possible, in order to get corrected if needed.

cyccommute 

The hub is free to roll, yes. The hub isn’t free to turn relative to the spokes. Applying force to the wheel through the brake system, whether disc or rim, put stress on the spokes. The hub shell (which the spokes are attached to) tries to turn relative to the spokes but the hub is locked in place. The amount of force on the spoke bed is the same in both cases. A disc doesn’t generate more force at the spoke bed than a rim brake. Therefore there is no reason for a limitation on using rim brake rims with a disc hub...aesthetics aside

cyccommute 

Both transfer the braking force to the fork. The only difference is one of symmetry. All of the force on the disc is transferred to the left fork leg. The load at the spoke bed...remember this is about the rim not the brake...is the same in each case. Rim brake rims aren’t inherently weaker than disc rims.

Your test wouldn't test the rim, it would test spokes. The only way to test the rim would be a longevity test. How many years do you have to dedicate to the test?