2014 too slow? (or not, as the case may be)

[Artur Craft]

Of note, my original comparison was not with a 2003 car, or a 2008 car, it was was with a 1990 car. The idea that a modern car can set very similar lap times to a 1990 car with significantly less power and skinnier tyres, but at the same time not have more downforce is frankly moronic. It's clear simply from the fact that lap times are basically the same, but straight line speeds are lower that the car must be faster in the corners, and therefore must be generating more downforce.

As far as 2003 goes, I'd bet on a 2013 car having more downforce than a 2003 one, but it would probably be pretty damn close. The reason I'd bet on this is simple – in 2003 radilon and blanchimont were corners that would regularly chuck drivers into the barriers because they did not have sufficient downforce to make the corner. That is not true any more, despite speeds being roughly the same. I suspect in 2014 that we'll be back to 2003 style lifting through there.

That's the problem with your reasoning. Your understading of how this work is incredibly poor and naive and would make me think you're a layman, albeit I know you're not.

What makes a car quicker in a corner is grip, and only grip. Downforce increase grip but it's not the only thing. Tyre compounds produces huge differences in grip and I see you obviously are not much aware of the evolution of tyre compounds over the years.

If you look into Senna qualifying lap from Suzuka in 1991(it was a P2 time that was 0,4s slower than team mate Berger on sector 1, so the limits of the car were even higher than those I'm gonna mention), the speed on the Esses are a match for Suzuka's 2012 qualifying laps of Alonso and Vettel(directly from speedmeter). At the hairpin(sector 2, just 2 back then, Senna was 0,2s faster than Berger so likely at the limit here), and despite wider qualifying tyres, Senna corners with just 85% of the Red Bull's speed.(Senna's min speed is 66kmh, Vettel's 71. During the cornering Senna's average speed is around 70kmh, while Vettel's over 80kmh)

So, the tyres give recent cars more grip, not the downforce. And the tyres compensates and even surpass the downforce deficit in many cases.

Plus, another wrong issue that you posted is regarding early 90's cars's powers. In 1990, the engine figures were this:
Renault : 660HP
Ferrari : 680HP
Honda : 690HP

And because they had fat tyres, it gave a lot more drag which implies that their's straight line speeds were lower, not greater, contrary to what you said

The "radilon and blanchimont" mention one is also not a good argument. Nobody needed to lift in any of these corners, or Eau Rouge.

Martin Brundle said on ITV, during Suzuka 2003 qualifying, that Allan Mcnich(Toyota's test driver back then)told him Toyota was achieving 5Gs on the Esses(using grooved tyres)

At the same corners, 2011-2013 cars reach no more than 3.9Gs

You'd lose. P. Symonds in Amus interview said that current era cars (-2013) downforce levels exceed the ones from 2008 and Marussia has more downforce than 2003 Renault and would be winning championships back then (he saw some numbers too ). Sentiment to "good old times" is one thing - technical development reality - something different.

http://translate.google.com/translate?h ... 36558.html

Firstly, thanks very much for the link

I think it's pretty likely the 2013 cars had more downforce than 2008s. The diffusers from either year are very small compared to the diffusers of 95-2004, and even more to the 1994 one. Ride height also is the same so from 95 onwards, nobody gains time on that. In 2005-2008, the RW had more width but were run shallower because the FW gave relative poor downforce being so high. Despite the 2009-2013 RWs being narrower, they have bigger span than the 1995-2008 ones.

Also, with the exhaust blown diffuser, you can run more rake and seal the side of diffuser, effectively increasing it's size a bit. Diffuser-wise, 2011-2013(EBD generation)cars had advantage over 2008. The RW is 0,75m wide versus the previous 1m, but they didn't lose 75% of RW downforce because it's run higher now and with bigger span(taking some surface area back). FW-wise, 2011-2013 cars had way more downforce than 2008 ones. Despite the neutral central zone, it was 1,8m wide, low(which improves downforce by a very big margin. For instance, depending on the case, a 10mm lower ride height can improve downforce by upwards of 50%) and with very steep angles.

When they started the 2009 season, they were around 85% of the 2008 downforce levels. With development of 5 years and the implement of devices such as EBD, it's no surprise they surpass the 2008 downforce values.

The rake and EBD might take them equal to 2003 diffuser levels and as FW gave more df last year, I believe Red Bull, Mercedes could surpass Renault 2003. Marussia seems a bit of a stretch and Symmonds have done such things before.

Did you realize he doens't mention the 1993 cars, only the 2003 and 1983 ones?

I doubt that. Much of the performance deficit is masked behind the grooved tyres.

I think many people don't make the difference between more downforce and more advanced downforce. The way the RB9 creates downforce is of course much more complex and technology behind it a lot more advanced, but in 1998 they had more means for "mass-downforce creation": bigger diffusers, bigger bargeboards, bigger rear wings, lower front wings. All in all they also were much more efficient.

Yes, you understand it.

Wind tunnel and CFD technology advanced quite a lot and as a consequence, aerodynamicists can develop their car's shapes much quicker and precisely.

There is no doubt that they "squeeze" way more downforce out the regulations now than they did back in 94. But because there was much more freedom back then, they still had more downforce.

With the values of a contemporary tyre that I have seen and the downforce of a 94 car, they would be able to corner at 7G. And if you gave 1994 aero regulations to Adrian Newey, for instance, today, he could improve that downforce considerably and we could see potentianly +8G

PS: Of course, this is all saying the max force cars would theoretically get with that combination. In reality, they couldn't reach that because at 6G, the F1 cars are already not far from rolling over/flipping

[Artur Craft]

If you look into Senna qualifying lap from Suzuka in 1991(it was a P2 time that was 0,4s slower than team mate Berger on sector 1, so the limits of the car were even higher than those I'm gonna mention), the speed on the Esses are a match for Suzuka's 2012 qualifying laps of Alonso and Vettel(directly from speedmeter). At the hairpin(sector 2, just 2 back then, Senna was 0,2s faster than Berger so likely at the limit here), and despite wider qualifying tyres, Senna corners with just 85% of the Red Bull's speed.(Senna's min speed is 66kmh, Vettel's 71. During the cornering Senna's average speed is around 70kmh, while Vettel's over 80kmh)

So, the tyres give recent cars more grip, not the downforce. And the tyres compensates and even surpass the downforce deficit in many cases.

Plus, another wrong issue that you posted is regarding early 90's cars's powers. In 1990, the engine figures were this:
Renault : 660HP
Ferrari : 680HP
Honda : 690HP

And because they had fat tyres, it gave a lot more drag which implies that their's straight line speeds were lower, not greater, contrary to what you said

Just to bring some screenshots regarding this: (Senna's qualifying lap which gave him P2 versus Alonso's 2012 Q3 lap versus Vettel's 2012 pole)

On the Esses:


On Hairpin:


Top Speed, firstly before Spoon and then on back straight before 130R(no speedmeter on front straight on Senna's video):


Once again, sorry for taking this thread to this. I intend to stop with it in this post

On topic, 2014 cars will be some 15% slower on corners(just a guess to be confirmed on Melbourne) but their torque, less drag and supposedly more total power will make them not much bellow in laptime. At least not on tracks with long straights and not many high speed demanding corners.

What I wonder and ask you guys is how much do you expect the fixed gear ratio to influence on performance? Will we see cars performing relatively better in some circuits than on anothers due to it as well?

[acosmichippo]

fixed gears, but they also have an additional gear to play with.

[Juzh]

Also, with the exhaust blown diffuser, you can run more rake and seal the side of diffuser, effectively increasing it's size a bit. Diffuser-wise, 2011-2013(EBD generation)cars had advantage over 2008.

2010 semi-blown double deckers had an advantage over all of these versions. RB6 was the only car ever to take turn 8 in turkey and turn 9 in barcelona flat. Copse was also flat but some cars managed to so in 2008 as well (RB4 was one of them). They did this on a much harder bridgestones compared to pirelli.
RB6 also owns RB7,8 and 9 in all DF heavy stuff. It was 0,4s faster in S1 in suzuka than RB9 despite track being washed away on saturday by a hurricane and then quali being held on sunday on a green, drying track. Difference is even more than that compared to RB7 and 8.
Example of DF advantage on RB6 compared to overall fastest car of 2012, mp4-27:
[youtube]http://www.youtube.com/watch?v=aywDRf_Dm5A[/youtube]
Hamilton is clearly much faster on the straights, yet looses all advantage in 1 corner. 1!
2013 is much better in that regard, but still long way off.

There is no doubt that they "squeeze" way more downforce out the regulations now than they did back in 94. But because there was much more freedom back then, they still had more downforce.

They had even more leeway in 84 and 74 so they must had more DF in those years, right?

PS: Of course, this is all saying the max force cars would theoretically get with that combination. In reality, they couldn't reach that because at 6G, the F1 cars are already not far from rolling over/flipping

Why would it flip over? I don't understand this. Center of gravity is too low to roll on a flat surface imo.

[turbof1]

They had even more leeway in 84 and 74 so they must had more DF in those years, right?

They made more potentional. They issue is that back then they didn't had proper tools to get to micro aerodynamics. They might have heard for instance about vortices back then, but they for sure couldn't have known that those actually have benefits. Something like a boundary layer was very poorly understood.

Tighter regulations drove aero technology. They are now much higher up the S curve then back then.

[Juzh]

They had even more leeway in 84 and 74 so they must had more DF in those years, right?

They made more potentional. They issue is that back then they didn't had proper tools to get to micro aerodynamics. They might have heard for instance about vortices back then, but they for sure couldn't have known that those actually have benefits. Something like a boundary layer was very poorly understood.

Tighter regulations drove aero technology. They are now much higher up the S curve then back then.

Of course, I was being sarcastic

[beelsebob]

They had even more leeway in 84 and 74 so they must had more DF in those years, right?

They made more potentional. They issue is that back then they didn't had proper tools to get to micro aerodynamics. They might have heard for instance about vortices back then, but they for sure couldn't have known that those actually have benefits. Something like a boundary layer was very poorly understood.

Tighter regulations drove aero technology. They are now much higher up the S curve then back then.

They didn't in 94 really either. 94 was still the era of flat, or one dimensionally curved aluminium surfaces for the most part. There were no complex CF shapes to be had really.

[Artur Craft]

They had even more leeway in 84 and 74 so they must had more DF in those years, right?

Why would it flip over? I don't understand this. Center of gravity is too low to roll on a flat surface imo.

I'm not aware of much difference from 84 to 94. The flat bottom rule and ban of skirts was introduced very early in the 80's.

I saw a tv segment with Ross Brawn in 86 and he said cars back then were not that far from the ground effect ones downforce-wise. Williams made public the downforce of their 81 ground effect car and the 94 one had 95% of it.

But the 81 used small wings and most of it's downforce came from underbody(they lost 70% of all downforce when removed skirts- that was said on same video) while the 94 had a big diffuser and big wings to compensate for the flat bottom

Because CoG is low, the force needed to roll the car will be bigger, but it will eventually do it. I based that considering that at around 6G and with F1car's CoG height and track, it would have already 100% of the load on the outer wheels

They didn't in 94 really either. 94 was still the era of flat, or one dimensionally curved aluminium surfaces for the most part. There were no complex CF shapes to be had really.

yeah, if you compare FW shapes from then and from now, it's absurd the difference. The refinement extracts quite more from what the regulations restrict them into, but there is a physical limit to that. Curiously, the RWs back them were not very different to contemporary ones(end plates apart).

It would be so nice if they revealed aero figure from past generation cars. I mean, what's the problem? why the secrecy? The public knowing the numbers of 2008 or 2003 cars wouldn't jeopardize them in anyway. But maybe I'm missing something here.

On Topic, I wonder if Mercedes, or any other, will low Rosberg's time(from last week) in these next 4 days.

If somebody lows Rosberg's pole from last year, we will see tones of posts of enthusiasm of "how fast 2014 will be" but I will conservatively wait after Melbourne and Sepang to really have an clear idea of how these cars are gonna stand in relation to recent past.

Testings are a very different thing, the track get's a lot more rubber and the different temperature have quite an impact. I remember some years ago that Ricciardo lowered Vettel's pole in Abu Dhabi by 2,5s, just a couple of days after it was settled in a GP weekend. Maybe they used a different track configuration back then and I don't remember?

[Stradivarius]

Because CoG is low, the force needed to roll the car will be bigger, but it will eventually do it. I based that considering that at around 6G and with F1car's CoG height and track, it would have already 100% of the load on the outer wheels

I don't understand how you think this can happen. I appreciate that the sideway's forces from the wheels will set up a bending moment that tries to flip the car over. But the downforce itself will counteract this bending moment, as I think it is fair to assume that the downforce is pretty much symmetrically distributed. Today we don't see cars flipping over, because the friction coefficient of the wheels is too small compared to the ratio of the car's width to the height of the car's center of mass. So if you increase the downforce, you will be able to increase the grip and thus increase the bending moment that tries to flip the car over. But the increased downforce itself will counteract this bending moment accordingly.

My understanding of the situation relies on the following equations which I believe to be correct for a car running on the limit of grip through a turn. (Please correct me if you think I am making any false assumptions.)

Fs = µ*Fd (1)

Fs - the total sideway's force acting on the wheels
Fd - the total downforce (including gravity)
µ - the friction coefficient of the tires


M1 = Fs*h (2)

M1 - the bending moment trying to flip the car over
h - the height of the car's center of mass above the ground


M2 = min[M1; Fd*w/2] (3)

M2 - the bending moment from the downforce, preventing the car from flipping over. This will always be equal to M1 until the car flips over, when it is limited to Fd*w/2. So M2 is the smallest of the two.

w - the car's width


If we combine equations 1 and 2, we get: M1 = µ*h*Fd

Since today's cars do not flip over before they start to slide or spinn, we can conclude that M2 = M1, which in turn means that M1 is smaller than Fd*w/2. So we have:

M1 < Fd*w/2 which is equivalent to µ*h*Fd < Fd*w/2

We see that the total downforce Fd appears on both sides in this inequality, so we can simply divide by Fd on both sides and we see that the criterion of the car flipping over is independent of the downforce.

For the car to flip over, the following criterion needs to be met:

2*µ > w/h

As long as this criterion is not met, the car will not flip over, no matter how high the downforce levels get.

[turbof1]

They had even more leeway in 84 and 74 so they must had more DF in those years, right?

They made more potentional. They issue is that back then they didn't had proper tools to get to micro aerodynamics. They might have heard for instance about vortices back then, but they for sure couldn't have known that those actually have benefits. Something like a boundary layer was very poorly understood.

Tighter regulations drove aero technology. They are now much higher up the S curve then back then.

They didn't in 94 really either. 94 was still the era of flat, or one dimensionally curved aluminium surfaces for the most part. There were no complex CF shapes to be had really.

Hmmm, yes they still had relative simple wings, yet did have much more knowledge on the aero side. Also they had way more powerful engines that could handle the much higher drag. They were at that point higher up the S-curve. Actually you could say that huge gains were made between 84 and 94. They did have a clear understanding about ground effect for instance, and the importance of having cars low to the ground.

I think a very clear distinction should be made between the real performance and the true potential at a given time. Aero in 74 is comparable to fire and humans: they barely started to know how to work with it.

[Artur Craft]

I don't understand how you think this can happen. I appreciate that the sideway's forces from the wheels will set up a bending moment that tries to flip the car over. But the downforce itself will counteract this bending moment, as I think it is fair to assume that the downforce is pretty much symmetrically distributed. Today we don't see cars flipping over, because the friction coefficient of the wheels is too small compared to the ratio of the car's width to the height of the car's center of mass. So if you increase the downforce, you will be able to increase the grip and thus increase the bending moment that tries to flip the car over. But the increased downforce itself will counteract this bending moment accordingly.

My understanding of the situation relies on the following equations which I believe to be correct for a car running on the limit of grip through a turn. (Please correct me if you think I am making any false assumptions.)




M1 = Fs*h (2)

M1 - the bending moment trying to flip the car over
h - the height of the car's center of mass above the ground


M2 = min[M1; Fd*w/2] (3)

M2 - the bending moment from the downforce, preventing the car from flipping over. This will always be equal to M1 until the car flips over, when it is limited to Fd*w/2. So M2 is the smallest of the two.

w - the car's width


If we combine equations 1 and 2, we get: M1 = µ*h*Fd

Since today's cars do not flip over before they start to slide or spinn, we can conclude that M2 = M1, which in turn means that M1 is smaller than Fd*w/2. So we have:

M1 < Fd*w/2 which is equivalent to µ*h*Fd < Fd*w/2

We see that the total downforce Fd appears on both sides in this inequality, so we can simply divide by Fd on both sides and we see that the criterion of the car flipping over is independent of the downforce.

For the car to flip over, the following criterion needs to be met:

2*µ > w/h

As long as this criterion is not met, the car will not flip over, no matter how high the downforce levels get.

Hmm, you brought a very interesting point.

After thinking about it, wouldn't M2(which resists to flipping force) be neutralized by the other half?

I mean, in the other side of the axles there is also a moment with the same magnitude as M2, but countering it.

I think the total vertical force(aero downforce + weight)moments cancels themselves out and as you increase downforce, you increase sideforce as an effect and that will keep transfering more and more to the outside untill it reaches a point of flipping it.

When you increase the lateral force, it has the same effect, numerically, as if you increased CoG height or decreased the car's track(width from left to right wheel)

percentual load transfer from one side to the other is = vl*a*h/w
vl= total vertical load(weight + dynamic aero downforce)
a= lateral acceleration or longitudinal acceleration
h= CoG height
w= track for lateral or wheelbase for longitudinal acceleration

Assuming an f1 car has h between 25 to 30cm and considering their track is 180cm, it follows that a 100% load transfer, ie, all load on the outer wheels, will happen with a= 7,2Gs for h= 25cm and a=6Gs for h=30cm

[Stradivarius]

Hmm, you brought a very interesting point.

After thinking about it, wouldn't M2(which resists to flipping force) be neutralized by the other half?

I mean, in the other side of the axles there is also a moment with the same magnitude as M2, but countering it.

I think the total vertical force(aero downforce + weight)moments cancels themselves out and as you increase downforce, you increase sideforce as an effect and that will keep transfering more and more to the outside untill it reaches a point of flipping it.

When you increase the lateral force, it has the same effect, numerically, as if you increased CoG height or decreased the car's track(width from left to right wheel)

percentual load transfer from one side to the other is = vl*a*h/w
vl= total vertical load(weight + dynamic aero downforce)
a= lateral acceleration or longitudinal acceleration
h= CoG height
w= track for lateral or wheelbase for longitudinal acceleration

Assuming an f1 car has h between 25 to 30cm and considering their track is 180cm, it follows that a 100% load transfer, ie, all load on the outer wheels, will happen with a= 7,2Gs for h= 25cm and a=6Gs for h=30cm

I think the critical issue is regarding the sentence I highlighted in the quote. There is no neutralization from the other half since the load is shifted away from this half. Ultimately, at the limit, there is no force on the inner wheels at all. If both sides of the car neutralized each other, the car would flip immediately when it started turning, as there would be a net bending moment trying to flip it. The only reason why the car doesn't flip is that the sum of all bending moments is zero. If you have no sideway's force, as is the case when the car is traveling in a straight line or standing still, then you are right, the bending moment contribution from one side of the car will be neutralized by the other half. This is easy to imagine, as the weight is equally distributed on both sides. But this means that M1 = 0. M1, I defined as the resulting bending moment from the sideways forces acting on the wheels when the car is running through a corner. When M1 is non-zero, the wheight is no longer equally distributed between the two sides of the car.

Just to be clear, I defined M2 to be the sum of the contributions from both sides of the car, so if one half neutralizes the other half, M2 is zero. If you apply a small sideways force, what actually happens is that the normal forces on the wheels on one side of the car will increase slightly, while the normal forces on the wheels on the oposite side of the car will decrease accordingly. The result is a bending moment M2 (resulting from the uneven weight distribution) which will cancel the bending moment M1 (resulting from the sideway's force). M2 will always be equal to M1 in magnitude until there is no more weight to shift, i.e. all the weight is shifted to one side of the car and the wheels on the oposite side of the car are no longer in contact with the ground. This is when the car flips over. However, for this to happen, you need sufficient friction from the wheels, or the car will simply slide instead of flipping. And that is not case, as I think I have demonstrated above.

In your calculation with h = 25 cm and w = 180 cm, you seem to have forgotten to divide by 2, since the center of gravity is located at the center of the car, and thus only 90 cm away from the outer wheels. In that case I understand how you are reasoning, as 7.2 is the ratio of w to h. You are comparing the sideways acceleration 7.2 g to the acceleration of gravity, which is g. (As I said, it should really say 3.6 g as it is the ratio of w/2 to h that matters here.) But you are not accounting for the fact that increased downforce also means increased load on the wheels. The car flips over when you have no more load to shift to the outer wheels. But as the downforce increases, you increase the ammount of load you are able to shift to the outer wheels. To lift the outer wheels, you need to overcome a much larger load than just half the wheight of the car, you have to overcome both gravity and aerodynamic downforce.

Maybe it is easier to think about the first step of the process when the car flips: For the inner wheels to lift up the first 10 millimeters, the centerplane of the car needs to lift up 5 mm. On the centerplane of the car you have downforce from the front wing, from the rear wing and from the "underwing" made very efficient by the diffuser. So there you have forces directed downwards, but the car is moving upwards. So of course, you need to overcome more than just gravity. Try to make a sketch where you draw in all the forces involved. Then try to double the downforce. You will see that all other forces involved will double as well. If the initial resulting bending moment is zero, it will still be zero after you double all contributions. Or if you tripple them or scale them by a factor of 10 if you like. 2*0 = 0

[Artur Craft]

Sorry for this post

[Artur Craft]

I revised some calculations and I see that you're right because:

Going through the equation: Load transfer = ( Lateral acceleration(in Gs) x Vertical Load x Height of CG ) / Track width

It's needed half the vertical load to be transferred in order to have 100% load on outside wheels. Or:
Lateral acceleration x Height of CG ) / Track width=0,5

Putting a guess value for CoG height:
Lateral acceleration(G)= 0,5 x 180cm/30cm = 3Gs

If the cars were to flip, it wouldn't take much more than 3Gs to do so, if they could peak at 5,5Gs in the past, then I was wrong

Thinking about it, I saw what was wrong with me The flipping is around the outer wheels and not around the CoG. Almost all of the vehicle's vertical load will act against lifting the body around the outer wheels.

M2s, as I had defined them(not your definition) will not cancel each other as the flipping is not around the CoG. Virtually all of body's vertical load(weight and downforce) will work to counter the sideways force rolling moment as it attempts to flip the body around the outer wheels.

If the CoM and center of pressure(due to aerodynamics) is at half the car's width, then indeed, the counter rolling moment will be M2= VL*TW/2

VL= total vertical load, TW= track's width

if not, it will be where the center of vertical load is in the track's width

The initial set of equations you posted seems correct to me, now.

How stupid of me, sorry I kept thinking about it regarding forces to flip around the CoG(the ground doesn't allow it, obviously)

[Stradivarius]

Well, to start, the load transfer, afaik, is that way, ie, the whole track length, not half of it because it's not related to a moment contribution from one side

Then, on the first bold part, I think we are still not speaking the same language.

The vectorial sum between all vertical forces due to donwforce and weight, that could cause a rolling(no matter to which side) moment is always 0(in a symmetrical design) regardless if the car is on a straight line, standstill or during a turn maneuver.

I should probably have specified this instead of relying on conventional terminology, that when I write "normal force" I do not mean all vertical forces, I only mean the force acting on an object due to contact. In this particular case, the vertical forces acting on the car would be the normal forces (the contact forces from the ground acting on the wheels normal to the ground, i.e. not friction) and the aerodynamic downforce in addition to gravity. This seems to have caused some confusion for which I apologize.

Half the load(weight + downforce) is on the left side, and the other half on the other, previous to the car start turning. As the tyres induces sideways force on the car, it will cause a rolling moment as it's produced away from the CoG horizontal line(which I guessed to be around 25cm above the ground. btw, it's also away from CoG vertical line but the front and rear moments cancel each other out in equilibrium).

The bigger the sideways force, the bigger the rolling moment will be. There is no M2, the way you defined it. There is no counter resulting moment that originates once the car starts turning.

If there is no counter resulting moment, how do you explain that the cars do not flip over today? If there is no counter resulting moment, only the slightest movement of the steering wheel even at low speed would cause the car to flip over, if there was not counter resulting bending moment.

There is just the sideways force applied away from CoG line that is trying to roll the vehicle outwards of the turn. The total vertical force(downforce + weight) on the inner side is virtually the same as before(half the total vertical force). The tyre load is being transferred to the outside tyres as the moment is already starting to roll the body.

If you assume the same vertical force on the inner side, how can the car ever flip over? When the car flips over, there can be no load on the inner wheels. How do you explain that?

You claimed M2 to be "resulting from the uneven weight distribution". Firstly, there is no uneven weight distribution, but uneven load distribution. The outer tyres will be more and more pressed onto the tarmac as the sideways force increases.

First of all, this is exactly what is commonly refered to as the "normal force" i.e. the forces on the tires from the tarmac which acts normal to the surface, and not tangentially along the the surface (friction). Secondly, what is the difference between weight and load in this case? This is nitpicking anyway, but it is quite common to use the expression weight distribution in this context, as it is less ambiguous than load distribution. For example, you will read many places the weight transfers forward during braking. This does not mean that any mass is moving relatively to the car, it refers to the loads. You can find an example of the usage here (third sentence of the penultimate section: "wheight transfers under braking").

On the second bold part. First of all, the rolling will lift the inner wheels not the outside, as you wrote. When you increase downforce, you have more load to transfer but you also have more force transferring the load. That's why when you increase lateral acceleration(commonly expressed in G units) you keep transferring more and more load percentually or proportionally to the outer wheels.

On the third par, you corrected yourself with the inner wheels lifting, that's a good start.

Again I apologize, but it seems you understood it was an accidental error and not a misjudgement from my side. Just to confirm: The load will transfer outwards away from the inner wheels and if the friction is high enough, the inner wheels will lift when the sideway's load is sufficiently large.

What you are not grasping here is that the "rolling moment" is not being countered by the downforce here.

Again, the sum of vertical forces cancels when it comes to rolling moment.

The vertical forces(weight + downforce) acting on the inner part of the body would cause a moment. However, there is the same amount of vertical forces(talking about a symmetrical design) acting on the outside of the vehicle's body and this forces will cancel the inner part moment.

The moment from one side will always cancel the moment from the other, regardless if there is only weight acting on it or downforce as well, or if the car is turning or not. The only way to have a sum different from 0 is if you design a car where one side of the bodywork(let's say the left) will generate more downforce than the other. Then, if you take a left-hand turn, you will have more vertical force in the inner side(in this case the left) than on the outer. That will lead to a counter moment.

For a car that does not flip over, you can basically choose any point you like and claim that the sum of all moments relative to this point must be equal to zero, due to conservation of angular momentum. If you choose this point to be in the middle of the car (not displaced either left or right from the center plane), then gravity and aerodynamic downforce will not contribute directly at all. But it will increase the normal forces acting on the wheels, and thus contribute indirectly. If you instead select a point at the center of pressure of the outer wheel (assume front an rear axle the similar), then the contact force on the outer wheels do not contribute directly, but the downforce does. We are free to do so, as there is neither any rotation around this point and hence, the sum of moments must be zero there as well. In that case, it is easy to determine the correct "flip condition" as we basically only need to consider the (sum of gravitational and aerodynamic) and the centrifugal* force ( *assuming an inertial reference frame that follows the car, for simplicity), which both act in the center plane of the car. In the critical case, the inner wheels are loosing contact with the ground, so there is no load and hence no moment contribution from any of the wheels. We then arrive at the same condition as before. Fs*h must be greater than Fd*w/2 for the car to flip over. There is no other forces involved that don't act through the contact point, which in this case is chosen as the point of rotation.

If you like, we can also consider the forces around the center of mass. The sum of all vertical loads must be zero, since the car is not taking off into the air or penetrating the ground, and the vertical load on the inner wheels is still zero, since they are about to lift off the ground. That means all the vertical load is taken by the outer wheels, displaced w/2 from the center plane, and the sideway's force Fs still contributes with a moment equal to Fs*h. So we still reach the same condition: Fs*h > Fd*w/2.

I am not sure if this is really on topic, but at least it is a technical discussion.

Edit: It took me a while to write this post and you have posted again in the mean time. It seems we pretty much agree on the conclusion. I still want to leave my reply here, as I think it clariys some important things: It doesn't matter which point around which you consider the flip to happen. The laws concerning conservation of angular momentum apply to any rotation point, so it is not wrong to choose the center of mass in such an evaluation.