From AutoBild.de.. the gallery has a bunch of pictures, and there was German sentences to give a description of each picture underneath. Being bored, I ran them through a translator.. and after reading all of them, I got some great info!

The brake performance is to mark optimum with less than 36 meters from 100 km/h.

For us Americans, that conversion runs to 118.11 feet! For comparison, a BMW 330 stops in 117 feet, a BMW M3 in 112 feet, Audi A4 in 118 feet, G35 in 129 feet.

Edit: Those stopping distances are from 60-0 MILES per hour. Since the conversion on 100km/h is around 62 miles, that second of stopping distance could be made up and we'd be beating the 330 as well :P Either way, the car stops might quick!

We are in good company! :D

I should get bored more often :)

That's pretty decent. :) I expected no less. It's almost as good as the MR2 Spyder, a car with probably the best brakes around in the sub-$30k price range. I've personally experienced the unbelievable stopping power of these brakes on the track. You can move your braking points so far back it's not even funny -- scares passengers though. ;) The car hauls down from high speed in what seems like a second. Really amazing. Motorweek recorded an average of 94ft from 60-0 mph! But then Road & Track lists it as 115ft.

weather conditions, track conditions, and tires can cause a HUGE discrepancy in the braking distances. To be as accurate as possible, all compared cars should be tested in the same conditions, though not necessarily the same size wheel and tire.

Very true. Tires and ambient temperature play a huge role. That said the 8 looks like a stopper! :)

yes!! now, i can't WAIT for the comparo tests... it's possible the 3 will be unseated as the sedan braking champion of the world!! :D

For comparison, the 93-95 FD was about 110 feet, so the RX-8 is only slightly behind in that regard, which is very good.

I wonder how the handling will measure up... so far it's looking like the 8 will be slightly slower than an FD around a track, but slightly slower than a FD is still quite fast - keep in mind that in 1993 the NSX and Porsche 911 Turbo were also slightly slower than an FD on a track. :)

Originally posted by Hercules


For us Americans, that conversion runs to 118.11 feet! For comparison, a BMW 330 stops in 117 feet, a BMW M3 in 112 feet, Audi A4 in 118 feet, G35 in 129 feet.

Edit: Those stopping distances are from 60-0 MILES per hour. Since the conversion on 100km/h is around 62 miles[/B]


For any math geniouses: please convert this measurement (62-0) down to 60-0. I think we all will be suprised. I could probably find some comparisons but would really like to know how to do it with an equation. The surprise will come from the statistic for cars doing 70-0 is easily double than that same car on the same day doing 60-0.

Originally posted by SPDFRK



For any math geniouses: please convert this measurement (62-0) down to 60-0. I think we all will be suprised. I could probably find some comparisons but would really like to know how to do it with an equation. The surprise will come from the statistic for cars doing 70-0 is easily double than that same car on the same day doing 60-0.
Done. I did it this way:

I set up a ratio of 36 over 62, = x over 60. Then I cross multiplied, and divided to get x. I got X = to ~34.838. That's meters. I then converted meters to feet, and that comes to:

114.29790026201 feet :) Not too shabby!

Correct me if I'm wrong but I plugged in 70 to see what came up because I felt your method was clearly too simple and I got 134ft which I certainly hope was true but I think you have missed a few laws of physics.

Now after thinking for a second, with this limited info this question is impossible to answer. There are just too many variables to come up with a good educated answer for the 70-0 distance.

Originally posted by SPDFRK
Now after thinking for a second, with this limited info this question is impossible to answer. There are just too many variables to come up with a good educated answer for the 70-0 distance.
Well I did the calculation from 62-60.... going to 70, yes there will be many variables to account for. It's just a matter of testing it and seeing what happens in the real world :)

I'll wait till C&D get their hands on it :D

Originally posted by Hercules

Done. I did it this way:

I set up a ratio of 36 over 62, = x over 60. Then I cross multiplied, and divided to get x. I got X = to ~34.838. That's meters. I then converted meters to feet, and that comes to:

114.29790026201 feet :) Not too shabby!

A simpler way to expain what Herc did is to take the distance of 62-0 and divide by 62.

Then multipy the result by 60. For example:

118.11 ft / 62 mph = 1.905 * 60 mph = 114.3 ft

This is flawed however because it assumes that it will take will take 1.905 ft to slow down every mph.

As an example, if you are going 1mph and you stomped on the brake, would you go almost 2 feet before stopping?

What if you were going 100 mph and you stomped on the brake? You would certainly go more than 2 feet before dropping to 99mph. The time to slow down may be close to linear but the distance will not be. That would explain why the stopping distance for cars doing 70-0 is easily double than that same car on the same day doing 60-0.

The good news in all of this is that since the distance does not drop linearly with increased speed, we can expect the 60 - 0 stopping distance to be even less than 114 ft. :o

Brian

Hercules - it's not a linear equation so your calculation is not correct.

For example a linear braking behaviour would suggest that a car that can brake from 60-0 in 120' would therefore brake 70-0 in 140', but in reality it's more like 170'.

To calculate it mathematically is probably impossible - too many variables (e.g. as well as braking force (and all it's variables) there is drag force (continuously changing with speed^2) and others).

I would guess that if 62.5-0 is 118' then 60-0 is definitely less than the linear solution of 114', maybe around 110'.

I conclude from the given information. that it will be less then 118, and likely less then 117 for 60-0 :)

Hercules may be close, another way to guesstimate it would be to use some physics and find the average braking force using F = ma for the known quanities, and then apply them to 60. This may be faily accurate assuming the ABS kept the braking force constant during a full run. There will be variations every run, so best you will ever get is an average anyway.

Originally posted by R.Cade
I conclude from the given information. that it will be less then 118, and likely less then 117 for 60-0 :)

Hercules may be close, another way to guesstimate it would be to use some physics and find the average braking force using F = ma for the known quanities, and then apply them to 60. This may be faily accurate assuming the ABS kept the braking force constant during a full run. There will be variations every run, so best you will ever get is an average anyway.
Yea, right now we're just ballparking so I don't think it matters too much until we get solid numbers from some reputable magazines :)

110.93 ft!!! :D

Originally posted by Buger
110.93 ft!!! :D
WOW!

Where did u get that number from ? :D Let's just keep making them up!

14 feet!!!!! :D

I actually used 114 ft when I should have used 118.11 in the original equation. The answer I got is 110.67. All of you math people, please correct my calculations if I made any errors. I like trying to apply my mind but it has been quite awhile since I've really done any math.

The answer can be calculated by assuming that the time to slow down 1mph is the linear (not the distance). I used x as the variable for the time to slow down 1mph.

Speed * time = distance

62x + 61x + 60x + ... 3x + 2x + x = 118.11ft

63 * 31 = 1953

1953 mph (x) = 118.11 ft

1953 mph = 2864.4 ft/sec

2864.4 ft/sec (x) = 118.11 ft

x = .041234 sec (note the dist units cancel)

Assuming the linear rate of braking, it will take approx .041234 seconds to slow down 1 mph.

We now need to subtract 62x + 61 x distance from 118.11 ft

123mph (x) = 180.4 ft/sec (x)
180.4 ft/sec (x) = dist from 62mph to 60mph
180.4 ft/sec (.041234) = dist from 62mph to 60mph
dist from 62mph to 60mph = 7.4386

118.11ft - 7.4386ft = 110.67ft

Your guess of 14 ft was alittle low Herc. :D

Brian

Originally posted by Buger
Your guess of 14 ft was alittle low Herc. :D

Brian
I stand by my fake number.

:cool:

I will put on some sticky Hoosiers and stop under 100 feet. So there...

Originally posted by randyc
I will put on some sticky Hoosiers and stop under 100 feet. So there...
OH yea!???!?!?

I'll stop under 5 feet from 10 mph! :D

use s = 1/2 * a * t^2 (1)
and v = a * t <=> t = v/a (2)

(2) in (1): s = 1/2 v^2/a <=> a= v^2/(2s)

v is speed, s is stopping distance

100 km/h = 62,1 miles/h = 27,8 m/s
96,6 km/h = 60 miles/h = 26,8 m/s
112,7 km/h = 70 miles/h = 31,3 m/s

a = 27,8^2/(2 * 36) m/s^2 = 10,7 m/s^2
(a is approximately constant)

stopping distance at 60 miles/h:
s = v^2/(2a) = 33,6 m = 110 ft

stopping distance at 70 miles/h:
s = v^2/(2a) = 45,8 m = 150 ft

I doubt the 36 m from 100 km/h are a realistic value.

And who said that math wasn't practical? :) Nice work oecher, I assume those are various forms of motion equations (physics).

The 118.11ft (36m) from 100 km/h does seem a bit low. If that is assumed, math can show that the 60-0 distance will be approx 110ft. To compare this, see the below top cars in 60-0 braking:

2004 Mazda rx-7.....................110ft????
2002 BMW M3...........................112ft
2002 Lexus IS 300.....................113ft
2002 Lexus GS..........................114ft
2001 BMW M5...........................116ft
2002 Chevrolet Corvette Z06.......116ft
2002 Acura NSX........................117ft

:o

The direct quote from the Autobild caption is below:

Dynamische Optik: flacher Vorderwagen, große Räder, ausgestellte Kotflügel. Die Bremsleistung soll mit weniger als 36 Metern aus 100 km/h Bestwert markieren.

Babelfish.Altavista.com translated this to:

Dynamic optics: flat front car, large wheels, issued fenders. The brake performance is to mark optimum with less than 36 meters from 100 km/h.

Brian

Hey - I guessed 110' first!

Actually Newtons equations of motion only apply to CONSTANT acceleration or deceleration. They are:
v=u+at
s=ut+(1/2*at^2)
v^2=u^2+2as

where
u=initial velocity
v=final velocity
a=CONSTANT acceleration
s=distance
t=time


I did quite a few years of Applied Mathematics and Engineering at University and to really calculate this is MUCH tougher than that - definitely a 2nd or 3rd order differential. Even the simple formula F=m*a is in reality F=m*dv/dt or F=m*d^2s/dt^2 - luckily we can ignore changes in mass due to relativistic effects :) ). Then there are unknown variables like coefficient of friction for the surfaces (tires and road), possible stiction (even with ABS) etc.

For example you also need to allow for the constantly changing drag force:
F=0.5*rho*Cd*A*v^2
= 0.5*density of air*drag coefficient*cross-sectional area* velocity squared
That velocity squared is a killer at high speed as we all know (else all cars geared correctly could do 150mph+ easy).

Also the braking behaviour may be changing (non-linear) as the brakes heat up between the start of the test and the end etc. Same goes for tires.

The above will have minor effects both for and against deceleration, so I think 110'-113' is about right (on that day, on that track with the same car).

Originally posted by pelucidor

Even the simple formula F=m*a is in reality F=m*dv/dt or F=m*d^2s/dt^2 - luckily we can ignore changes in mass due to relativistic effects :) ). Then there are unknown variables like coefficient of friction for the surfaces (tires and road), possible stiction (even with ABS) etc.


My thoughts exactly! :confused:

Originally posted by pelucidor
- luckily we can ignore changes in mass due to relativistic effects :)

....well I don't know.....the RX-8 should be a pretty fast car....

:D

I just re-read my post and I must apologize most profusely for the nerd-speak. I am not a geek. I am not a geek. I am not a geek. I am not ...

What I wanna know pelucidor is whether you needed to consult a book to write that stuff up or did that come to you off the top of yer head??!! I am a mechanical engineer and I had to check one of my many textbooks to make sure that is correct. If you used a book, good for you! If not, and since I'm an engineer and didn't know that that easily does that make me stupid? :eek: :D

I wrote it off the top of my head (I am not a geek!). But that's about all I remember of my 5 years at college doing maths and engineering. I hated mechanical engineering (expecially thermo) but in the first year I had a lot of fun computationally designing a gearbox based on combustion pressure numbers (designed for maximum speed and acceleration naturally :)). I still remember some of those equations (from 16 years ago) because it was the only fun thing I ever did in class. I really wanted to study theoretical physics but my parent's were against it (any Indian would know there are sadly only two things you can study at University - medicine or engineering).

@pelucidor
Imho the only thing that has considerable influence is the drag force. And compared with the breaking force (approx. 13-15 kN), it (approx. < 1 kN at 60mph) is neglectable.

If we presume a is constant (I know it isn't), dv/dt or d^2s/d^2t is also constant.

I'm not sure, what you want to tell (probably my English is too bad), but I think you agree, the results are acceptable.

I agree - around 110'-112' is realistic. Which is pretty spectacular by any standards. The drag-force (which slightly aids a car stopping from the higher speed) is almost negligible as it only changes the stopping distance during the 2.5mph delta between 62.5mph (100kmph) and 60mph - so minimal impact (1-2 feet?). I was just trying to make the point that an accurate calculation is very difficult (if not impossible based on the few variables we know).

A more important question for those intending to track the RX-8 is how the brakes will fade under heavy use. For example the IS300 has a great one-time (or even 10-time) 60-0 stopping distance of 113' but (according to people on is300.net) the stock brakes fade too much during heavy track use.

DISCLAIMER - any engineering knowledge I have is purely theoretical (and I've forgotten almost everything anyway). In fact I probably couldn't change a flat tyre if I had to (never had to yet). That is why I leave all my cars completely stock (even the crappy BOSE stereo will be staying). I left the engineering field a soon as I got my degree as I thought it was too hard and carried too much responsibility. In my final year at college a professor (who was a consultant on an investigation into a prototype helicopter crash - rotor design problem) asked the class if ANYONE would fly in a vehicle that they had helped to design - nobody put their hands up. Funnily enough my first ever job offer was to design the canopy of the European Jet Fighter for British Aerospace, which I declined. I look up to real engineers like tallguylehigh. Now if I make a mistake I only screw up your bank accounts ;)

Wow pelucidor nice offer on the British Aerospace. I am working for the Department of Defense right now and kind of enjoy it, it holds a certain element of duty that I appreciate. :) If you decided not to do engineering, what field did you go into. Engineer's fit in anywhere, because, ahem, we are just so awesome :o lol

Why IT of course, like everyone else. I worked as a developer (fun), then networking (fun), then consulting (lots of international travel, usually fun) and eventually product management (eh) and operations management (blah). Currently I work on banking systems and services that we run for several hundred small banks in the USA. Your bank details are safe with us - really. Bwahahahahaha.

Originally posted by oecher
@pelucidor
Imho the only thing that has considerable influence is the drag force. And compared with the breaking force (approx. 13-15 kN), it (approx. < 1 kN at 60mph) is neglectable.

The coefficient of the friction between the tires and the road surface overrules all of that, even the rotors/calipers. Want to stop faster? Put higher grip tires on and brake on grippy concrete.

We are not talking about how to stop faster. We are trying to figure out what the 60-0mph distance is based on a 100-0kph (62.5-0mph) distance of 118'. The car, driver, tyres, conditions, etc are all constant. The general agreement is 60-0mph in about 110' or slightly more.

Ok, here's my attempt (I’m bored).

At its most basic level, braking is the conversion of kinetic energy into heat. Therefore, the braking distance is proportional to the amount of kinetic energy the brake has to dissipate. So the ratio between braking distance at different speed is the same ratio as their respective kinetic energy. D1/D2 = KE1/KE2.

The equation for kinetic energy is ½MV^2 where M = mass and V = velocity. Since we are calculating ratio, we can eliminate the mass and ½. We are left with just V^2. So the ratio for two braking distance is equal to the ratio of the velocity square, V1^2 / V2^2.

V1 = 100kph = 62.14mph
V2 = 60.00mph

V1^2/V2^2 = 1.073

36m/D2 = 1.073

D2 = 33.566m = 110.13ft

As for the validity of this method, here are some empirical data from Road & Track magazine. They tested braking distance at both 60mph and 80mph. Based on the 60mph and distance, the number in braket is the calculated braking distance at 80mph by the above method.

V1^2/V2^2 = 60^2/80^2 = .5625 = (D1 @ 60mph)/ (D2 @ 80mph)
Solve for (D2 @ 80mph)
(D2 @ 80mph) = (D1 @60mph)/ .5625 = (D1 @ 60mph)* 1.77778

Acura NSX
D1 @ 60mph = 117ft
D2 @ 80mph = 215ft [est. 208ft]

Chevrolet Corvette Z06
D1 @ 60mph = 118ft
D2 @ 80mph = 212ft [est. 210ft]

Nissan 350Z
D1 @ 60mph = 119ft
D2 @ 80mph = 217ft [est. 212ft]

Subaru WRX
D1 @ 60mph = 138ft
D2 @ 80mph = 247ft [est. 245ft]


As you can see, the estimated distance is close and always shorter than actual measured distance. This occurred since we have assumed that the dissipation of kinetic energy by the brake would be constant regardless of the amount of energy. Of course, this is not true in real life as the efficiency of the brake decrease as the amount of heats it dissipate increase.

Alright, now everyone is as bored as I am.