ACC Blog Aris Drives

Very good stuff Aris, just love to watch amd learn! Thank you very much mate

 

I found this comment on Youtube , it is interesting and a good laugh at same time....made me think about tyre physics though. check it out :


" Gt cars run mostly on front bumpstop is total BS. No race car runs on bumpstop even F1.....with WAYY more downforce than GT3 !! The hype talk on how GT3 cars aero is so big deal etc is totally BS as well. What another thing is that ACC generates unrealistic low speed grip from aero. These 1.5 tonnes of vehicles couldn't possibly be affected by aero at speeds below 120km/h but in ACC they do ! Not that aero cannot have effects at these speeds but if you take into account how much downforce they produce and how much they weigh you can see how ACC aerodynamic physics are exaggerated. Remember why they tell you to run on bumpstop because that exaggerated aero shifts the balance way to forward with way to much magnitude which is not possible with given GT3 car design. They somehow managed to create mechanical grip even though front tyres are basically running on bumpstops. Not to mention other things that are completely wrong in physics of ACC. for example how oversteer is only dependent on wheel spin rather than load sensitivity,tyre friction circle etc. This is why cars in ACC understeer hard on fast corner exit (BEFORE you say there us aero on fast corners to help i have to remind you that even if aero is there if tyre can only do 2 Gs the only one direction then no matter what the hell you do you CANNOT go beyond it) As long as there is no wheels spin there is grip even if rear wheel is running zero camber or reaching temperatures of sun or even when they are way overloaded......as long as there is no wheel spin there is no tyre slip.....TOTAL BS. You can watch real life onboard and see how real GT3 cars are sliding little bit on fast cornering (for example look at spa puohon and T16 IRL and compare it to ACC) but in ACC they understeer, as there not enough torque to wheel spin....which is completely wrong as you can still lose traction at rear by other reason (overloading,temperatures,oversteery setup like stiff rear ARB ) "

 

https://www.facebook.com/Assetto.Corsa/photos/a.340573865968938/5375675735792034/

 

So much horse**** in one post, I've never seen it in such quantities before.
Do you think that aero load shifting is the only aero affect of a diving front end under braking?

Your jibber really shows you know nothing. You also say if a tyre has only 2g of grip, thats all it will produce. What crap.

GT3 cars do run close to the bumps and with very good reason.
Even the Mercedes AMG suspension geometry has lots of anti dive in its design.

I suggest you go play iracing or forza etc if your after simpler challenges in physics modelling.
ACC is very advanced in this area, and you have demonstrated that you haven't a clue what you are talking about.

 

@GONKO It's not his jibber, he said he found it on Youtube in the comment section under a video.

 

LOL

 

Ah OK, I thought it was his muck. Apologies to @A-Jin

 

ye no worries mate mistakes happen : )
I found that YT comment somewhat interesting not because i fully agree with it but , about tyres I sort of agree (sry not a native speaker can't think of correct word.)
What he said about oversteer in ACC ,which I too always felt bit off .
As many of you I too watch onboards of GT cars. Sometimes i see some car handling traits which basically never happen in ACC, So that YT comment made me think. This was "new" way of thinking for me.
I don't have simulator knowledge, all I do is drive in sim and observation IRL , no harm intended.

BTW @GONKO what do you mean by this ? Even I think of this same way i.e if tyre can only do 2Gs then it will never go above it am I wrong ?

 

Depending on the load on the tire. More downforce equals more traction which results in more g.
But what is meant by this statement is: If a tire is capable of delivering 2g, it is always in one direction only. If you have lateral 2g then there is nothing left for longitudinal (brake/accelerate) force. You need to have less than lateral 2g to be able to have some leftover to brake. The less the lateral force the more you can use longitudinal force and vice versa.

 

Yes this is exactly what I think as well. IMO this is also what that YT comment meant. So this is correct right ?
If tyre is rated 2G then it cannot magically with the help of downforce do 3G right ?

I think @GONKO is denying that, hence maybe i'm wrong.

 

It's not "rated", just the laws of friction. The utoob comment was probably from a troll which has been banned from these forums and hate ACC.

 

hm? Of course, as I said:

The equation for a tire of delivering up to 2G is always taken into account the downforce is staying the same.
Let's imagine a card box for moving, where you put stuff in...
When it is standing on the carpet empty you can push it around with ease.
Now increase the downforce by putting some books into the box, it will get harder to push the box around.
Now fill up that box with more books, you will find it very hard to push that box.

The same applies to the tire.
You have an amount of friction, which holds the tire from sliding off the track, combined with speed in a turn it generates a centrifugal force. That centrifugal force lets you feel g-force.
When adding downforce (pressure to the ground) you increase the friction, which gives you the ability to have a higher speed in the turn, which raises the centrifugal force and lets you feel more g-force as before (as long as the tire holds the line).
That friction is the same when braking and it goes along the same equation. More downforce equals shorter braking distance, equals higher g-force during braking.


But this is not was I was talking about in my last post.
There I was talking about the g-force is a vector. A vector is a force in a direction.
The tire (with a specific friction coefficient, e.g. amount x of downforce) is capable of generating a g-force in a direction (= vector) of a.
When braking in a straight line, there is no force to the side (ax = 0) and all of a can be used to brake (ay = 100). Roughly speaking...
When braking in a light turn (e.g. 1g), there is force sideways (ax = 50) and force longitudinal (ay = 50), which results in max g-force left over for braking is 1g, because both together are the limit of 2g (of that tire with that x amount of downforce which results in that amount of friction).

 

That is right, only slight correction is that you need to use vector addition for the lateral and longitudinal forces.
So, if your tyre can cope with 2g and you have a lateral load of 1g the potential left for longitudinal load equals to sqrt(2^2 - 1^2)g = 1,73g.

The circle of forces is an ideal model though, so real tyres usually show an elipse instead of a perfect circle and are capable of transmitting slightly more longitudinal forces than lateral forces.

 

@C#27 you're absolutely right. Wanted that explanation of this highly complicated topic to be as simple as possible and messed up the equation itself. Thanks for correcting.

 

Very cool stuff guys thanks.

 

Today: "Spa 12Hours RWB International - The race with Turn1!"!

 

Good stuff, really nice listening to Aris talk and race with the pros.