# Does heavier car need more aerodynamic downforce in turn than lighter car for same cornering speed?

When a car turns, the frictional force $$F_f = (mg\mu + A\mu)$$, where A is aerodynamic downforce and $$\mu$$ the friction coefficient, needs to match the centrifugal force $$F_c = m \times V^2/r$$. If the car's mass increases by 20%, both the frictional and centrifugal forces will increase by the same percentage. Does heavier car need more aerodynamic downforce in turn than lighter car for same cornering speed?

(Assumption is that car produce downforce,I know that car can make a turn without any downforce.)

• In general, a car does not need aerodynamic forces to turn. Commented Nov 2, 2023 at 12:52
• After checking the comment of @JAlex, I realised that I had misunderstood the question, so I delete my answer.
– Alex
Commented Nov 2, 2023 at 13:28
• @JAlex in general, all cars produce non zero aerodynamic force, lift or downforce...car is not symetric object Commented Nov 2, 2023 at 16:45
• @user628075 - yes, but they do not need downforce or lift to turn. Mechanical grip suffices. Commented Nov 8, 2023 at 14:52

Yes, a heavier car would typically need more downforce than a lighter car to make a turn at the same speed.

Note that the centrifugal force $$F_c$$ is proportional to its mass and an increase in the car's mass leads to a corresponding increase in $$F_c$$. To avoid skidding, the frictional force $$F_f$$ must compensate accordingly. If only the car's weight increases and the downforce remains constant, the total frictional force might be insufficient. Consequently, for the heavier car to turn at the same speed as before, an increase in the aerodynamic downforce $$A$$ is required.

• You say No, but the answer is actually Yes you need more aero force to contribute the same centripetal acceleration. Commented Nov 2, 2023 at 12:54
• @Alex I edit my question in text because it is opposite from what title ask(just confuse)...so you just change No into Yes and your answer is correct. Commented Nov 2, 2023 at 22:03
• @user628075 done it!
– Alex
Commented Nov 2, 2023 at 22:15

Yes.

Given a turn of fixed radius $$r$$, and a required corner speed of $$V$$, such that the centripetal acceleration is $$a_c = \frac{V^2}{r}$$, you can back solve for any additional downforce $$A$$ needed to sustain this turn, if the mechanical grip is $$\mu$$, from $$a=F/m$$

$$\frac{V^2}{r} = \frac{ \mu ( m g + A) }{m}$$

with solution

$$A = m \left( \frac{V^2}{\mu \,r} - g \right)$$

which indeed shows the proportionality with mass $$m$$, given all else remains the same.

• If car has more weight at rear axle that mean we must produce more downforce at rear axle? Commented Nov 6, 2023 at 15:17
• Definitely yes, Commented Nov 8, 2023 at 14:51