Front tyres locked, otherwise he would have spun. He clearly didn't know what the hell he was doing - to lock the tyres for that distance and still not figure out what was happening. No damn business driving a Miata, let alone any type of Ferrari, especially with no ABS.
i'm thinking by that second pic at the 75m mark his anus would have had more foot pounds of torque than the testarossa.
That's a very good first approximation, but by assuming m is the same on both sides of the equation you have implicitly assumed either 100% weight transfer to front tires and they are skidding, or that all 4 tires locked and are all skidding so can ignore weight transfer effects. Since only the front tires locked, a better approximation would have been to assign mass and CF values for fronts and rears. Assuming typical brake biase of 60% front 40% rear and .... Ahh... memories of freshman physics.
m is the same on both sides by definition, there is only 1 car, it has one mass which is supplying the normal force for the friction and the momentum/kinetic energy, so it drops out. Your point about weight bias really isnt valid in that the total mass still has to represented regardless of which axle it ends up acting on. What is valid is that the F front and rear are probably need slightly different math since the fronts where locked and the rears where not .that complicates it quite a bit. The front math is then as I wrote, just NCf. The rear is not locked due to brake bias differences, so Cf is not the limiting factor .but because the wheels were still turning, the Cf is closer to 1.0 than 0.8, so that probably cancels out the bias setting difference and then some. I dont think there is enough data to get much closer, maybe alter the Cf value if there is some test data available, that is really the biggest guess I mean assumption, I made, but I think its close.
It should be easy enough for someone to do 130 kmh in their TR and stomp the brakes. At 85 mph I would bet that it takes MUCH less distance to slow to 45 kmh (about 30 mph) than 109 meters. BT
No, m is not the same on both sides of the equation. True there is only one car with total mass m, but total weight (m x g) is not acting on one tire, which is what you have done in treating m as a point mass. It is distributed on 4 tires. So the weight transfer towards the front during braking will directly affect the frictional force generated by the front tires versus the unloaded rear tires. Cf will also be different on skidding front tires versus rolling rear tires. It is not my intention to argue with you nor nit pick your equation. As I said in my previous post, your estimate is a very good first approximation. For a slightly more accurate estimate, I suggested taking weight transfer into consideration, but even then we are still talking about using statics equations here. For accurate computer simulation, we will need to write the full dynamics equations, and take into account not only linear but rotational inertia, damping, etc.. Now that's complicated and will require a lot of testing to determine the simulation model parameters. Back in the late eighties I spent two years modelling the dynamics behviour of helicopters on rolling ship decks. We started with a statics simulation model, which we ultimately proved to be completely wrong once we looked at the dynamic behaviours.
pv908----there are some VERY interesting thoughts here, about 90% are so far out in left field--an accurate speed would never be figured out. If you want me to work the formulas, send me a PM.
I understand your point, it's just not relevant unless I used a different Cf front/rear. As long as Cf is assumed constant, the weight could be all on 1 tire or distributed in any way among the 4, it all works out the same. The assumption I made was the fronts are locked and Cf for the sliding tire is 0.8. I also assumed that although Cf in the rear is higher because the tires are not lock. The fact that they are not locked says they have probably not reached their limit either, and 0.8 would be a fair approximation or what the effective Cf is .and make my math way, way easier
as a matter of relativity.... international soccer pitches and american football fields are 109m long. the Menkaure pyramid (smallest of the 3 pyramids of Giza) is 109 metres across on each side. the SS Yongala which sank in 1911 was 109 metres long. the world's highest swing (in nz) is 109 metres off the ground. during the swing you get to 150kph. 109 metres is about 30-35 storeys up in an office/condo tower.
Makes you wonder how a car can skid the entire length of a soccer pitch! That Testrossa must have been moving at a f*cken high rate of knots alright! Almost seems unbelievable how the driver was doing in the vincinty of 200km/h(would have to be close to that), then locked the wheels up and skidded perfectly straight, without deviating for 109m!!!! How long was this street again?
Sorry guys, I'm I missing something. I don't see any signs of a wreck on the road. It looks more like a "burn out" to me.
You must have missed post 8 where he says the Testarossa hit a stationary trailer double parked. But it seems that the wreck/s are gone. The Police probably would have marked it afterwards.
The car is not wrecked, luckily. Parts bill currently around $39K AUD with some more to go. The bonnet is 10 grand alone. Not sure of finished quote. Insurance is taking care of that. The road is probably about 900 meters long after coming out of some nice curves. Could do 100 Kmh around the curves if pushed. The road was marked out by the Police 24 hrs after the accident for some reason. It has been off the road for 7 weeks+ and the parts are still not in the country yet. This was a very sad engine clean after 2 years of getting the car right.
Farrrrrrrkkkkkk! $39K and insurance is covering it after the Police report says there was a 109m long skid mark? I find that very hard to believe mate. Insurance company's really delve into this sort of stuff, especially since there is a Ferrari involved and the quote so far costs nearly as much as Ford XR6. Whats the speed limit on that street. 50km/h or 60km/h.? What the f*ck was your mate doing.......trying to break a record for speeding down that particular street?
cosmetic only then? 45kph can pack some wallop, but if it was a jims mowing trailer (mate of mine is a mr clip ) then that might have absorbed more of the inertia than your car did. did it push the trailer far from it's original position? how was 45kph derived as the terminal velocity?
insurance most likely meaning they'll chase the money from the playful party if the owner's black-and-white not at fault. as goober says "you fook it you fix it". never drive faster than you can afford to. (try not to anyhoo )
The only problem i see here is.....that the insurance company looks at the playful party and see's that they cannot come to the party, then owner has to come to the party. Then..........with massive skid marks like that, maybe the insurance company may not come to the party for the owner of the Trossa.
