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G-force Question


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#1 ilious

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Posted 25 February 2003 - 03:06

Every circuit has corners rated at different g's

For example 2g or 3g corner

Please give an explanation for this difference of rates.

What is the reason each corner has a different g rate?

Where do the corners depend on, in order to have different g rates?

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#2 DEVO

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Posted 25 February 2003 - 03:46

my best guess is that it's the maximum amount of lateral forces sustained before breaking the cohesion of friction of the tires on the tarmac of an F1 car.... in other words measure the g-force within an F1 car while the car is going through the corner as fast as it can.

#3 furious_camel

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Posted 25 February 2003 - 06:56

Originally posted by ilious
Every circuit has corners rated at different g's

For example 2g or 3g corner

Please give an explanation for this difference of rates.

What is the reason each corner has a different g rate?

Where do the corners depend on, in order to have different g rates?


When cornering, you have to change the direction of your velocity-vector. This can be seen as accelerating in a horizontal plane perpendicular to your current velocity. Depending on how 'fast' you have to change directions, this will result in different accelerations. By dividing the required accelaration by 'g', you get expressions like '2g', 2.4g' etc...

I'm sorry for eventual faults in my English, but the explantion should be correct in technical terms.


Sincerely,

Jurn De Cleyn
student 4th year civil engineer - option chemical engineering techniques

#4 Froilan_G

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Posted 25 February 2003 - 07:06

For a car to corner optimally, you'd think that in every corner the G-forces would remain the same right? As this is dependent on the tyre coefficient of friction, and you definitely want to be on the peak in every corner.

However, when aerodynamic downloads come into play, then the max. G-force might be bigger in a corner, because then the tyres can take more.

La Source at Spa and Blanchimont at Spa would have different max. G-forces because Blanchimont has a bigger corner radius and is therefore way quicker. I would however, see no reason why the La Source hairpin and Loews Hairpin at Monaco would differ much in max. G-forces.

#5 DOHC

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Posted 25 February 2003 - 10:54

Originally posted by ilious
Every circuit has corners rated at different g's

For example 2g or 3g corner

Please give an explanation for this difference of rates.

What is the reason each corner has a different g rate?

Where do the corners depend on, in order to have different g rates?


In a fast corner the car generates more downforce.

A simple model for the g-force of the corner, if you're on the limit, is

mu*(m*g + d)/m,

where

mu = friction coefficient
m = car's mass
g = gravity acceleration (9.81 m/s^2)
d = downforce

If you'd have no downforce, d=0 and the g-force of the corner becomes mu*g.

But in a simple model, downforce is proportional to the square of the speed, d=k*v^2 (with k being a constant), so the g-force rating becomes

mu*(m*g + k*v^2)/m.

In a high speed corner then, v is large, so the g-force rating of that corner is much higher than for a slow corner with v small.

At high speed, the downforce d can be twice or three times as large as m*g. Let's say d=3*m*g. The g-force rating is then

mu*(m*g + 3*m*g)/m = 4*mu*g,

and with a friction coefficient of mu=1 you get a 4*g rating.

#6 DOHC

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Posted 25 February 2003 - 11:25

As an addendum...

Given that you travel at speed v, the expression mu*(m*g + k*v^2)/m is the highest lateral g-force you can sustain at that speed.

Also, the constant k can be viewed as depending on the aero setup. WIth a high downforce setup, you have a bigger value for k, with a low downforce setup, k has a smaller value.

#7 Wuzak

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Posted 26 February 2003 - 11:25

That explains how much side grip is available to the tyres given a certain speed, but does not show why there are different speeds through corners.

The lateral force on a car going at velocity V around a corner of radius R is given by:

F = m*V^2/R - the centrifugal force on the car.

This is balanced by the sideways friction force of F = mu * Force Normal

or F = k*V^2 + m*g (as has been shown before).

If the centrifugal force exceeds the friction force, the car will not be able to maintain its trajectory.

The friction force, and downforce equations are simplifications.

#8 DOHC

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Posted 26 February 2003 - 20:22

Right, that's a good point.

We can also add a comment on "flat-out" corners. Flat-out doesn't necessarily mean that you go at top speed, only that you don't have to lift or brake to make the corner.

A corner of radius r is "flat out" at speed v in this simple model if the centrifugal force m*v^2/r is smaller than the available lateral grip force mu*(m*g + k*v^2). This gives the condition

v^2/r < mu*g + mu*k*v^2/m.

Because we at least have the grip generated by downforce available, the corner is always flat out if we drop the mu*g term above. This gives a simplified sufficient condition,

v^2/r < mu*k*v^2/m,

which is the same as

r > m/(mu*k).

So if the radius of the corner is greater than car mass divided by friction coefficient mu times downforce constant k, the corner is flat-out.

Maybe not a very important comment, but anyway.