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93 500 sel - 95 s500
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Discussion Starter · #1 ·
I will be needing new pads and discs soon. Do I have the option of switching to a cross-drilled or slotted rotor and ceramic pads?

My google search turned up allot of ebay results but I am not comfortable buying there.

1. Do any of you run this kind of set up? (drilled rotors with ceramic pads?)
2. If so where can I get up-graded rotors and pads from?

Quality is my top concern.

Thanks
Kevin

ps I got a quote from Movit Brakes USA, these sure are sexy but being a student out of my price range...


Front 370mm w/6 Piston calipers: $4681 US
Rear 342mm w/ 4 piston Calipers: $3399 US
 

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2014 CLS63S SB
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Movits prices are about spot on for a brake upgrade of that order. I too suffered sticker shock the first time I ever looked at aftermarket brakes. Before you do though you may want to look at the pros and cons.
Drilled/slotted discs are reported to be noisy. I don't know about the drilled but can certainly comment on the slotted, (6 grooves). I've run them a couple of times and I tell you the grumble they make got right on my tits.

Oh, one other thing about brake upgrades. When you have the shiny new stoppers on all is well until you get a flat..................and try to put the 15" spare on the hub and find that the brakes will not allow you to.
 

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'14 E350, '07 CLK350, '01 SLK320
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Do a search on this forum too. It was discussed not too long ago.

IIRC, the Ate slotted rotors and the Akebono ceramic pads were rated favorably.
 

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Discussion Starter · #4 · (Edited)
I never would have thought slotted or drilled rotors would have been noisy. That is something I DEFINITELY do not want. Thanks Gumsie.

Yea I should have done a search first but Im on here sooooooo much and I couldnt remember ever reading such a post, I must have missed it.
I will now do some research on Ate slotted rotors and the Akebono ceramic pads.
Thanks bobs
 

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1973 W115 200 /8
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If quality is of importense.. Why not go for original Mercedes-Benz parts ?
The original brakes are big enough for your ride..
 

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93 500 sel - 95 s500
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Discussion Starter · #6 ·
If quality is of importense.. Why not go for original Mercedes-Benz parts ?
The original brakes are big enough for your ride..
My main reason for not wanting going with OEM parts is I am sick of cleaning my wheels of break dust every 2 days.
Also I thought there might be a manufacturer of drilled or grooved rotors that were of the same or highier quality then the OEM's.
Plus if I could slightly upgrade my stopping power from OEM I would like that also.

HOWEVER

I have been doing some reading and have seen that some people have found ceramic pads to not stop as well as OEM pads. Any opinions?
 

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1993 300SEL (Sold) 2007 X5
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I'm pretty sure you can run a ceramic pad on the stock rotor. Also I too think you're wasting your time getting aftermarket rotors you don't really need the extra stopping power unless you are modding the car quite a bit.
 

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Discussion Starter · #8 ·
Yea I think running ceramic pads on stock rotors is what I will end up doing.
However if if I can find drilled rotors I like I would want to run them. Even though I am not modding the car I feel adding breaking power is always a positive, sometimes 6 inches of distance can save thousands or dollars or even life.

Im still confused as to how ceramic pads might not grip as hard as regular semi-metalics? It it because the compound is harder? Or do they have to have heat built up in them to perform well?

Anyone here with ceramic pads that could give me their $.02?
 

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I never would have thought slotted or drilled rotors would have been noisy. That is something I DEFINITELY do not want.
If possible get a ride in a car, any car which has them and see for yourself. Some may not be bothered by the noise, it's just that I was. I don't know how much effect the number of grooves had as I only tried the six slot discs.
 

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1999 S500 1999 S420 1994 S320 1994 E320
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Zimmerman Cross Drilled Sport rotors/Axxis Ultra Ceramic pads/HD Bilsteins F/R

2 years no problem,less brake dust and good stopping power.



On my E320 running Ultra Ceramic and stock ATE rotors :thumbsup:
 

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Plus if I could slightly upgrade my stopping power from OEM I would like that also.
Actually I'm not so sure you'll have increased stopping power form those brakes. Remember you'll actually have less surface area working for you.
The only time drilled brakes will help you is because the allow the build up of gas to escape so there is less float from the pad, but his only happens to any significant amount when the pad is getting really hot, that is after repeated heavy stops. There are also reports of drilled discs cracking when overheating but then standard ones warp so the pint is moot.
Slotted discs apparently de-glaze the pads for you, but again this only happens when the pads overheat after even more repeated heavy stops.

Increased stopping power can really only be achieved with either better fluid/cooling, thicker/bigger discs and better pistons. Unlike other areas on the car where you can make reasonable gains from not that much money, you really need to spend hard to get reasonable improvement.
 

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1993 W140 500SEL Black/Black, 12014 Ford Fiesta ST Blk/Blk
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If possible get a ride in a car, any car which has them and see for yourself. Some may not be bothered by the noise, it's just that I was. I don't know how much effect the number of grooves had as I only tried the six slot discs.
My father had them on an older honda:rolleyes: They made a fluttering sound as you stopped from the air being trapped then released from the slots(OEM size), very strange....rotors got fouled up very quickly and developed a pulse within months, specific application and usage I suppose, a ratio of how fast the rotor can cool to it's entire mass etc. The drilled holes, an apparent lack of material as opposed to a bigger thicker aftermarket rotor that can deal with more heat initially IMO.
 

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W140 Mercedes 500 SE, 1992, European, 440.000 km
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Actually I'm not so sure you'll have increased stopping power form those brakes. Remember you'll actually have less surface area working for you.

there is an equatation in mechanical engineering which says that friction force does NOT depend on the surface of friction, so the area of the brake rotor or pads does not influence the braking performance directly. the later depends on A) friction coefficient between pads and rotor, and B) normal (perpendicular) force of the pad to the rotor, which is a function of pressing the brake pedal ...

I would say that a good feature of a drilled disc is larger area which results in better drain of heat (positive) and second, such discs should behave better in rain because water (if any) can escape into the holes .... thus the friction between pads and rotor is closer to that one in dry weather.
 

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there is an equatation in mechanical engineering which says that friction force does NOT depend on the surface of friction, so the area of the brake rotor or pads does not influence the braking performance directly. the later depends on A) friction coefficient between pads and rotor, and B) normal (perpendicular) force of the pad to the rotor, which is a function of pressing the brake pedal ...

I would say that a good feature of a drilled disc is larger area which results in better drain of heat (positive) and second, such discs should behave better in rain because water (if any) can escape into the holes .... thus the friction between pads and rotor is closer to that one in dry weather.
Sure, the coefficient of friction remains constant. But just the coefficient, I think that the energy created between two surfaces of for arguments sake 1sq foot will still be less than that created between two surfaces with the same pressure applied between them of say a square mile area.
But yes, take your point about water.
 

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Sure, the coefficient of friction remains constant. But just the coefficient, I think that the energy created between two surfaces of for arguments sake 1sq foot will still be less than that created between two surfaces with the same pressure applied between them of say a square mile area.
But yes, take your point about water.
:):):) One of the three newton's law indirectly says that input energy (influenced by driver's foot) is proportional to the output energy (available directly on discs), so we cannot expect wonders. Input energy is the energy which is actually delivered by the driver's foot. So if you want more energy on discs you also must press with larger force onto the brake pedal. What I want to say is that "energy logic" is not correct one to discuss about the role of the surface area of the discs. the energy approach cannot reveal the role of disc surface.

The energy is defined as a force multiplied with distance of traveling or displacement. The later is not dependent on the disc's shape ... namely the distance is a distance which is made by a disc during braking manouver (rotational moving results in distance). So the only parameter which remains in equation for energy is the force. It is called friction force, and as i wrote, it is defined as a product between a normal force on the discs (i.e., force of pads onto discs) and coefficient of friction between the pads and disc. coefficient of friction is a constant, and the normal force depends on the driver's activity of course. i think you mix force and pressure. indeed at smaller disc (or drilled disc) the pressure of pads onto disc is larger in comparsion to a larger disc (or non-drilled disc) but this has nothing to do with a force or braking performance.

Of course there is one big disadvantage of smaller disc area: the result is relatively large pad wear, because the pressure between the pads and the disc is larger in case of smaller disc area ... and due to this fact the pad (and disc) wear is relatively large. Pressure = force/surface. At a constant force (driver's foot) smaller surface gives higher pressure. Simpe as a bean :)

So, your example is not correct. In both cases (1 sq foot or 1 sq mile) the created or spent energy will be the same. because energy is never correlated with the surface of acting, it is related only to the acting force and displacement of the objects.
 

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. Pressure = force/surface. At a constant force (driver's foot) smaller surface gives higher pressure. Simpe as a bean :)

So, your example is not correct. In both cases (1 sq foot or 1 sq mile) the created or spent energy will be the same. because energy is never correlated with the surface of acting, it is related only to the acting force and displacement of the objects.
Yeah I know that, but consider this:
The rotor is there to generate a retarding torque which is what slows the vehicle. That torque is related to the brake system thus;
T=FR where F is friction, (constant), and R is the effective radius or moment of the rotor.
So from your statement are not saying that, In order to generate the same force, if you increase the surface area you can reduce the pressure?
 

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Yeah I know that, but consider this:
The rotor is there to generate a retarding torque which is what slows the vehicle. That torque is related to the brake system thus;
T=FR where F is friction, (constant), and R is the effective radius or moment of the rotor.
So from your statement are not saying that, In order to generate the same force, if you increase the surface area you can reduce the pressure?
No no no:)

It is vectors, do not forget. Thus, friction force acts in plane of discs, so let's say that it points in direction of driving (when wheels are straight). and this force (or momentum if we take radius of disc into consideration) is the one which stops the car, like you stated.

The force of the pads onto discs is perpendicular to the above friction force, do not forget this !!:):) So in this moment we cannot mix the normal force (pads onto discs) and the force which stops the vehicle. these are two completely different forces in terms of mathematic.

Now it comes the interesting part (I am sure you know it): the normal force, so the one which is acting perpendicular to the discs, depends on the area of the piston in the braking caliper (jaws). more precisel, we must consider the diameter of the braking line (usually very very small) and the area of that piston which is usually much bigger than the area of the line (metal or/and rubber pipe).

So, for a relatively large movement of a braking pedal we obtain actually zero movement of braking piston. We transform the movement of a brake fluid in the line into hydrostatic pressure onto the braking piston. So, the perpendicular force (F=pressure x area) is large (because the area of the piston is so much bigger than the area of the line). the point is that the pressure inside brake system is a constant. And force = pressure x area, thus we obtain relatively high perpendicular force onto discs after applying the brake pedal.

In another words, the perpendicular force onto discs (which determines the braking force in direction of wheels) does not depend on the area of pads or disc, but it depends on the area of the piston which is in the braking caliper. I think you did not pay any attention on that fact. I hope it is clear now.
 

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In another words, the perpendicular force onto discs (which determines the braking force in direction of wheels) does not depend on the area of pads or disc, but it depends on the area of the piston which is in the braking caliper. I think you did not pay any attention on that fact. I hope it is clear now.
I think we are coming at this same point from two different angles-this is how I see it. Yes of course you'll increase the piston area but nobody I know ever fits bigger discs without fitting bigger calipers generally with larger/more pistons or vice versa.

Ignoring uncontrollable losses in the braking system, mechanical force generated by the caliper will be equal to:
Fcal=Pcal x Acal.
Where like you said, F=Force and A is the effective area. All multiplied by two as there are two calipers giving the clamping force. So you can see how varying the surface area affects the braking.
Now the product of that clamping force, and the coefficient of friction is equal to the the overall frictional force or opposing forces generated in the discs/pads, and then the torque generated by the rotor becomes the product of that Frictional force and the effective radius.
I had an article somewhere that dealt with it. If I can find it I'll post it.

Still, one thing we can surely see the same way is that if the OP wants to improve his braking performance he needs a combination of;
Extremely fat wallet.
Thicker/bigger and more rigid discs and calipers.
Rigid brake lines.
Higher boiling point brake fluid, and improved cooling, (drilling ironically).
He also needs to pay heed to the fact that if he modifies the front brakes too much he changes the effective brake balance.
Also remember the spare wheel thing.
.....and maybe a couple of other things that escape me now.
 

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Where like you said, F=Force and A is the effective area. All multiplied by two as there are two calipers giving the clamping force. So you can see how varying the surface area affects the braking.
Only surface area of the piston affects the force (normal one and therefore also the frictional force which means braking performance in our case). So, it is only the piston's area not the pads' or disc's area. you stated in the beginning of discussion that due to holes in the disc, this one will have smaller surface and consequently braking performance will be worse. But this is not true :). Please do not mix surface of the disc with surface of the piston in the caliper, because these are two completely different things. In another words, the magnitude of the braking force onto the disc (or energy or whatever you want ...) is determined inside of the braking caliper. Once this force is applied to the brake disc, the story is practically finished. Only the coefficient of friction affects how this normal force onto the disc will affect frictional force which lies in the plane of disc.

before you answer this, please study theory of friction and you will notice that area of bodies in contact does not appear in any equation. This should persuade you if I am not able to :)

I agree with the rest (about rigid lines, etc...).
 
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