Cursed brakes

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Comments

  • konadawg wrote:
    I used to think that, but then said to myself, this could not be right

    You were right the first time.
    konadawg wrote:
    Should I be doing something better with my time?

    Note to self: never buy any brakes designed by konadawg...
  • The Rookie
    The Rookie Posts: 27,812
    Just to be clear, most oil based products have an auto-ignition temperature of between 250-330C, so get it that hot and the rubbish will burn off.

    When we fitted hydraulics (used) to the lads bike the pads were contaminated, just so we could decide if he wanted to keep them (and therefore worth buying new pads for) we plonked them in a jam jar of petrol (outside the house) for a couple of hours then baked them in the oven (no flame!) for 30 mins on max and then I roughened the surface with sandpaper.

    When tested they worked fine, so well we've never got around to fitting the new pads we then ordered!
    Currently riding a Whyte T130C, X0 drivetrain, Magura Trail brakes converted to mixed wheel size (homebuilt wheels) with 140mm Fox 34 Rhythm and RP23 suspension. 12.2Kg.
  • YeehaaMcgee
    YeehaaMcgee Posts: 5,740
    edited September 2012
    I've said it before, and here it is again... You may have a very different opinion on what defines "fine" to me.

    By the way, that's not a dig at anyone, it just that I've ridden a ton of bikes which had brakes that "worked perfectly", or were "fine", only to find out that they were what I'd call "absurdly innefective".
    Judgment on such things is coloured by what experiences you've had of brakes, and I tend to be a stickler for having my brakes set up and working properly.
  • I've said it before, and here it is again... You may have a very different opinion on what defines "fine" to me.

    TBH I think I'd go with a £5 replacement pair for what is arguably the most important item on a bike.
  • The Rookie
    The Rookie Posts: 27,812
    As the lad could lift the back wheel easily enough he figured there was no point having any more braking (if they even gave any).....but yup, each to there own, he's a poor student!
    Currently riding a Whyte T130C, X0 drivetrain, Magura Trail brakes converted to mixed wheel size (homebuilt wheels) with 140mm Fox 34 Rhythm and RP23 suspension. 12.2Kg.
  • Heat and noise.
    If you were to immedaitely and suddenly stop the wheel from rotating, then the job of dissipating your kinetic energy is passed on to your tyres, which will invariably fail, causing you to skid.
    Skidding is far less efficient than not skidding, so not much heat energy is generated, and so it takes longer to slow down.

    I did mention having cogs for wheels/tyres, working in a rack, in which case, there would be no slippage, the stopping would be instantaneous.

    The kinetic energy practically in it's entirety less minor losses would be transferred to Earth.

    Think of two snooker balls, one hits the other, the other shoots off, almost all of the energy is transferred. It's just the scale that is different, of course.

    Certainly heat is generated when a brake is used, but that's because of the method used to slow down the rotation, and I don't think that all the energy is converted into heat - just some of it. (ignore noise, keep things simple!)

    Re skidding/not skidding, it's all to do with static friction being higher than dynamic friction, not with how much heat is generated.
    Giant Reign X1
  • If an object was to stop immediately, then the energy is dissipated in the deformation of that object. It has to go somewhere.
    Deforming an object takes a lot of energy since it requires the reorganisation of molecular bonds. There would also be a hell of a lot of noise. Have you ever seen a truck crashing, for example?
  • deveng
    deveng Posts: 132
    konadawg wrote:
    Very nearly right. It's almost just semantics.
    Disc brakes work by converting movement into heat. Heat is not a by-product, it's the actual goal of their function.

    I used to think that, but then said to myself, this could not be right i.e. not all the energy is converted into heat, what if, rather than a brake pad rubbing against the disc, one would have a bolt went through it and that immediately stopped the wheel's rotation, and the wheel was a cogwheel on a rack, so the deceleration was just about instantaneous. So no heat.

    Energy can't be created nor destroyed - and equal and opposite reaction and all that - I figure that when one accelerates, the entire planet reacts by "relatively" accelerating in the opposite direction, and when one brakes, it accelerates in the same direction... Of course the masses are incomparable, so the planet's reaction is utterly, utterly, unfathomably and unmeasurably infinitesimal...

    Should I be doing something better with my time?

    Friction brakes work in exactly as expected: they convert kinetic energy into heat by means of friction. The example you give is pointless: you're describing an over constrained system, so you've made a pretty big leap in logic to come to the conclusion"the deceleration was just about instantaneous. So no heat."

    If you were to have this perfect system in which the tyres couldn't brake traction and the wheels stop "instantaneously" then you'd convert kinetic energy into heat energy by placing the bike frame under a great deal of tension/compression. The wheels may stop instantly (they wouldn't but assume they would), but you still have to dissipate the stored kinetic energy, which would be in the form of heat by doing work putting the frame in tensile/compressive load.
    Wilier Izoard XP 105
    Saracen Zen 1
    SRAM/RaceFace 1x10
  • deveng wrote:
    Friction brakes work in exactly as expected: they convert kinetic energy into heat by means of friction. The example you give is pointless: you're describing an over constrained system, so you've made a pretty big leap in logic to come to the conclusion"the deceleration was just about instantaneous. So no heat."

    If you were to have this perfect system in which the tyres couldn't brake traction and the wheels stop "instantaneously" then you'd convert kinetic energy into heat energy by placing the bike frame under a great deal of tension/compression. The wheels may stop instantly (they wouldn't but assume they would), but you still have to dissipate the stored kinetic energy, which would be in the form of heat by doing work putting the frame in tensile/compressive load.

    I don't understand why you seem to be claiming that I am claiming that I said that no heat is generated when brakes are applied - I'm not. The examples given - imperfect as they were - were just that, examples. Certainly one does not deny that a snooker ball that has been struck by another departs at almost the same velocity as the striking ball. The striking ball has stopped completely. If all it's energy was converted into heat then where is the struck ball getting it's kinetic energy from?

    But I also don't understand why you are insisting that all the kinetic energy is converted into heat (again, forget noise) and none of it is transferred to Earth.

    Equal and opposite reaction. The Earth is slowing the bike and rider down, hence some/a lot/an unknown amount of energy is being transferred to it, and likewise, some is being converted into heat. If there were no Earth i.e. the bike were in outer space then one could apply the brakes till ET's call finally gets through, yet the bike would not decelerate.

    The ratio of energy transferred to Earth and that converted to heat? I don't know, I suppose it all depends. If one drags the brakes and decelerates slowly then the pads are likely to generate quite a lot of heat, but if one grabs a fistful of brake and stops as near instantaneously as possible then likely not that much.

    The above is about the upper limit of my recollection of A level physics... :mrgreen:
    Giant Reign X1
  • konadawg wrote:
    I don't understand why you seem to be claiming that I am claiming that I said that no heat is generated when brakes are applied - I'm not. The examples given - imperfect as they were - were just that, examples. Certainly one does not deny that a snooker ball that has been struck by another departs at almost the same velocity as the striking ball. The striking ball has stopped completely. If all it's energy was converted into heat then where is the struck ball getting it's kinetic energy from?
    Er, there's no heat generated there, because the energy from one ball is transferred into another.

    I've already explained what happens when a body comes to a sudden instantaneous stop. Check back a bit.

    I can be a volatile person at times, but on this occasion I'm not trying to be, I can't actually think of a polite way to say this...
    Your knowledge of physics/mechanics is so misguided that it baffles the mind on where to begin correcting you, I've not encountered someone with such a comprehensive misunderstanding of the topic they're trying to discuss, and I'm at a loss on how to deal with it.
  • Snooker balls colliding will generate heat, small amout yes but heat all the same.
  • Pretty much any interaction between more than one of anything will generate heat.
  • deveng
    deveng Posts: 132
    konadawg wrote:
    deveng wrote:
    Friction brakes work in exactly as expected: they convert kinetic energy into heat by means of friction. The example you give is pointless: you're describing an over constrained system, so you've made a pretty big leap in logic to come to the conclusion"the deceleration was just about instantaneous. So no heat."

    If you were to have this perfect system in which the tyres couldn't brake traction and the wheels stop "instantaneously" then you'd convert kinetic energy into heat energy by placing the bike frame under a great deal of tension/compression. The wheels may stop instantly (they wouldn't but assume they would), but you still have to dissipate the stored kinetic energy, which would be in the form of heat by doing work putting the frame in tensile/compressive load.

    I don't understand why you seem to be claiming that I am claiming that I said that no heat is generated when brakes are applied - I'm not. The examples given - imperfect as they were - were just that, examples. Certainly one does not deny that a snooker ball that has been struck by another departs at almost the same velocity as the striking ball. The striking ball has stopped completely. If all it's energy was converted into heat then where is the struck ball getting it's kinetic energy from?

    But I also don't understand why you are insisting that all the kinetic energy is converted into heat (again, forget noise) and none of it is transferred to Earth.

    Equal and opposite reaction. The Earth is slowing the bike and rider down, hence some/a lot/an unknown amount of energy is being transferred to it, and likewise, some is being converted into heat. If there were no Earth i.e. the bike were in outer space then one could apply the brakes till ET's call finally gets through, yet the bike would not decelerate.

    The ratio of energy transferred to Earth and that converted to heat? I don't know, I suppose it all depends. If one drags the brakes and decelerates slowly then the pads are likely to generate quite a lot of heat, but if one grabs a fistful of brake and stops as near instantaneously as possible then likely not that much.

    The above is about the upper limit of my recollection of A level physics... :mrgreen:

    You're confusing conservation of momentum with conservation of energy.

    The snooker ball example is not an example of friction brakes, whereas a bike slowly doing using the brake is an example of friction brakes.

    Yes, there is an equal but opposite reaction between the tyre and the earth; but this is not what dissipates the bike and riders kinetic energy.

    Here's the science behind friction brakes to explain where the energy goes:

    force is applied to the pads. coefficient of friction (based on pad surface area and pad compound) * force applied to pad * radius at which the centre of the pad acts = braking torque. When you transfer torque between objects rotating at different speeds the energy dissipated as heat is equal to the torque applied * difference in rotational speeds.

    Because the caliper is static the difference in rotational speeds is equal to the speed at which the front wheel rotates, which means there is no useful work done, as the useful work is equal to torque*output speed. The caliper is not rotating, so the useful work = torque * 0, so 0. If there is no useful work being done then all the energy is being transferred to heat.


    The above is some of my recollection of 2 years developing regenerative braking systems...
    Wilier Izoard XP 105
    Saracen Zen 1
    SRAM/RaceFace 1x10
  • cooldad
    cooldad Posts: 32,599
    deveng wrote:
    konadawg wrote:
    deveng wrote:
    Stuff.

    I don't understand

    No you don't...
    Simplified for konadawg
    I don't do smileys.

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