# How many Gs would a driver of the Bugatti Veyron experience on the Ehra-Lessien track when cornering before record attempt?

Apologies if too specific.

Watched a documentary National Geographic Megafactories Bugatti Veyron

I told an colleague (engineer) about the need to warm the car up and then unleash it on the long stretch to be able to make it to 405km/h and mused that the curve before the straight probably had to be taken at 250km/h to do so. He immediately replied that the G-force of going around a(ny) curve at 250km/h would kill the driver... I find that very hard to believe but do not have the skills nor knowledge to refute it. It would be cool to get him this app http://www.dynolicious.com/ and a trip in a Veyron ;)

Here is the Ehra-Lessien track and here is the actual curve where some of the clever people here could glean the radius - I think 350m

Can someone show some math and a graph or so?

I have been looking at http://en.wikipedia.org/wiki/Formula_One_car#Lateral_force

Turn 8 at the Istanbul Park circuit, a 190° relatively tight 4-apex corner, in which the cars maintain speeds between 265 and 285 km/h (165 and 177 mph) (in 2006) and experience between 4.5g and 5.5g for 7 seconds—the longest sustained hard cornering in Formula 1.

which supports my claim - but would love some more information on this specific matter

Here is a link on what an Apex means in motoring
And here are some formulas

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Neglecting friction, the force experienced is the Centrifugal Force $F=\frac{mv^2}{r}$ (it would be less if you included friction since the car actually slips) vectorially added to the orthogonal gravitational force $F_g=mg$, i.e. $F = m\sqrt{\left(\frac{v^2}r\right)^2 + g^2}$ where $g = 9.81 \frac{m}{s^2}$1. Divide this by $F_g$ to obtain a result in Gs. Remember to use SI-units, i.e. divide a km/h speed by 3.6 (1000 km/m / 3600 s/h) to obtain m/s.