Aston Martin wants hyper-car to be faster than F1 cars

[Cold Fussion]

So even the amr pro version is still 100 km/h slower through copse than an F1 car.

[Andres125sx]

Nice, great video

So even the amr pro version is still 100 km/h slower through copse than an F1 car.

He says 100km/h, but also that they were doing 200 when one of the laps they never go below 207 (just watched it at 0.25 speed to not miss any number in the speedometer), and also that in F1 they´re doing 280-290, so 100km/h was just a round number, real difference even if we take 290 for F1, was 83.

Also, Nico said going with a passenger makes a huge difference, and also that they had some understeer due to a not perfect aero balance wich should be adjusted. So the car can go faster through copse, no idea how much but real difference could be around 70 km/h. Still a huge difference, but not that high


I´ve just performed a search to compare with different categories, and found this 6h of Silverstone, with the Audi R18 doing at the only clean lap I found (4:18), exactly 200. Does this mean the AMR pro is faster through fast corners than LMP1 cars?



Maybe it was just a slow corner, if someone find some more evidences in any direction, please share

[Cold Fussion]

6th gear entry and exit for an lmp2 car in 2014. The speed is very hard to make out but I would guess around ~220km/h through copse for this lmp2 car at least.

[Mchamilton]

Random comparison, James rossiter did a 48.1 on a slightly damp cold day in an Honda RA107, minimum speed through copse was 111mph. Nico was doing like 47.3s in the AMR pro but apexed copse at 130mph ish.

[Cold Fussion]

So from this video, the Valkyrie and AMR Pro do not share a chassis, and the AMR Pro is 250kg lighter (so essentially hitting the original 1000kg target). 250kg seems like a lot to remove from the Valkyrie, while seemingly being physically larger. I wonder if this basically confirms the AMR Pro is using a standard LMP2 chassis as was rumored here previous.

[Holm86]

https://www.youtube.com/watch?v=iNQPNtMYITk

So from this video, the Valkyrie and AMR Pro do not share a chassis, and the AMR Pro is 250kg lighter (so essentially hitting the original 1000kg target). 250kg seems like a lot to remove from the Valkyrie, while seemingly being physically larger. I wonder if this basically confirms the AMR Pro is using a standard LMP2 chassis as was rumored here previous.

Just watched Hagerty's video on it, they say it's "only" 1270kg, that's quite a lot more that the initial target, and didn't they say something last year about them hitting a weight of 1100kg? Where does all that extra weight come from? Sound insulation?? Or have they had to redo the chassis design to live up to crash regulations??

[Fluido]

The other problem is that a road car needs to produce MORE downforce than an F1 car to equal it's cornering performance because:

1. It weighs more
2. It's tyres have a lower coefficient of friction

A 700kg F1 car needs 2100kgf of cornering force to make 3G. A 1500kg road car needs 4500kgf. Essentially you need a downforce to weight ratio of at least 1.5 at cornering speeds. The mythical McLaren F1 has a downforce to weight ratio of around 0.4... So they need to be at least three times better than that.

I did a calc earlier today that a 1000kg car needs a Cz of around 8.0 to hit 3G. An F1 car has a Cz of around 2.0-3.0.

A fan car would do the job but the problem is always going to be the tyres. Nothing exists even close to what is required.

Why heavier car need more downforce for same cornering force?
if you increase car mass by 50%, friction force(centripetal force) increase by 50% and centrifugal force rise by 50%.
Ff=mgx friction coeff.
Fc=mxV^2/r

Can you explain that?

[gruntguru]

Why heavier car need more downforce for same cornering force?
if you increase car mass by 50%, friction force(centripetal force) increase by 50% and centrifugal force rise by 50%.
Ff=mgx friction coeff.
Fc=mxV^2/r

If the car already has DF = X x m x g:

Normal force = (X+1) x m x g
Centripetal force = m x a (a = centripetal acceleration)
Friction coefficient required = centripetal force/normal force = (m x a)/((X+1) x m x g) = a/((X+1) x g)

Note this is independent of mass. So to achieve the same centripetal acceleration with an increase in mass, the downforce must be a fixed percentage of m ie if you double mass you must also double DF.

[Fluido]

Why heavier car need more downforce for same cornering force?
if you increase car mass by 50%, friction force(centripetal force) increase by 50% and centrifugal force rise by 50%.
Ff=mgx friction coeff.
Fc=mxV^2/r

If the car already has DF = X x m x g:

Normal force = (X+1) x m x g
Centripetal force = m x a (a = centripetal acceleration)
Friction coefficient required = centripetal force/normal force = (m x a)/((X+1) x m x g) = a/((X+1) x g)

Note this is independent of mass. So to achieve the same centripetal acceleration with an increase in mass, the downforce must be a fixed percentage of m ie if you double mass you must also double DF.

Df=downforce
Ff=Fc
mgμ + Dfμ = Fc

3+2=5.......only mass double.....................3x2 +2= 5x2......................8≠10..... Ff < Fc
3+2=5.......double mass and downforce....... 3x2 + 2x2=5x2......10=10 ..... FF = Fc

Yes I must multiply both numbers on left side with 2...

[Andres125sx]

The other problem is that a road car needs to produce MORE downforce than an F1 car to equal it's cornering performance because:

1. It weighs more
2. It's tyres have a lower coefficient of friction

A 700kg F1 car needs 2100kgf of cornering force to make 3G. A 1500kg road car needs 4500kgf. Essentially you need a downforce to weight ratio of at least 1.5 at cornering speeds. The mythical McLaren F1 has a downforce to weight ratio of around 0.4... So they need to be at least three times better than that.

I did a calc earlier today that a 1000kg car needs a Cz of around 8.0 to hit 3G. An F1 car has a Cz of around 2.0-3.0.

A fan car would do the job but the problem is always going to be the tyres. Nothing exists even close to what is required.

Why heavier car need more downforce for same cornering force?
if you increase car mass by 50%, friction force(centripetal force) increase by 50% and centrifugal force rise by 50%.
Ff=mgx friction coeff.
Fc=mxV^2/r

Can you explain that?

I´m construction engineer so this surpass my knownledge, but I understand from your reply an increase in mass should not reduce cornering speed, since centrifugal force and friction force increase proportionally. Did I get it right?

Then what´s the reason lighter cars are faster at corners?

[Fluido]

I´m construction engineer so this surpass my knownledge, but I understand from your reply an increase in mass should not reduce cornering speed, since centrifugal force and friction force increase proportionally. Did I get it right?

Then what´s the reason lighter cars are faster at corners?

IF car produce zero dowforce or lift,than yes forces increase proportionally, so no reduced cornering speed.
But if car produce non zero downforce than, heavier car has reduced cornering speed.
Any car must produce non zero aerodynamic force, because car is not symetrical object.

Indeed this is basic mathematics as I explain in post above, you can see if you only increase mass in equation, that friction force become smaller than centrifugal force, friction force is centripetal force, force that keep car moving in circle.

[Andres125sx]

I´m construction engineer so this surpass my knownledge, but I understand from your reply an increase in mass should not reduce cornering speed, since centrifugal force and friction force increase proportionally. Did I get it right?

Then what´s the reason lighter cars are faster at corners?

IF car produce zero dowforce or lift,than yes forces increase proportionally, so no reduced cornering speed.
But if car produce non zero downforce than, heavier car has reduced cornering speed.
Any car must produce non zero aerodynamic force, because car is not symetrical object.

Indeed this is basic mathematics as I explain in post above, you can see if you only increase mass in equation, that friction force become smaller than centrifugal force, friction force is centripetal force, force that keep car moving in circle.

That´s exactly the reason for my question, as it looks contradictory with real life where lighter vehicles are faster at corners

Do you mean on a vacuum environement that would be true, but aerodynamics (even on production cars) causing some lift are the cause for this?

[Fluido]

I´m construction engineer so this surpass my knownledge, but I understand from your reply an increase in mass should not reduce cornering speed, since centrifugal force and friction force increase proportionally. Did I get it right?

Then what´s the reason lighter cars are faster at corners?

IF car produce zero dowforce or lift,than yes forces increase proportionally, so no reduced cornering speed.
But if car produce non zero downforce than, heavier car has reduced cornering speed.
Any car must produce non zero aerodynamic force, because car is not symetrical object.

Indeed this is basic mathematics as I explain in post above, you can see if you only increase mass in equation, that friction force become smaller than centrifugal force, friction force is centripetal force, force that keep car moving in circle.

That´s exactly the reason for my question, as it looks contradictory with real life where lighter vehicles are faster at corners

Do you mean on a vacuum environement that would be true, but aerodynamics (even on production cars) causing some lift are the cause for this?

yes in vacuum heavy and light car have same max cornering speed

[Cold Fussion]

That´s exactly the reason for my question, as it looks contradictory with real life where lighter vehicles are faster at corners

Do you mean on a vacuum environement that would be true, but aerodynamics (even on production cars) causing some lift are the cause for this?

Newtonian friction is F=m*μ (assuming no downforce and level ground) which is all well and good, but you also need to consider the tyre friction coefficient is a function of the normal force and a whole bunch of other variables (of which i have zero knowledge), and this relationship is not necessarily linear with normal force. The Newtonian model of friction is ultimately way too simplified for tyre friction physics.