From: John_H on
D Walford wrote:
>On 28/05/2010 7:42 PM, John_H wrote:
>>
>> The stiletto example also supports the case that contact area doesn't
>> matter. In fact it's pressure (force per unit area) that generates
>> the friction but if you double the contact area you halve the pressure
>> so the end result is the same (area cancels out when you do the sums)
>
>Which is what I said, the brake with the smaller contact area with the
>same pressure applied will be more affective since the "force per unit
>area" is greater.

And what I said is that force per unit area (pressure) cancels out, so
the contact area doesn't matter. I can't make it any simpler than
that.

The stiletto provides a good example of that fact, it's just that
you're still not seeing it correctly. My last reply to Noddy might
help, or you could take another look at what Sylvia has to say about
it.

--
John H
From: John_H on
Noddy wrote:
>
>However, I'd add that in increasing the size of the pad you reduce the
>pressure between the pad and rotor for a given amount of force holding the
>two together. The amount of braking force applied to the rotor will be
>exactly the same as you mentioned, but the pressure between the rotor and
>pad will reduce as the size of the pad goes up, and in direct proportion to
>the size of the pad since pressure equals force divided by the area of
>contact. If you were to keep the pressure between the pad and rotor the same
>for both large and small pads, then the friction applied by the larger pad
>would be greater and have a stronger braking effect.

Which suggests that you haven't quite got it yet, but we're getting
close! :)

Since doubling the surface area would halve the pressure, pressure
cancels out of the equation (Coulomb's) and you're left with the
applied force as the only relevant factor.

You could work through the example using either pressure of force and
you'll get the same answer (as I've just done on he back of an
envelope to satisfy myself I've got it right). Put in a value for the
force you can apply with your finger (say 50kg) and a value for COF
(0.3 is around the mark for brakes). The arithmetic is simple so I
won't bother to reproduce it here.

Other way to look at is from the conservation of energy perspective.
Since you're doing work when you apply the finger you expect to get
back exactly what you put in (though not necessarily in the same
form). In the example you'll get very, very close to the same amount
back as braking force (just as you do in the real world).

The rest is lost as heat of course, but changing the size of the pad
won't significantly affect the operating temperatures... ie the same
amount of heat energy will be lost in both cases.

--
John H
From: John_H on
Jason James wrote:
>"John_H" <john4721(a)inbox.com> wrote in message
>news:2e0vv55ltgpqhvvffi3o7o4lsle8btobem(a)4ax.com...
>> Jason James wrote:
>>>"Noddy" <me(a)home.com> wrote in message
>>>news:4bff0e6c$0$11949$c30e37c6(a)exi-reader.telstra.net...
>>>> "John_H" <john4721(a)inbox.com> wrote in message
>>>>>
>>>>> What I'm disputing is the existence of any credible theory that says
>>>>> the increased surface area due to scoring affects the performance in
>>>>> any way.
>>>>
>>>> I also tend to agree, however the theory that increased contact via
>>>> larger
>>>> surface area seems to make sense.
>>>
>>>Except, the angled sides of scoring dont present the pad with as an
>>>effective friction surface as parts of the disc surface that are parallel
>>>with the pad surface. Make sense? :-)
>>
>> No. Because the contact area is irrelevant to their performance, so
>> is the shape. Two sheets of corrugated iron will behave exactly the
>> same as two flat sheets if you were to slide them apart under the same
>> load.
>
>Yup,..the force or work, provided by the hydaulic system (from
>Master-cylinder) couldn't care less what the force/work dissappation is
>engineered like,...however there is one exception? IE if the disc was
>covered in grease?

There are numerous exceptions, and the one you've given is called
*lubricated friction*, which is why wet brakes don't work nearly as
well as dry ones... because the COF is effectively reduced.

Contact area still doesn't make a hoot of difference, except you'd
expect smaller pads to dispel the liquid (normally water) quicker
because of the higher pressure they generate (which doesn't translate
to a greater braking force once the dry COF is restored).

--
John H
From: John_H on
Brad wrote:
>
>I can see some points but not all.
>
>More brake pad / rotor contact means that there is less force (Pedal
>pressure) required to slow a vehicle.

No it doesn't. The same force applied to pads (and hence the pedal)
will provide exactly the same braking force.

> More often than not effective braking
>especially in motor is to do with heat dispersal. Larger surface area of the
>rotor gives greater dispersal of heat.

The surface area of the rotor is a separate issue to the contact area
of the friction surfaces. The pads only contact a small proportion of
the rotor surface at any instant.

Heat loss through the pads is SFA (relatively speaking) otherwise the
brake fluid would boil in the calipers. The contact area will
therefore make little or no measurable difference. For all practical
purposes rotor temperature (and hence heat loss) will be a near as
damnit to the same whatever the contact area.

--
John H
From: Noddy on

"John_H" <john4721(a)inbox.com> wrote in message
news:26c006tu1svsaurhqmhmm9ak63i29037j4(a)4ax.com...

> Which suggests that you haven't quite got it yet, but we're getting
> close! :)

Lol :)

> Since doubling the surface area would halve the pressure, pressure
> cancels out of the equation (Coulomb's) and you're left with the
> applied force as the only relevant factor.

Well, you are, but I was merely pointing out that while the applied force
pushing the pad against the rotor might be the same, the pressure between
the rotor and pad will be different. If you were to keep the pressure
between the *pad and the rotor* the same, which would require an increase in
the amount of applied force pushing them together, then more friction would
be generated with the larger pad. Of course, if you increased the applied
force on the smaller pad to the same level you'd generate more friction as
well, but then the pressure between the pad and rotor would be different
again.

The point I'm trying to make is that while I agree with you that changing
the size of the contact area doesn't affect the coefficient of friction and
won't change the braking ability for the same amount of applied force, it
*does* change the pressure between the pad and rotor which effectively makes
the smaller pad work harder than a larger one.

--
Regards,
Noddy.