Can differing brake pads affect disc tempartures
10-27-2009, 02:23 AM
#11
Originally Posted by lower,Oct 27 2009, 08:53 AM
Don't forget that you continue to apply pressure through the brake pedal whilst you're braking and that pressure is magnified by the brake servo.
As i said in a previous post, you get heat generated purely from the friction between the pads and the disc which would occur even if the car didn't slow down as would happen with glazed pads.
10-27-2009, 03:32 AM
#12
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To what extent does the shifting of weight onto the front wheels play a part? For example, under hard braking there is a more pronounced weight transfer that presumably acts to increase the grip of the front tyres. Under gentle braking, there is a less pronounced weight transfer.
I'm not sure of the physics here, but does this mean that if you can brake later and thereby increase the grip at the front, are you not able to exert a greater braking force than would otherwise be possible and thus benefit pads that have a higher coefficient of friction?
I'm not sure of the physics here, but does this mean that if you can brake later and thereby increase the grip at the front, are you not able to exert a greater braking force than would otherwise be possible and thus benefit pads that have a higher coefficient of friction?
10-27-2009, 03:45 AM
#14
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Originally Posted by Dembo,Oct 27 2009, 11:23 AM
I guess the question is does the chemical or other reaction going on between pad and disc produce heat in itself?
I have set greenstuff pads on fire on my first Civic. If they are on fire then the pad material is undergoing a reduction of enthapy by means of oxidisation - endothermic reaction.
However, the catalyst for this was extreme heat.
Oh, and when I have been mentioning tyres as the limiting factor, I am talking in terms of ultimate force, not repeated braking - I thought everyone would have grasped that, but apparently not
I mentioned about repeated braking in my other post regarding higher temps for greater thermodynamic transfer.
Mans,
Yes, braking harder does mean more grip at the front.
However, it's a progressive scale and is paradoxical upto a cetain point
e.g
The harder you brake, the more weight you transfer forwards, the harder you can brake, the more weight you transfer forward, the harder....you gett he picture.
Up until you:
a) Run out of brake friction by means of insufficient caliper pressure
b) Pad vapourisation - brake fade
c) Tyre grip is relinquished by terms of the pressure physically breaking down the rubber.
10-27-2009, 04:02 AM
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Originally Posted by Hypersonik,Oct 27 2009, 11:45 AM
Mans,
However, it's a progressive scale and is paradoxical upto a cetain point
e.g
The harder you brake, the more weight you transfer forwards, the harder you can brake, the more weight you transfer forward, the harder....you gett he picture.
However, it's a progressive scale and is paradoxical upto a cetain point
e.g
The harder you brake, the more weight you transfer forwards, the harder you can brake, the more weight you transfer forward, the harder....you gett he picture.
10-27-2009, 04:35 AM
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Originally Posted by lower,Oct 27 2009, 12:39 PM
that is a good point.
Braking hard with higher friction pads certainly stands the car on its nose so must increase the pressure of the tyres into the road and therefore the friction.
The other thing i've not got my head round totally is that braking just before the lock up point gives greater deceleration than locking all 4 wheels.
It's the slip angle that's the important factor with generating grip from a tyre; I think locking the wheels overloads the slip threshold and thus somehow the grip falls off dramatically resulting in the tyre sliding across the tarmac.
10-27-2009, 04:43 AM
#18
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Originally Posted by Mans Best,Oct 27 2009, 01:02 PM
I have a feeling this is a power equation, so it's going to be exponential and not progressive, but I know what you mean.
Think Young's modulus
10-27-2009, 05:24 AM
#20
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Originally Posted by Mans Best,Oct 27 2009, 01:53 PM
Isn't deceleration a power equation?
As the rubber deforms into the road, it bonds with it.
There is only so much force the bond will take before it breaks.
Once one bond breaks, you will then have a cascade failure (at the molecular level) which WOULD be exponential
Have a think, but I need to go to bed now as I have my last nightshift to get up for in a few hours!!