F-150 brakes

On 06/07/2011 07:10 PM, Pete C. wrote:
>
> 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.

standard brake lines are closer to 5000psi actual. if you got a
blow-out it was because of condition issues, not pressure issues.


>
>>
>>>
>>>>
>>> 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.

not even that is right.

flow rate = pressure change / flow resistance is analogous to current =
voltage change / resistance.

and

pressure = energy / volume is analogous to voltage = energy / charge

but it's still 5th grade. sorry.


>
>>
>>>
>>> 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.

fluids have considerably more mass than air. that means more inertia.


>
>>
>>>
>>>>
>>>
>>> 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.

still missing the point.


>
>>
>>>
>>>>
>>>
>>> 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.

you evidently still haven't read the cite - it is /specifically/ about
hydraulics and pascal's law.


> 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.

wow. /please/ read the cite. please.


>
>>
>>>
>>> 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?

"the" application? singular? hardly any pwm apps use or are reliant on
filtering. if you think you're witnessing "low pass filtering" when
you're witnessing pwm, you're absolutely not understanding the words.
see below.


>
>>
>>>
>>>>
>>>
>>> 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.

ok, there's primary three levels of knowledge.

1. what you know.
2. what you know you don't know.
3. what you don't know you don't know. [think about it]

you're in the third on this one. which is bad because the information
is here to at least take you to #2 - if you wanted it to.

<snipping the rest because it makes no sense>


--
nomina rutrum rutrum

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

> > 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
>
> is this a coulomb friction coefficient ed? yes or no. [careful, that's a
> loaded question.]


I understand that tires don't act according to the "perfect" coulomb
friction theory. But the the fricitional force available is still a function
of the normal force on the tire. It is just that instead of:

F(f) = N*cf it is something like F(f) = (N**x)*cf

whe

F(f) = maximum frictional force available
N = Normal force on the tire
cf = coefficient of friction
x = a factor somewhere in the range of 0.7 to 0.9

But this does not affect the main point - if part of your load is on a
trailer without brakes, then you have less normal force on the vehicle's
tires, and therefore less maximum braking force available. If you have less
maximum braking force available, you can't slow the load as quickly. I know
it is a lot more complicated when you start accounting for weight transfer,
and other factors but I am certain that if you have two otherwise identical
vehicles, one with all the load in the bed and one with the same load on a
trailer that does not have trailer brakes, the one with the load in the bed
can stop shorter.

Ed

> > 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.
>
>
> >
> > Ed