Furniture
Well our countries are perhaps the most efficient, hard working or the cleanest.Originally Posted by LeonOfTheDead
But we're certainly passionate about our things and have pride and honor high in our list, amonsgt other things.
Other countries may be better objectively, but I can't help thinking that our lives wouldn't be better there.
Lack of charisma can be fatal.
Visca Catalunya!
Fanatic
Among passion, I would say a sort of characteristic "artistic" touch is something quite distinctive in what we do.
I would like, and I eventually I will, try to live somewhere else, not only as a comparative experiment, but also to gain some direct information about what makes latins different from say Americans or English people.
I mean, I know why Alfa is Italian and couldn't have been say Japanese or German, and I would like too know really why say Jaguar is English, beyond stereotypes and common beliefs.
KFL Racing Enterprises - Kicking your ass since 2008
*cough* http://theitalianjunkyard.blogspot.com/ *cough*
Enthusiast
I love the Ariel Atom and Caterhams.Just because it isn't cushy enough for you doesn't mean you need to hate it.
The first time I saw this car was at Geneva two years ago. My friend's girlfriend saw it and instantly proclaimed it the ugliest car she'd ever seen. Then we saw what it was called and we couldn't stop laughing. I don't care how aerodynamic the car is, I don't care how fast it is, I wouldn't be caught dead in one. And finally, I love all cars, but I'm a GT car guy.
"The Metric System is the tool of the Devil! My car gets 40 Rods to the Hogshead and that's the ways I likes it!" -Grandpa Simpson
Furniture
Fair enough.
Novice
Hi, first post. Sorry to necropost this topic, but I can't create new ones and didn't want to be completely off-topic somewhere else.
I've seen it time and again on car page after car page the claim that you can drive this car upside-down in a tunnel. I'm here to prove mathematically that this is rubbish. I recently posted this on the talk page for the Gumpert Apollo page on Wikipedia. I'm going on estimated figures, so if anyone has the proper figures I'd be glad to redo the math with them.
300 kilos of downforce more than the weight of the car means 300 kilos of normal force between tire and ceiling. Your typical supercar generates about 1.2 g in the average corner. F=ma, F=1100kg*(1.2*10m/s^2)=13,200N. If we assume that in a 3rd gear corner it generates more than 100kg of downforce (not an outrageous claim at all), and realize that tires on ceiling tile have much less grip than tires on asphalt, the tires are gonna generate less than 20% of that figure, probably quite a lot less, but being generous, let's say 3000N of grip. Aerodynamic drag is calculated as F=.5*p*v^2*Cd*A. p (mass density) in 20-degree air is about 1.2kg/m^3, Cd (coefficient of drag) for the Gumpert is supposedly about .43, v=220mph=98m/s, A (frontal area) is greater than 2m^2 (about 2 meters wide a 1 meter tall). F=.5*1.2*(98m/s)^2*.43*2=5000N of aerodynamic drag pushing it back.
edit: I just realized I didn't make it clear- the force of friction in the tires has to completely overcome the aerodynamic drag. If it can't, the tires will be unable to propel the car forward, meaning the speed will drop below the necessary threshold to overcome its own weight with aerodynamics.
Last edited by kickflipthecat; 02-22-2009 at 07:10 AM.
Fanatic
afaik, that claim regards the overall downforce generated at its top speed, or probably at speeds higher than 300 km/h.
I don't see why you should take into account the tires' friction on cornering (not a part of the claim) or the drag, which would be meaningless regardless of the car going upside down or not.
you get about 1500 kg of downforce at those speeds afaik, and 1100 of weight, bot acting, theoretically, to the CG, and those are the only forces acting in a vertical direction.
regardless of which is the drag value, it would be compared to the thrust provided by the engine, which are the two forces acting in an horizontal and longitudinal direction.
while cornering, an inertial force, the centrifuge one, would be generated, and again it would be applied to the CG. if the value of the grip of the tires is enough to balance it, the car would steer in the direction required by the driver, while if not, there would be oversteer.
you are right saying that the cornering capability of the car would be highly affected driving upside down, because there would be only 400 (or 300 using your figure) kg pushing the car to the surface it would be going on.
since the force the tires can generated is equal to their friction coefficient, multiplied by the force pushing the car normally to the surface, you would end with a much lower value of grip/force.
that would only affect the direction of the car, not its speed though (apart from generating more friction reducing the real speed, but we don't have data on that), so it would be still possible to run upside down but avoiding close corners.
KFL Racing Enterprises - Kicking your ass since 2008
*cough* http://theitalianjunkyard.blogspot.com/ *cough*
Furniture
Thing is, who's crazy enough to drive any car upside down in a tunnel?
Top Gear's actually right about that one detail.
Novice
The grip in the tires doesn't have anything to do with cornering, and I realize that my math is flawed. Here's what I'm basically saying- the amount of force the tires are capable of exerting on the car in any direction is 3000N, including forwards, when the car is driving upside-down at its top speed. If you look at a free-body diagram of the car going upside-down, you've got a downward force of 1100 kg = 10,780 N, a upwards aerodynamic force of 1500 kg = 14,700 N. Because the roof of the tunnel prevents the car from accelerating upwards, the downwards normal force is 3,920 N.
Now, look at the forward and backwards forces on the car. The normal force between tires and tunnel means a net forward frictional force that can be no greater than 3,000 N (this assumes all the grip is concentrated in the driven tires, which is not true, there's actually much less than 3,000 N). This is the only forward force acting on the car. The backwards force, caused by aerodynamics, is roughly 5,000 N. Because of the 2,000 N net force pushing the car backwards, the car will slow down, drop below 190 mph, and fall to the ground.
All the power of the engine HAS TO GO THROUGH THE TIRES! The reason a Veyron accelerates from 0-60 in 2.5 seconds is because 4WD allows it to put all of its torque on the road. Aerodynamic drag pushes the car directly backwards and the engine cannot overcome it if the tires aren't grippy enough to overcome the drag. This is why the SSC Aero got wheelspin at 190 mph when they first tried to break the record.
Last edited by kickflipthecat; 02-25-2009 at 04:43 AM.
Fanatic
I see your point now, still I disagree.
apart from the fact tires react in different ways if we are considering forces acting on them transversally or longitudinally, and even forward or backward, because of the design of their surface, we could say so:
-vertically, the car is pushed by 300/400 kg of force against the ceiling while driving upside down
-with 300/400 kg of force the tires' ability to put the torque to the ground is affected due to a low normal force to combine with the friction coefficient
-assuming the car is traveling at a certain speed, constantly, it means that longitudinally the car is in an equilibrium status, with a longitudinal net force equal to zero. it means that the aerodynamic force (A) generated by the air resistance is equal to the generated by the engine and transmitted to the ground by the tires (E), minus the loss (F) with the ground/surface.
now, if E-F<A, the car simply never reached that speed.
in order for the car to accelerate until that speed is that during all the time E-F>A and E-F=A when the car is going constantly at that speed.
now, I don't know if your value of A=5.000 N is correct, but let's assume it is, and let's assume it's at the top speed.
it would mean that, in standard conditions, at top speed, the complex of the engine-tires can push the car forward with a certain force less the friction loss, with the car being pushed to the ground by about 25.000 N of force.
now we don't know anything about E and F.
what we know about F is that's equal to the "friction coefficient" * "normal force against the road" = "friction coefficient" * 25.000 N
About E: at top speed, the car is likely going to use it's peak of power, about 800 hp (600 Kw, Sport version), at about 7.500 rpm (785 rad/s) or so. that means the engine is developing about 765 Nm of torque.
Considering the car has a actual sequential gearbox, and it's a performance oriented low-volume car, I would say the transmission loss is of about the 20%, maybe even less.
So at the wheels we have about 612 Nm, and with a 18" rim + tire (about 27 cm of radius) we obtain a force acting on the ground of about 2.300 N.
now, we have to subtract the value of the friction force generated by the ground, so the final force pushing the car forward is surely lesser than 2.300 N.
what we have here is A = E-F = 2.300 N - (X * 25.000 N)
as you can see A can't be 5.000 N, but also A < 2.300 N.
I don't have a value of X = "friction coefficient" though, but it can't be negative either.
with the car upside down, we have about 3.500 N pushing the car against the ceiling, at top speed.
the engine is generating the same power-->torque--> force to the ground.
the friction force is equal to "normal force pushing the car against the ceiling" * "friction coefficient" = 3.500 * X, assuming a similar coefficient (it could be both better or worst).
Now
- E = 2.300 N provided by the engine
- F = 3500 N * X of friction provided by the ground
- A < 2.300 N
with the same X, we have an about 7 times smaller F, therefore (E - F) is higher than A, and the car slightly accelerate until A becomes equal.
theoretically, the car could start spinning, so the force moving the car forward would decrease, and the car would regain traction.
still, the margins are very small, so it's an unstable system, and probably the smaller ceiling imperfection or something else. would result in the car falling down.
now, check my whole post, I did it from scratch without a book or notes, so it could be wrong somewhere.
About the fact of the Veyron accelerating so fast:
4wd allows to distribute traction trough the 4 wheels, instead of just 2, so it means each tire has to sustain half the force, or the car can put to the ground twice the force. theoretically.
Considering even the Apollo Sport, with 800 hp, can accelerate in 3 seconds from 0 to 60, it's evident the relationship isn't that correct.
an 4wd system has theoretically another 50% of additional transmission loss, besides it also add quite some weight, it also require bigger front wheels (theoretically even smaller rear ones, but since such cars are usually using mostly the rear ones, they have to be still quite large) which therefore have a bigger intertia. also, accelerating 4 wheels instead of 2 means you are facing 4 bodies with their own intertias and not just 2.
KFL Racing Enterprises - Kicking your ass since 2008
*cough* http://theitalianjunkyard.blogspot.com/ *cough*
Enthusiast
Gumpert Apollo Speed #2
Furniture
I think that color scheme may actually make it worse...
Lack of charisma can be fatal.
Visca Catalunya!
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