Bicycle Mechanics - Are disc brakes dangerous?

My wife is considering a bike with disc brakes. While poking around on-line I found this website:

http://www.ne.jp/asahi/julesandjames/home/disk_and_quick_release/

It details how disc brakes can actually cause a front wheel to suddenly come out of the drop-outs and cause a serious crash.

His reasoning looks OK, but there are thousands of bikes out there with discs. Part of me thinks that if this were a serious problem, we'd hear more about it.

Should we steer clear of discs?

Thanks

1. The issue is real.

2. I'm not ready to say "Steer clear of disc brakes yet."

3. If you have disc brakes, do everything right. Don't file the lawyer tabs off of your fork. If you have a quick release, be sure that you use it correctly. Avoid quick releases that have an external cam mechanism that you can see.

4. Thru axles on mountain bike forks. Why do you suppose they suddenly have become so popular?

If the front drop-outs face slightly forward then its impossible for the wheel to pop out. My forks were designed that way, but I'm not sure standard forks are.

If the bike is a recent design and designed to use disc brakes it is likely the front dropouts are angled properly to prevent wheel ejection. Ask the dealer about this before buying.

Also, as recommended DO NOT fill off the lawyers lips and be sure the quick release is properly tightened.

Disc brakes are not dangerous. They actually work extremely well. While the writer of the article has somewhat of a valid argument, it's no where near as big of a "sky is falling" scenario as he makes it seem.

The odds say the dropouts won't face forward. It's okay. I will agree with Retro in that you should use an internal-cam skewer (like the kind Shimano makes), and make sure you get them on tight.

I've been riding a custom 29"er with a rigid fork (and no lawyer tabs) and disc brakes for quite some time with no problems.

I have feeling that people with disc brakes (me included) tend to bust many more spokes than someone running V's as well. The forces put on the spokes and rim are tremendous compared to rim brakes in which the spokes only have to support the riders weight. At the same time, it probably keeps the rim so tight and rigid during braking it may even be stronger during. Not sure.

Don't forget people should check the quick releases before every ride. I don't think you'll have any problems if you make sure the quick release tight.

If you set up a rim brake incorrectly or let it get very worn out it can move into the spokes stopping the wheel, and throwing the rider. Are rim brakes dangerous?

the title of this thread should ask

are poorly set up bicycles dangerous ?

yes they are.


disc brakes are not dangerous.

The force on the QR is exactly the same with disc-brakes as it is with rim-brakes; they both push the hub & axle backwards while the fork & frame tries to continue forward. This real problem is people not tightening their quick-releases correctly. An age-old problem with bikes and people not knowing how to use their equipment properly.

Heck, I've worked in a bike-shop for 10-years and I made an idiotic mistake this last weekend. I swapped in a new wheel and it turned out the rim was 2mm smaller in diameter than the Mavic rim I had on there before (Mavics tend to be oversized). So... I didn't check my brake-pads and they were now riding 1mm too high... Guess what? In less than 30-minutes later, I had a flat-tyre... Am I gonna blame the shop that sold me the new rim? Or Mavic for making their rims slightly too large? Or sue Shimano for making brakes that don't automatically adjust themselves to center the pads on the rims?

i'm not dead yet.


This real problem is people not tightening their quick-releases correctly. An age-old problem with bikes and people not knowing how to use their equipment properly.

+1

whatever happened to checking one's equipment out before using it, whether a bike or car or whatever?

i check any quick releases before riding, no matter what. i also walk around my car before leaving home. maybe i'm just an anal-retentive schmuck, but like i said, i'm not dead yet.

Are disk brakes dangerous?

well that makes me think of my grandfather; he had low blood pressure;
which he accepted as better than having none at all.

Peter

The force on the QR is exactly the same with disc-brakes as it is with rim-brakes; they both push the hub & axle backwards while the fork & frame tries to continue forward.

Not quite correct, disc brakes (as commonly positioned on the fork) tend to push the axle downwards, as noted in the earlier link and unlike rim brakes.

Not quite correct, disc brakes (as commonly positioned on the fork) tend to push the axle downwards, as noted in the earlier link and unlike rim brakes.

I think so too. I think that the wheel tries to rotate around the caliper when you put on the brakes.

This is true of either brake design. As viewed from the left side of the bike, the wheel will want to rotate CCW around the point where the pad grips the rim or disk. The rim brakes are also disk brakes, folks, using the rim as the disk. So if you move the pad (caliper) about, you can get the wheel to want to rotate up down, for or aft. It's a product of the position of the pad, not the brake type. Geeze, didn't you guyz have spiralgraphs? :D

This is true of either brake design. As viewed from the left side of the bike, the wheel will want to rotate CCW around the point where the pad grips the rim or disk. The rim brakes are also disk brakes, folks, using the rim as the disk. So if you move the pad (caliper) about, you can get the wheel to want to rotate up down, for or aft. It's a product of the position of the pad, not the brake type. Geeze, didn't you guyz have spiralgraphs? :D

Rim brakes and discs grip the disc (be it the rotor or the rim) at different radial points, and disc brakes exert a much stronger force than rim brakes. Rim brakes are not capable of producing enough force to eject a wheel and don't force the axle downward. Discs are and do.

That doesn't mean that it's a big risk if you take proper precautions. According to the linked article, disc brakes certainly ARE capable of ejecting a wheel even with a properly-tightened QR. Still, the liklihood of this happening is very small if you follow the proper precautions. The issue is real, but I don't think you're taking your life in your hands with discs, though, especially for road use.

Seems like a logical solution is a slightly redesigned disk break where the pistons are on the other side of the radial axis, so instead of applying downward force to the wheel it will have an upward force... thus making it even harder to lose a wheel.

Not quite correct, disc brakes (as commonly positioned on the fork) tend to push the axle downwards, as noted in the earlier link and unlike rim brakes.That's true, depending upon the location of the caliper itself.


Seems like a logical solution is a slightly redesigned disk break where the pistons are on the other side of the radial axis, so instead of applying downward force to the wheel it will have an upward force... thus making it even harder to lose a wheel.Yup, personally, I would mount it right behind or in front of the fork-tube like a motorcycle, rather than low and 90-degrees to it. This would tend to push the axle longitudinally rearward instead of down.

Funny they didn't do it that way right off the bat. Being a victom of the wheelie of death as a kid, watching that front wheel roll away as I rode a wheelie, I have a great deal of respect for front wheel security! I'd still be riding that wheelie if I could. I almost cleared the handlebars. I remember my brothers laughing as they drug me off the street by my feet, all racked up. Bastards.

This is more of a problem when you put bigger rotors on bikes with QR wheels. That's why you see so many thru-axle 8" rotor combo's.

OK, so we got the Novara Safari my wife was interested in. Couldn't see one in person, ordering was the only way to get one. The fork has vertical drop-outs, so it looks like Novara didn't design the fork to be safer with disc brakes.

That said, what I am taking from most of the replies is that if the front quick-release is checked regularly then it shouldn't be a problem.

Would putting a zip tie or hose-clamp on the quick release further increase the safety? It wouldn't be able to loosen much if the lever is basically locked against the fork. Does this sound reasonable? Thanks

Just how much downward force is generated?Anyone have any idea? If I pull the lever with 5 lbs of force, does that mean 30 lbs of brake force. I not sure what the mechanical advantage is.30 lbs on a 6" disc is just 15lbft-no way will that lift 1/2 me and the bike(100 lbs). Pretty sure it would lock the wheel and you would skid anyway.
I just completely loosened the QR on my Manitou forked NRS and did a bit of violent braking.No F_ _ _ing way is that wheel escaping downward. There is about 100 lbs of bike and me holding the fork down on the wheel axel. Even if the opening was straight down(it isn't, it is canted about 10-15 degrees forward) there is no way it was moving downward.
Now since this is a suspended bike, and I was eyeballing the front wheel, it doesn't tell me much about the rear. The weight shift increases force on the front,and decreases it on the rear. Of course, this also makes the wheel more likely to lose traction and lockup.
I maybe could see a rear wheel jumping out if the QR was completely loose, and you were heading downhill.I still think you would lose traction and skid-lockup the wheel 1st.
The caliper is actually holding the wheel in place to some extent. To jump out the wheel has to pivot against the clamping force of the caliper-rotor-even if the "hole" is straight down.
Now a wheel can always "bounce" out of a soft QR ,but that is unrelated to braking.
Why doesn't someone turn their bike upside down, loosen the rear QR,spin the wheel, and hit the brakes??Hmmm-just noticed that my rear caliper is above the rotor-not much downward force there.
This sounds like BS to me. The front wheels are bouncing out.I don't know if any rear wheels have the caliper positioned so that the force would be downward??
Disc brakes aren't dangerous-other than the increased risk of lockup in poor traction.Thanks,Charlie

take a look at the website in the initial post, he gives some ideas about the forces generated. There seems to be a lot of evidence that disk brakes can and do cause the front axle to slip--the question is how much and how frequently and is there anything that can be done with completely vertical drop-outs. I agree that it seems like it shouldn't have enough force, but apparently it does. And even if the caliper is holding it in place, on the next bump it could, as you say, pop out if the QR is loose.

I read and eyeballed his math. He is missing something, but I'm not sure what. His claim is that for a ~200 lb bike and rider(90kg) the ejections forces are 800 lbs more than the "stay in forces"(weight of bike rider). He said the net ejection force was about 1800 Newtons(I think that is about 400 lbs or so) per dropout, or 800 lbs total for a .6 G braking(stopping at about a rate of 18ft/sec/sec0 or going from 18ft/sec to Zero in one second(this is about 12 miles/hr to zero in one second). I did about that with a completely loose QR, and I felt no "lift" at all while riding the bike.
I just flipped my bike over. With the QR completely off, if I spin the wheel about 12mph(3 rps) and give a brisk squeeze, it will eject the wheel, pushing it maybe 5" up. If I give a really violent quick squeeze, it will push it upward about 3/4th of the way out of the dropout before it is seized in place.
If the QR are on at all they tend to stop it before it jumps out because they are in the countersunk "bores" and the "catch" on the way up and out.
I put a glove on and tried to get a feel for the ejection force by putting my hand on the top of the rotor-I really couldn't get much feel for it. Now an 800lb push/impULse should push the wheel up farther than 5-6 inches.
My dropout is pretty much straight up and down.
He might be underestimating/undercalculating the force/wt of the rider/bike since it will actually be moving-have some momentum- because of the wt shift of the suspension bike.I'm also pretty sure he is ignoring the "clamping" effect. Luck,Charlie

My Fox fork owners manual requires that the thickness of the drop-outs be measured every six months or 100 hours. That's to measure the wear on the lips that surround the skewer contact areas. They require that you replace the lower tubes if the lips get too thin. Smart those Fox folks. My wife's 7 year old RockShok has the same lip design.

In something like 4 years of using Avid mechanicals on three bikes, I've never had a skewer, external cam or internal cam, unscrew one iota. I ride a lot on steep single-track in N Georgia, E Tenn., and western N Carolina and some in Moab. After each ride I remove the front wheels to haul the bikes, so I would notice. Also, long before the skewer would release enough to lose a wheel, the wheel would have to wobble in the drop-outs. You'd notice it. Those lips are very pronounced.

These lips are very beefy and are obviously designed to counter the force from the disc brakes. Those of us who remove our wheels to transport the bike might be wise to loosen the Skewers a few extra turns to reduce the wear on the lips due to the frequent wheel removal/installation. Also measuring the drop-out thickness is a very good idea. The lower fork material on Fox forks is very soft and wears quickly.

Though the article claims that the fork manufacturers have ignored the problem, that appears to not be the case. Also, those pictures showing the bent fork just demonstrates that the fork was too weak in the first place to support a disc brake set-up. Poor design.

I have a ti cyclocross frame/fork for my road bike. The fork is designed for disc's and it's massive carbon fiber. It weighs at least twice what a typical road fork weighs and it's twice as thick. I run V's on it for weight savings.

A final point on loosening skewers-- I brake the Ned Overend way. I stay off the brakes until i
the need to scrub off a lot of speed than brake hard. That should loosen those skewers if they they are prone to do that and they haven't yet.

Al

I can't see the wheel managing to get past the countersunk bores(lips) very easily even if the skewers are kinda loose. With the skewers removed, the wheel would eject fairly easily when the bike was upside down.With the skewers loose,but touching the dropouts, the wheel would lift, but stop at the top of the countersunk(lips). I'm just not sure just how much a short push of 800 lbs is worth, and if the net push is as high as he showed.
When I did real braking, I didn't even feel the wheel push down and "tap" the bottom of the lips. I was probably going ~15 mpg, and gave a reasonably good squeeze.
If your bike has these countersunk bores(I'm guessing they are the lawyer lips some refer to), you have to have an extremely loose QR to get it to jump out under braking. So loose that it will jump out just from the normal bumps.

Just tighten your QR, and check it before every ride-good brakes will save your hide..Luck,Charlie

How about painting a small line on the screw end of the QR, and then you could look down as you ride along, and get an idea if it's unscrewing.

I read and eyeballed his math. He is missing something, but I'm not sure what.Here's what he's missing:

- weight over front-wheel from rider+bike. There is already a static force pushing the fork down over the axle. Any ejection-forces from braking must overcome the rider's weight first with no QR installed. Then add additional friction f=mu from the clamping of the QR.

- longitudinal weight-transfer to the front-wheel under deceleration: F= dG +Bh/wR where d= weight-distribution, G=static total weight, B=braking force, h=COG height, w=wheelbase, R=weight over rear-wheel

- his torque-ratio equation D = 540 * r2 / r1 showing ejection force at the drop-out is also reversed. It should be D = 540 * r1 / r2 which shows the correct leverage between disc-diameter and wheel-diameter (the larger the disc, the more braking force and more ejection force at the axle).

You'll notice that when recalculating to account for this revised equation that the ejection forces are much lower. But they do increase with increasing disc-diameter as one would expect since braking forces goes up as well.

But the equation is set up all wrong to begin with anyway. Calculating downward force at the axle's irrelevant because the wheel's being pushed down into the ground, it's not going anywhere unless you're airborne. In which case, you've got no braking-force because the wheel's not being pushed back by the ground.

The correct modeling equation would actually leave the wheel stationary and calculate forces pushing up on the fork. The part that pushes the fork upwards is actually the caliper. So he needs to come up with a new model that shows the upward push of the caliper. Here again, we see that a design which puts the caliper near the top of the rotor directly behind or in front of the fork like on a motorcycle would be better.

I read and eyeballed his math. He is missing something, but I'm not sure what. His claim is that for a ~200 lb bike and rider(90kg) the ejections forces are 800 lbs more than the "stay in forces"(weight of bike rider). He said the net ejection force was about 1800 Newtons(I think that is about 400 lbs or so) per dropout, or 800 lbs total for a .6 G braking(stopping at about a rate of 18ft/sec/sec0 or going from 18ft/sec to Zero in one second(this is about 12 miles/hr to zero in one second). I did about that with a completely loose QR, and I felt no "lift" at all while riding the bike.
I just flipped my bike over. With the QR completely off, if I spin the wheel about 12mph(3 rps) and give a brisk squeeze, it will eject the wheel, pushing it maybe 5" up. If I give a really violent quick squeeze, it will push it upward about 3/4th of the way out of the dropout before it is seized in place.
If the QR are on at all they tend to stop it before it jumps out because they are in the countersunk "bores" and the "catch" on the way up and out.
I put a glove on and tried to get a feel for the ejection force by putting my hand on the top of the rotor-I really couldn't get much feel for it. Now an 800lb push/impULse should push the wheel up farther than 5-6 inches.
My dropout is pretty much straight up and down.
He might be underestimating/undercalculating the force/wt of the rider/bike since it will actually be moving-have some momentum- because of the wt shift of the suspension bike.I'm also pretty sure he is ignoring the "clamping" effect. Luck,Charlie

This is a pretty poor experiment. Your bike is upside down, with no moving weight on the wheel. The wheel doesn't have a whole lot of kinetic energy 6 inches out from the hub. On a moving bike with the weight of a rider, all of that kinetic energy needs to be transferred to the wheel and dissipated as heat. In order to do that in an acceptable distance, your disc brakes need to exert a hell of a lot more force than the kind of force you're seeing with the wheel spinning freely. I'm not saying that the opposite conclusion - that disc brakes will easily eject the wheel - is true because of this, but this little experiment does absolutely nothing to resolve the matter either way. I trust the figures used by the guy who wrote that article a lot more than the mono-buttocked touchy-feely guesswork here.

As I said, I don't think that disc brakes are dangerous, but the fact that they are capable of ejecting a front wheel is documented, and people should be aware of it. Bad enough to not tighten your QR adequately on a bike with rim brakes. With discs, you are in for some serious pain if you don't take the time to do it right.

grolby, so I'm not going to get a call from the Nobel Committee? Guess I should cancel the flight to Stockholm!
Now, I didn't claim it was anything more than a rough and ready test to get a feel for what is happening.
You are wrong of course, the peak force that the caliper can exert has absolutely nothing to do with what the bike weighs. It is strictly how hard I squeeze the lever-nothing else. Now, how long it takes to stop the wheel has everything to do with the weight/mass/speed of the rider/bike. My little experiment gives an idea of the peak force; it probably underestimates the feel/upward push by a little because the clamping stops the unladen wheels so quickly.The spit out force-not net spit out,but just the spit out force-is independent of the weight-for a given squeeze and wheel speed because if the bike is heavier the decelleration will be less,if it is just a wheel-like my upside bike the decelleration will be much,much higher than a laden upright bike. The weight and decelleration are probably inversely proportional-10x the wt= 1/10the the decel for a given wheel speed and clamping force/torque sorta. I use clamping force to mean the whole force/torque deal which is unchanged in our experiments.
I don't know what you are referring to as 6".Nothing in this has anything to do with 6" out from the hub.The disc is 165mm- or about 3" out-the tire is maybe 13" out?
I didn't notice anyone else doing any rough and ready experiments-I did two. If this BS was true-I knew/felt it wasn't-then my wheel should have been spit out from beneath me when I rode with the very loose skewer. I don't mean that the skewer wasn't tightly clamped-it was literally 3mm from seated on both sides of the dropout.The wheel would kind of flop till it got up to speed.
DannoXYZ- I'm not really up to the math/physics-36 years since I've done any, but what you say seems to agree with what I noticed.If you are right, then his first figure of 2400N(ABOUT 1000LBS) is actually 118N or about 45lbs. Much,much less than would be required to spit a wheel out from under a 200lb bike rider(with the QR completely unclamped).This is what I "felt" or actually didn't feel or see, when I rode with the unclamped wheel. Thanks,Charlie
PS-grolby-don't believe everything you read! Test it when you can! Of course, I didn't just immediately go 15mph and clamp on the brakes(I have occasionally been wrong); I did 5mph with gently braking and slowly(over about 3 minutes you can do about 10 brake runs) worked up to the grand finale.
My final proof- all those folks who haven't managed to spit their wheels out.If his numbers/reasoning were correct, almost every lightly clamped wheel would be spit out.They would be littering the landscape!!

I suspect the numbers of wheel pop-outs due to disc brakes are relatively minor compared to normal wheels falling out due to untightened QR skewers. Even percentage-wise relative to total numbers of disc-brake wheels, the failure-rate's probably lower than QR problems on normal wheels.

Whoah there, phoebeisis. I didn't mean to sound quite so hostile. Sorry. And you're right, of course. The weight on the bike won't effect the maximum force at the caliper.

Now, my physics knowledge is fairly rudimentary at best, so forgive me if I'm totally on the wrong track here, but doesn't the momentum (or kinetic energy) of the bike and rider have some effect on the kinetic energy or angular momentum of the wheel, and isn't that going to cause a greater amount of ejection force as the caliper grips the disc? Eh, maybe not - it's sounding pretty stupid as I write it. Let me think about this a different way.

Okay, so when you brake a free-spinning wheel, it takes a tiny fraction of the force required to brake that same wheel with a bike and rider behind it moving at 15 MPH. The rider is therefore going to squeeze that lever a lot harder to bring the whole package to a short stop, and will have to squeeze it for a greater amount of time in order to stop. With a free-spinning wheel, it's likely that the wheel has stopped spinning long before you reach peak force in your squeeze of the lever, and long before the small amount of force produced in that time could cause the wheel to eject very far. Is that making sense? I'm trying to make a more coherent criticism of the experiment than I did before, without being a jerk about it. As before, I just don't think it really establishes anything either way. Of course, if my assumptions here are wrong, then I'm wrong too, and sorry for opening my big fat mouth about something that I don't understand as completely as you guys.

Grolby,I was a bit over the top also-sorry.
You are right; a lot more energy must be bled off a bike and rider than a spinning wheel- lot more kinetic energy as you say.
You are also right that the wheel "acts' like it is much heavier than it is because it is being"grabbed" by the contact patch and twisted. Right again, that the heavier the bike/rider that harder the "grab".
I'm not really up to the Physics anymore, but if the weight of the bike and rider mattered(all other things being equal) then a fat person would be more likely to spit a wheel off than a skinny person-a heavier bike more likely to lose a wheel than a light bike. Yes, this isn't proof. It is intuition, and intuition is frequently wrong.
I think that the reason that the weight isn't critical is that the decelleration will be greater for the light bike than for the heavy bike.(for the same caliper/rotor/lever). The increased energy /momentum of the heavy system is exactly offset by the decreased decelleration because of that weight-energy-momentum.
I think the grab of the contact patch-friction- pulling the wheel around is always "less" that the weight/mass of the rider/bike because the coefficient of friction of rubber is usually than one. The contact patch can never push as hard as the weight of a person can push because it can never get a 100% "grip".You grab too much brake, and the wheel skids.If the friction coefficient stays below one, then the the weight pushing down always wins. The exception to this is if you lose contact with the road-bounce- while braking-the wheel will be pushed down against air, and it can squirt out.
Rubber against asphalt for really good rubber, and really clean good asphalt can sometimes be greater than one, but I doubt our bike tires ever are over 1.
I've talked in a circle-I think the different decellerations of the light vs the heavy are why the weight doesn't matter much-the clamping power is the important thing-not the weight.Luck,Charlie
PS-Sorry I was so snappish.
PPS I pretty sure he has ignored the fact that the riders "weight" will actually be more than 200 lbs because of the weight shift from the braking; his weight is actually going to be "thrown" downward, so it will be more than 200 lbs.
Ignoring the clamping of the QR it is being held in place by
1)Riders weight.
2)The increase in the riders weight by virture of the forward weight transfer.
3) The actual "clamping" in place by the caliper-to jump out it is having to pivot thru a clamped caliper-pad.
4)Even if the Dropout is straight down, the wheel/axle will have to pivot downward and it will hit the rear lower part of the dropout-lotta friction as the axel has to scrape down past it.
I'm pretty sure he has only attempted to account for the riders weight. If he flipped the ratio as our fellow poster says, he has completely fouled up his calculation of the ejection force.
PPPS- I re eyeballed his math. Danno is certainly correct. His ejection force equation has the ejection force increasing as the disc gets smaller!!
D=540 WR/DR WHERE WR is wheel radius and DR is wheel radius. He used other letters R2/R1 but this is more clear. If the wheel radius is 10 and the disc radius is 2 then the answer is 5x540. If the wheel radius is 10 and the disc radius is 1(half as big) then the answer is 10x540-TWICE AS MUCH FORCE FROM A DISC HALF THE SIZE!! Danno is right-this is gibberish BS. No wonder he got that huge number-about 1000 lbs as the ejection force and a net force of 400 lbs(1000 newtons). The 1000 lbs(2400 Newtons) should have been less than 1/20th of the 2400 Newtons-about 40-50 lbs-not 1000 lbs.
He used 15"/3" or 5 when he should have used 3/15 or 1/5. I rounded all the numbers to make it easier for me( 15" is wheel radius 3" is disc radius-yeah the disc is bigger and the wheel smaller, but I wanted whole numbers).
Thanks Danno and Grolby.

Fascinating theory. I'm not sure we have a proven case of the problem. What concerns me about the theory are two points: a) The axle repeatedly moves up and down and has only a racheting affect on the quick-release nut; b) The quick release can become loose enough that upon application of violent force, it's forced past the lawyer tabs, but its not so loose that the rider notices anything wrong.

Why would the up and down motion only loosen the nut?

Why wouldn't the rider notice a loose axle?

How can it be proven that a braking related wheel ejection was caused by nut loosened when its axle oscillated up and down?

Without answering these questions, the theory is a bit loose. I'm no disk fan - don't have them - don't need them. But none of this seems to pass the smell test yet.

My front quick release came loose during a 6 hr race. I did crash, and fail to check it after the crash. However, I find it very unlikely that it came loose during the fall. When it came loose, I heard a loud thud when I squeezed the brakes. I had no idea what it was, but it was wrong enough for an immediate dismount, and I found that the quick release had come loose. Sure enough, whenever I hit the brakes, the wheel was beating against the 'lawyer tab.' I'm surely grateful, that those tabs are there.