I have enjoyed these threads and emails. I'm an experienced street rider but am new to the track. Having attended the level 1 class late October and a track day in November (pitting next to Will and Keith), I've spent a great deal of time thinking about high performance cornering lately.
The basic physics of cornering are very interesting.
Lateral Acceleration (LA) is a result of combining two factors. Speed (V) and turn Radius ®.
LA= V^2 / R
Divide by 32 to express LA as a magnitude of gravity (G)
Hence: a turn with a radius of 190' taken at 60 MPH looks like this.
60MPH = 88 Ft. Per. Sec. (MPH * 1.4666667)
88^2 / 190 = 7744 / 190 = 40.758
divide by 32 = 1.274 G
So tires in such a turn will have to support 1.274 times the total weight of the bike and rider to a direct side load.
The lean angle for such a turn is just as cut and dried.
lean in Deg. = (aTan(V^2 / (32 * R)))
so... (aTan(88^2 / (32 * 190))) = (aTan(7744 / 6080)) = (aTan * 1.274) = 51.86 Deg. lean.
The important thing to realize here is that any speed / radius turn combination that results in a lean of 51.86 Deg. will have a lateral acceleration of 1.274 G and vice versa. The lateral acceleration and the lean angle go up in a direct relationship.
Since there is only so much grip available, at some point LA will exceed friction and the bike will slide. Slide too much and you will low side. The optimal lean angle would be just at the point where the tires start to slide.
When you consider these concepts with the goal of getting around a corner as quickly as possible you realize that the ideal turn would be made with the bike at the optimal lean angle as much as possible. To get around the entire track as quickly as possible you would adjust your speed for each turn to hit that same optimal lean angle every time. Slower in tighter corners, faster in wider corners. We all know this intuitively; the math allows us to understand this exactly.
If you don't trail brake and hold a constant speed around the turn, your line will be a constant radius. If you trail brake into the turn and accelerate out, the radius of your line will tighten as you slow down and widen as you accelerate if you keep a constant lean angle. Looking back at Keith's experiments with turns on the no BS bike we see that if he established a lean angle and added speed the turn radius would increase. That could be perceived as a "straightening up" of the bike. I would like to conduct a no BS turn experiment on the flat open "skid pad" area. I'd establish a constant speed / radius / lean angle and ride around in a circle until a stable turning state was achieved then slowly change speed to observe the effect of changing only one variable. It may take a while riding in circles to accomplish a good experiment. Doing this at different speeds, radius and lean angles then changing speed up and down would be interesting. Conducting such experiments on the track introduces too many variables like camber and rise.
BTW (I've ridden up to 4 miles of gently winding road on my BMW without touching the bars at about 40 MPH) But this is a discussion for another day.
Back to trail braking. One concept that I'm having trouble understanding is this, as I watch GP racers, they corner with remarkably consistent lean angles. This makes sense with regard to the physics mentioned above. What's strange to me is that they seem to be braking well into the corner. They do this even when riding alone when the only thing that matters is lap time. It seems to me that if they had enough grip to corner AND brake at the same time that they could do less braking and use that grip for turning by having a higher corner speed (higher lean angle)
Obviously whatever Mr. Rossi is doing is damn near perfection. If there were a faster way around a track, he would find it.
So my question about trail braking is this, exactly how much braking are they doing while at max. lean and why is it fast?
My suspicion is that they approach the turn with 99% of grip being used for straight line braking. And as they tip in they come off the brakes at the same rate that lateral acceleration consumes grip. As they hit max lean angle 95% of grip is being used for lateral acceleration and the brakes are using less than 5% of the available traction. They may use the first half or more of the turn to release the last 5% of brake.
I've seen telemetry from fast riders. It seemed that they spent the entire corner decelerating to mid corner and accelerating from that point on. Such evidence would invalidate my theory. I'd love to see the telemetry from the top 10 GP racers! It would answer just about every question on this forum.
one possible factor to consider may be that lateral acceleration forces and braking forces apply themselves to the contact patch at right angles to each other and don?t simply add up on top of one another. For example if you are in a 45 deg. leaned turn, your lateral acceleration would be exactly 1G. This pushes directly sideways on the contact patch. Braking pushes directly back on the contact patch. This is 90 off from the side force of cornering. If you corner at 1G and bake simultaneously with say .5G the contact patch would not be holding 1.5 G, it would be holding 1.12 G.
If the tire could hold a maximum of 1.2 G before sliding, you would be within a safety margin for grip.
It is possible that adding a minor braking force at right angles to the cornering force yields gains over a technique that uses slightly higher corner speed and little or no braking in the turn. Perhaps due to a shorter line through the corner or something.
One last bit of math. The total G force on the bike and rider in a given turn is the square root of LA^2 + 1.
So that 60 MPH / 190' radius turn with a LA of 1.274G and a lean angle of 51.86 deg. would generate a total G force of 1.62 G.
1.274^2 + 1 = 1.623 + 1 = 2.623
The Sq. Rt. of 2.623 is 1.62
A lean angle of 60deg. will generate 2Gs on the bike and rider. With a LA of 1.73
Truly,
Scott@ltd-aero.com
Riding Fontana Sunday Dec. 14th
