I’ve read on certain manufacturers sites that a road disc fork cannot replicate the comfort of one designed for rim brakes due to the increased braking forces.
I can accept this.

I have read a thread on another forum which suggests that the differing positions of the application of braking force may have a bearing on this.

How so?
Surely a rim and disk brake both act on the wheel, which transfers the same braking force via the dropout?

If we could impart more force than the wheel or fork would tolerate, they would fail?

I’m sure it’s far more complicated than that, but does a disc fork need to be stiffer than a rim brake fork because of the position of the force applied, or just because of the increase in available power?

Just an idle question, ta.

I’d have thought the rim brake applied less force but is acting over an area further from the centre point.

A disc brake being closer to the centre of the wheel needs to apply more force.

Its location lower down the fork also means its further from the bulky uppers of the fork so they have to reinforce the skinny lower bits resulting in less flex and comfort.

Is the braking force transferred to the fork at the point of application, or at the dropout, or a proportion of both?

The force is applied at the location of the brake, so in different places for rim and disc brakes. Road forks tend to be wider at the top, presumeably meaning less reinforcing is required.

The braking force to slow the biggest lump on the bike, ie the rider would act through the tyre contact patch, the dropouts, fork legs and crown into the frame etc etc. The limiting factor is the tyrefriction, and as both disc and rim brakes will lock a wheel this is the maximum force that can be applied. The difference will be that there will need some support at the caliper mounting point so here may be a stiff section in the fork legs so taking out a bit of flex in that area. Nothing that a good designer could’nt work his/her way around though it may make a small difference.

Force is being applied at the end of the leg, rather than at the crown. Meaning braking has a lever the length of the fork leg, so the entire leg needs to be stronger throughout its’ length as a result.

honourablegeorge has it – disc brakes are force multipliers, braking at the same rate produces around 4x the force on the disc mount (ratio of disc rotor to wheel diameter), and most of the force is on one side.

So the fork has to be stiffer to compensate for braking forces, so it’s not so springy/comfy.

I remember seeing an explanation of this years ago which revolved around the leverage effect that stopping a rotating object causes but can’t find the article.

It basically drew a line from the axle to the tyre’s contact patch on the ground which was “lever 1” then another from the axle to the rim braking surface “lever 2” and a third from the axle to the centre point of a disc brake rotor, “lever 3”. It went on to show that the force on lever 1 (mass x velocity)needed to be counteracted to stop the bike.

Then it demonstrated that because 1 and 2 were fairly close in terms of length, the force (leverage) needed to slow the bike was slightly more than that being generated. However, because lever 3 (the disc brake) was so much shorter than 1, the actual force needed to stop was much, much greater.

The upshot was, because the ‘fulcrum point’ for all 3 was the axle and the only part of the bicycle that was connected to that was the fork, the fork itself needed to be considerably stronger and stiffer for the disc brake.

That all probably comes across as nonsense so I’ll keep looking for the article…..

Edit: Slow typing means that others have put it so much simpler 😳

So on a rim brake, roughly 75% of the braking force is at the point of application, 25% through out the rest of the fork toward the dropout?

Edit, asymmetry explains a lot.
A single disc on the centreline would be efficient.
I think the weight penalties of a Buell type system on a bicycle would be prohibitive however.

The hub centre steering you’d need to achieve it (presumably?) would be weighty, but glorious.
🙂

As per BenCooper; it’s the lever and twisting effects of the disc that require the beefed up fork.

So on a rim brake, roughly 75% of the braking force is at the point of application, 25% through out the rest of the fork toward the dropout?

The forces on brake mounts and dropouts from braking are equal and opposite*. The fork mount is pushed forwards, the dropouts backwards, rotating around their common centre.

They’re much further from that common centre with rim brakes, and equal on both sides.

*This is the bit that people forgot, which led to those problems of hubs being ejected from disc forks.

I’m fine with the equal and opposite thing.

So if a disc is employed, the lower leg /droput of one fork has to deal with 75% of the braking force, instead of half of the 25% applied by a rim brake?
Roughly?

Yes, it’s hard to say exactly because it depends on hub stiffness and the like – how much leverage is transmitted through to the right leg, etc. It also depends which force you’re looking at – there’s the force on the mounts and dropouts, then there’s the force trying to bend the forks backwards at the crown.

The force at the crown, assuming the same rate of deceleration, is the same. That’s just a horizontal force on the axle.

Bigger rotors = skinnier forks then 😉

Bigger rotors = skinnier forks then

You could have two rotors roughly the same diameter as the rim and then place a single calliper at the the fork crown to deal with the asymmetry problem 🙂

Perfect!

🙂

I had a British Eagle gas pipe road bike in the 80’s with forks that curled like a Turk’s boot.
Comfiest thing I’ve ever ridden, but a bit odd at high speed, a bike that needed it’s own lane at times.
😀

Mac, call Honda.
One single, central disc though.

I’ll fight Eric Buell for the patent rights, you think of a good name.

Bigger rotors = skinnier forks then

only if the disc caliper for a larger rotor is bolted to the fork further from the axle (which it isn’t, i think, unless they do with the flat mount standard?)

A larger rotor reduces the force anyway (ratio of rim to rotor) but yes, it’s then levering off the mount then that introduces lots of new interesting problems.

There is of course a solution to the problem of single-sided braking forces with disc brakes:

[url=https://flic.kr/p/ee8bre]Custom Bromptons 2[/url] by Ben Cooper, on Flickr

But the unsprung/sprung weight ratio goes up hugely.

Is the weight saving in the swingarm/fork not negated by the disk & caliper weight?

Looks well Bol hough.

This was on a Brompton – not such an issue with that 😉

Using smaller, lighter wheels would increase the proportional detrimental 8)effect of a two disc system re sprung/unsprung weight surely?

Surely being a Brompton makes it worse?

The fact that it’s a Brompton makes it worse?

Oh yes, it was ridiculous. That customer probably died. But he paid me, so all is good.

I’m not having a go, Ben, it’s a genuine thing of beauty and a true object of desire.
I love the madness required to do this.
😀

How did you square the physics with the design?

I do lots of Brompton disc forks anyway – the only complication really was getting a hub with double disc mounts. The rest of it was simple – the callipers are on opposite sides, no-one makes mirror image callipers.

With small wheels, the ratio of wheel size to rotor size is smaller, so much less force on the forks anyway. Plus shorter forks, so much less leverage on the crown and steerer.

I’m not having a go, Ben

Didn’t think you were 😉

So would this have practical benefits on a bigger wheeled bike?

Smaller discs and callipers, fork legs designed to deal with 25% less force?

Weight and feel benefits re servicing?

Just pondering..

I’d think that with double 200mm+ disks, although the theoretical force would be less to come to a stop gradually, who ever actually sticks to that, what would actually happen is people would brake harder and the crown would break. I know the limiting factor is the tyres friction, but you have to really try to get anywhere near that in a straight line on dry tarmac on a normal bike.

So all we have to do is to design a central, between spoke, integral disc/caliper/hub/axle system?

Same time tomorrow?

So, if we had discs that were about 700mm and had the caliper near the top of the fork, maybe in the middle to balance forces, then we could have comfier forks. There must be a solution in there somewhere.