70sSanO

My wife has a 20h radial laced front wheel. Are you willing to try a hard stop with a disc brake from 40mph going downhill?

John

tomtomtom123

Ok just for fun I tried calculating the additional load on the spokes from braking.

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

Let's assume you travel at 39.6km/h, which is 11m/s, and slow to 0m/s in 3 seconds at a constant rate of deceleration. That is 11/3 = 3.7m/s2. But this is like extreme deceleration. (Maybe it could go up to 4.5m/s2) (edit: wikipedia says front brake can apply up to 0.5G of deceleration, or approx 5m/s2)
https://en.wikipedia.org/wiki/Bicycl...und_conditions

F=ma

Mass of 100kg for a heavy rider and heavy bike .

F=3.7x100= 370N against the direction of travel

Assume you brake hard to almost lock/lift the rear wheel and 100% of the deceleration force goes to the front wheel. (edit: or you simply only use the front brake)

I don't know how to convert the force needed for deceleration into the braking force. But if we assume the force of deceleration is the same unit of F in my see-saw diagram (this assumption could be wrong), then the spokes behind the axle will have a 2F = 740N additional radial load when braking with rim brakes. 740/9.8 = 76kgf

Assume 6 (12) spokes on average take this load equally, although I could be wrong and the load is distributed to more spokes.
76/612 = 13kgf 6kgf additional radial load per spoke when braking with rim brake.

With disc brake, the radial load behind the axle is 2.5 times greater than with rim brake = 13x2.5= 32.5 kgf additional radial load per spoke.

i previously thought the ratio of disc to spoke hole flange was 2 but it's maybe 3. (150mm disc pad diameter vs 50mm spoke hole flange circle diameter)
Let's assume that load from torque from disk brake is 4F x 3 ratio / 8 trailing spokes x 90% disc side = 500N
500/9.8= 51kgf additional torque load per trailing spoke on disc side

So disc brake would apply 32.5+51= 83.5 kgf to the trailing spoke behind the axle.

Rim brake would be 13kgf. 6.3kgf That's 6.4 8 times greater for the disc brake. It's more than the 4.3 that I initially thought because of the change of 2 to 3 ratio of the disc to spoke hole circle diameter, and average of 4 spokes to 6 12 spokes that take the radial load behind the axle.
​​​​​​
the actual amount of force that I calculate could be very wrong depending on what is the correct way to convert the deceleration force to the braking force, and the average number of spokes that take the radial load when braking (I could maybe be off by a factor of 2 to 4). But the 4 to 6 8 times of the extra load on spokes from disc brake compared to rim brake should be (maybe) close.

(edit: This paper says that 80kg rider/bike applying enough force before rollover would need 400N of rim braking force. The explanation for disc braking force is confusing because I guess they're German and English is not their native language, but it seems either 3.5 times of rim braking force 400x3.5= 1400N or 1666N/400N = 4.2 times of rim brake force. I estimated 4 times, which gives the approximately 2.5 times of additional radial load on the spokes behind the axle, as shown in the see-saw diagram, not including the torque load from the disc brake on the trailing spokes, so maybe my calculations are somewhat realistic. But if 12 spokes on average take the load equally instead of 6 [assuming the 370N example with 4x disc braking force], then the total additional braking load for rim brakes would be 6kgf radial per spoke [behind the axle], and for disc brakes would be 15radial+51torque= 66kgf per spoke [trailing spoke behind the axle on disc side], or 11 times greater)
https://www.sensorprod.com/news/whit...tb-2010-03.pdf

Bike Gremlin

Mostly an addition to the quoted post:
At the cost of stating the obvious - think it's worth noting (correct me if I'm wrong):
Total (aggregate) spoke tension in a bicycle wheel doesn't increase when forces are applied. For that to happen, forces would have to try to expand the rim's circumference (or diameter, if they are symmetrically being applied around the whole rim) - acting in a direction from the hub, towards the rim, i.e. outwards.

But that is never the case. Forces only act laterally, or towards the inside of the wheel - towards the hub.

So, while some particular spokes do get increased tension, other spokes get loosened.

With disc brakes braking: every 2nd spoke (pulling ones) get increased tension, while "pushing" ones get decreased tension. It is more-less evenly distributed around the whole wheel, so particular spoke tension change is not that great. How much? I have no way of measuring that.

With rim brakes braking: rearwards facing spokes get increased tension, while front facing spokes get decreased tension. Jobst Brandt says it goes up to 5% of tension change. And, while it does get distributed over both trailing and leading spokes, it is limited to a certain portion of the rim, hence it could be argued that the tension change spread is not evenly distributed around the whole wheel.

Is that more stressful for the rim? I think it is. Is it more stressful for the spokes? Possibly.

What kills spokes? Fatigue - they are a lot stronger than any rim, or hub flange can handle. But, apart from something getting stuck into them, when they break, they break from fatigue - caused by a huge number of cyclic changes in tension - loosening, then re-tightening.
What kills the rims? Apart from big potholes (and highly over-inflated tyres), it's either spokes being too tight (damaging the nipple to rim interface), or changes in spoke tension, causing material fatigue, but apart from some weak rims, spokes are usually the first to give in this case.

cyccommute 

“4 times“ what additional load? 4 times over the load of a rim brake which you’ve said doesn’t exist?

Rather than use a poor analogy, how about using the wheel. Consider the rim brake example below. The green arrow is the forward momentum of the bicycle acting through the hub. The yellow arrow is the braking force acting against the forward momentum. The red arrow is the rotation of the hub and the blue arrow is the action of the hub on a spoke. (Each spoke would have it’s own arrow.) During braking, the hub is pulling the spoke down in a circle. The spoke should continue in a straight line but the hub is pulling it in a circle. That’s torque, plain and simple.

The amount of force that is applied to the spoke is proportional to the force used to stop (or slow) the forward motion of the bike. The force is also limited to the force used to stop the bike. That force is the same independent of the type of brake used. Bicycles can’t develop much more than about 0.5g of deceleration. Multiply that times the mass of the rider/bike and you have the maximum force that can be applied.



Now compare the above to a hub mounted disc. The same parameters...forward momentum, braking force and torque on the spokes... applies. I can’t see where there is any increase on the force on the spokes. The application of the force is in a different location but the forces work the same way. The hub is pulling the spokes around just like the hub in the rim brake example is.



You keep saying that a rim brake doesn’t impart torque onto the hub through the spokes but that simply isn’t true. The brakes don’t impart torque onto the axle but they most certainly impart torque onto the hub shell. The hub shell pulls on the spokes when the bike moves forward and the rim pulls on the spokes when the brakes are applied. If they didn’t impart torque onto the hub shell, the wheel would never stop.

Now the reason that I say there is little to no difference is because the only force that we can work with is the braking force. The braking force is the same in each system because the mass and acceleration is the same. If you are going to have 4 times as much stress on the spokes of the wheel, you have to have 4 times the braking force. Disc brakes don’t provide 4 times the deceleration of rim brakes.

Wait a minute. I seem to recall someone saying “By how much extra force? You figure it out, or ask the manufacturer who has to have calculated this.” Now it’s 4.3 times which is a pretty accurate “estimate” based on something that has assumptions associated with it.

cyccommute 

Kinda. It’s not the fork, per say, but the fork tips where the hub axle is attached. And it’s not really even the fork that is doing the work. It’s the contact patch.
The braking force goes from the fork tip through the hub through the spokes to the ground.

Technically a rim brake is a disc brake. The rotor is just larger and wider. In a rim brake case, the forces go through the spokes at the caliper through the hub through the fork tips through the lower spokes to the ground.

No. Look at my picture of a rim brake given above. The brake force in the rim brake example is applied to the top of the wheel. The brake forces of a disc are applied at the mid-point of the wheel. the rim brake acts across the entire diameter of the wheel. But what you have to consider is that the wheel is a whole structure. Force applied at one point is shared around the wheel. The leading and trailing spokes don’t act independently of each other nor do the spokes in the upper part of the wheel act independently of each other. In braking, the spokes tighten and loosen because the spoke isn’t a lever arm.

In some spokes. It decreases tension in others. And that is for both types of brakes.

Braking force creates tangential compression in the rim in both instances.

Why would the path be shorter in the case of the rim brake? The rim brake acts on the rim at the top of the wheel above the contact patch. That force has to translate through the wheel to the contact patch about 30 inches away (for a 622mm rim). Applying the brake applies friction to the rim but the real work of stopping is done by the tire in both braking systems. Brandt says in your quote that the tension is higher in the front half of a wheel and tighter in the rear. Remember that the wheel is turning so each spoke is constantly undergoing a swing from looser to tighter and back. This happens constantly. This means that the rim is undergoing a constant change in force. I’m not sure that you should think of it as a vertical line but more like a slopped line that bifurcates the wheel.
Not necessarily. It’s complicated. There are a whole bunch of factors to consider. Most of this discussion has been of the “how many angels can dance on a pinhead” variety. The main takeaway from Brandt should be the final sentence in the quote you provided. My takeaway from this statement is that the braking forces are of such small magnitude that they won’t matter. Just riding the bicycle puts more stress on the wheels than braking does. That answers the question of “Why would a rim say ‘for rim brakes only’?” There is no good reason. A rim that is strong enough for someone to ride on will be strong enough to withstand the braking forces...independent of the kind of brake.

billridesbikes

This is wrong, there is no unequal forces on some spokes and not others. For a rim brake under hard braking, say 400N applied by the brake shoes you have a rearward force of 200N pushing the rear of the rim and compressing it and an equal force pulling on the front of the rim. This causes all the spokes in the front half the wheel to be looser and the ones in the rear to tighten by the same amount (equal and opposite). As the rim passes through the caliber/contact patch there is a transition of spoke tension as the spokes pass through the point of contact and there is no change in spoke tension at these two points.

For a disk brake there is a torsional load that is dynamic. In this case the hub has to twist to act against the wheel rotation and brake disk. Because spokes are not rigid they can't transmit toque by acting like levers with x amount of force they can only loosen or tighten. And this is how they transmit the twisting motion of the hub. The spoke pattern and flange diameter determine how much the spoke tension changes. For cross-laced wheels exactly half of the spokes will become tighter under hub torque and the other half will loosen equally. This is why disk brake wheels are cross-laced.

All in all, the differences on the rim under different braking should be small, maybe about the same as the change in tension from tire pressure (?). But if I was a rim maker and I hadn't tested a given rim with disk brakes under heavy loading I would probably put a label to only use it with rim brakes to be safe. Why add additional risk to your company for a condition that hasn't been tested?

AlmostTrick

Cyccommute. You continue to say there is no difference in forces between rim and disc brakes. (well, you have allowed for "a little" more force with discs a few times ) Yet you skirt the fact that discs can't run radial spokes. This is not a red herring, there must be logical reasons for it. The fashionistas would run it if worked. Obviously, forces at the hub are not anywhere near the same.

tomtomtom123

You are mixing up radial and torque loads caused by the braking force.

(edit: after thinking more, disc brakes don't create radial loads on the axle, so no radial loads get transmitted to the spokes)
Rim brake force transferred to the spokes: Radial load only
Disc brake force transferred to the spokes: Approximately 2.5 times the radial load, plus a torque load

Radial load from the rim brake force is perpendicular to the axle. (edit: radial load on the axle will be approximately 2 times the rim brake force, if ignoring the effect of the tire diameter) This in turn becomes approximately 2.5 times the radial load on the axle for disc brake wheels than rim brake wheels, which gets transmitted only to the spokes behind the axle that are exiting in the opposite direction of the radial load (this is because the radial load on the axle is the sum of the braking force plus the road friction, as shown in the see-saw diagram). (For a rim brake) The (key) spoke that eventually moves directly behind the axle perpendicularly to the radial load gets the most amount of this load, and the adjacent spokes pick up less of it for each consecutive spoke further from this first (key) spoke until you reach 0 load at 90 degrees. To make the calculations simpler I disregard the differences in distribution of the radial load among the adjacent spokes to the key spoke, and assume that the maximum load on that first (key) spoke will be the same as the radial load divided equally between 12 spokes (I initially thought 4 spokes, then 6 spokes, but 12 spokes seem to be more correct when you look at this wheel simulator which shows around +4kgf for the key spoke on an example 32 spoke rim with an 80kgf radial load on the axle https://bicyclewheel.info/wheel-simulator/).

This radial load is affected by the factors of the tire diameter, rim brake sidewall diameter, and disc brake rotor diameter. If you have a larger disc rotor, the difference in radial load compared to rim brake wheels will decrease. So you'll have to make assumptions about the sizes (I assume the 2 are equal). The radial load will also be affected by the angle of the fork to the ground, because the brake force is perpendicular to the fork, but for simplicity, I assume that the brake force is parallel to the ground.

Torque load from the brake force to the spokes only occurs if the 2 bodies where the 2 ends of the spokes (hub flange and rim) have force applied tangentially in opposite directions around the center of rotation (axle). With a rim brake wheel, brake force is applied to the rim to counter the road friction on the tire tread. The hub is free to rotate, therefore there is no torque load transmitted to the spokes (except for the insignificant amount of friction from the bearings in the hub).

Torque load from the brake force is transmitted to the trailing spokes with disc brakes, because the disc brakes are applying a tangential force to the hub flange, which is in the opposite direction to the tangential road friction on the tire/rim. Disc brakes will create somewhere around 3.5 to 4 times (assume 4) the brake force than rim brakes because the disc brakes are 3.5 to 4 times closer to the axle than the rim brakes, in proportion to the radius of the tire (assuming 160mm disc rotor and ISO 622mm rim, 32 spokes). Torque load will be applied to all trailing spokes (there are 16 on a 32 spoke wheel), but almost all will be to the trailing spokes on the disc side (there are 8 on a 32 spoke wheel, assume 90% load), and only some to the drive side (assume 10%). Torque load on the trailing spokes from the brake force will be a multiple of the proportion of the diameter of the disc rotor to the diameter of the effective hub flange spoke hole circle (which depends on the exit angle of the spokes, which needs a lot of assumptions for many factors). Because every wheel will have very different dimensions, I assume the disc brake pads are clamping at 150mm diameter on the rotor, and the effective hub flange spoke hole circle is 50mm. That makes the torque load from a disc brake transmitted to the trailing spokes a multiple of 150/50 = 3 times the disc brake force. So then if the disc brake force is 4 times the rim brake force, then torque load on the trailing spokes from disc brake force transmitted to all the trailing spokes will be 3 x 4 = 12 times the rim brake force.

Then you will have to divide the radial load and the torque load each separately among the affected spokes.

With my previous example of 370N rim brake force, assuming (tire has no thickness and) the radial load is divided equally to 12 spokes (for simplicity), then a 6.3kgf from the radial load would be applied to the key spoke by the rim brake force.

A disc brake force producing an equivalent deceleration, with all the previous assumptions on sizes of components and 32 spokes, would produce a 15kgf (from radial load) + 51kgf (from torque load) = 66kgf on the key trailing spoke on the disc side.

edit - with all the assumptions: F = rim brake force
rim brake key spoke load from rim brake force (radial load) = 2F/12 = 0.167F
disc brake spoke load from disc brake force (torque load) = 12F/8x0.9 = 1.35F
1.35/0.167 = 8.1
disc brake spoke load = 8.1 x rim brake key spoke load

edit of edit: https://www.sheldonbrown.com/torque-spoking.html
For disc brake torque, the sheldon brown article ignores the disc rotor diameter, and simply estimates the ratio of effective hub flange spoke hole circle diameter to be 1/12 of the rim diameter. I still got 12F with my assumptions, so the proportion is the same.
However, the sheldon brown article divides the torque load by all 32 spokes to calculate the increase in tension of the trailing spokes. I assumed 90% of the torque went to the 8 trailing spokes on the disc side. So my result of 51kgf would be (32/8x0.9) = 3.6 times greater than what the sheldon brown article came up with = 51/3.6 = 14.2kgf. That would be 14.2/6.3 = 2.3 times greater than the radial spoke load from rim brakes, compared to the 8.1 times that I came up with. I don't know which method is correct. For 120kgf pretension, the rim brake would add 5% extra load to the spokes, and disc brakes would add extra 12%.

This is only the braking force. This does not include the pretension of the spokes and the force of gravity on the mass of the bike and rider.

I've already admitted that I don't know how to correctly translate some of the forces and I've made a lot of assumptions, so the proportions and amounts that I've calculated could be very wrong. But I think 2 things are clear, even with the incorrect assumptions, which are: 1) disc brakes create more load on spokes than rim brakes do, and 2) rim brakes do not transmit torque to the spokes.

Your force diagrams are very wrong. You are mixing the forces on the fork with the internal forces of the wheel. What we are discussing is the wheel as a closed system.

grayEZrider

I recall making a lot of money truing the early spoke/disc wheels on motorcycles. The first ones sported a single disc in front of the fork leg. First improvement was stiffer forks to combat the twisting problem. Second was moving the disc behind the fork. Finally came DUAL discs to actually solve it. A few lighter bikes retained a single disc but by then most wheels left spokes behind for cast wheels.
I suspect some CYA lawyer wanted the sticker because the guys who built the rims were not always the guys who built the wheels.

tomtomtom123

After trying harder to look up more information, I still couldn't figure out exactly how to translate the external forces into internal forces within the closed wheel system. One video showing the external forces that would act outside of the closed wheel system:

My see-saw diagram (and all the previous calculations) with internal radial load on the spokes behind the axle caused by the braking force may be flawed, because the force from the ground in my diagram may be completely different from how I've shown it, and the diagram may only work if the wheel was not rotating (brake locked), just before the instantaneous point of skidding without rolling over. The diagram may not be valid if the wheel is rotating. I really don't know and I'm not going to continue searching for an answer. A mechanical engineer would probably know.

However, if the diagram is valid, then the additional radial load for rim brake wheels would be 2 times the rim brake force (when ignoring tire thickness and assuming rim brake radius is the same as wheel radius, and that rim brake force is parallel to the ground). I changed my mind a few times about whether there would also be radial load from disc brakes, but if the above case were true, then I now think that there would be 4.1 times the additional radial load from disc brakes compared to rim brakes (assuming that the disc brakes are located 1/4 the radius of the wheel from the center of rotation, and that the brake pads are located behind the fork and tangentially 90 degrees vertically from the ground [sqrt(4^2+1^2)]=4.1).

On the other hand, if my diagram is wrong and there are no internal radial loads on spokes caused by the braking force, then there would only be an external radial load on the spokes, caused by the combination of the force of gravity on the mass of the rider/bike and the force of deceleration at the center of mass, like this diagram from Wikipedia, and would be equal for both rim brake spokes and disc brake spokes:
https://en.wikipedia.org/wiki/Bicycl...und_conditions


It is definitely true that rim brakes do not transmit torque load from braking force through the spokes. Only disc brakes do, as previously explained in past posts.

The rim brake force (when using only the front wheel brake) will be equal to the force needed to decelerate the rider/bike, assuming maximum of half the force of gravity, and rim brake diameter is the same as wheel diameter, and ignoring tire thickness, air resistance, energy loss from deformation of tire, mass of the wheel, and any inclines),
F=ma = 100kg x (0.5x9.8) = 490N
(although the video above or the other videos in the series seem to show that the deceleration force can be reduced by the friction on the tire, but I don't know how to apply this)


Disc brake force will definitely be 4 times the rim brake force (assuming the above and the disc brake is located 1/4 the distance of the wheel from the center of rotation),
490N x 4 = 1960N

Torque load on the spokes from disc brake force will definitely be 12 times the rim brake force (assuming the above and the effective hub flange spoke hole circle is 1/12 the diameter of the wheel),
490N x 12 = 5880N

I thought 90% of the torque load would be divided among the 8 trailing spokes on the disc side (of 32 spoke wheel, 10% to the 8 trailing spokes on the drive side, and an opposite negative load would apply to the leading spokes).
The sheldon brown article says the torque load would be divided by the total number of spokes (32) and all the 16 trailing spokes would increase their tension by that amount while the 16 leading spokes would decrease their tension by the same amount.
(Ignoring the lateral exit angle of the spokes from the hub flange)
My way: 5880N / 8 x 0.9 = 661.5N per 8 trailing spoke on disc side
Sheldon way: 5880N / 32 = 183.75N per all 16 trailing spokes on both sides
Difference of 3.6 times

The original point I was trying to make wasn't to calculate the exact forces, because I don't know how to correctly calculate them. My point was that:
1) disc brakes apply more load to spokes than rim brakes.
2) disc brakes transfer torque load through the spokes, while rim brakes do not. This is why full radial spoke lacing is not compatible with disc brakes.

How much greater are the loads on disc brake spokes than rim brake spokes, and would it require a stronger spoke hole bed in the rim? I don't know how to correctly calculate the difference in the amount of loads, but you can look at my attempts at trying.
.

holytrousers

Thank you for this extensive response, and it's really cool that you take the time to reply to every post. That quote from Jobst Brandt's book was about pedaling torque as well as hub brakes , which are far less powerful than disc brakes, isn't that right ? I feel that the problem is that you are maintaining all the time the view that a wheel is one unit : As pointed out by tomtom123 you should try to see what's happening inside the wheel : even though the overall system of forces is the same (more or less) , inside the wheel there is a huge difference.
I would like you to consider a thought experiment for a while : if you had two wheels with 3 spokes each, one had disc brakes and the other rim brakes, what would happen if you hit the brakes ? Would both wheels behave the same way ?
I think the additional torque made by the disc brakes would be clearly visible and the additional tensions transmitted to the rim would clearly indicate there is a difference in between the two designs.
I believe the difference could be significant enough to require a stronger spoke bed design.

70sSanO

What about... drum brakes?

John

ShannonM

No, but that exact same rim,. with 12 more holes in it, and laced in a cross-3 pattern?

No problem.

That exact same rim, with the same number of spokes, but laced cross-2?

It's a dumb way to build a wheel, but it wouldn't be unsafe.

The sticker is pure, unadulterated, ass-covering lawyerese. (Which would be very, very unlikely to work in an actual court, unless the plaintiffs had Vinny for a lawyer.)

--Shannon

billridesbikes

Generally manufacturers are assumed to know more about their product than customers. Manufacturers also have a “duty to warn” if there is danger to the user that is not obvious. However, you’re correct that the manufacturer could still be liable in a court because warning labels are highly subjective. If a customer built a rim with this warning using disk brakes and the rim failed under heavy braking because the spokes pulled out of the spoke bed killing the customer, would a jury regard the warning “Use with rim brakes only” as sufficient? Hard to say.

I notice a lot of comments are dismissive of the warning “ har har har, dumdum lawyer talk”. But most likely the warning was added by the product engineering team, which set the product specs, not the lawyers. The lawyers just jazz up and check the language and spelling the engineers supply.

When I was part of a group that made OEM electronic components the engineers would determine the use conditions (e.g. “not for use in medical equipment”, etc.) Legal only reviews these, they don’t add them on their own.

conspiratemus1

You are correct. Braking of all vehicles (except for aircraft and spacecraft that use thrust reversers)* is achieved by converting the “organized” kinetic energy of the moving vehicle into the “disorganized” kinetic energy of molecules, which we recognize as heat. But the work of doing this transformation involves a ******ing force acting through a distance. What contributors to this thread are trying to accomplish is to understand the manner in which this force acts, in different modes of braking, whether at the rim, a disc, air resistance, or scuffing your sneaker sole along the pavement.

Edit: The censor deleted my perfectly correct physics term for slowing or decelerating because it’s also that old playground taunt that starts with re and ends with tard.

Edit:* brain cramp: of course reverse thrusters dissipate heat, too. Duh.

conspiratemus1

No fair. The rim with 12 more holes drilled in it is no longer “that exact same rim”. Laced into a 32-spoke wheel it might not warrant the sticker. But the original rim with 20 spokes cannot be made into a 32-spoke wheel, and so might warrant the sticker. A good lawyer would have you cornered right there.

You say the sticker is “pure . . . lawyerese”. Yet you say in the same breath that it wouldn’t work in an in an “actual” court (as opposed to a make-believe one?), so no asses are actually covered. Which is it? Or can we just rely on your demonstration that you don’t know what you’re talking about?

Moe Zhoost

I've used them. I like them. The rims did not have a label saying "for drum brakes only".

sweeks

Same principle as discs, no?
Just less force!

Bike Gremlin

I generally agree with your thoughts in this topic (and in general - I appreciate your knowledge and experience).
However, mostly out of curiosity, I will start with a disagreement. Interested to hear your thoughts on the topic.

With rim brakes, force is transferred by the front side spokes mainly loosing tension (spokes roughly at 3 o'clock looking from the right side of the bicycle). And those at the rear (at 9 o'clock) gaining tension.
Radially laced spokes are OK when handling such, non torque loads (not a native speaker, correct any terms if I used them incorrectly). That is why one can more easily "get away" with using radial spoking with rim brakes.

With disc brakes, braking force creates torque. Cross spoked wheels are a lot better at handling that.
Still - why I agree with your general conclusion: this torque is transferred by all the spokes in the wheel. Leading ones gain some tension, while trailing ones loose some tension. So I suppose there's less stress on both the spokes and the rim - because the load is more equally spread.

Quote from Jobst Brandt's "The Bicycle Wheel" on radially spoked wheels:
"Radial spokes carry loads just as well as crossed spokes, but they cannot transmit
torque. They transmit torque only after the hub rotates ahead of the rim,
making the spokes no longer truly radial. This rotation produces a small
tangential offset, or lever, on which spoke tension can act to produce torque.
This lever is the distance between the axis of the hub and the extended axis of
the no-longer-radial spokes. The driving torque is the product of this small
offset and the tension of all the spokes.
In a radial rear wheel the windup that occurs while riding is small (less than two
degrees). However, this motion increases spoke fatigue, and spoke rotation in
the flange causes wear. As radial spokes wind up under torque, they become
appreciably tighter causing high rim stress and, in some instances, flange or rim
failure. Looser spoking would reduce windup induced tension, but it would also
reduce wheel strength.
Even though they transmit no torque, front wheels should not be spoked radially
because high radial stress can cause fatigue failure of their flanges. The spoke
holes of aluminum alloy hubs can break out causing wheel collapse. Flange
fatigue takes time, so these failures do not occur immediately. Some lightweight
hubs carry specific warnings against radial spoking.
Radial spoking has no aerodynamic advantage over other patterns because near
the rim, where the spokes produce the greatest drag, they occupy exactly the
same positions, regardless of pattern. At the rim, spokes arrive alternately from
the left and right sides and do not draft one another. Without resorting to disk
wheels or flat spokes, the best way to reduce drag is to use fewer spokes. Flat
spokes have their own problems that are described elsewhere."

Same book, on braking force transfer when using rim brakes:
"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)."

Het Volk

I guess I look at this this way, and that we are trying to solve for two problems:

(a) Do disk brakes exert more stress on spokes than rim brakes to stop the wheel

(b) Do disk brakes exert more torsional stress on spoke given the applying of the pressure one one side of the bike.


Question 1: Spoke Stress

As for the first question, and I am waffling between whether there is a difference or not. If ultimately, you assume the wheel is as one unit, and the spokes carry the load regardless, then the question to me is whether this load is higher or lower for each brake type.


In a rim brake, the rim is the object with pressure to slow the wheel, and Sheldon Brown assumes that the rest of the wheel is along for the ride. Others have noted that this might be incorrect, as the hub still wants to move forward, and the spokes need to exert force on the hub as well to slow the wheel down. However, and this is where I waffle.....if this is true, and the spokes either have to slow the hub (rim brakes) or slow the rim (disc), does one create greater stress than the other on the spokes, given the diameter of the object the spokes are being asked to slow down. I keep coming back to the fact there is some concept of leverage, and that asking spokes to direct the braking force from a larger object (rim) to the smaller object (hub) is less stressful (i.e.- a rim brake) than asking a spoke to exert that same energy from a smaller object (hub) to a larger object (the rim).


Isn't there greater centrifugal force at the rim when asking the spokes to slow down the rim (disc braking system) than the spoke being asked to slow down the hub from the rim (rim brake system)?


Question 2: Uni-Directional Braking Force

I know fork flex on disc brakes are an issue, and this raises the question whether there is also more torsional tension on the spokes from the side the disc brake is positioned. It seems to me, that this is like minimal at best, since the hub likely absorbs almost all of this stress, and applies the same (or almost the same) tension on the spokes, and so this is not in the equation, even if the fork is flexing.

Bike Gremlin

As for the spoke load - I did quote Jobst Brandt who did some measurements from what I understood. And explained how and why, with rim brakes, the load is carried by just a few spokes.
With disc brakes - the load is carried by all the spokes. Even the ones on the side opposite from the disc - modern hubs are rather stiff. Also, the rim doesn't get deformed, as when using rim brakes - the load gets a lot more evenly spread around the whole rim.

This is my poor production, Tarzan English and rather amateur approach, but I did measure both the right and the left hand side spokes taking the driving torque. I suppose it is similar with disc brake load (only in reverse direction, with trailing spokes coming less tight):

Het Volk

Follow-up question: Based on your comment, rim brakes are tougher on spokes....yet, rim brakes can use radial lacing while you cannot use such lacing with disc brakes, meaning there are forces on spokes for disc brakes not in existence on rim brakes, and this gets back at what is the different force with disc brakes requiring a more bulked up wheel.


Something also that I am curious about is the fact that rims for mountain bike and many cross frames use beefier rims, which also help stiffen the wheel. Using light weight aluminum rims (Open Pros) are even more reliant on the spokes to maintain a true wheel shape, which if disc brakes do put more strain on the wheel, is something I am not sure we have enough data yet as to how this impacts the rims.


That being said, with rim brakes ultimately destroying the rim anyways, this may be offsetting issues for road rim longevity.

Bike Gremlin

I discussed (explained?) that in my previous replies (radial vs crossed lacing) - see post #96 .

Rims could easily be replaced when done, if only manufacturers didn't make small changes in ERD and discontinued tried and tested models - which they often do.

Het Volk

So maybe the question is more down to this: Ignoring the tension changes between spokes, is the torque introduced from disc brakes, more destructive to spokes, even if the load factor on each spoke (tensioning and untensioning) is more evenly distributed on a disc brake. Which of course, leads to the question, is the torque on beefed-up, modern hubs, really so prevalent for disc hub, since the hub should, in theory, soak up most of the torque?

Het Volk

It could be a combination of things, but generally speaking, depending on the model year of the rim, there could very well have been a decision by the Company that it was not sure how its rim would hold up under the loads when braking under rim brakes, and even if engineering felt it was safe if properly installed, in the event someone were to die or get seriously injured by a failed wheel, this would provide additional protection in the event they were sued. I mean, I am guessing that Mavic has worked on testing how their road rims work with disc brakes, but a smaller company may not have the bandwidth to get comfortable, and when designed, there was never a thought that these would be used for anything other than rim brakes.

As someone who works in the corporate world, and the litigious nature of society (especially as we deregulate, making caveat emptor more and more a theme) the companies protect against the black swan event.