Relationship between downforce and cornering speed (simplified)?

First, we need to clarify one assumption:

When you say:  "Lets say without any downforce, a 1000kg car is able to take a corner (radius 10m) at 10m/s. ", you need to keep in mind WHY the car is able to take the corner at that velocity.  You don't mention anything about a banked turn, so let's assume the roadway is flat.  Otherwise, the banking could be providing the centripetal force.

In an unbanked turn, the only force available to provide the necessary centripetal force is roadway friction.  This frictional force is proportional to the down-directed (normal) force Fn, but it is not equal to it.  That's where the coefficient of friction ? enters in.

If we ignore "downforce" (i.e., that caused by aerodynamic modifications -- ref. 1), then the only source of Fn is the weight of the car:

(1)  Fc  =  ? * m * g  =  ? * Fn  [Fc is centripetal force]

and the following must hold:

(2)  ? * Fn must be  >  m * v^2 / r.

If you add 5000N of normal force (downforce), then

(3)  Fn'  =  m * g  +  5000

and (2) becomes

(4)  ? * (Fn') must be  >  m * v^2 / r.

So you only get an additional (? * 5000 N) of centripetal force to play with.

If the turn is banked, then you need to consider also the horizontal component of the force normal to the bank.
See ref. 2 for more info on that situation.

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