why should he get a payout – he should check his bike before use (and during if need be). Duty if care and all that is bollox, taking responsibility for yourself is where it should be at!

why should he get a payout – he should check his bike before use (and during if need be). Duty if care and all that is bollox, taking responsibility for yourself is where it should be at.

Not necessarily, remember the Alan Ide case re defective handlebars. I also know for a fact that Halfords, Evans amd Cycle surgery have paid out before a case comes to court re shoddy bike builds where parts have catastrophically failed after a few rides due to shoddy workmanship

I understand that the vibration can undo the QR, to expand on my point, a QR needs to move a fair bit to get past the tabs, even slightly loose cones are noticeable, i just find it hard to believe that you wouldn’t notice that the QR was loosening before the point where it became so loose that the wheel was able to come out.

At the end of the day the QR was undone the wheel did come out, whether it was “user error” or a mechanical defect is something i don’t think will ever be known.

mrmo – and at the heart of the case it doesnt matter how the QR came loose, only that the fork design made it possible for the wheel to forcibly eject under braking. The QR question is a red herring.

If the experiement or the accident was repeatable and mathematically statistically tested and significantly proved dangerous then yes the manufacturer would be at fault.

I think he would have noticed a loose wheel within 5 secs of riding one.

Downhill I would never use a quick release e.g. whistler. Bolts or 20mm-which is what it was designed for.

Materials can fail though and there have been many recalls on shimano QRs!

Most of us haven’t got a law degree or mechanical physics Phd.

Best we can do is learn and check out QRs before and during check points of trails. 😈

James annan calculates a force od 1800N and cannondale give almost twice that figure for the force acting to push the wheel out of the dropout. Easily enough to lift the weight of the rider and bike. I can’t see any flaw in either set of calculations

Even if these calculations of force are correct (and the fact that 2 people have produced results that differ by a factor of 2 suggests that there’s a lack of agreement on how to do this), in what direction are they acting?

Even in a ‘vertical’ dropout fork, the slot is typically in-line with the fork leg (Annan’s involvement/interest originated in his custom fork with a backwards sloping dropout…). Given that the caliper is typically mounted on the back of the fork leg the force is going to be acting at around 20deg away from the perpendicular to the dropout. Although part of the force is acting downwards, more of it is acting into the back of the dropout.

In both cases the force quoted is the resultant force in the direction of the dropout. the difference is due to a difference in max deceleration used to calculate the force

stoner, i know the case revolves around the angle and placement of the dropout, and how the braking force acts so that it pushes the wheel out. Just unable to get my head round how the QR got so loose as to be able to come out without the rider noticing?

My experience is that when a wheel is no longer secure it is noticeable.

Fox paying out suggests they weren’t confident of winning, which implies that there might be an issue, it might need a certain set of conditions to have been met, but there is something about dropout design and disk brakes that could be an issue.

Just unable to get my head round how the QR got so loose as to be able to come out without the rider noticing?

Have you ridden The Gap? It’s quite bumpy.

Which I’d have thought would make it all the more obvious when something wasn’t right on the bike in question…

I find this debate very depressing and can only wish Russ well and hope that the outcome helps ease his situation in any way possible.

In both cases the force quoted is the resultant force in the direction of the dropout. the difference is due to a difference in max deceleration used to calculate the force

OK; but presumably we are saying that this directional force is in the first instance created as a result of the interaction between the pads and disc rotor, creating a pivot point. A component of this force is then acting in a particular direction, in line with the dropout. This component will, clearly, be smaller in magnitude than the original force. So… if the pad/disc interface is sufficient to cause the larger decelerative force in the first place, how does the (smaller) component of this force become sufficient to break that same interface? Particularly when that directional force component must also overcome the clamping force of the QR at the same time?

Or am I missing something?

leverage ratios M

leverage ratios M

In what sense?

James Annan describes it better than I can.. Its to do with how far from the effective pivot the force is applied.

In Annan’s own ‘analysis’ he states that the force at the disc is 2460N, whilst the force at the dropout is 2000N. If my understanding of his argument is correct, he is stating that this 2000N force, created as a result of the interaction between the pad and disc, is sufficient to overcome that interaction.

So effect is bigger than cause?