Brake bits
- Thread starter DavidL
- Start date
mik_ok
New member
Always worth a phone call to Eurocarparts Leeds branch. If you want a discount see THIS THREAD
I'd insist on Sebro discs as opposed to Zimmermanns - I had warping problems with two sets of front Zimm's (rears have been fine) ~ Sebro's now on and they are fine.
I'd insist on Sebro discs as opposed to Zimmermanns - I had warping problems with two sets of front Zimm's (rears have been fine) ~ Sebro's now on and they are fine.
Paul Fraser
New member
I cannot second the above suggestion too much: you will pay _much_ less using them than anyone else.
sawood12
New member
That's odd - last time I checked with the OPC they wanted about £80 per disk inc. vat and the guy said he'd be able to give me a discount "at least equivalent to the vat". I went with non-OEM disks last time round but in future I think i'll be paying the little bit extra for the OEM disks. You don't seem to be saving a significant amount with non-OEM disks irrespective of where you get them from.
Well I have a Porsche stock list with prices dated 24.7.06 which gives retail prices as £107.75 each for the fronts and £110.98 each for the rears. Add on the VAT and your're over £125 for each. My discs fit '87 model year onwards cars, I think the earlier discs are a fair bit cheaper. Just to stir it up a bit Porsche reviewed their parts prices on the 1st August (mainly upwards I suspect).
Just checked my previous post, the part number for the rears should end 03 (not 08 as posted), typo on my part.
Just checked my previous post, the part number for the rears should end 03 (not 08 as posted), typo on my part.
sawood12
New member
Mybe it's an individual dealer thing then - for example we all know that Exeter OPC can often do good deals on various items for forum members. It is definately worth shopping round the dealers as they are independant businesses at the end of the day and can charge what they like irrespective of the official Porsche price list as long as they can cut a profit. When they are charging upto and over £1k for brake disc and pad replacement on 993's and as long as people are still prepared to blindly take their cars to their OPC for these jobs and pay the premium, it easily funds £40 or so off the price of a disc or two for people like us that run the older cars. They know too well that we will happily go elsewhere for the business if they can't be flexable on prices.
I went from the 250 Turbo "reasonably big blacks" to 928GTS "big blacks" with 964 Turbo "drilled" disks. To be honest I did it because they are a direct bolt-on and I never want to run out of brakes. They are awesome, but I imagine that's more to do with new fluid, new calipers and the Performance Friction 97 pads I use as with the upgrade itself. What I can say is that they have never so much as hinted at a drop in performance after being used hard.
I have the Mo30 brakes which are the 928 s4 ones I believe. Also s/s braided lines, EBC Yellows front + greens rear, very expensive race brake fluid, and the 968cs underbody air deflectors. They have proved very good at stopping me most times but watching Boxters, Cayman's, 996/7's, Gt3's etc they seem to be on the brakes for a much, much shorter time which makes me wonder why? I am losing so much time under brakes compared. Some have mentioned changing the brake proportioning valve to alter the bias. Not sure this would achieve shorter duration though?
All of the cars you mention are mid or rear engined, which I guess may mean they can get more out of the rear brakes.
There was an interesting discussion on Titanic last week about this, but when it came ot an engineer and a physicist discussing it I got a bit lost. Can anyone else who may have understood it offer a precis for Patrick?
This is where I lost it:
The turning moment, or force as it appears when you specify the point of its reaction with the ground (front tyre contact patch), is proportional to:
· the height of the centre of gravity above the contact point,
· the mass of the car
· the size of deceleration (laymen talk about acceleration is g's. Engineers use m/s/s, the SI unit that fits all of the equations without conversion factors)
Height hardly changes, so you can largely ignore that. Mass never changes. But if you decelerate the car hard enough the force, expressed as a function of mass x deceleration, can increase way beyond the vehicle weight. Reaction force at the front contact point then equals the force arising from braking plus static front axle weight. If you look at the rear tyre in the above scenario the reaction force at point of contact with the ground equals rear axle weight minus the force from braking. If you brake hard enough you can get the braking force at the rear tyre to be greater than the static axle weight. You will have undoubtedly experienced this first hand when you were a child and you pulled very hard on the front brake of your bike, only to find the rear wheel lifting clear of the ground and you tumbling over the handlebars. Clearly this situation shows how the braking force can exceed the static axle weight. This applies to the front tyre as well as the rear. Think of the force applied by a decelerating bullet of 10grams hitting a person in the chest. Decelerate the mass hard enough and the force will knock you backwards several metres. The force is many times greater than the weight of the bullet held in your hand with gravity acting upon it. The simple model of grip is equal to vertical force applied x friction coefficient. This model says that contact area has no part to play in the grip force. However this model does not adequately describe our car and tyre scenario. You know that wider tyres provide more grip from your own experience. This is why in the case where deceleration is increased you reach a point where the ever increasing reaction force at the ground does not provide enough grip force to equal the deceleration force and you go into a skid. Now, if you put more rubber on the ground the point at which the tyre can't cope becomes higher. You are thinking that front axle weight increases by the same amount as rear axle weight decreases. So overall braking effect should stay the same Well that's roughly true but not exactly true. As you lighten the rear tyre load when braking the grip available decreases. So you can't apply as much deceleration force as you do with the front brake, or else you will lock up the rears. On top of this effect manufacturers further limit rear brake pressure by giving a strong bias to the front, for safety reasons (rear lockups result in spins). The rear brakes therefore do not act efficiently. The fronts get overloaded as they are doing all of the work and then they lock up. But if you could get the rears to do more of their share of the braking then the car will stop sooner. You could do this by fitting a more rear biased brake proportioning valve, or moving the centre of gravity lower. In fact if you could move the centre of gravity below the road when you brake the rear wheels will experience the highest loading compared to the fronts, obviously this will be a clever modification if anyone can achieve it J You could also get the rears brakes to do more work if you increase the static rear axle loading compared to the front. Rwemember at the rear reaction force equals axle load minus breaking force. This is why 911's brake so well. Lets imagine a 1 tonne and 2 tonne car both decelerate at the same rate. Because the 2 tonne car weighs more the force expreinced at the tyre/road interface will be greater, as must be the clamping force from the brake caliper. If you try both cars at higher and higher decelerations you will reach a point where the caliper clamping force needed for the 2 tonne car cannot be achieved. At this point the 2 tonne car will not be able to brake equally with the 1 tonne car. Or, despite the greater weight transfer of the 2 tonne car, the 2 tonne car wil lock the front wheels before the 1 tonne car.
There was an interesting discussion on Titanic last week about this, but when it came ot an engineer and a physicist discussing it I got a bit lost. Can anyone else who may have understood it offer a precis for Patrick?
This is where I lost it:
The turning moment, or force as it appears when you specify the point of its reaction with the ground (front tyre contact patch), is proportional to:
· the height of the centre of gravity above the contact point,
· the mass of the car
· the size of deceleration (laymen talk about acceleration is g's. Engineers use m/s/s, the SI unit that fits all of the equations without conversion factors)
Height hardly changes, so you can largely ignore that. Mass never changes. But if you decelerate the car hard enough the force, expressed as a function of mass x deceleration, can increase way beyond the vehicle weight. Reaction force at the front contact point then equals the force arising from braking plus static front axle weight. If you look at the rear tyre in the above scenario the reaction force at point of contact with the ground equals rear axle weight minus the force from braking. If you brake hard enough you can get the braking force at the rear tyre to be greater than the static axle weight. You will have undoubtedly experienced this first hand when you were a child and you pulled very hard on the front brake of your bike, only to find the rear wheel lifting clear of the ground and you tumbling over the handlebars. Clearly this situation shows how the braking force can exceed the static axle weight. This applies to the front tyre as well as the rear. Think of the force applied by a decelerating bullet of 10grams hitting a person in the chest. Decelerate the mass hard enough and the force will knock you backwards several metres. The force is many times greater than the weight of the bullet held in your hand with gravity acting upon it. The simple model of grip is equal to vertical force applied x friction coefficient. This model says that contact area has no part to play in the grip force. However this model does not adequately describe our car and tyre scenario. You know that wider tyres provide more grip from your own experience. This is why in the case where deceleration is increased you reach a point where the ever increasing reaction force at the ground does not provide enough grip force to equal the deceleration force and you go into a skid. Now, if you put more rubber on the ground the point at which the tyre can't cope becomes higher. You are thinking that front axle weight increases by the same amount as rear axle weight decreases. So overall braking effect should stay the same Well that's roughly true but not exactly true. As you lighten the rear tyre load when braking the grip available decreases. So you can't apply as much deceleration force as you do with the front brake, or else you will lock up the rears. On top of this effect manufacturers further limit rear brake pressure by giving a strong bias to the front, for safety reasons (rear lockups result in spins). The rear brakes therefore do not act efficiently. The fronts get overloaded as they are doing all of the work and then they lock up. But if you could get the rears to do more of their share of the braking then the car will stop sooner. You could do this by fitting a more rear biased brake proportioning valve, or moving the centre of gravity lower. In fact if you could move the centre of gravity below the road when you brake the rear wheels will experience the highest loading compared to the fronts, obviously this will be a clever modification if anyone can achieve it J You could also get the rears brakes to do more work if you increase the static rear axle loading compared to the front. Rwemember at the rear reaction force equals axle load minus breaking force. This is why 911's brake so well. Lets imagine a 1 tonne and 2 tonne car both decelerate at the same rate. Because the 2 tonne car weighs more the force expreinced at the tyre/road interface will be greater, as must be the clamping force from the brake caliper. If you try both cars at higher and higher decelerations you will reach a point where the caliper clamping force needed for the 2 tonne car cannot be achieved. At this point the 2 tonne car will not be able to brake equally with the 1 tonne car. Or, despite the greater weight transfer of the 2 tonne car, the 2 tonne car wil lock the front wheels before the 1 tonne car.
sawood12
New member
I would have thought that a mid engined car that has 50/50 weight distribution (i.e. a Boxter) wouldn't be better under braking than a front engined car with 50/50 weight distribution (i.e. a 944) as the weight on a boxter is all in the centre of the chassis between the axels which means the rear axel becomes more unloaded for the same braking force therefore reducing the rear wheels braking effect. Because a 944 achieves it's weight distribution by having a dirty great big lump of an engine over the front axel and a dirty great lump of a gearbox mounted directly over the rear axel means that for the same braking force there is less weight transfer as the gearbox is still over the rear axel no matter how hard you are braking therefore meaning the rear brakes are more effective. This is also basically what I thought the long techno blurb was saying in a much longer and protracted way!!
Having said that I can't think why a Boxter would be better under braking than a 944 - in fact I would have thought them to be roughly equal if not the 944 having the edge for the reasons i've suggested above. Could it be that the new generation of calipers are better than the previous ones - or the more modern disc/pad material combinations are better - i've noticed with newer cars i've owned that generally pads will last the same time as discs i.e. when you replace pads the discs also need replacing. With older cars (944 included) you can usually get a couple or three sets of pads to each disc set.
Having said that I can't think why a Boxter would be better under braking than a 944 - in fact I would have thought them to be roughly equal if not the 944 having the edge for the reasons i've suggested above. Could it be that the new generation of calipers are better than the previous ones - or the more modern disc/pad material combinations are better - i've noticed with newer cars i've owned that generally pads will last the same time as discs i.e. when you replace pads the discs also need replacing. With older cars (944 included) you can usually get a couple or three sets of pads to each disc set.
Hi Scott,
Just read through the technical post from Fen and the way I understand a Boxster could out brake a 944 is that more of the boxsters weight is away from the front of the car i.e not directly over the front axle. This reduces the static weight of the front axle slightly moving it closer to the rear axle. Then according to the formula above this would allow longer effective tyre contact for the rear brakes before the reaction force equals the static axle weight minus braking force. I would imagine that the Boxster has a slightly different brake bias valve because of this ? Any Boxster owners confirm this ?
Dave K.
Diver944
Active member
After the Titanic discussion a number of us went out with G-Tech accelerometers and did the 60-0mph brake test. Obviously there were lots of variables involved like different tyres, road surfaces, car types etc but the results ranged from a best of 120 feet to a worst of 144
The only way to answer this question properly is for a few of the Boxster guys to do the same thing. I could bring my accelerometer to Eynsham and we could get a few Boxsters to do the test in the campsite [
] (should raise a few eyebrows and guarantee a favoruable result for the 944 [])
The only way to answer this question properly is for a few of the Boxster guys to do the same thing. I could bring my accelerometer to Eynsham and we could get a few Boxsters to do the test in the campsite [
] (should raise a few eyebrows and guarantee a favoruable result for the 944 [])
sawood12
New member
Fen (or anyone else that know's the answer) - I'm off out to the states with work in Sept and have a funny feeling I will return with a shiny set of big blacks which work out very cheap in the states given they are the same in $ as they are in £ (or even a bit less). My current spare wheel is the full tyred 16"wheel which obviously wont fit around the big blacks. Are the spare wheels with the collapsable tyre 17"?
If the Boxster is better under braking than the 944 then I suspect it's probably more to do with the lower centre of gravity of the car due to the low-slung flat six engine and the substantially beefed up sill structure due to the car being a convertable.
If the Boxster is better under braking than the 944 then I suspect it's probably more to do with the lower centre of gravity of the car due to the low-slung flat six engine and the substantially beefed up sill structure due to the car being a convertable.
Mine is in the garage so I'd need to check, but I think it's still a 16" rim. I have a can of spray in both my 944 and my Roadster in lieu of a spare wheel.
The other option you have is as was the handbook recommendation for my 3.2 Carrera (albeit the other way round for that), namely if you have a front wheel puncture you take off a rear and fit that to the front then put the spare that you have on the back as it will clear the rear calliper. It means you will have a different sized wheel across each axle on both ends and that's of dubious legality in the UK I believe, but that's the case with any space saver as it only takes one axle to make you illegal.
The other option you have is as was the handbook recommendation for my 3.2 Carrera (albeit the other way round for that), namely if you have a front wheel puncture you take off a rear and fit that to the front then put the spare that you have on the back as it will clear the rear calliper. It means you will have a different sized wheel across each axle on both ends and that's of dubious legality in the UK I believe, but that's the case with any space saver as it only takes one axle to make you illegal.
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