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RX-8 stopping distance.. this just in!

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Old 11-07-2002 | 12:00 PM
  #26  
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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
Old 11-07-2002 | 12:04 PM
  #27  
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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 ...
Old 11-07-2002 | 12:13 PM
  #28  
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Talking Nerd Talk

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? :D
Old 11-07-2002 | 12:34 PM
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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).

Last edited by pelucidor; 11-07-2002 at 12:41 PM.
Old 11-07-2002 | 12:46 PM
  #30  
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@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.
Old 11-07-2002 | 01:28 PM
  #31  
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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

Last edited by pelucidor; 11-07-2002 at 01:36 PM.
Old 11-07-2002 | 01:50 PM
  #32  
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British Aerospace eh?

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
Old 11-07-2002 | 02:52 PM
  #33  
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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.
Old 11-07-2002 | 09:15 PM
  #34  
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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.
Old 11-07-2002 | 10:59 PM
  #35  
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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.
Old 11-08-2002 | 03:15 PM
  #36  
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My turn

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.
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