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Done with my Zee brakeset.. whats next?

toodles

ridiculously corgi proportioned
Aug 24, 2004
5,480
4,720
Australia
They're German brakes. Think about that for a second.
Not sure if serious hey. Hope are all made in the UK. Sure there's economies of scale, and the Trickstuffs are a better quality brake - but thats a hell of a premium.

Also, again we're digressing. The problem isn't that Trickstuffs are insanely expensive. Given their manufacturing size, batch volume and location, the price is merely a bit more than I'd expect. The real problem is that every company who could make a good brake is busy selling us new freehub standards, electronic shifting, multi-coloured rotors or some other non-essential rubbish instead of safe, consistent, powerful brakes.
 

jonKranked

Detective Dookie
Nov 10, 2005
85,573
24,192
media blackout
Not sure if serious hey. Hope are all made in the UK. Sure there's economies of scale, and the Trickstuffs are a better quality brake - but thats a hell of a premium.

Also, again we're digressing. The problem isn't that Trickstuffs are insanely expensive. Given their manufacturing size, batch volume and location, the price is merely a bit more than I'd expect. The real problem is that every company who could make a good brake is busy selling us new freehub standards, electronic shifting, multi-coloured rotors or some other non-essential rubbish instead of safe, consistent, powerful brakes.
Germans are, in general, excellent at manufacturing and engineering. I've run some German made manufacturing lines before. Equipment was top notch.
 
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freeriding

Monkey
Jun 5, 2011
138
1
The MTB industry these days keeps making wheel diameters bigger (cutting brake force proportionally),
This is a false statement since the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant.
 

toodles

ridiculously corgi proportioned
Aug 24, 2004
5,480
4,720
Australia
Germans are, in general, excellent at manufacturing and engineering. I've run some German made manufacturing lines before. Equipment was top notch.
Yeah I buy a fair bit of machinery from zee Germans at work, as well as from the UK, Italy and the US. The German stuff is nicely made thats for sure.

This is a false statement since the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant.
I think the point is that its much harder to lock a 29er wheel than a 26" wheel. The ratio of wheel size to rotor size is the issue.
 

freeriding

Monkey
Jun 5, 2011
138
1
I think the point is that its much harder to lock a 29er wheel than a 26" wheel. The ratio of wheel size to rotor size is the issue.
This is what I'm saying, this is a false statement. Again: the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant.
Same weight, same velocity 26 and 29 bikes, have equal kinetic Energy, thus will need the same Braking force. (equal rotor size)
 
This is what I'm saying, this is a false statement. Again: the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant.
Same weight, same velocity 26 and 29 bikes, have equal kinetic Energy, thus will need the same Braking force. (equal rotor size)
Um, no.
 

toodles

ridiculously corgi proportioned
Aug 24, 2004
5,480
4,720
Australia
This is what I'm saying, this is a false statement. Again: the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant.
Same weight, same velocity 26 and 29 bikes, have equal kinetic Energy, thus will need the same Braking force. (equal rotor size)
I don't have the big engineery words to explain this, but wouldn't the 29er wheel have more leverage over a 203mm rotor than a 27.5 wheel would? Plus a bigger footprint and more traction? The same reason they don't put 15" go-kart steering wheels on a tractor with no power steering. Larger wheel = more turny force stuff = harder to lock up.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,915
1,200
Same weight, same velocity 26 and 29 bikes, have equal kinetic Energy, thus will need the same Braking force. (equal rotor size)
The part you're missing is that to generate that "same braking force" you now need greater input force from the rider's finger, because you have a shorter moment arm between rotor and contact patch. The wheel has increased leverage over the brake, unless you increase rotor diameter proportional to wheel diameter.

You also have greater angular momentum from the larger wheel having greater mass further away from the center of rotation (in L=Iw, the moment of inertia I is higher for a 29" wheel of same components), so you don't actually have equal kinetic energy for equal velocity either.

@toodles / @johnbryanpeters are correct.
 

HAB

Chelsea from Seattle
Apr 28, 2007
11,580
2,006
Seattle
I don't want to spend $800 on a pair of brakes, but I'm fortunate enough to make enough money that I could, and I'm not entirely happy with any of the other options.

I'm thinking I'll probably get another pair of RORs though, since it doesn't sound like I could actually buy the Trickstuffs even if I wanted to.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,915
1,200
Hey HAB if you wanted to try something cheap, I scribbled this out for toodles from the spreadsheet just before from your numbers (just for interest since we'd ridden all these between us):

31.418, Tech3 / E4
32.084, Formula RO-R
35.052, Formula RO-R on Hayes Prime caliper
35.591, Tech3 / V4

You can get those Prime things cheap on ebay/CRC (sometimes), the expert and pro models have 26mm pistons. You sacrifice a bit of throw but they have slightly bigger pads than Formula and are roughly equal to the V4 for peak force (when combined with RO-R lever). I wouldn't put a bunch of money into it, but if you see them cheap somewhere, it's an option. I moved mine to my trailbike recently. Throw grows a bit because of greater rollback on the Hayes caliper but it's still decent (better than Saint etc). They stopped 650b on my DH bike noticeably more competently than the stock RO-R.

I also meant to mention, if you could get a longer lever blade machined for the RO-R (it's a pretty simple design), you and Flo33 could probably have some fun with that.
 

Cerberus75

Monkey
Feb 18, 2017
520
194
The part you're missing is that to generate that "same braking force" you now need greater input force from the rider's finger, because you have a shorter moment arm between rotor and contact patch. The wheel has increased leverage over the brake, unless you increase rotor diameter proportional to wheel diameter.

You also have greater angular momentum from the larger wheel having greater mass further away from the center of rotation (in L=Iw, the moment of inertia I is higher for a 29" wheel of same components), so you don't actually have equal kinetic energy for equal velocity either.

@toodles / @johnbryanpeters are correct.
Udi is correct in layman's terms you must slow down the momentum of the wheel before you slow the vehicle. A larger wheel will have more momentum to overcome.
 

StiHacka

Compensating for something
Jan 4, 2013
21,560
12,504
In hell. Welcome!
Udi is correct in layman's terms you must slow down the momentum of the wheel before you slow the vehicle. A larger wheel will have more momentum to overcome.
I am not sure this is very useful. What I find *moar* illustrative is inverting the problem and think about acceleration instead. For the same chain ring, which cassette sprocket requires the smallest force on pedals for instant acceleration from a full stop - 11T or a 50T?
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,915
1,200
Udi is correct in layman's terms you must slow down the momentum of the wheel before you slow the vehicle. A larger wheel will have more momentum to overcome.
That's not the key point though, even if the 29" wheel had identical angular momentum (plausible scenario, for example if you used a lighter rim and tyre than on the 27.5" bike), the leverage it generates over the rotor is still greater which means that the brake lever needs greater input force to stop the bike in an otherwise identical scenario.

It is this moment arm length differential which causes the loss of braking force, which I have shown elsewhere is ~4% when going from 26" to 27.5", and a further ~6% when going from 27.5" to 29".
 

Nick

My name is Nick
Sep 21, 2001
23,928
14,450
where the trails are
the front Saint m820 was acting up, getting soft once a month after 2 years or so of use.
quick call to my bestest LBS.
new complete m820 assembly shipped in 2 days, dropped off the old, picked up the new.

Just sayin' ...
 

Electric_City

Torture wrench
Apr 14, 2007
1,993
716
My Hayes HFX mags with a 22mm mount from 1998 still work. Lol! And here we are allowing The Industry™ to "justify" $1000 brake sets! Hahahahahahahhhaahahaaha!

I have to go over to PB or MTBR so I can come back to reality apparently! Lololololololol lol!
 

Sandwich

Pig my fish!
Staff member
May 23, 2002
21,031
5,921
borcester rhymes
wait what? There are plenty of still functional hayes brakes out there. They work, just not well.

I think what people are asking for is reliability + function + reasonable weight + low cost
 

HAB

Chelsea from Seattle
Apr 28, 2007
11,580
2,006
Seattle
Udi is correct in layman's terms you must slow down the momentum of the wheel before you slow the vehicle. A larger wheel will have more momentum to overcome.
It's true that a larger diameter wheel and tire of the same construction has more kinetic energy at a given linear velocity of the bike, both because it's heavier and because the mass is centralized further out, but that mass is a very small percentage of the total bike/rider system and isn't the most significant part of the equation.

A brake converts kinetic energy to heat. The force applied to the rotor is the clamping force of the brake times the coefficient of friction between the pads and the rotor. The power of the brake, the rate at which it converts kinetic energy to heat, is the product of this force and the velocity of the rotor brake track past the caliper. At a given linear speed of the bike/rider and a given rotor size, a bike with bigger wheels will have the rotor spinning more slowly, and therefore have less braking power.

@Udi I'm game to try some frankenbraking, but I need a new pair of brakes for one bike, not just to upgrade my RORs, so that unfortunately doesn't help much.
 

freeriding

Monkey
Jun 5, 2011
138
1
the larger wheel having greater mass further away from the center of rotation
I clarified that we have same weight, same velocity 26 and 29 bikes that have equal kinetic Energy -> which means that rotating mass will have to be the same.

It's true that a larger diameter wheel and tire of the same construction has more kinetic energy at a given linear velocity of the bike, both because it's heavier and because the mass is centralized further out,
This is a false statement since the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant. As far as heavier is concerned, i clarified that the rorating mass has to be the same.

The part you're missing is that to generate that "same braking force" you now need greater input force from the rider's finger, because you have a shorter moment arm between rotor and contact patch. The wheel has increased leverage over the brake, unless you increase rotor diameter proportional to wheel diameter.
After second thought you are correct, cause angular velocity ω comes into play. Torque acting on the entire surface of disc brake rotor (replace π with φ, cause brake pad is not occupying the entire rotor surface) : T = (2/3)*µ*P*π*ω*[r2^3 – r1^3]
µ - Coefficient of friction
P - Pressure applied on disc brake by the pad
r1 -Radius at inlet of rotor
r2 - Radius at outlet of rotor

As you can see, the torque apllied and thus the the work done per second by the pad on the rotor surface and which is equal to the heat generated (transformation of the kinetic energy), depends on the angular velocity, which is lower in a 29er wheel compared to a 26er.

Increasing the rotor radius, will counterbalance the lower angular velocity or increasing the pressure P.

In conclusion, between same weight, same brakes, same velocity 26 and 29 bikes that have equal kinetic Energy, the 29er will need more braking force in order to stop in the same given distance.

Cheers!

PS. took me some effort/time but this is the best way to explain it in a scientific way in terms of energy. :dirol:
 
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Cerberus75

Monkey
Feb 18, 2017
520
194
That's not the key point though, even if the 29" wheel had identical angular momentum (plausible scenario, for example if you used a lighter rim and tyre than on the 27.5" bike), the leverage it generates over the rotor is still greater which means that the brake lever needs greater input force to stop the bike in an otherwise identical scenario.

It is this moment arm length differential which causes the loss of braking force, which I have shown elsewhere is ~4% when going from 26" to 27.5", and a further ~6% when going from 27.5" to 29".
Trying to simplify it made it too simple. Correct even if the wheel weight was lighter it's like using a breaker bar for more leverage. The further out from the axle the more force the brakes and rotor.
 

HAB

Chelsea from Seattle
Apr 28, 2007
11,580
2,006
Seattle
This is a false statement since the kinetic Energy of a bike relates only to mass and velocity (After transforming the translational and rotational kinetic Energy equations), radius of wheel is irrelevant. As far as heavier is concerned, i clarified that the rorating mass has to be the same.
You're right, I didn't think it though enough. For equal weight wheels the radius cancels between the moment of inertia and the square of angular velocity, for a given bike speed.

I do think that's a poor assumption, because going to bigger wheels isn't going to let you use lighter duty parts. In any case, my main point that the reduced angular velocity of the rotor decreasing brake power on bigger wheels stands.
 

Cerberus75

Monkey
Feb 18, 2017
520
194
You're right, I didn't think it though enough. For equal weight wheels the radius cancels between the moment of inertia and the square of angular velocity, for a given bike speed.

I do think that's a poor assumption, because going to bigger wheels isn't going to let you use lighter duty parts. In any case, my main point that the reduced angular velocity of the rotor decreasing brake power on bigger wheels stands.
If the distance of the force is greater, and moving in an arc it will create a more leverage on the rotor. Breaker bars,longer wrench, wider handlebars ect all are doing the same thing a larger wheel will do.
 

freeriding

Monkey
Jun 5, 2011
138
1
I do think that's a poor assumption, because going to bigger wheels isn't going to let you use lighter duty parts.
It certainly can happen. Simple example: 26 tyres with super gravity casing and 29 tyres with snakeskin.

In any case, my main point that the reduced angular velocity of the rotor decreasing brake power on bigger wheels stands.
It stands, i explained it scientifically here :happydance:
 

kidwoo

Artisanal Tweet Curator
It certainly can happen. Simple example: 26 tyres with super gravity casing and 29 tyres with snakeskin.
I've spent a lot of time on the side of trails the last few years as friends of mine just HAD to try this approach out. At least theres the benefit of the comedy involved in watching 26" tubes go into 29" wheels.

If you needed tough 26 or 27.5 tires, you still need tough 29 tires.

Ask the syndicate team.
 

toodles

ridiculously corgi proportioned
Aug 24, 2004
5,480
4,720
Australia
It stands, i explained it scientifically here :happydance:
I think what you mean to say is that you realised where you'd gone wrong and conceded there. I'm no mechanical engineer, but even without the leverage difference acting on the rotor, the increased traction would mean that the brakes are going to have to be more powerful to lock the wheel. I'd certainly prefer my skidding tyres to be heating up than my brake pads.
 

toodles

ridiculously corgi proportioned
Aug 24, 2004
5,480
4,720
Australia
Just ride the fucking things and you'll be able to tell your brakes work worse on bigger wheels.
Kidwoo went full lizard and is pushing 29ers. You heard it here first.

Seriously though, good on the guys willing to nerd it out. Without them we'd all just be riding around with the vague suspicion shit isn't working correctly.
 

Electric_City

Torture wrench
Apr 14, 2007
1,993
716
Ridemonkey- We talk like we know our shit about every leverage ratio, oil viscosity, shimz stack and suspension curvature, but we don't really ride bikes.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,915
1,200
After second thought you are correct
So you realised you were wrong, then ate your words in a very roundabout way.

the 29er will need more braking force in order to stop in the same given distance.
This is what I said all along.

KE is only proportional if the moment of inertia (I) increase and angular velocity (ω) decrease exactly proportional to each other (on the larger wheel), which in reality is not always the case. Mass distribution affects (I), not just the mass itself. Thus even if you have two 29" wheels of identical mass, and identical (ω) for a fixed linear velocity - you could still have a different (I) and thus different total KE.

However I very clearly stated that it is primarily the lever arm difference between rotor radius and wheel outer radius that changes the braking input force required, regardless of the KE being the same or different.

It doesn't matter if you use the (r) or (ω) delta between wheel and rotor for calculation purposes, either way unless you increase the rotor radius proportional to wheel radius, you will lose braking force with a larger wheel.

This was a simple moment arm equation and you overcomplicated it.

My Hayes HFX mags with a 22mm mount from 1998 still work. Lol! And here we are allowing The Industry™ to "justify" $1000 brake sets! Hahahahahahahhhaahahaaha!
Yet somehow you can "justify" your expensive Hope V4 brakes instead of just using your HFX Mags?

I must say, many contestants are tied for this week's "Absolute muppet of Ridemonkey" award.
 

freeriding

Monkey
Jun 5, 2011
138
1
I explained it using a scientific way, based in energy terms. This is how things make perfect sense, when one abides to scientific ethics.
To use torque is not compatible with energy terms, which i had in mind. (conservation of energy)
The explanation to sum it up, was clearly fulfilling for me and this is why i posted it.
It has nothing to do with playing smart or better or anything.
Cheers! :neo:



PS. to admit a mistake and explaining it, is the way to behave, don't be rude :rockout:

So you realised you were wrong, then ate your words in a very roundabout way.
 
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Gary

"S" is for "neo-luddite"
Aug 27, 2002
7,541
5,472
UK
That guy who keeps coming in the bike shop you work in to *discuss* the subtle differences between the latest high end products. That guy you humour for hours on end. Partly because it's your job to. Partly because eventually he will pay for a product you'd probably never otherwise order into stock?


 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,915
1,200
I explained it using a scientific way, based in energy terms. This is how things make perfect sense, when one abides to scientific ethics.
To use torque is not compatible with energy terms, which i had in mind. (conservation of energy)
You implied that the various explanations you were already given (by numerous people) were not correct.
You are still implying that the energy balance is the "correct" way, with your BS about "scientific ethics".

In reality you can explain these things (equally correctly) with basic newtonian physics OR a lagrange-style energy balance - if you read my previous post you can see that both methods if performed properly will give the same result. You made a mistake here which is why we've wasted a page of discussion.

Just because you misunderstood a simple torque-based explanation (everyone else understood) doesn't make everyone else incorrect.

PS. to admit a mistake and explaining it, is the way to behave
I have all the respect in the world for someone who can admit a mistake.
None for someone who thinks they can but didn't really. :)
 

Jm_

sled dog's bollocks
Jan 14, 2002
18,855
9,560
AK
You implied that the various explanations you were already given (by numerous people) were not correct.
You are still implying that the energy balance is the "correct" way, with your BS about "scientific ethics".

It is true, the the light, the dark side, it must be all in balance. It is the energy between all living things. Use The Force to stop.
 

Udi

RM Chief Ornithologist
Mar 14, 2005
4,915
1,200
It is true, the the light, the dark side, it must be all in balance. It is the energy between all living things. Use The Force to stop.
It was all going well until you mentioned "The Force".
We mustn't explain anything using forces and torques, we must use only energy.

It goes against scientific ethics man.
 

freeriding

Monkey
Jun 5, 2011
138
1
Of course energy method is superior to elementary school torque and leverage. It is far more difficult to master and comprehend and can be used for far more complex problems.
Feel free to print my answer and study it.
Aaahhhh, Just kidding men:cool:
 

Electric_City

Torture wrench
Apr 14, 2007
1,993
716
you can "justify" your expensive Hope V4 brakes instead of just using your HFX Mags?

I must say, many contestants are tied for this week's "Absolute muppet of Ridemonkey" award.
Since you've only been talking about bikes online for the past few years, I'll explain it to you newbies.
Hayes HFX mags retailed for $300 a piece. Though sometimes you could find them for $250.

Being that the Hope V4's cost me about $185 a piece (not over $200 I recall) - and the fact that it's 20 years later- I'm not sure my "expensive" brakes are expensive at all. As a matter of fact, I never said they were expensive. Compared to brakes that don't work they might seem expensive, but that's it.

I'll let you claim your own reward this week. Congratulations! You win!