F-150 brakes

jim beam wrote:
>
> On 06/06/2011 05:49 PM, Pete C. wrote:
> >
> > jim beam wrote:
> >>
> >> On 06/06/2011 03:04 PM, Pete C. wrote:
> >>>
> >>> jim beam wrote:
> >>>>
> >>>> On 06/06/2011 01:05 PM, Pete C. wrote:
> >>>>>
> >>>>> jim beam wrote:
> >>>>>>
> >>>> <snip for brevity>
> >>>>
> >>>>>> there is no point in increasing the total piston areas. see below.
> >>>>>
> >>>>> Of course there is.
> >>>>
> >>>> ok, so we double the slave area, then we double the master area because
> >>>> we need to keep the pedal travel the same, then what have we achieved?
> >>>
> >>> Providing better distribution of the clamping force on the rotors.
> >>
> >> increasing piston size doesn't do that easily. piston count, as per
> >> your response to hls, is a better solution.
> >
> > Which is what I originally indicated about dual pistons.
> >
> >>
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>> If for example the F150 calipers had a
> >>>>>>>>> single 40mm dia piston and the F250+ calipers had dual 40mm dia pistons,
> >>>>>>>>> the applied clamping force on the rotor would double given the same
> >>>>>>>>> hydraulic pressure.
> >>>>>>>>
> >>>>>>>> if the master piston diameter stays the same. but because of the
> >>>>>>>> leverage ratio, and the limited amount of brake pedal travel available,
> >>>>>>>> ratios typically remain close. [given a long enough pedal travel, you
> >>>>>>>> could dispense with brake boosters.]
> >>>>>>>
> >>>>>>> Larger brake boosters (or hydroboost) allow for larger master cylinder
> >>>>>>> diameters increasing volume without increasing pedal travel.
> >>>>>>
> >>>>>> - to keep the relative ratios fairly constant.
> >>>>>>
> >>>>>> by definition, if the slave area doubles, the master must also double or
> >>>>>> the pedal travel will double. the latter is clearly impracticable -
> >>>>>> there is very little latitude to vary these ratios, and certainly not by
> >>>>>> a factor of two.
> >>>>>
> >>>>> The master can double as readily as the slaves can.
> >>>>
> >>>> indeed, but to what purpose? provided ratios remain the same, there are
> >>>> significant reasons not to do so - do you know why?
> >>>
> >>> Increasing the piston sizes allows more force to be transmitted at a
> >>> given hydraulic pressure, much like increasing voltage allows more power
> >>> to be transmitted at a given current.
> >>
> >> it only transmits more pressure if the ratio between master/slave
> >> changes. if that doesn't change, and as we've discussed repeatedly
> >> above there are substantial limitations to that, then it can't.
> >
> > Sorry, you are very much mistaken.
> >
> > Example with some arbitrary values:
> >
> > 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
> > applied to a 1 sq in slave cylinder = 100# force on the caliper.
> >
> > 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
> > applied to a 2 sq in slave cylinder = 200# force on the calipers.
> >
> > Twice the force transmitted, at the same hydraulic pressure. Obviously
> > master cylinder bores are much smaller than slave cylinder bores, but
> > that doesn't change the fact that increasing both bores allows more
> > force to be transmitted without increasing hydraulic pressure.
> >
> > Simple 1:2 ratio:
> >
> > 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
> > applied to a 2 sq in slave cylinder = 200# force on the caliper.
> >
> > 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
> > applied to a 4 sq in slave cylinder = 400# force on the calipers.
>
> you're not reading what i said. if you want to double the output force
> on an hydraulic system, you can increase the piston size, but you will
> double the input travel. this is immutable and there is no way around it.

You're not reading what I said. If you want to transmit more force
without increasing the hydraulic pressure, you increase the bore on both
sides of the equation which increases fluid flow without changing travel
and allows you to transmit more force at the same pressure. This is
little different than stepping up voltages in order to transmit more
power at lower currents.

>
> >
> >>
> >>> Keeping pressures within the
> >>> limits of standard components is beneficial.
> >>
> >> yup. that's one.
> >>
> >>> The larger volume of fluid
> >>> to be displaced won't have any notable effect since the fluid is
> >>> essentially non compressible, this isn't like air brakes and the reasons
> >>> for relay valves.
> >>
> >> right. so what else? there's one particular practical issue that's
> >> very important. apart from cost, weight, etc.
> >>
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>> The displacement and required fluid of course would
> >>>>>>>>> also double.
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> and as for "weight improving traction" [this is going to be a classic] -
> >>>>>>>>>> why exactly does increasing weight increase stopping distance ed? go
> >>>>>>>>>> on, give it a shot.
> >>>>>>>>>
> >>>>>>>>> The short answer is that increasing weight does not automatically
> >>>>>>>>> increase stopping distance. There is something of a dip in the stopping
> >>>>>>>>> distance where the stopping distance decreases as weight is added up to
> >>>>>>>>> a point before the stopping distance begins to increase again as further
> >>>>>>>>> weight is added.
> >>>>>>>>>
> >>>>>>>>> What happens is that an axle that is too lightly loaded (such as the
> >>>>>>>>> rear in an unloaded pickup) is not able to apply the full available
> >>>>>>>>> braking force without wheel lockup so that the effective braking
> >>>>>>>>> capacity is limited by the lack of traction. As weight is added,
> >>>>>>>>> traction improves allowing more brake force to be applied without
> >>>>>>>>> lockup, resulting in shorter stopping distances. Once there is enough
> >>>>>>>>> weigh to provide sufficient traction to match the maximum braking force
> >>>>>>>>> available, that will be the shorted stopping distance in the graph.
> >>>>>>>>> Additional weight past this point will again increase the stopping
> >>>>>>>>> distance.
> >>>>>>>>
> >>>>>>>> that's a function of two things: tire contact area and control. if
> >>>>>>>> tires are over-inflated relative to load, contact area is reduced.
> >>>>>>>> that's introducing an additional variable rather than talking braking
> >>>>>>>> physics. same for control - if it's too cheap or ineffective to not be
> >>>>>>>> proportioning relative to weight, again, the solution is not to load
> >>>>>>>> more weight, it's to exercise better control.
> >>>>>>>
> >>>>>>> It has nothing to do with control,
> >>>>>>
> >>>>>> then, with respect, you don't understand what "control" means.
> >>>>>
> >>>>> Of course I do.
> >>>>
> >>>> let's come back to that...
> >>>>
> >>>>>
> >>>>>>
> >>>>>>> it has everything to do with a light
> >>>>>>> weight which provides insufficient traction weight on the axle to give
> >>>>>>> the traction necessary to utilize the brakes full capacity.
> >>>>>>
> >>>>>> that's adding weight to increase tire contact area, not braking capacity.
> >>>>>
> >>>>> Added weight increases traction (both tire contact area and tire contact
> >>>>> pressure) which allows the use of more of the brakes capacity.
> >>>>> Insufficient traction prevents the use of the brakes full capacity.
> >>>>
> >>>> the brakes' "full capacity" is locked solid. what you really mean is to
> >>>> the full capacity of the tire traction. tire compounds being constant,
> >>>> traction is maximized by maximizing contact area. in that regard, you
> >>>> should lower your tire pressure before you increase weight.
> >>>
> >>> Full braking capacity is not "locked solid", that is a static state.
> >>
> >> sorry, it is. if the brake can't lock the wheel, it's ineffective.
> >
> > A locked brake is dissipating no more energy than an un-applied brake.
>
> right, but i didn't say that.

You said that full braking capacity was a locked state, which it is not.
Braking is dissipating energy, a locked state is "holding", not
"braking". A brake will hold when it has dissipated all of the energy,
or caused another component to dissipate that energy such as locking the
wheel and letting the tire skid.

>
> >
> >>
> >>> Full braking capacity is the maximum energy dissipation rate the brakes
> >>> are capable of while the rotors are still turning.
> >>
> >> and dissipation is caused by? and what energy are we dissipating?
> >
> > Brake friction converting the momentum of the vehicle (and thus rotation
> > of the brake rotor) into heat.
>
> not the momentum, the kinetic energy. momentum is m.v. ke is 1/2.m.v^2

Sorry, not a physics major.

>
> >
> >>
> >>>
> >>> You need to increase tire pressure as you increase the loading on them
> >>> (per tire manufacturers).
> >>
> >> and [theoretically] decrease it as you reduce the load again! there are
> >> practical limitations of course, but anyone who goes off-road and
> >> scrambles loose surfaces knows this first hand.
> >
> > I don't find many loose surfaces around here, I'm quite good at getting
> > stuck in teflon-like clay mud around here. Airing down doesn't help much
> > on that.
> >
> >>
> >>>
> >>>>
> >>>>>
> >>>>>
> >>>>>>
> >>>>>>> Proportioning and ABS both work to limit the applied brake force to a
> >>>>>>> level that the traction can support, in effect reducing the braking
> >>>>>>> capacity.
> >>>>>>
> >>>>>> they don't "decrease" braking capacity, they decrease skidding.
> >>>>>> skidding isn't braking capacity.
> >>>>>
> >>>>> They decrease the braking force applied and thus the percentage of the
> >>>>> braking capacity available in order to prevent skidding when there is
> >>>>> insufficient traction.
> >>>>
> >>>> abs doesn't decrease pressure, it pulses it so that high pressure is
> >>>> applied intermittently [in electronics, it's called "pwm", pulse width
> >>>> modulation].
> >>>
> >>> ABS PWM relies on the time constant of the pistons and calipers to
> >>> average the PWM to a lower effective pressure.
> >>
> >> no, the pressure starts high, just like a non-abs brake, then gets cut
> >> by the abs system momentarily before reapplication. it doesn't
> >> "average" a low pressure, it averages a lower braking force.
> >
> > Semantics, now you're relying on the elasticity of the tire tread to
> > average the near lockup of the rotor.
>
> i'm not, the abs system is. and it's not relying on the tire
> elasticity, it's relying on the physical inertia of the moving parts.

It's a complex system, multiple components are performing that
averaging.

>
> >
> >>
> >>> The same in electronics
> >>> where PWM signals are averaged by a low pass filter such as a capacitor,
> >>> or by rotor inertia in a PWM motor drive.
> >>
> >> you misunderstand pwm. put a capacitor across the poles of a pwm-driven
> >> dc motor and tell me what happens. seriously.
> >
> > I don't misunderstand PWM and I said nothing about putting a capacitor
> > across a motor, though that is common for RFI suppression.
>
> oh, please. and don't waffle, do what i said so you can see what happens.

I clearly said PWM signals are averaged by a low pass filter such as a
capacitor, or by rotor inertia in a PWM motor drive. Perhaps you are
misreading and thinking the capacitor applied to a motor drive, it did
not, that was other PWM applications. Capacitors are still common across
motors for RFI suppression however.

>
> >
> >>
> >>>
> >>>>
> >>>> proportioning valves decrease pressure, but as above, pressure over the
> >>>> locking value is ineffective. proportioning, done right, /increases/
> >>>> braking capacity.
> >>>
> >>> Again, these devices are limiting the brake force to match the available
> >>> traction, so yes, they are reducing the braking capacity from it's
> >>> maximum value.
> >>
> >> preventing lock-up is increasing braking capacity, not reducing it. by
> >> definition.
> >
> > No, it is limiting the braking capacity to match the available traction,
>
> which is /increasing/ the braking capacity if it's preventing lock-up -
> skidding tires don't offer as much traction.

No, it's *reducing* braking capacity to match available traction
capacity. If the traction was up to par the brakes would be operating at
full capacity. There is no way around it, ABS or proportioning causes
the brakes to be applied at less than full capacity.

>
> > which is often less than the maximum capacity of the braking system.
>
> no, the "less than" is the applied hydraulic force relative to the
> maximum, not the braking force.

The brake is the pads and the rotor, if the braking force has to be
reduced to prevent a failure outside of the brake, i.e. a traction
failure of the tire, it is still reduced braking force.

>
> >
> >>
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>> When added weight adds traction, more of the available braking
> >>>>>>> force can be utilized.
> >>>>>>
> >>>>>> KE = 1/2 M.V^2
> >>>>>>
> >>>>>> if you add weight, you increase the amount of braking required.
> >>>>>
> >>>>> When that increased weigh has also provided the traction necessary to
> >>>>> utilize more of your existing braking capacity it does not increase your
> >>>>> stopping distance. When you already have sufficient traction to utilize
> >>>>> your full braking capacity, additional weight increases stopping
> >>>>> distance.
> >>>>
> >>>> which is a matter of control. when i say you don't understand control,
> >>>> that's not a pejorative, it's just that you're not fully aware of all
> >>>> the factors. if you're talking a big rig trailer that is skidding and
> >>>> not stopping when unladen, that's because of poor control and excess
> >>>> tire pressure relative to the load. if the brake
> >>>> anti-lock/proportioning system was better, it wouldn't skid. if tire
> >>>> pressures were actively managed, as some modern vehicles now are, again,
> >>>> it wouldn't skid. both are control issues, not "insufficient weight"
> >>>> issues.
> >>>
> >>> Again, it's not at all about control. ABS and proportioning maintain
> >>> control by reducing the braking forces to less than the maximum values
> >>> which the brakes can sustain, the values that would cause a skid due to
> >>> insufficient traction.
> >>
> >> it's not about control but abs/proportioning maintain control??? you're
> >> confused.
> >
> > Nope, you're confused. If you do not have the traction to allow the use
> > of the brakes at their maximum capacity, you are operating at *reduced*
> > braking, even if you want to twist semantics to claim that ABS or
> > proportioning are maximizing the (available before skidding) braking.
>
> abs' sole function is CONTROL.
> http://en.wikipedia.org/wiki/Control_theory
>
> it is CONTROL that allows you to maximize braking without skidding.

It is the control that automatically reduces braking force to prevent
skidding. This is not any different than traction control which reduces
engine power to prevent skidding from excess drive power,
"anti-braking".

>
> >
> >>
> >> any brake that works properly should be able to lock its wheel.
> >
> > The subject of quite a bit of debate, since a locked wheel with the
> > vehicle moving is not a desired condition.
>
> any "debate" is whether a driver can control maximum braking better than
> a machine, not whether a braking system should be powerful.

No, I've read plenty of debate over whether the ability to lock a wheel
at speed is a relevant qualification for a brake system, since such a
condition is not desirable.

>
> >
> >> whatever conditions that wheel is experiencing. ensuring it /doesn't/
> >> lock is absolutely /all/ about control.
> >
> > Yep, ABS is certainly about control, but that control is all about
> > limiting the brakes to less than their maximum braking capacity to
> > compensate for inadequate traction.
>
> no, it's to maximize available braking and CONTROL. this is why abs was
> invented - drivers are typically BAD at CONTROL.

No it's about limiting braking to match the available traction thus
aiding a poor driver in maintaining control.

>
> >
> >>
> >>> A vehicle will not be able to achieve the maximum
> >>> braking that the brakes are capable of unless it has sufficient load to
> >>> achieve enough traction.
> >>
> >> nope. see above.
> >
> > Sorry, inadequate traction = less than the brakes maximum braking
> > capacity.
>
> no, CONTROL

Sorry, in conditions of inadequate traction such as an unloaded pickup
truck, whether controlled by a competent driver or automatic by ABS, the
braking force is reduced to match the available traction.

>
> >
> >>
> >>> This is why for many vehicles the stopping
> >>> distance decreases as you add cargo weight up until you reach maximum
> >>> traction at which point further weight will increase stopping distance.
> >>
> >> nope, it's because the control systems don't work well at low loads.
> >
> > Nope, it's because an inadequately loaded axle has little traction and
> > the brakes on that axle can provide little braking capacity as a result.
>
> no. and you contradict yourself below.

No, and there is no contradiction below.

>
> >
> > The control systems also suck eggs when such a lightly loaded vehicle
> > with stiff suspension hits a slight bump while braking moderately hard.
> > This is / was a significant safety hazard with some trucks, including
> > ones I owned, where you had to watch the road surface and stop braking
> > at each bump, lets the POS ABS freak out and cause you to loose all
> > braking and greatly increase your stopping distance. Many NHTSA incident
> > reports are directly attributed to this ABS fault.
>
> so we're back right where we started - if you're experiencing poor
> braking, it's because of poor control, not poor physics.

Nope, in the specific case I cited above, it is a defective ABS system
that is presenting a hazard by not being properly matched to the vehicle
and/or having inadequate sensors such that it could differentiate the
momentary wheel lock due to the tire coming off the ground in a bump
from an actual traction loss / skid event. When that defective ABS
system saw that fraction of a second lock when the wheel was off the
ground, it went into full panic mode causing a near total loss of
braking for an extended period of time which in many cases caused
accidents that would not have occurred had the system worked properly or
not been operating.

On 06/06/2011 08:50 PM, Pete C. wrote:
>
> jim beam wrote:
>>
>> On 06/06/2011 05:49 PM, Pete C. wrote:
>>>
>>> jim beam wrote:
>>>>
>>>> On 06/06/2011 03:04 PM, Pete C. wrote:
>>>>>
>>>>> jim beam wrote:
>>>>>>
>>>>>> On 06/06/2011 01:05 PM, Pete C. wrote:
>>>>>>>
>>>>>>> jim beam wrote:
>>>>>>>>
>>>>>> <snip for brevity>
>>>>>>
>>>>>>>> there is no point in increasing the total piston areas. see below.
>>>>>>>
>>>>>>> Of course there is.
>>>>>>
>>>>>> ok, so we double the slave area, then we double the master area because
>>>>>> we need to keep the pedal travel the same, then what have we achieved?
>>>>>
>>>>> Providing better distribution of the clamping force on the rotors.
>>>>
>>>> increasing piston size doesn't do that easily. piston count, as per
>>>> your response to hls, is a better solution.
>>>
>>> Which is what I originally indicated about dual pistons.
>>>
>>>>
>>>>>
>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> If for example the F150 calipers had a
>>>>>>>>>>> single 40mm dia piston and the F250+ calipers had dual 40mm dia pistons,
>>>>>>>>>>> the applied clamping force on the rotor would double given the same
>>>>>>>>>>> hydraulic pressure.
>>>>>>>>>>
>>>>>>>>>> if the master piston diameter stays the same. but because of the
>>>>>>>>>> leverage ratio, and the limited amount of brake pedal travel available,
>>>>>>>>>> ratios typically remain close. [given a long enough pedal travel, you
>>>>>>>>>> could dispense with brake boosters.]
>>>>>>>>>
>>>>>>>>> Larger brake boosters (or hydroboost) allow for larger master cylinder
>>>>>>>>> diameters increasing volume without increasing pedal travel.
>>>>>>>>
>>>>>>>> - to keep the relative ratios fairly constant.
>>>>>>>>
>>>>>>>> by definition, if the slave area doubles, the master must also double or
>>>>>>>> the pedal travel will double. the latter is clearly impracticable -
>>>>>>>> there is very little latitude to vary these ratios, and certainly not by
>>>>>>>> a factor of two.
>>>>>>>
>>>>>>> The master can double as readily as the slaves can.
>>>>>>
>>>>>> indeed, but to what purpose? provided ratios remain the same, there are
>>>>>> significant reasons not to do so - do you know why?
>>>>>
>>>>> Increasing the piston sizes allows more force to be transmitted at a
>>>>> given hydraulic pressure, much like increasing voltage allows more power
>>>>> to be transmitted at a given current.
>>>>
>>>> it only transmits more pressure if the ratio between master/slave
>>>> changes. if that doesn't change, and as we've discussed repeatedly
>>>> above there are substantial limitations to that, then it can't.
>>>
>>> Sorry, you are very much mistaken.
>>>
>>> Example with some arbitrary values:
>>>
>>> 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
>>> applied to a 1 sq in slave cylinder = 100# force on the caliper.
>>>
>>> 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
>>> applied to a 2 sq in slave cylinder = 200# force on the calipers.
>>>
>>> Twice the force transmitted, at the same hydraulic pressure. Obviously
>>> master cylinder bores are much smaller than slave cylinder bores, but
>>> that doesn't change the fact that increasing both bores allows more
>>> force to be transmitted without increasing hydraulic pressure.
>>>
>>> Simple 1:2 ratio:
>>>
>>> 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
>>> applied to a 2 sq in slave cylinder = 200# force on the caliper.
>>>
>>> 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
>>> applied to a 4 sq in slave cylinder = 400# force on the calipers.
>>
>> you're not reading what i said. if you want to double the output force
>> on an hydraulic system, you can increase the piston size, but you will
>> double the input travel. this is immutable and there is no way around it.
>
> You're not reading what I said. If you want to transmit more force
> without increasing the hydraulic pressure, you increase the bore on both
> sides of the equation which increases fluid flow without changing travel
> and allows you to transmit more force at the same pressure.

absolutely incorrect. if the bore area ratios on both sides of the
equation are the same, the force ratios are the same. by definition.

<http://www.grc.nasa.gov/WWW/K-12/WindTunnel/Activities/Pascals_principle.html>

you can only "transmit" more output force with the same input force if
you monkey with the ratio, and that's impracticable.


> This is
> little different than stepping up voltages in order to transmit more
> power at lower currents.

that's your analogy, not mine. and they're not equivalent.


>
>>
>>>
>>>>
>>>>> Keeping pressures within the
>>>>> limits of standard components is beneficial.
>>>>
>>>> yup. that's one.
>>>>
>>>>> The larger volume of fluid
>>>>> to be displaced won't have any notable effect since the fluid is
>>>>> essentially non compressible, this isn't like air brakes and the reasons
>>>>> for relay valves.
>>>>
>>>> right. so what else? there's one particular practical issue that's
>>>> very important. apart from cost, weight, etc.

you didn't go here, so i'll tell you - it's brake line internal
diameter. they need to fine enough for the surface tension of air
bubbles to occupy the whole tube, not simply float through it. without
that, brake systems are next to impossible to bleed.


>>>>
>>>>>
>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>>> The displacement and required fluid of course would
>>>>>>>>>>> also double.
>>>>>>>>>>>
>>>>>>>>>>>>
>>>>>>>>>>>> and as for "weight improving traction" [this is going to be a classic] -
>>>>>>>>>>>> why exactly does increasing weight increase stopping distance ed? go
>>>>>>>>>>>> on, give it a shot.
>>>>>>>>>>>
>>>>>>>>>>> The short answer is that increasing weight does not automatically
>>>>>>>>>>> increase stopping distance. There is something of a dip in the stopping
>>>>>>>>>>> distance where the stopping distance decreases as weight is added up to
>>>>>>>>>>> a point before the stopping distance begins to increase again as further
>>>>>>>>>>> weight is added.
>>>>>>>>>>>
>>>>>>>>>>> What happens is that an axle that is too lightly loaded (such as the
>>>>>>>>>>> rear in an unloaded pickup) is not able to apply the full available
>>>>>>>>>>> braking force without wheel lockup so that the effective braking
>>>>>>>>>>> capacity is limited by the lack of traction. As weight is added,
>>>>>>>>>>> traction improves allowing more brake force to be applied without
>>>>>>>>>>> lockup, resulting in shorter stopping distances. Once there is enough
>>>>>>>>>>> weigh to provide sufficient traction to match the maximum braking force
>>>>>>>>>>> available, that will be the shorted stopping distance in the graph.
>>>>>>>>>>> Additional weight past this point will again increase the stopping
>>>>>>>>>>> distance.
>>>>>>>>>>
>>>>>>>>>> that's a function of two things: tire contact area and control. if
>>>>>>>>>> tires are over-inflated relative to load, contact area is reduced.
>>>>>>>>>> that's introducing an additional variable rather than talking braking
>>>>>>>>>> physics. same for control - if it's too cheap or ineffective to not be
>>>>>>>>>> proportioning relative to weight, again, the solution is not to load
>>>>>>>>>> more weight, it's to exercise better control.
>>>>>>>>>
>>>>>>>>> It has nothing to do with control,
>>>>>>>>
>>>>>>>> then, with respect, you don't understand what "control" means.
>>>>>>>
>>>>>>> Of course I do.
>>>>>>
>>>>>> let's come back to that...
>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>>> it has everything to do with a light
>>>>>>>>> weight which provides insufficient traction weight on the axle to give
>>>>>>>>> the traction necessary to utilize the brakes full capacity.
>>>>>>>>
>>>>>>>> that's adding weight to increase tire contact area, not braking capacity.
>>>>>>>
>>>>>>> Added weight increases traction (both tire contact area and tire contact
>>>>>>> pressure) which allows the use of more of the brakes capacity.
>>>>>>> Insufficient traction prevents the use of the brakes full capacity.
>>>>>>
>>>>>> the brakes' "full capacity" is locked solid. what you really mean is to
>>>>>> the full capacity of the tire traction. tire compounds being constant,
>>>>>> traction is maximized by maximizing contact area. in that regard, you
>>>>>> should lower your tire pressure before you increase weight.
>>>>>
>>>>> Full braking capacity is not "locked solid", that is a static state.
>>>>
>>>> sorry, it is. if the brake can't lock the wheel, it's ineffective.
>>>
>>> A locked brake is dissipating no more energy than an un-applied brake.
>>
>> right, but i didn't say that.
>
> You said that full braking capacity was a locked state, which it is not.

i didn't say "any locked state". i was talking about the locking of a
brake after passing through peak threshold. and that varies with the
energy of the vehicle.


> Braking is dissipating energy, a locked state is "holding", not
> "braking".

no, in order to lock, it has to pass through the peak threshold.


> A brake will hold when it has dissipated all of the energy,

its /local/ energy.


> or caused another component to dissipate that energy such as locking the
> wheel and letting the tire skid.

red herring.


>
>>
>>>
>>>>
>>>>> Full braking capacity is the maximum energy dissipation rate the brakes
>>>>> are capable of while the rotors are still turning.
>>>>
>>>> and dissipation is caused by? and what energy are we dissipating?
>>>
>>> Brake friction converting the momentum of the vehicle (and thus rotation
>>> of the brake rotor) into heat.
>>
>> not the momentum, the kinetic energy. momentum is m.v. ke is 1/2.m.v^2
>
> Sorry, not a physics major.

mmm. so why are we arguing? i'm not trying to tell you you're wrong,
i'm trying to tell you how to get it right.


>
>>
>>>
>>>>
>>>>>
>>>>> You need to increase tire pressure as you increase the loading on them
>>>>> (per tire manufacturers).
>>>>
>>>> and [theoretically] decrease it as you reduce the load again! there are
>>>> practical limitations of course, but anyone who goes off-road and
>>>> scrambles loose surfaces knows this first hand.
>>>
>>> I don't find many loose surfaces around here, I'm quite good at getting
>>> stuck in teflon-like clay mud around here. Airing down doesn't help much
>>> on that.
>>>
>>>>
>>>>>
>>>>>>
>>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>>> Proportioning and ABS both work to limit the applied brake force to a
>>>>>>>>> level that the traction can support, in effect reducing the braking
>>>>>>>>> capacity.
>>>>>>>>
>>>>>>>> they don't "decrease" braking capacity, they decrease skidding.
>>>>>>>> skidding isn't braking capacity.
>>>>>>>
>>>>>>> They decrease the braking force applied and thus the percentage of the
>>>>>>> braking capacity available in order to prevent skidding when there is
>>>>>>> insufficient traction.
>>>>>>
>>>>>> abs doesn't decrease pressure, it pulses it so that high pressure is
>>>>>> applied intermittently [in electronics, it's called "pwm", pulse width
>>>>>> modulation].
>>>>>
>>>>> ABS PWM relies on the time constant of the pistons and calipers to
>>>>> average the PWM to a lower effective pressure.
>>>>
>>>> no, the pressure starts high, just like a non-abs brake, then gets cut
>>>> by the abs system momentarily before reapplication. it doesn't
>>>> "average" a low pressure, it averages a lower braking force.
>>>
>>> Semantics, now you're relying on the elasticity of the tire tread to
>>> average the near lockup of the rotor.
>>
>> i'm not, the abs system is. and it's not relying on the tire
>> elasticity, it's relying on the physical inertia of the moving parts.
>
> It's a complex system, multiple components are performing that
> averaging.

any "complex system" comprises simple components. figure out the simple
stuff and you can get a handle on the complex.


>
>>
>>>
>>>>
>>>>> The same in electronics
>>>>> where PWM signals are averaged by a low pass filter such as a capacitor,
>>>>> or by rotor inertia in a PWM motor drive.
>>>>
>>>> you misunderstand pwm. put a capacitor across the poles of a pwm-driven
>>>> dc motor and tell me what happens. seriously.
>>>
>>> I don't misunderstand PWM and I said nothing about putting a capacitor
>>> across a motor, though that is common for RFI suppression.
>>
>> oh, please. and don't waffle, do what i said so you can see what happens.
>
> I clearly said PWM signals are averaged by a low pass filter such as a
> capacitor, or by rotor inertia in a PWM motor drive. Perhaps you are
> misreading and thinking the capacitor applied to a motor drive, it did
> not, that was other PWM applications. Capacitors are still common across
> motors for RFI suppression however.

but pwm signals are /not/ averaged through a low pass filter - unless
you have some kind of funky audio application, and even then, the people
that say it is probably don't understand what's really going on. the
reason i said to use the capacitor on a pwm controlled motor is so
you'll see why capacitor "averaging" is untrue.


>
>>
>>>
>>>>
>>>>>
>>>>>>
>>>>>> proportioning valves decrease pressure, but as above, pressure over the
>>>>>> locking value is ineffective. proportioning, done right, /increases/
>>>>>> braking capacity.
>>>>>
>>>>> Again, these devices are limiting the brake force to match the available
>>>>> traction, so yes, they are reducing the braking capacity from it's
>>>>> maximum value.
>>>>
>>>> preventing lock-up is increasing braking capacity, not reducing it. by
>>>> definition.
>>>
>>> No, it is limiting the braking capacity to match the available traction,
>>
>> which is /increasing/ the braking capacity if it's preventing lock-up -
>> skidding tires don't offer as much traction.
>
> No, it's *reducing* braking capacity to match available traction
> capacity.

this is becoming pointless. i'm saying the flag is red white and blue,
you're saying it's purple green and black.


> If the traction was up to par the brakes would be operating at
> full capacity.

there is no such thing as "par" on braking - it constantly changes per
conditions. control, whether from the driver or from an abs system,
matches the two.


> There is no way around it, ABS or proportioning causes
> the brakes to be applied at less than full capacity.

but at maximum braking [i.e stopping] capacity!!!


>
>>
>>> which is often less than the maximum capacity of the braking system.
>>
>> no, the "less than" is the applied hydraulic force relative to the
>> maximum, not the braking force.
>
> The brake is the pads and the rotor, if the braking force has to be
> reduced to prevent a failure outside of the brake, i.e. a traction
> failure of the tire, it is still reduced braking force.

but it's increasing braking! again.


>
>>
>>>
>>>>
>>>>>
>>>>>>
>>>>>>>
>>>>>>>>
>>>>>>>>> When added weight adds traction, more of the available braking
>>>>>>>>> force can be utilized.
>>>>>>>>
>>>>>>>> KE = 1/2 M.V^2
>>>>>>>>
>>>>>>>> if you add weight, you increase the amount of braking required.
>>>>>>>
>>>>>>> When that increased weigh has also provided the traction necessary to
>>>>>>> utilize more of your existing braking capacity it does not increase your
>>>>>>> stopping distance. When you already have sufficient traction to utilize
>>>>>>> your full braking capacity, additional weight increases stopping
>>>>>>> distance.
>>>>>>
>>>>>> which is a matter of control. when i say you don't understand control,
>>>>>> that's not a pejorative, it's just that you're not fully aware of all
>>>>>> the factors. if you're talking a big rig trailer that is skidding and
>>>>>> not stopping when unladen, that's because of poor control and excess
>>>>>> tire pressure relative to the load. if the brake
>>>>>> anti-lock/proportioning system was better, it wouldn't skid. if tire
>>>>>> pressures were actively managed, as some modern vehicles now are, again,
>>>>>> it wouldn't skid. both are control issues, not "insufficient weight"
>>>>>> issues.
>>>>>
>>>>> Again, it's not at all about control. ABS and proportioning maintain
>>>>> control by reducing the braking forces to less than the maximum values
>>>>> which the brakes can sustain, the values that would cause a skid due to
>>>>> insufficient traction.
>>>>
>>>> it's not about control but abs/proportioning maintain control??? you're
>>>> confused.
>>>
>>> Nope, you're confused. If you do not have the traction to allow the use
>>> of the brakes at their maximum capacity, you are operating at *reduced*
>>> braking, even if you want to twist semantics to claim that ABS or
>>> proportioning are maximizing the (available before skidding) braking.
>>
>> abs' sole function is CONTROL.
>> http://en.wikipedia.org/wiki/Control_theory
>>
>> it is CONTROL that allows you to maximize braking without skidding.
>
> It is the control that automatically reduces braking force to prevent
> skidding.

increasing braking.


> This is not any different than traction control which reduces
> engine power to prevent skidding from excess drive power,
> "anti-braking".

increasing traction where it would otherwise be lost.


>
>>
>>>
>>>>
>>>> any brake that works properly should be able to lock its wheel.
>>>
>>> The subject of quite a bit of debate, since a locked wheel with the
>>> vehicle moving is not a desired condition.
>>
>> any "debate" is whether a driver can control maximum braking better than
>> a machine, not whether a braking system should be powerful.
>
> No, I've read plenty of debate over whether the ability to lock a wheel
> at speed is a relevant qualification for a brake system, since such a
> condition is not desirable.

exercising it and having capacity for it are two totally different
things. you need excess braking capacity for safety because, as any
truck driver should know, if you experience overheating and fade, you'll
need it.


>
>>
>>>
>>>> whatever conditions that wheel is experiencing. ensuring it /doesn't/
>>>> lock is absolutely /all/ about control.
>>>
>>> Yep, ABS is certainly about control, but that control is all about
>>> limiting the brakes to less than their maximum braking capacity to
>>> compensate for inadequate traction.
>>
>> no, it's to maximize available braking and CONTROL. this is why abs was
>> invented - drivers are typically BAD at CONTROL.
>
> No it's about limiting braking to match the available traction thus
> aiding a poor driver in maintaining control.

by increasing braking...


>
>>
>>>
>>>>
>>>>> A vehicle will not be able to achieve the maximum
>>>>> braking that the brakes are capable of unless it has sufficient load to
>>>>> achieve enough traction.
>>>>
>>>> nope. see above.
>>>
>>> Sorry, inadequate traction = less than the brakes maximum braking
>>> capacity.
>>
>> no, CONTROL
>
> Sorry, in conditions of inadequate traction such as an unloaded pickup
> truck, whether controlled by a competent driver or automatic by ABS, the
> braking force is reduced to match the available traction.

no, it's limited by poor control. the driver doesn't control the
proportioning, the manufacturer does. and i'll bet he's also not
controlling the tire pressure. his only bet is if there's an abs system
on there, and most production versions are cheap enough that at low
loads/low tire contact, they don't modulate [CONTROL] finely enough.


>
>>
>>>
>>>>
>>>>> This is why for many vehicles the stopping
>>>>> distance decreases as you add cargo weight up until you reach maximum
>>>>> traction at which point further weight will increase stopping distance.
>>>>
>>>> nope, it's because the control systems don't work well at low loads.
>>>
>>> Nope, it's because an inadequately loaded axle has little traction and
>>> the brakes on that axle can provide little braking capacity as a result.
>>
>> no. and you contradict yourself below.
>
> No, and there is no contradiction below.

there is. but you need to understand to understand.


>
>>
>>>
>>> The control systems also suck eggs when such a lightly loaded vehicle
>>> with stiff suspension hits a slight bump while braking moderately hard.
>>> This is / was a significant safety hazard with some trucks, including
>>> ones I owned, where you had to watch the road surface and stop braking
>>> at each bump, lets the POS ABS freak out and cause you to loose all
>>> braking and greatly increase your stopping distance. Many NHTSA incident
>>> reports are directly attributed to this ABS fault.
>>
>> so we're back right where we started - if you're experiencing poor
>> braking, it's because of poor control, not poor physics.
>
> Nope, in the specific case I cited above, it is a defective ABS system
> that is presenting a hazard by not being properly matched to the vehicle
> and/or having inadequate sensors such that it could differentiate the
> momentary wheel lock due to the tire coming off the ground in a bump
> from an actual traction loss / skid event. When that defective ABS
> system saw that fraction of a second lock when the wheel was off the
> ground, it went into full panic mode causing a near total loss of
> braking for an extended period of time which in many cases caused
> accidents that would not have occurred had the system worked properly or
> not been operating.

a "defective abs" is neither the basis on which to argue or the
foundation of a subject understanding. it's simply a defective abs.


--
nomina rutrum rutrum

"Pete C." > wrote in message news:4ded507c$0$32686>
> The main advantage of multi piston calipers is to better distribute the
> clamping force across the pads. A caliper piston can't really be lager
> diameter than the pad is wide, and most pads are not round which leaves
> the overhanging areas not backed by the piston to flex and be less
> effective. Multiple pistons can better cover the area of the pad and
> allow the whole pad area to work more evenly.

That makes sense. I suppose the common use of single piston calipers
is just a concession to lowering costs.

"jim beam" > wrote in message
t...

>>>>> getting back to the point, any vehicle with a cargo bed should be able
>>>>> to stop a load. on a hill. more than once. whether that load is on
>>>>> the bed or following behind is irrelevant.
>>>>
>>>> Not irrelevant. A load in the bed increases the tire traction available
>>>> for braking. The same load on a trailer without trialer brakes doesn't
>>>> load the tow vehicle tires nearly as much (maybe 20% as much if you
>>>> have
>>>> the tongue weight right). Therfore, you have less tire traction
>>>> available
>>>> to stop the combination. If you don't have anti-lock brakes, then you
>>>> can
>>>> easily lock the rear brakes on the tow vehicle when trying to stop the
>>>> vehicle plus trailer (with the dire consequences that follow). If you
>>>> do
>>>> have antilock brakes, then you can't lock the rear wheels, but you also
>>>> won't have as much braking force available (comparded to having the
>>>> load
>>>> in the bed). And then there is the whole problem with jack-knifing a
>>>> trailer with no trialer brakes. Certainly stopping a vehicle with a
>>>> trailer without trailer brakes is much different than stopping a
>>>> vehicle
>>>> with the load in the bed.
>>>>
>>>> I looked into brakes of the current F150 some more. It seems Ford
>>>> claims
>>>> they have a dual diaphram brake booster and electronic brake force
>>>> distribution. Neither claim is made for the SuperDuties. Not sure of
>>>> the
>>>> implications. The biggest difference at the wheel end is that the
>>>> SuperDuties have dual piston rear calipers, the F150's have single
>>>> piston
>>>> rear calipers. Brake disc sizes are similar, total swept area is
>>>> similar.
>>>> I really don't see any reason to think that the F150 brakes are
>>>> inadequate if you tow trailers that don't exceed the rated GCVW.
>>>>
>>>> Ed
>>>
>>> Assuming both rotors have the same swept area, and that
>>> the pads are of the same size and material, would you
>>> expect the addition of a second piston alone to have a
>>> major effect on the stopping power? Pressure applied
>>> to the pads should be the same whether you use one or
>>> two pistons. You dont double the pressure when you
>>> apply pressure to both sides, versus deadheading the
>>> same pressure on one side only.
>>
>> In this context, the dual pistons are on the same side of the caliper.
>> The
>> brakes are still sliding caliper types, just with two pistons side by
>> side
>> on one side of the caliper. I haven't found the actual piston sizes, but
>> I
>> assume the F250+ brakes can apply significantly more pressure on the rear
>> pads than the F150 single piston calipers
>>
>> Ed
>>
>>
>
> you, of all people ed, shouldn't "assume" a damned thing.
>
> piston count does not "increase" pressure. it can improve braking because
> mechanical elasticity becomes less of a concern, and thus you can have a
> more effective brake for less weight [and money].

The total force exterted on a brake pad is a function of piston area exposed
to the brake fluid times the brake fluid pressure. Brake caliper pistons are
usually (always?) round. So the maximum area of the piston is limited by the
diameter of piston that will fit between the wheel hub and the inside of the
actual wheel (along with other factors....). Two pistons side by side can
have more piston area than the largestest possible single piston that weill
fit in the space (even if they are slightly samller in diameter than the
single piston). Plus they better spread the load over the pad. I used the
word "assume" becasue I don't know the relative sizes of the F150 and F250
pistons. It is possible that the F150 has one large piston that has as much
or even more area than the two pistons used by the F250 calipers - but I
doubt it. Putting it simply - Two 1.8" diamter pistons (side by side in a
caliper) can exert more force on the brake pad than a single 2" diamter
piston and they better spread the load on the pad (in this case about 60%
more force).

> and as for "weight improving traction" [this is going to be a classic] -
> why exactly does increasing weight increase stopping distance ed? go on,
> give it a shot.

My original comment was only in the context of carrying a load on a trailer
(without trailer brakes) versus carrying a load in the bed of the truck. You
do understand that right? Most modern vehicle have brakes powerful enough to
lock the tires under most conditions at least on the first stop (except that
most vehicles with ABS prevent this from happening). So for most vehicles,
the making braking force is limited by the tire / road friction available.
This maximum braking force is a function of the frictional coefficent
between the tire and the road, the tires contact patch, the normal force
pressing the tires down on the road and other minor factors. The amount of
braking force needed to stop the vehicle is a function of the mass to be
stopped and the desired deceleration rate.With a load in the back of the
truck, the normal force applied to the tires is directly related to the mass
to be stopped - right? If you have more weight in the bed, you have a higher
normal force on the tires and therefore you have a higher maximum braking
force available. When you have a trailer, without trailer brakes, the
situation is different. The load on the trailer only partially contributes
to the normal force on the tires that provide the braking - depending on how
the tailer load is distributed, maybe only 10% or 20% . The load on the
trailer still contributes 100% to the mass to be stopped. So although you
have the same total mass to be stopped with part of the load on the trailer
instead of in the bed of the truck, you have less maximum braking force
available to stop it. Therefore, it is possible (very likely) you cannot
stop the vehcile plus trailer as qucikly as you could stop the vehicle alone
if the weight was all carried in the bed of the single vehicle. OK?

Ed

"hls" > wrote in message
...
>
> "Pete C." > wrote in message news:4ded507c$0$32686>
>> The main advantage of multi piston calipers is to better distribute the
>> clamping force across the pads. A caliper piston can't really be lager
>> diameter than the pad is wide, and most pads are not round which leaves
>> the overhanging areas not backed by the piston to flex and be less
>> effective. Multiple pistons can better cover the area of the pad and
>> allow the whole pad area to work more evenly.
>
> That makes sense. I suppose the common use of single piston calipers
> is just a concession to lowering costs.

More pistons have more seals to leak, pistons can freeze or bind
individually, you have to properly port fluid to all pistons, etc.

If one piston provides enough force to lock the brakes under all conditions,
then one piston is best. KISS.

Ed

jim beam wrote:
>
> On 06/06/2011 08:50 PM, Pete C. wrote:
> >
> > jim beam wrote:
> >>
> >> On 06/06/2011 05:49 PM, Pete C. wrote:
> >>>
> >>> jim beam wrote:
> >>>>
> >>>> On 06/06/2011 03:04 PM, Pete C. wrote:
> >>>>>
> >>>>> jim beam wrote:
> >>>>>>
> >>>>>> On 06/06/2011 01:05 PM, Pete C. wrote:
> >>>>>>>
> >>>>>>> jim beam wrote:
> >>>>>>>>
> >>>>>> <snip for brevity>
> >>>>>>
> >>>>>>>> there is no point in increasing the total piston areas. see below.
> >>>>>>>
> >>>>>>> Of course there is.
> >>>>>>
> >>>>>> ok, so we double the slave area, then we double the master area because
> >>>>>> we need to keep the pedal travel the same, then what have we achieved?
> >>>>>
> >>>>> Providing better distribution of the clamping force on the rotors.
> >>>>
> >>>> increasing piston size doesn't do that easily. piston count, as per
> >>>> your response to hls, is a better solution.
> >>>
> >>> Which is what I originally indicated about dual pistons.
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>>> If for example the F150 calipers had a
> >>>>>>>>>>> single 40mm dia piston and the F250+ calipers had dual 40mm dia pistons,
> >>>>>>>>>>> the applied clamping force on the rotor would double given the same
> >>>>>>>>>>> hydraulic pressure.
> >>>>>>>>>>
> >>>>>>>>>> if the master piston diameter stays the same. but because of the
> >>>>>>>>>> leverage ratio, and the limited amount of brake pedal travel available,
> >>>>>>>>>> ratios typically remain close. [given a long enough pedal travel, you
> >>>>>>>>>> could dispense with brake boosters.]
> >>>>>>>>>
> >>>>>>>>> Larger brake boosters (or hydroboost) allow for larger master cylinder
> >>>>>>>>> diameters increasing volume without increasing pedal travel.
> >>>>>>>>
> >>>>>>>> - to keep the relative ratios fairly constant.
> >>>>>>>>
> >>>>>>>> by definition, if the slave area doubles, the master must also double or
> >>>>>>>> the pedal travel will double. the latter is clearly impracticable -
> >>>>>>>> there is very little latitude to vary these ratios, and certainly not by
> >>>>>>>> a factor of two.
> >>>>>>>
> >>>>>>> The master can double as readily as the slaves can.
> >>>>>>
> >>>>>> indeed, but to what purpose? provided ratios remain the same, there are
> >>>>>> significant reasons not to do so - do you know why?
> >>>>>
> >>>>> Increasing the piston sizes allows more force to be transmitted at a
> >>>>> given hydraulic pressure, much like increasing voltage allows more power
> >>>>> to be transmitted at a given current.
> >>>>
> >>>> it only transmits more pressure if the ratio between master/slave
> >>>> changes. if that doesn't change, and as we've discussed repeatedly
> >>>> above there are substantial limitations to that, then it can't.
> >>>
> >>> Sorry, you are very much mistaken.
> >>>
> >>> Example with some arbitrary values:
> >>>
> >>> 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
> >>> applied to a 1 sq in slave cylinder = 100# force on the caliper.
> >>>
> >>> 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
> >>> applied to a 2 sq in slave cylinder = 200# force on the calipers.
> >>>
> >>> Twice the force transmitted, at the same hydraulic pressure. Obviously
> >>> master cylinder bores are much smaller than slave cylinder bores, but
> >>> that doesn't change the fact that increasing both bores allows more
> >>> force to be transmitted without increasing hydraulic pressure.
> >>>
> >>> Simple 1:2 ratio:
> >>>
> >>> 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
> >>> applied to a 2 sq in slave cylinder = 200# force on the caliper.
> >>>
> >>> 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
> >>> applied to a 4 sq in slave cylinder = 400# force on the calipers.
> >>
> >> you're not reading what i said. if you want to double the output force
> >> on an hydraulic system, you can increase the piston size, but you will
> >> double the input travel. this is immutable and there is no way around it.
> >
> > You're not reading what I said. If you want to transmit more force
> > without increasing the hydraulic pressure, you increase the bore on both
> > sides of the equation which increases fluid flow without changing travel
> > and allows you to transmit more force at the same pressure.
>
> absolutely incorrect. if the bore area ratios on both sides of the
> equation are the same, the force ratios are the same. by definition.
>
> <http://www.grc.nasa.gov/WWW/K-12/WindTunnel/Activities/Pascals_principle.html>
>
> you can only "transmit" more output force with the same input force if
> you monkey with the ratio, and that's impracticable.

Sorry, you are absolutely incorrect. I said nothing about changing the
ratio, I said transmit increased force without changing the fluid
pressure. If you want to transmit more force without changing the bore
size on both ends you will have to increase the hydraulic pressure.

100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
applied to a 1 sq in slave cylinder = 100# force on the caliper.

200# of force applied to a 1 sq in master cylinder = 200 PSI fluid,
applied to a 1 sq in slave cylinder = 200# force on the calipers.

200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
applied to a 2 sq in slave cylinder = 200# force on the calipers.

As you can clearly see, increasing the bores on both sides allows more
force to be transmitted (not multiplied) without increasing the line
pressure.

>
> > This is
> > little different than stepping up voltages in order to transmit more
> > power at lower currents.
>
> that's your analogy, not mine. and they're not equivalent.

They are exactly equivalent, this is fact.

>
> >
> >>
> >>>
> >>>>
> >>>>> Keeping pressures within the
> >>>>> limits of standard components is beneficial.
> >>>>
> >>>> yup. that's one.
> >>>>
> >>>>> The larger volume of fluid
> >>>>> to be displaced won't have any notable effect since the fluid is
> >>>>> essentially non compressible, this isn't like air brakes and the reasons
> >>>>> for relay valves.
> >>>>
> >>>> right. so what else? there's one particular practical issue that's
> >>>> very important. apart from cost, weight, etc.
>
> you didn't go here, so i'll tell you - it's brake line internal
> diameter. they need to fine enough for the surface tension of air
> bubbles to occupy the whole tube, not simply float through it. without
> that, brake systems are next to impossible to bleed.

And this relates to the topic how? Moving more fluid due to larger bore
sizes doesn't automatically equate to larger line diameter, we aren't
moving 4' stroke cylinders on an excavator here, the total volume is
still very low.

>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>>> The displacement and required fluid of course would
> >>>>>>>>>>> also double.
> >>>>>>>>>>>
> >>>>>>>>>>>>
> >>>>>>>>>>>> and as for "weight improving traction" [this is going to be a classic] -
> >>>>>>>>>>>> why exactly does increasing weight increase stopping distance ed? go
> >>>>>>>>>>>> on, give it a shot.
> >>>>>>>>>>>
> >>>>>>>>>>> The short answer is that increasing weight does not automatically
> >>>>>>>>>>> increase stopping distance. There is something of a dip in the stopping
> >>>>>>>>>>> distance where the stopping distance decreases as weight is added up to
> >>>>>>>>>>> a point before the stopping distance begins to increase again as further
> >>>>>>>>>>> weight is added.
> >>>>>>>>>>>
> >>>>>>>>>>> What happens is that an axle that is too lightly loaded (such as the
> >>>>>>>>>>> rear in an unloaded pickup) is not able to apply the full available
> >>>>>>>>>>> braking force without wheel lockup so that the effective braking
> >>>>>>>>>>> capacity is limited by the lack of traction. As weight is added,
> >>>>>>>>>>> traction improves allowing more brake force to be applied without
> >>>>>>>>>>> lockup, resulting in shorter stopping distances. Once there is enough
> >>>>>>>>>>> weigh to provide sufficient traction to match the maximum braking force
> >>>>>>>>>>> available, that will be the shorted stopping distance in the graph.
> >>>>>>>>>>> Additional weight past this point will again increase the stopping
> >>>>>>>>>>> distance.
> >>>>>>>>>>
> >>>>>>>>>> that's a function of two things: tire contact area and control. if
> >>>>>>>>>> tires are over-inflated relative to load, contact area is reduced.
> >>>>>>>>>> that's introducing an additional variable rather than talking braking
> >>>>>>>>>> physics. same for control - if it's too cheap or ineffective to not be
> >>>>>>>>>> proportioning relative to weight, again, the solution is not to load
> >>>>>>>>>> more weight, it's to exercise better control.
> >>>>>>>>>
> >>>>>>>>> It has nothing to do with control,
> >>>>>>>>
> >>>>>>>> then, with respect, you don't understand what "control" means.
> >>>>>>>
> >>>>>>> Of course I do.
> >>>>>>
> >>>>>> let's come back to that...
> >>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>> it has everything to do with a light
> >>>>>>>>> weight which provides insufficient traction weight on the axle to give
> >>>>>>>>> the traction necessary to utilize the brakes full capacity.
> >>>>>>>>
> >>>>>>>> that's adding weight to increase tire contact area, not braking capacity.
> >>>>>>>
> >>>>>>> Added weight increases traction (both tire contact area and tire contact
> >>>>>>> pressure) which allows the use of more of the brakes capacity.
> >>>>>>> Insufficient traction prevents the use of the brakes full capacity.
> >>>>>>
> >>>>>> the brakes' "full capacity" is locked solid. what you really mean is to
> >>>>>> the full capacity of the tire traction. tire compounds being constant,
> >>>>>> traction is maximized by maximizing contact area. in that regard, you
> >>>>>> should lower your tire pressure before you increase weight.
> >>>>>
> >>>>> Full braking capacity is not "locked solid", that is a static state.
> >>>>
> >>>> sorry, it is. if the brake can't lock the wheel, it's ineffective.
> >>>
> >>> A locked brake is dissipating no more energy than an un-applied brake.
> >>
> >> right, but i didn't say that.
> >
> > You said that full braking capacity was a locked state, which it is not.
>
> i didn't say "any locked state". i was talking about the locking of a
> brake after passing through peak threshold. and that varies with the
> energy of the vehicle.

And as soon as it is locked, it is "holding", not "braking" and energy
dissipation (if any remains) has moved elsewhere, such as grinding tread
off tires.

>
> > Braking is dissipating energy, a locked state is "holding", not
> > "braking".
>
> no, in order to lock, it has to pass through the peak threshold.

Pass through and no longer be "braking".

>
> > A brake will hold when it has dissipated all of the energy,
>
> its /local/ energy.
>
> > or caused another component to dissipate that energy such as locking the
> > wheel and letting the tire skid.
>
> red herring.
>
> >
> >>
> >>>
> >>>>
> >>>>> Full braking capacity is the maximum energy dissipation rate the brakes
> >>>>> are capable of while the rotors are still turning.
> >>>>
> >>>> and dissipation is caused by? and what energy are we dissipating?
> >>>
> >>> Brake friction converting the momentum of the vehicle (and thus rotation
> >>> of the brake rotor) into heat.
> >>
> >> not the momentum, the kinetic energy. momentum is m.v. ke is 1/2.m.v^2
> >
> > Sorry, not a physics major.
>
> mmm. so why are we arguing? i'm not trying to tell you you're wrong,
> i'm trying to tell you how to get it right.

Then why are you making factually false claims such as those around
hydraulic line pressure vs. bore size and force transmitted?

>
> >
> >>
> >>>
> >>>>
> >>>>>
> >>>>> You need to increase tire pressure as you increase the loading on them
> >>>>> (per tire manufacturers).
> >>>>
> >>>> and [theoretically] decrease it as you reduce the load again! there are
> >>>> practical limitations of course, but anyone who goes off-road and
> >>>> scrambles loose surfaces knows this first hand.
> >>>
> >>> I don't find many loose surfaces around here, I'm quite good at getting
> >>> stuck in teflon-like clay mud around here. Airing down doesn't help much
> >>> on that.
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>> Proportioning and ABS both work to limit the applied brake force to a
> >>>>>>>>> level that the traction can support, in effect reducing the braking
> >>>>>>>>> capacity.
> >>>>>>>>
> >>>>>>>> they don't "decrease" braking capacity, they decrease skidding.
> >>>>>>>> skidding isn't braking capacity.
> >>>>>>>
> >>>>>>> They decrease the braking force applied and thus the percentage of the
> >>>>>>> braking capacity available in order to prevent skidding when there is
> >>>>>>> insufficient traction.
> >>>>>>
> >>>>>> abs doesn't decrease pressure, it pulses it so that high pressure is
> >>>>>> applied intermittently [in electronics, it's called "pwm", pulse width
> >>>>>> modulation].
> >>>>>
> >>>>> ABS PWM relies on the time constant of the pistons and calipers to
> >>>>> average the PWM to a lower effective pressure.
> >>>>
> >>>> no, the pressure starts high, just like a non-abs brake, then gets cut
> >>>> by the abs system momentarily before reapplication. it doesn't
> >>>> "average" a low pressure, it averages a lower braking force.
> >>>
> >>> Semantics, now you're relying on the elasticity of the tire tread to
> >>> average the near lockup of the rotor.
> >>
> >> i'm not, the abs system is. and it's not relying on the tire
> >> elasticity, it's relying on the physical inertia of the moving parts.
> >
> > It's a complex system, multiple components are performing that
> > averaging.
>
> any "complex system" comprises simple components. figure out the simple
> stuff and you can get a handle on the complex.
>
> >
> >>
> >>>
> >>>>
> >>>>> The same in electronics
> >>>>> where PWM signals are averaged by a low pass filter such as a capacitor,
> >>>>> or by rotor inertia in a PWM motor drive.
> >>>>
> >>>> you misunderstand pwm. put a capacitor across the poles of a pwm-driven
> >>>> dc motor and tell me what happens. seriously.
> >>>
> >>> I don't misunderstand PWM and I said nothing about putting a capacitor
> >>> across a motor, though that is common for RFI suppression.
> >>
> >> oh, please. and don't waffle, do what i said so you can see what happens.
> >
> > I clearly said PWM signals are averaged by a low pass filter such as a
> > capacitor, or by rotor inertia in a PWM motor drive. Perhaps you are
> > misreading and thinking the capacitor applied to a motor drive, it did
> > not, that was other PWM applications. Capacitors are still common across
> > motors for RFI suppression however.
>
> but pwm signals are /not/ averaged through a low pass filter - unless
> you have some kind of funky audio application, and even then, the people
> that say it is probably don't understand what's really going on. the
> reason i said to use the capacitor on a pwm controlled motor is so
> you'll see why capacitor "averaging" is untrue.

Give me a PWM application where the end result is not averaged by some
form of low pass filter. Even PWM drive for lighting dimming relies on
human eyes as the low pass filter.

>
> >
> >>
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>> proportioning valves decrease pressure, but as above, pressure over the
> >>>>>> locking value is ineffective. proportioning, done right, /increases/
> >>>>>> braking capacity.
> >>>>>
> >>>>> Again, these devices are limiting the brake force to match the available
> >>>>> traction, so yes, they are reducing the braking capacity from it's
> >>>>> maximum value.
> >>>>
> >>>> preventing lock-up is increasing braking capacity, not reducing it. by
> >>>> definition.
> >>>
> >>> No, it is limiting the braking capacity to match the available traction,
> >>
> >> which is /increasing/ the braking capacity if it's preventing lock-up -
> >> skidding tires don't offer as much traction.
> >
> > No, it's *reducing* braking capacity to match available traction
> > capacity.
>
> this is becoming pointless. i'm saying the flag is red white and blue,
> you're saying it's purple green and black.

Yep it's pointless. You don't want to admit that inadequate traction
forces a limitation in brake capacity to prevent skidding.

>
> > If the traction was up to par the brakes would be operating at
> > full capacity.
>
> there is no such thing as "par" on braking - it constantly changes per
> conditions. control, whether from the driver or from an abs system,
> matches the two.

I said "par" on traction. If traction was always up to "par", the
braking would not have to be reduced from maximum.

>
> > There is no way around it, ABS or proportioning causes
> > the brakes to be applied at less than full capacity.
>
> but at maximum braking [i.e stopping] capacity!!!

Braking reduced to maximum traction capacity.

>
> >
> >>
> >>> which is often less than the maximum capacity of the braking system.
> >>
> >> no, the "less than" is the applied hydraulic force relative to the
> >> maximum, not the braking force.
> >
> > The brake is the pads and the rotor, if the braking force has to be
> > reduced to prevent a failure outside of the brake, i.e. a traction
> > failure of the tire, it is still reduced braking force.
>
> but it's increasing braking! again.

No, it's reduced braking to match available traction.

>
> >
> >>
> >>>
> >>>>
> >>>>>
> >>>>>>
> >>>>>>>
> >>>>>>>>
> >>>>>>>>> When added weight adds traction, more of the available braking
> >>>>>>>>> force can be utilized.
> >>>>>>>>
> >>>>>>>> KE = 1/2 M.V^2
> >>>>>>>>
> >>>>>>>> if you add weight, you increase the amount of braking required.
> >>>>>>>
> >>>>>>> When that increased weigh has also provided the traction necessary to
> >>>>>>> utilize more of your existing braking capacity it does not increase your
> >>>>>>> stopping distance. When you already have sufficient traction to utilize
> >>>>>>> your full braking capacity, additional weight increases stopping
> >>>>>>> distance.
> >>>>>>
> >>>>>> which is a matter of control. when i say you don't understand control,
> >>>>>> that's not a pejorative, it's just that you're not fully aware of all
> >>>>>> the factors. if you're talking a big rig trailer that is skidding and
> >>>>>> not stopping when unladen, that's because of poor control and excess
> >>>>>> tire pressure relative to the load. if the brake
> >>>>>> anti-lock/proportioning system was better, it wouldn't skid. if tire
> >>>>>> pressures were actively managed, as some modern vehicles now are, again,
> >>>>>> it wouldn't skid. both are control issues, not "insufficient weight"
> >>>>>> issues.
> >>>>>
> >>>>> Again, it's not at all about control. ABS and proportioning maintain
> >>>>> control by reducing the braking forces to less than the maximum values
> >>>>> which the brakes can sustain, the values that would cause a skid due to
> >>>>> insufficient traction.
> >>>>
> >>>> it's not about control but abs/proportioning maintain control??? you're
> >>>> confused.
> >>>
> >>> Nope, you're confused. If you do not have the traction to allow the use
> >>> of the brakes at their maximum capacity, you are operating at *reduced*
> >>> braking, even if you want to twist semantics to claim that ABS or
> >>> proportioning are maximizing the (available before skidding) braking.
> >>
> >> abs' sole function is CONTROL.
> >> http://en.wikipedia.org/wiki/Control_theory
> >>
> >> it is CONTROL that allows you to maximize braking without skidding.
> >
> > It is the control that automatically reduces braking force to prevent
> > skidding.
>
> increasing braking.

Reduced.

>
> > This is not any different than traction control which reduces
> > engine power to prevent skidding from excess drive power,
> > "anti-braking".
>
> increasing traction where it would otherwise be lost.

Preventing the loss of traction does not equate to increasing traction.
The available traction does not change at all, engine power is simply
reduced to prevent exceeding the limits of the traction. Increasing
traction would require adding a traction modifier such as spraying
racing "traction compound" on the road ahead of the tire.

>
> >
> >>
> >>>
> >>>>
> >>>> any brake that works properly should be able to lock its wheel.
> >>>
> >>> The subject of quite a bit of debate, since a locked wheel with the
> >>> vehicle moving is not a desired condition.
> >>
> >> any "debate" is whether a driver can control maximum braking better than
> >> a machine, not whether a braking system should be powerful.
> >
> > No, I've read plenty of debate over whether the ability to lock a wheel
> > at speed is a relevant qualification for a brake system, since such a
> > condition is not desirable.
>
> exercising it and having capacity for it are two totally different
> things. you need excess braking capacity for safety because, as any
> truck driver should know, if you experience overheating and fade, you'll
> need it.

And a runaway ramp...

>
> >
> >>
> >>>
> >>>> whatever conditions that wheel is experiencing. ensuring it /doesn't/
> >>>> lock is absolutely /all/ about control.
> >>>
> >>> Yep, ABS is certainly about control, but that control is all about
> >>> limiting the brakes to less than their maximum braking capacity to
> >>> compensate for inadequate traction.
> >>
> >> no, it's to maximize available braking and CONTROL. this is why abs was
> >> invented - drivers are typically BAD at CONTROL.
> >
> > No it's about limiting braking to match the available traction thus
> > aiding a poor driver in maintaining control.
>
> by increasing braking...

By limiting braking to match available traction.

>
> >
> >>
> >>>
> >>>>
> >>>>> A vehicle will not be able to achieve the maximum
> >>>>> braking that the brakes are capable of unless it has sufficient load to
> >>>>> achieve enough traction.
> >>>>
> >>>> nope. see above.
> >>>
> >>> Sorry, inadequate traction = less than the brakes maximum braking
> >>> capacity.
> >>
> >> no, CONTROL
> >
> > Sorry, in conditions of inadequate traction such as an unloaded pickup
> > truck, whether controlled by a competent driver or automatic by ABS, the
> > braking force is reduced to match the available traction.
>
> no, it's limited by poor control. the driver doesn't control the
> proportioning, the manufacturer does. and i'll bet he's also not
> controlling the tire pressure. his only bet is if there's an abs system
> on there, and most production versions are cheap enough that at low
> loads/low tire contact, they don't modulate [CONTROL] finely enough.

Braking force is still reduced to match available traction.

>
> >
> >>
> >>>
> >>>>
> >>>>> This is why for many vehicles the stopping
> >>>>> distance decreases as you add cargo weight up until you reach maximum
> >>>>> traction at which point further weight will increase stopping distance.
> >>>>
> >>>> nope, it's because the control systems don't work well at low loads.
> >>>
> >>> Nope, it's because an inadequately loaded axle has little traction and
> >>> the brakes on that axle can provide little braking capacity as a result.
> >>
> >> no. and you contradict yourself below.
> >
> > No, and there is no contradiction below.
>
> there is. but you need to understand to understand.

Nope, there isn't.

>
> >
> >>
> >>>
> >>> The control systems also suck eggs when such a lightly loaded vehicle
> >>> with stiff suspension hits a slight bump while braking moderately hard.
> >>> This is / was a significant safety hazard with some trucks, including
> >>> ones I owned, where you had to watch the road surface and stop braking
> >>> at each bump, lets the POS ABS freak out and cause you to loose all
> >>> braking and greatly increase your stopping distance. Many NHTSA incident
> >>> reports are directly attributed to this ABS fault.
> >>
> >> so we're back right where we started - if you're experiencing poor
> >> braking, it's because of poor control, not poor physics.
> >
> > Nope, in the specific case I cited above, it is a defective ABS system
> > that is presenting a hazard by not being properly matched to the vehicle
> > and/or having inadequate sensors such that it could differentiate the
> > momentary wheel lock due to the tire coming off the ground in a bump
> > from an actual traction loss / skid event. When that defective ABS
> > system saw that fraction of a second lock when the wheel was off the
> > ground, it went into full panic mode causing a near total loss of
> > braking for an extended period of time which in many cases caused
> > accidents that would not have occurred had the system worked properly or
> > not been operating.
>
> a "defective abs" is neither the basis on which to argue or the
> foundation of a subject understanding. it's simply a defective abs.

It points out that those marvelous control systems are not as infallible
as many believe.

On 06/07/2011 06:52 AM, Pete C. wrote:
>
<extensive snippage for brevity>

>
> Sorry, you are absolutely incorrect. I said nothing about changing the
> ratio,

but you should have. because you kept missing the fact that you can't
get "more out" without there being some kind of "more in".


> I said transmit increased force without changing the fluid
> pressure.

see above.


> If you want to transmit more force without changing the bore
> size on both ends you will have to increase the hydraulic pressure.

well, you got that bit right at least. and this is precisely what we
see in practice with brake boosters.


>
> 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
> applied to a 1 sq in slave cylinder = 100# force on the caliper.
>
> 200# of force applied to a 1 sq in master cylinder = 200 PSI fluid,
> applied to a 1 sq in slave cylinder = 200# force on the calipers.
>
> 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
> applied to a 2 sq in slave cylinder = 200# force on the calipers.
>
> As you can clearly see, increasing the bores on both sides allows more
> force to be transmitted (not multiplied) without increasing the line
> pressure.

so where does this #200 input come from??? this is why i keep coming
back to ratios - there is ALWAYS a trade!

and worrying about keeping line pressure in the hundreds is pointless
when standard brake lines take over 1000 psi.


>
>>
> They are exactly equivalent, this is fact.

no it's not, it's what you tell 5th graders!


>
> And this relates to the topic how? Moving more fluid due to larger bore
> sizes doesn't automatically equate to larger line diameter, we aren't
> moving 4' stroke cylinders on an excavator here, the total volume is
> still very low.

if you increase fluid flow and you don't increase tube bore, you get
flow resistance and thus response lag. kind of important on a braking
system.


>
>>
>
> And as soon as it is locked, it is "holding", not "braking" and energy
> dissipation (if any remains) has moved elsewhere, such as grinding tread
> off tires.

missing the point.


>
>>
>
> Pass through and no longer be "braking".

missing the point.


>
>>
>
> Then why are you making factually false claims such as those around
> hydraulic line pressure vs. bore size and force transmitted?

again:
<http://www.grc.nasa.gov/WWW/K-12/WindTunnel/Activities/Pascals_principle.html>

read the cite and look for the magic phrases "velocity ratio" and
"mechanical advantage". they are inextricably linked. if you think i'm
making false claims, you're also saying nasa have it wrong.


>
> Give me a PWM application where the end result is not averaged by some
> form of low pass filter. Even PWM drive for lighting dimming relies on
> human eyes as the low pass filter.

then "low pass filter" is simply a buzzword to you - you don't actually
understand it.


>
>>
>
> Yep it's pointless. You don't want to admit that inadequate traction
> forces a limitation in brake capacity to prevent skidding.

sorry, but you're asking me to deny reality. i'm sure you're sincere,
but you're underinformed, and there's no reason to think others are
bull****ting you. similarly, it's no reason for you to ignore that
CONTROL is the missing piece of your jigsaw.


>
>>
>
> I said "par" on traction. If traction was always up to "par", the
> braking would not have to be reduced from maximum.

of course it would if you don't want to lock tires - what a bizarre concept!


>
>>
>
> Braking reduced to maximum traction capacity.

uh, that's increased to maximum traction capacity since skidding is a
lower amount.


>
>
> No, it's reduced braking to match available traction.

see above.


>
> Reduced.

this is schoolyard. what's next - is your brother bigger than my brother?


>
>>
>
> Preventing the loss of traction does not equate to increasing traction.

yes it does. by definition.


> The available traction does not change at all,

it sure does.


> engine power is simply
> reduced to prevent exceeding the limits of the traction.

see?!!


> Increasing
> traction would require adding a traction modifier such as spraying
> racing "traction compound" on the road ahead of the tire.

lost it.


--
nomina rutrum rutrum

jim beam wrote:
>
> On 06/07/2011 06:52 AM, Pete C. wrote:
> >
> <extensive snippage for brevity>
>
> >
> > Sorry, you are absolutely incorrect. I said nothing about changing the
> > ratio,
>
> but you should have. because you kept missing the fact that you can't
> get "more out" without there being some kind of "more in".

Hydroboost works nicely for the "more in" part.

>
> > I said transmit increased force without changing the fluid
> > pressure.
>
> see above.
>
> > If you want to transmit more force without changing the bore
> > size on both ends you will have to increase the hydraulic pressure.
>
> well, you got that bit right at least. and this is precisely what we
> see in practice with brake boosters.
>
> >
> > 100# of force applied to a 1 sq in master cylinder = 100 PSI fluid,
> > applied to a 1 sq in slave cylinder = 100# force on the caliper.
> >
> > 200# of force applied to a 1 sq in master cylinder = 200 PSI fluid,
> > applied to a 1 sq in slave cylinder = 200# force on the calipers.
> >
> > 200# of force applied to a 2 sq in master cylinder = 100 PSI fluid,
> > applied to a 2 sq in slave cylinder = 200# force on the calipers.
> >
> > As you can clearly see, increasing the bores on both sides allows more
> > force to be transmitted (not multiplied) without increasing the line
> > pressure.
>
> so where does this #200 input come from??? this is why i keep coming
> back to ratios - there is ALWAYS a trade!

I get the increased input from increased boost.

>
> and worrying about keeping line pressure in the hundreds is pointless
> when standard brake lines take over 1000 psi.

I've seen some brake lines blow out, I like headroom in the ratings.

>
> >
> >>
> > They are exactly equivalent, this is fact.
>
> no it's not, it's what you tell 5th graders!

Yes, they are exactly equivalent in this context. Force = W, bore = V,
line pressure = A.

>
> >
> > And this relates to the topic how? Moving more fluid due to larger bore
> > sizes doesn't automatically equate to larger line diameter, we aren't
> > moving 4' stroke cylinders on an excavator here, the total volume is
> > still very low.
>
> if you increase fluid flow and you don't increase tube bore, you get
> flow resistance and thus response lag. kind of important on a braking
> system.

Lag doesn't seem to be a big deal on an air brake system, so I'm not
going to get overly concerned about a tiny lag in a hydraulic brake
system.

>
> >
> >>
> >
> > And as soon as it is locked, it is "holding", not "braking" and energy
> > dissipation (if any remains) has moved elsewhere, such as grinding tread
> > off tires.
>
> missing the point.

Not at all, if the brake is locked it isn't doing any work dissipating
energy.

>
> >
> >>
> >
> > Pass through and no longer be "braking".
>
> missing the point.

Nope.

>
> >
> >>
> >
> > Then why are you making factually false claims such as those around
> > hydraulic line pressure vs. bore size and force transmitted?
>
> again:
> <http://www.grc.nasa.gov/WWW/K-12/WindTunnel/Activities/Pascals_principle.html>
>
> read the cite and look for the magic phrases "velocity ratio" and
> "mechanical advantage". they are inextricably linked. if you think i'm
> making false claims, you're also saying nasa have it wrong.

Sorry, we are talking hydraulics, not wind tunnels. Applied force x
Piston bore area = hydraulic pressure and conversely, hydraulic pressure
x piston bore area = applied force. Velocity and mechanical advantage
have no relevance in that equation.

>
> >
> > Give me a PWM application where the end result is not averaged by some
> > form of low pass filter. Even PWM drive for lighting dimming relies on
> > human eyes as the low pass filter.
>
> then "low pass filter" is simply a buzzword to you - you don't actually
> understand it.

So what is the PWM application that isn't relying on some form of low
pass filtering?

>
> >
> >>
> >
> > Yep it's pointless. You don't want to admit that inadequate traction
> > forces a limitation in brake capacity to prevent skidding.
>
> sorry, but you're asking me to deny reality. i'm sure you're sincere,
> but you're underinformed, and there's no reason to think others are
> bull****ting you. similarly, it's no reason for you to ignore that
> CONTROL is the missing piece of your jigsaw.

There is nothing missing, if traction is inadequate such as due to
insufficient axle weight, braking capacity is reduced.

>
> >
> >>
> >
> > I said "par" on traction. If traction was always up to "par", the
> > braking would not have to be reduced from maximum.
>
> of course it would if you don't want to lock tires - what a bizarre concept!

Nothing bizarre about it, try reading back in the thread as you seem to
have lost the context of what we are discussing.

>
> >
> >>
> >
> > Braking reduced to maximum traction capacity.
>
> uh, that's increased to maximum traction capacity since skidding is a
> lower amount.

Braking reduced to maximum traction capacity. This is a fact, there is
no way around it, if you increase braking, you will loose traction.

>
> >
> >
> > No, it's reduced braking to match available traction.
>
> see above.

Deny facts all you want, you wont change them.

>
> >
> > Reduced.
>
> this is schoolyard. what's next - is your brother bigger than my brother?

I don't have a brother.

>
> >
> >>
> >
> > Preventing the loss of traction does not equate to increasing traction.
>
> yes it does. by definition.

No, it does not. The amount of traction available does not change. No
amount of brake "control" will give you more traction on the ground, all
that control will do is help you avoid exceeding that available
traction.

>
> > The available traction does not change at all,
>
> it sure does.

Does not.

>
> > engine power is simply
> > reduced to prevent exceeding the limits of the traction.
>
> see?!!

Engine power reduced or braking power reduced, both to prevent the force
applied to the wheels either accelerating or decelerating from exceeding
the available traction.

>
> > Increasing
> > traction would require adding a traction modifier such as spraying
> > racing "traction compound" on the road ahead of the tire.
>
> lost it.

Nope. No amount of "control" will change the available traction, it only
helps you stay within that limit by *reducing* the traction demand by
either *reducing* engine power, or *reducing* braking as appropriate.

jim beam wrote:
>
> > between the tire and the road, the tires contact patch, the normal force
> > pressing the tires down on the road and other minor factors. The amount of
> > braking force needed to stop the vehicle is a function of the mass to be
> > stopped and the desired deceleration rate.With a load in the back of the
> > truck, the normal force applied to the tires is directly related to the mass
> > to be stopped - right? If you have more weight in the bed, you have a higher
> > normal force on the tires and therefore you have a higher maximum braking
> > force available. When you have a trailer, without trailer brakes, the
> > situation is different. The load on the trailer only partially contributes
> > to the normal force on the tires that provide the braking - depending on how
> > the tailer load is distributed, maybe only 10% or 20% . The load on the
> > trailer still contributes 100% to the mass to be stopped. So although you
> > have the same total mass to be stopped with part of the load on the trailer
> > instead of in the bed of the truck, you have less maximum braking force
> > available to stop it. Therefore, it is possible (very likely) you cannot
> > stop the vehcile plus trailer as qucikly as you could stop the vehicle alone
> > if the weight was all carried in the bed of the single vehicle. OK?
>
> no, not ok. just like pete c, what you're experiencing is not "better
> braking", it's better control. apparently this is a really difficult
> concept.

What you're experiencing is neither "better braking", nor "better
control", it's "better traction" which translates into "better braking"
by allowing more brake force without loosing traction.

On 06/07/2011 07:16 PM, Pete C. wrote:
>
> jim beam wrote:
>>
>>> between the tire and the road, the tires contact patch, the normal force
>>> pressing the tires down on the road and other minor factors. The amount of
>>> braking force needed to stop the vehicle is a function of the mass to be
>>> stopped and the desired deceleration rate.With a load in the back of the
>>> truck, the normal force applied to the tires is directly related to the mass
>>> to be stopped - right? If you have more weight in the bed, you have a higher
>>> normal force on the tires and therefore you have a higher maximum braking
>>> force available. When you have a trailer, without trailer brakes, the
>>> situation is different. The load on the trailer only partially contributes
>>> to the normal force on the tires that provide the braking - depending on how
>>> the tailer load is distributed, maybe only 10% or 20% . The load on the
>>> trailer still contributes 100% to the mass to be stopped. So although you
>>> have the same total mass to be stopped with part of the load on the trailer
>>> instead of in the bed of the truck, you have less maximum braking force
>>> available to stop it. Therefore, it is possible (very likely) you cannot
>>> stop the vehcile plus trailer as qucikly as you could stop the vehicle alone
>>> if the weight was all carried in the bed of the single vehicle. OK?
>>
>> no, not ok. just like pete c, what you're experiencing is not "better
>> braking", it's better control. apparently this is a really difficult
>> concept.
>
> What you're experiencing is neither "better braking", nor "better
> control", it's "better traction" which translates into "better braking"
> by allowing more brake force without loosing traction.

are you done? because you're no closer to reality now than you were
when we started this.


--
nomina rutrum rutrum