The Perfect Corner Line (Answer)

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:eek: CAUTION: Read on only if you are good at math! :eek:

Racing is about feel. You don't have time to compute things in your head and execute them perfectly. Cornering is best learned through practice. It takes experience and timing to know when and how to get way down on the rumble strips or to cut across the dirt.

With that in mind...step back from reality and into theory. Picture this...a right hand 90º turn. If you were to drive from the right side of the approach straight, around the turn exactly on the right hand edge, and exit on the far right of the exit straight (inside line), your turn radius would be (r1). If you were to drive from the far left of the approach straight, around the very outside edge of the turn and exit on the far left side of the exit straght (outside line), your turn radius would be (r2). If you've been following so far, you will understand that r2 - r1 = the width of the track

We all know that the fastest line through a corner is outside-inside-outside...so concider a perfect line. You approach from as far to the left side as you can...touch the inside of the corner at exactly the middle...and exit to the far left of the exit straight, all in a perect, smooth, 90º arc. Beautiful! Basically, you are following the path of the largest semi-circle that will fit in the width of the track and touch the outside, inside, and then outside again.

The question is...what is the radius (r3) of you perfect line. It obviously depends on (r1) and (r2), and the width (r2 - r1). Is it able to be determined given just (r1), the inside line, and (r2) the outside line? The answer is yes. There is a formula to determine r3 given r1 and r2. Where can you find it? I don't know, it has almost no use and (like i said) racing is about feel, not math. I determined the formula (out of sheer boredom)...but I am not gonna show it right away. If you like challenges, try to figure it out. The formula should express r3 in terms of r1 and r2. I would give out a prize to the first to get it, but I really have no way to give anything out. You get the satisfaction of knowing you did it though. :lol:

For those of you brave enough to try to figure it out, when it drives you :mad: crazy :mad: , remember the first paragraph of this post.
 
looks like nobody wants to try it...let me clear things up a bit:

Perfect Line Diagram

r1 is the radius of the inside of the turn
r2 is the radius of the outside of the turn
r3 is the radius of the perfect line

the formula should show r3 in terms of r1 and r2
(ex: r3 = r1 + r2) <<< I wish it were that simple!!!

i bet nobody tries...oh well...i will wind up posting it in a few hours anyway.
 
Hey thats what i was gonna say js2.

The outside-inside-outside cornering technique is the method I always use for cornering. Example: Tokyo first turn....

What u exactly said was explained on the last pages of the GT3 manual under Skip Barber training.
 
You're saying the 'turn radius.' I'm not familiar with this term. Would you please explain to me what this means. I'm sure I could come up with something close if I studied it and looked at it, but I dont want to educate myself wrongly.

Thank you.
 
i know this stuff is in the back of the manual, that is where i got the idea. and i know math is not for racing. this is a challenge for anyone who wants one...if not, no big deal, just ignore it. the idea is to have a formula to find the turn radius of the ideal line given the turn radius of the inside and outside lines.

Presarc: turn radius would the radius of an imaginary circle that your turn is part of (i.e. if you make a 90° right hand turn, the turn radius would be the radius of the circle you would make if you continued to turn a full 360° instead of just 90°...for more clarification go here:
gt3a-spec.tripod.com/perfectline.html

just don't scroll down if you are planning to try it yourself...the answer is at the bottom.

anyone who wants the answer can also go to the link above...it is explained in detail. anyone who doesn't care...keep on not caring.
 
Interesting, but this is a bit too idealized. First, those equations only work for a 90 degree turn. For a 180 degree turn, for example, r3 would just equal r2. From some quick math, we can see that r3=r2/(1-cos(THETA/2)) where THETA is the angle, if r1=0. if r1!=0, then r3=(r2 - (r1*cos(THETA/2))) / (1 - cos(THETA/2)). Yes, I am an engineering major and yes, there is an easier way to do this than the way shown on the website.

Also, this doesn't take into account the fact that you can gain time by braking into the corner and accelerating out of it. This equation assumes that you slow down to cornering speed while driving in a straight line, then turn at that exact speed, and not accelerating until you have finished cornering and are again driving in a straight line. I think the GT1 manual had diagrams explaining this. However it would be difficult to find this equation as the perfect line is not a circle arc, and depends on the length of the straights before and after the corner.
 
I understand shiet about what you guys are talking about but i know the fastest way from point a to point b is a straight line so if you were going around a 90degree you would need 3 points one for where to start your turn in and one being as close to the tip of the 90 degree corner as posible and last being where you will end up with out hiting the outside wall. some im going to say 270 degrees with 45 degrees on each side of the 270 > and i have no clue what i am saying never mind anything i have said except for the straight line cuz thats cool and true. oh well haha
 
That would be true if you didn't have to slow down for the corner, except that if the inside were curved, you would have to follow it in a curved line for a while. Basically, if you don't have to slow down for a corner (and it doesn't slow you down), think of where a string would go if it was stretched around the track and that is the fastest line.

What we're talking about is the way to get the fastest cornering speed if you do have to slow down for the corner. In this case, since your tires can only produce a certain g-force (which is acceleration), and a=(v^2)/r or v=sqrt(r*a) where v is the speed, r is the radius, and a is the acceleration or g-force; it is clear that the larger the radius, the larger the speed. So we are trying to find the largest possible radius for a certain turn.
 
I'd say it depends on the corner? :curvyrd:

Some corners you best to do Wide In >> Apex >> Wide Out

Other times >> Very Very Wide and far in to the corner >> Good Turn in >> Full power straight over the apex and Exit :exitonly:

I also find the situation your in can make a big difference. If you've got a good lead you can afford to go slow in and fast out and hold Race Position :burnout:

But if there's a guy on you bumper :car::car: you'll need to keep your speed constant :speedlmt: through the racing line in order to defend your position, or of course you can do slow in fast out but be prepared to re-pass the guy who most likely passes you on the Entrance? :car::gcar:

Can you see what I'm getting at? :smilewink
 
thanks for the input, by-tor. I knew 180° was r2 = r3 and that my formula does not account for change in speed during the turn, but I had never even concidered non-90° turns besides that because it moves from basic trig equations to more advance trig stuff (I would GUESS that you used the law of cosines: cos C = (A^2 + B^2 - C^2) / 2AB in some way to determine your formula. However, I am not an engineering major nor do I remember much from trigonometry, so if you could confirm that I would like to try understand it better. anyway, I'm glad a few people appreciate the challenge of applied knowledge.
 
(please note that I have not taken geometry yet...) Actually, I thought that it would be 2/3(r2) + 1/3 (r1) = r3. It just seems like the answer would be close to that. Or maybe it depends on what r2 - r1 equals? (something along the lines of [(r1)-(r2)+(r1)(r2)^2=(r3)] or something like that. Please remember I haven't taken this kind of math yet, and probably won't for another few years. But then, something comes to mind: r3 cannot exist (as that at you are turning at a different radius at any point in the turn). Also, it is different from car to car (it makes sense that an f1's r3 would be less than an Escudo's r3). so, it really has no answer. (just thinking aloud).
 
Originally posted by Jpec07
But then, something comes to mind: r3 cannot exist (as that at you are turning at a different radius at any point in the turn).
Click to expand...
This problem assumes that the turn is circular (which it usually isn't), and therefore the r3 would be constant. Actually an r3 does exist even for a noncircular turn for each point in the turn. However, you are right that there would be no overall radius.
Originally posted by Jpec07
Also, it is different from car to car (it makes sense that an f1's r3 would be less than an Escudo's r3)
Click to expand...
Actually, the handling limit of the car would not affect r3 in this idealized case. the escudo would take the corner at a slower speed, but would follow the same line (and therefore have the same r3).

Stand by for an explanation of how I solved the equation for any angle, ZZII.
 
perfectline.jpg


Ok, as you can probably see, this picture is color coded (sorta). Blue=r1, green=r2, red=r3, and purple=THETA(actually THETA/2). Point A is the center point of the inside and outside lines of the turn (r1 and r2). Point B is the center point of the racing line (r3). Point C is the point directly above Point B, and directly to the right of Point A, such that angle ACB is a right angle.

From the right side of the diagram it can be seen that BC=r3-r2. We can also see that AB=r3-r1 from the middle (line through the apex). From trig, we know that since triangle ABC is a right triangle, BC=AB*cos(THETA/2) (adjacent leg equals the hypontenuse times the cosine of the angle). So we have these three equations:

BC=AB*cos(THETA/2) ____ BC=r3-r2 ____ AB=r3-r1

Substitution yields:

r3 - r2 = (r3 - r1)*cos(THETA/2)
r3 - r2 = r3*cos(THETA/2) - r1*cos(THETA/2)
r3 - r3*cos(THETA/2) = r2 - r1*cos(THETA/2)
r3*(1 - cos(THETA/2)) = r2 - r1*cos(THETA/2)

and finally:

r3 = (r2 - r1*cos(THETA/2)) / (1 - cos(THETA/2))

By the way, this equation coincides with yours (did you write the website?) at 90 degrees, further proving that both are right.

edit: the purple circle with a line through it in the diagram is a theta symbol. For some reason it came across as a "è" when I tried to put it in the text, hence all the "THETA"s.
 
This math stuff :yuck:
I find that if I first do the simple math...
If the number of braincells is exceeded by the number of variables, and there is a graph involved...
don't mess with it.

Besides my favorite physics lab is as close as the nearest pool table.
 
very nice by-tor.

yes that was my website, but i am taking it down now. u win the war of the perfect racing line. i understand your method now and it's amazing that it is actually simpler than mine and yet it can be applied to turns of any angle.

i drove myself crazy with the 90° turn so much that i didn't even concider any other turns. had i tried to come up with something, it may have taken me a while and some review of trig, but i think i might have been able to do it. that's not important...u got to it first. congratulations to by-tor, the master of the perfect racing line. thanx for showing us how u did it.

now that i have exhausted my mathematical ability, i am going back to my original theory...turning is about feel. The more you do it, the more you can sense the perfect line through any given turn.

this will be my last post in this thread, congrats to the master of the race line. ;)
 
Originally posted by ZZII
very nice by-tor. u win the war of the perfect racing line.
Click to expand...
The thing you're missing, though, is that a constant turn of the largest possible radius is NOT the perfect racing line for a given corner. The perfect racing line is to turn as tightly as possible at the beginning of the curve, without early-apexing. This allows you to get back on the gas sooner because the tighter initial radius buys you a much wider exit radius. The earlier you begin accelerating, the faster you will exit the corner, and the faster you will be travelling at the other end of the next straight - both of which can mean big improvements in your lap times.
 

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