Do you guys know the cost of the tires on the new roadster? The veyron can do 250mph+ and those tires cost $40k. I was just curious if roadsters will be that much also.
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Tire price for 2020 roadster?
- Thread starter MaxPowers
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If you search this forum for the word “tires” you will find that the tires on the red Roadster prototype (the only one seen driving on public roads so far, I believe) are Michelin Pilot Sport Cup 2 tires, 325/30ZR21 rear and 295/35ZR20 front. You can search a site like tirerack.com to find current pricing. Far less than $40K. 
WaitingForY
Banned
If you search this forum for the word “tires” you will find that the tires on the red Roadster prototype (the only one seen driving on public roads so far, I believe) are Michelin Pilot Sport Cup 2 tires, 325/30ZR21 rear and 295/35ZR20 front. You can search a site like tirerack.com to find current pricing. Far less than $40K.
Yep, those are a lot cheaper.
No no, It's the tires for Bugatti Veyron that cost $40k. Maybe just the special tires if you want to go 250mph.
I remember reading somewhere that the tires were expensive because they had to have them at a certain exact pressure (something like 3 decimal places) psi to ensure safety at high speed.
Also, I ready that they were glued on or something similar, requiring special process, labor, blah blah blah
There's probably a sign in the service center that says "whatever the actual price of service is, increase by $39,500"
Also, I ready that they were glued on or something similar, requiring special process, labor, blah blah blah
There's probably a sign in the service center that says "whatever the actual price of service is, increase by $39,500"
Wow, so much misinformation I'm I going Forest Whitaker
Nothing is reading to 3 decimal places in a standard shop and air leaks naturally through the rubber much faster than that, they're aired up like any other tire.
They aren't glued on, they use the Michelien PAX system which is basically a fancy run flat process that has a special rim and inner ring. The tires themselves are Power sport 2 base with specific compounds. The Tires themselves are 17K and most of that comes from the rear sizes which are 365mm wide. Normal Super car tires in the 325-345 range will be over $1,000 each.
CarlK
Active Member
Yes and those tires will last about 10 minutes at top speed on the Veyron. Cup 2 on the Roadster prototype is much cheaper than that although it is not known whether you car drive at 250 mph with that.
The Veyron tires were just stupid, they went with a non-standard rim geometry requiring one of a kind equipment to mount and a unique size of Pilot Sport 2s. Their cost isn't representative of tires for any other modern hypercar, even ones that go 250mph. Bugatti went back to "normal" tires for the Chiron.
It was the special process of the rim that required the special tire on the bugatti. They're putting normal tires on the next car that discount tire could change if you wanted them to, and they're testing those tires at 260ish right now. Konnesegg set the 285mph record on their stock michelen power sport 2s. Although "stock" may still mean special model for that car.
The tires are experiencing over 3000 Gs at 250mph. That takes some special engineering to keep them from disintegrating. I do not expect the car to reach that top speed with its standard tires it's driving around now. But we'll see.
I just hope Tesla also realises that a sports car, super car, hyper car, halo car and whatever new category this is in, is not just about top speed and straight lines (like the Model S kinda is). It needs to corner as fast as electricity, to quote Jeremy Clarkson
I just hope Tesla also realises that a sports car, super car, hyper car, halo car and whatever new category this is in, is not just about top speed and straight lines (like the Model S kinda is). It needs to corner as fast as electricity, to quote Jeremy Clarkson
The tires are experiencing over 3000 Gs at 250mph. That takes some special engineering to keep them from disintegrating. I do not expect the car to reach that top speed with its standard tires it's driving around now. But we'll see.
I just hope Tesla also realises that a sports car, super car, hyper car, halo car and whatever new category this is in, is not just about top speed and straight lines (like the Model S kinda is). It needs to corner as fast as electricity, to quote Jeremy Clarkson
Total curiosity question:
f=mv^2/r or m=f/(v^2/r)
250 MPH = 112 m/s
Tire radius (325/30R21) is 0.3642
Force is 3,000G = 29,400N
How did you settle on 0.85kg for a tire mass?
Edit, it's /r not *r, numbers are right though
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Got this from several documentaries and car reporting websites. One of them quotes "Imagine the engineering that went into the tyres alone? Michelin had to develop a compound that can deal with 261mph any time the owner of the car wishes. At full speed the edge of the tyre is subject to 3,800 G's, but equally this car can be driven as easily as any grand tourer and so the tyres need to also be capable of displacing standing water, complying to road noise regulations and coping with the lateral loads of a car that is over twice as powerful as a Ferrari 488 GTB." But there are many more that quote the 3000+ number
It's a more complex integral calculation than this because the mass of the tyre is not all at a single point on the outside diameter. So you can't treat it like a point mass moving in a circular path.
Oh, i understand that, but the use of Gs as the unit means acceleration is being expressed, and on a quickly rotating tire in a straight line, the major acceleration of note is centripetal. The stresses in the sidewall and tread would be a force in Newtons, which would then require a mass to convert to Gravitys. So you would need a physical crosssection or such to take the mass of to convert those forces to Gs.
Running though the calculations with more coffee, I realized I forgot to cancel out mass when going from Newtons to Gs.
f=m*v^2/r (centripetal)
f=ma
so
ma=m*v^2/r
or
a=v^2/r
In other words, acceleration is independent of mass
v=112m/s (250MPH)
r=0.38m (325/30/R21)
a=34,295m/s^2
1G=9.8m/s^2
a=3,500 G
All makes sense now.
OK, now I'm really confused by this. I was assuming (with my limited knowledge) that they were talking about centrifugal forces, maybe instead of outwards only with the combination of the forwards directed rotational vector. But you're talking about centripedal force, which is a totally different direction.Oh, i understand that, but the use of Gs as the unit means acceleration is being expressed, and on a quickly rotating tire in a straight line, the major acceleration of note is centripetal. The stresses in the sidewall and tread would be a force in Newtons, which would then require a mass to convert to Gravitys. So you would need a physical crosssection or such to take the mass of to convert those forces to Gs.
Running though the calculations with more coffee, I realized I forgot to cancel out mass when going from Newtons to Gs.
f=m*v^2/r (centripetal)
f=ma
so
ma=m*v^2/r
or
a=v^2/r
In other words, acceleration is independent of mass
v=112m/s (250MPH)
r=0.38m (325/30/R21)
a=34,295m/s^2
1G=9.8m/s^2
a=3,500 G
All makes sense now.
They are linked things.
Centripetal is the force/ acceleration pulling toward the center (and amount of force the tire needs to exert to stay intact). Centrifugal is a perceived force you would feel being inside the tire against the tread, it is based on your inertia being reacted against by the tread wall.
So riding in a car though a curve, centripetal force is need to curve, but centrifugal is what you feel as you are pushed against the side of the car (you body tries going straight).
Centrifugal force - Wikipedia
Centripetal force - Wikipedia
Mostly:
Top fuel dragsters rely on ballooning to provide extra effective gearing.
Going oval is bad for handling though...
