tzrider

Another factor I haven't seen mentioned is that as you lean over, the final drive ratio changes. the difference between upright and fully leaned is the equivalent of half a downshift. This puts the engine in a different part of the powerband and can alter the effect a given amount of roll-on has, seeming to amplify the torque as the lean angle increases.

If we're going to use the admittedly artificial construct of "adding 1 mph" in a turn, regardless of speed, a turn that is twice as fast as another must be more than four times as long before you see a bigger lap time reduction in that turn over the slower one. That said, there is another piece we haven't yet touched on: Even if the elapsed time difference in 50/51 vs. 100/101 is the same (.09 seconds in one of our examples), the distance between you and a slower competitor at the exit will be much greater in the fast turn. In a slow turn, it may be less than a bike length, where it may be a couple of bike lengths in the fast turn even though the time gain over your competitor is the same. It can be the difference between completing a pass (and taking the line) or not. I do think this is a better way of looking at it, as it probably fits better with how riders really assess speed in corners of different sizes. In this case, giving priority to the fast turns is the clear winner.

Matt, I should mention that I made a typing error in my example above, having stated the dimensions for the second turn were the same as the first. They are not and are corrected in that post. I followed your math, but don't see that it supports your quote above, as stated. Taken literally, we could simplify the example to remove the turn and say we're riding the same 1,000 feet of straight road. If we make a pair of passes, one at 50 and the other at 51 mph, the difference in the time it takes to cover the 1,000 feet is .27 seconds. If we make a pair of passes at 100 and 101 mph, the elapsed time difference is .07 seconds. For your statement to be true, the distance covered must be proportionally larger. Your math allows for this, but the opening statement doesn't seem to.

I ran some calculations to compare turns of different sizes, but with similar configurations. To pick a couple out of many examples, my turns are 90º and the line is constant radius. I decided that the limiting factor on how fast the rider can go is traction, rather than bravery, so I determined a target speed based on a known lateral acceleration value. Assumption: The bike and surface will support about .8g lateral acceleration (± a bit) Turn 1: Radius = 150' Turn length for 90º at this radius = 235.6 feet Lap 1 speed = 42.25 mph Lap 1 lateral accel = .8g Lap 1 elapsed time = 3.8 seconds Lap 2 speed = 43.25 mph (1 mph increase) Lap 2 lateral accel = .84g Lap 2 elapsed time = 3.71 seconds Elapsed time difference = .09 seconds Turn 2: Radius = 600' Turn length for 90º at this radius = 942.48 feet Lap 1 speed = 84.5 mph Lap 1 lateral accel = .8g Lap 1 elapsed time = 7.6 seconds Lap 2 speed = 85.5 mph (1 mph increase) Lap 2 lateral accel = .82g Lap 2 elapsed time = 7.52 seconds Elapsed time difference = .09 seconds The above comparison shows the break even point in the turn length: For a fast turn that is twice as fast as a slow one, the fast turn must be 4 times longer than the slow turn before a 1 mph increase will result in the same elapsed time reduction in the turn. Even this is a bit deceiving though, as a 1 mph increase over a greater distance will have a greater affect on overall lap time because the bigger turn is a larger percentage of the overall lap distance. In other words, the fast turn might only be 3 times larger and the difference in elapsed time through the turn won't be as great as in the slow turn, but it will still reduce the overall lap time by a greater amount. In my example above, if we were talking about two turns at a track the size of Laguna Seca (11,816 feet long), the first turn is 2% of the track, where the second is 8% of the track. A .09 second gain over 8% of the track is more meaningful than a .09 second gain over 2% of the track. A proportionate speed increase in a faster corner is not 1 mph, but something more than that. In my example above, a 1 mph increase in the slow turn produced a .04g increase in lateral load. To get to the same increase in lateral load in the fast turn would require a 2 mph increase. While lateral load and the sensations it produces aren't all there is to our sense of speed, it's a big part of it, once in the turn. While I can't prove it, I would guess that most riders have a harder time distinguishing a 1 mph difference on the approach to a 100 mph corner than they would approaching a 25 mph one.

T, you pretty well got to the heart of the matter. While it's possible to find examples of fast corners and slow ones of equal length, it's not common and might be close to nonexistent on a given racetrack. A faster corner will have a larger radius, and if the turn configuration is similar (both 90º, for example), the distance covered from entry to exit will be longer. So, we know that if the corners are the same length (but clearly different geometry), picking up 1 mph in the 50 mph corner is more significant than gaining 1 mph in the 100 mph corner. If you do the math in Keith's ToTW example, he's right and we gain more for a 1 mph increase in the fast turn. Interestingly, both of these examples assume we are dealing with two turns where one can be ridden at twice the speed of the other. How much longer must the 2x faster corner be before we break even on the elapsed time gain? The fast turn must be about 4 times longer before we break even on the time gained by a 1 mph increase. This turns out to be roughly the ratio you get if you assume both turns are constant radius 90º's, and we begin with a pace in each that results in .9g of lateral acceleration (humor me). A constant radius 90º arc that produces .9g at 45 mph is 235 feet long; a 90º that gives the same .9g at 90 mph is 943 feet, for a distance ratio of 1:4. Could be we're over thinking it, but this does put the rule of thumb to focus on fast turns into some context. As long as we're talking about 1 mph increases, the faster turn must be a lot longer. That said, is a 1 mph difference in speed as significant to a rider in a 100 mph turn as in a 25 mph one? Does adding 1 mph in each case have equivalent effects on the sensations in the turn and the rider's state of mind?

You may have heard from any number of places that a 1 mph speed gain in a fast turn makes a bigger difference in your overall lap time than a 1 mph gain in a slow turn would. We recently received an email from a rider who read the chapter in Twist II that discusses this (page 26, Heading "100ths"), felt that he understood it, but then applied some math and found that his math didn't seem to agree with what the book said. Here is a paraphrase of what he said: He then asks what he may be doing wrong. On the surface, his math does seem to confirm the above: Scenario 1 - 50 to 51 mph Lap 1 Turn Length = 300 feet Speed = 50 mph (73.3 feet/sec) Elapsed Time In Turn: 4.09 seconds Lap 2 Turn Length = 300 feet Speed = 51 mph (74.8 feet/sec) Elapsed Time In Turn: 4.01 seconds Difference in elapsed time = 0.08 seconds Scenario 2 - 100 to 101 mph Lap 1 Turn Length = 300 feet Speed = 100 mph (146.7 feet/sec) Elapsed Time In Turn: 2.05 seconds Lap 2 Turn Length = 300 feet Speed = 101 mph (74.8 feet/sec) Elapsed Time In Turn: 2.03 seconds Difference in elapsed time = 0.02 seconds So, what gives? Is this rider making an error or has he shown that going 1 mph faster in a slow turn makes a bigger difference than going 1mph faster in a fast turn?

I think there may be pictures...

It occurs to me that after a little run-in with a couple of White Russians a few years back, a good name for Cobie would be Sergei.

Yeah, with traction control!

^^ We've got to stop over-inflating his tires.

It's not real until I see it on my leathers, jacket and cone.

This brings back other memories, like when Paul e-braked the rental car and you went into the dashboard.

Does me too! Might have to bring her out of retirement...

For that matter, we could call him "Rip," which also carries the suggestion that he's fast.

Pete used to think it was great sport to fart near the rest of us. I don't know that he ever actively tried to get others to smell it, like by letting an SBD and then saying, "Anybody smell weed?" Still, he was prolific and sometimes in enclosed spaces. Pepe is a Looney Tunes skunk character:

Speaking of Pete, here's an alternate for him: "Pepe le Peu "

I'm confused...

Tim = "Chops" Dylan = "Bob" Cobie = "Magoo"

Tim, I hope you're not really gone from the conversation. In case you're not, good for you for trying this out. A comment or two about your findings: If all you did was turn the bars, I would have to agree, but... You accompanied turning the bars with a body weight shift. Did you have a chance to try turning the bars without shifting your body? If you do that, I think you will have a revelation.

I won't obfuscate the issue. If you do these two things and ONLY these two things, your bike will lean left. Period. Try it. Sit perfectly still on the bike as you do it. I should stop right here to avoid tangents that make this more confusing than it is. You asked me direct questions though and I'll respond. The other thing that makes the wheel go right is it's pointed to the right and it's rolling. If it has traction, it's gonna go right. I'm not disputing that the wheel will go right; I'm disputing what the rest of the bike will do in response. That is true, the wheel does lean slightly right. It doesn't mean the *bike* also leans. In this narrative, you have not indicated when the bike itself leans. You have accounted for changing the direction the front tire is going, but are drawing a wrong conclusion about what the bike itself will do in reaction to that. If you literally did ONLY the things you have written above, your front tire would veer right and your motorcycle would lean left. If your motorcycle is leaning right, you are doing something other than what you've written here. The front wheel of a bike with no rake or trail would have no tendency to center itself on the path of travel and would be miserably hard to control. But if you could control it, such a contraption would also countersteer. No need to refute these things; they are true. The statements stop short of describing what the motorcycle does in response.

On the definition of countersteering it does sound like we have a logjam. Here's a related question: When the bike is leaned over and you need to quickly stand it back up to vertical, how do you do that? I'm curious given your current view how you decide whether to countersteer or not. At what speed does countersteering begin to work?

I define countersteering as turning the front wheel the opposite direction of a lean angle change that I want to cause in the bike. My definition and yours above are not the same. Also, by steering the bike's front wheel to the right, I am not making a tighter right turn, I am relaxing the lean angle so it's more appropriate for the arc I'm already on.

Here is a single frame from just after the 10 second mark where the rider is in the process of steering the bike to rider's left. Note the ski tip pointing to rider's right, as well as the angle of the lower triple clamp in relation to the body of the bike. I have placed a red circle around a small black vertical feature sticking out of the top of the ski. That feature is a grab loop, which can be found in the center of most snowmobile skis. The ski is pointing to the rider's right.

Let me make a couple of distinctions to keep the discussion as clear as possible. Pressing the inside bar harder makes the bike change lean angle faster. Pressing the inside bar farther produces a greater change in the bike's lean angle. Applying that to your request above, I would restate the request: "So now explain how pressing the bar farther produces greater lean angle even after the turn is initiated." That's the request I'll answer and if I've misunderstood you, I trust you'll say so. I'm still kind of asking you to take this on faith, but here's my assertion: A motorcycle changes its lean angle to the opposite direction the rider steers the tires. The lean angle change results when there is a change in the lateral relationship between the path the contact patches are following and the path the bike's center of mass (CoM) is following. If the motorcycle is already leaned into a righthand turn and you want to make it lean more, you steer the front tire to the left. The tires take a path that diverges from the path the CoM is on, which is another way of saying that you're steering the tires farther out from under the CoM. The bike leans over more. Here's a little bit of a whacky example, but if you were to try it (and were really coordinated), very similar principles would apply. Let's say you're walking straight ahead and balancing an eight foot pole vertically in the palm of your left hand. Your hand is directly in front of your shoulder. You decide to turn gradually to the right. As you began to turn, you would need to move the hand you're balancing the pole on a little to the left to keep it in balance. This is an exact analogy to steering the tires of a motorcycle left to initiate a lean angle change to the right. Let's now say you want to tighten the righthand arc you're walking on. As you tighten the arc, you must move your hand a bit more to the left to keep the pole in balance. Now you want to walk straight again. As you straighten out your path, you must now move your hand to the right to restore the pole to vertical. All of these hand movements are similar to the lateral movements of your tires in relation to your motorcycle's CoM as you enter and exit turns. In the walking example above, your body provides a fixed reference from which you can more easily tell that you are moving your hand in and out. When you are riding or watching someone else ride, there is no similar reference, which makes the observation so much more difficult. Referring back to the video you saw of the guy countersteering abruptly, we can see that the bike rolls around a point somewhere above the tires as the tires swerve to the left. It's easy enough to observe this when the rider turns the bike quickly but it's more difficult to observe when the rider turns the bike slowly. When turning slowly, the bike will begin to enter an arc the instant it goes off vertical, even though the rider is still adding lean angle. Since the bike begins to turn, it's not as easy to observe the point around which the bike rolls over and it's also not as easy to see the tires veer out from under the bike's CoM. The faster a rider steers the bike, the better you can see the tires moving laterally to the outside of the CoM's path of travel because there is more movement over a given period of time. Here's my thinking on that: Let's say the turn above is a 60 mph corner. At that speed the bike is covering 88 feet per second. The rider is about to steer the bike to a 35 degree lean angle. The height of the CoM of the rider and the bike is 21 inches, so this means the tires would need to deflect out from under the CoM a total of 12 inches during the time it takes the rider to steer. Note that this does not mean the tires will move 12 inches to the right before the bike ever begins to turn left, as the bike will bend into the arc the instant it leans over. However, if you're looking at the bike head on at the turn entrance, you might see the tires veer to the outside of the turn as the rider initiates countersteering, as we did in that video clip. Two Scenarios Scenario 1: If the rider takes 2.5 seconds to lean the bike from vertical to 35 degrees, he will cover 220 feet before he reaches full lean and in the first 20 feet of travel the front tire will deflect 1.1 inches to the outside of the bike's CoM. In this first twenty feet, the rider has achieved only about 3 degrees of lean angle and the bike has barely entered an arc. If you look closely, you might detect the tires veering to the outside of the turn, but the amount of movement would be very small. Once the bike did bend into the turn, observing the tires veering outside the current path of travel of the bike's CoM is much harder because your viewing angle changes. Scenario 2: If the rider takes three quarters of a second to lean the bike from vertical to 35 degrees, he will cover 66 feet before he reaches full lean and in the first 20 feet of travel the front tire will deflect 3.7 inches to the outside. In this first twenty feet, the rider will achieve 11 degrees of lean angle and the bike will turn into the arc more quickly. You should be able to detect the tires veering to the outside of the turn, as they should visibly move at least three inches to the outside, even taking into account that the bike will begin to turn. I had previously said: You replied: Here is our disconnect: In the example of trying to catch a fall to the right, I am already in a right turn and find that I am leaning too far. By steering into the turn, I am not attempting to negotiate a right turn, I am attempting to produce a lean angle change to the left of where I am right now. If I am at 35 degrees to the right and feel like I'm about to fall into the turn, I may want to bring the bike to 25 degrees right, for a net change of 10 degrees left. Turning the front wheel to the right will accomplish that. That is countersteering. If the bike is moving forward, countersteering will work. Despite that statement, we are closer to agreeing than not on this point, at least if we can leave precession out of it for the moment. At parking lot speeds, you can initiate lean with upper body movement and then maintain balance by steering into the turn. You could also countersteer. The reason I belabor this point is that if you believe there is a speed below which countersteering does not work, it presents you with the problem of knowing what speed that is. It can result in confusion over which way to turn the bars in an emergency. I'm not gonna tell you that you must countersteer at five MPH, but I will insist that you can. This isn't a counter-rotating rotor, but how about video of a single track snow vehicle with a front ski? No precession in play here: At the 10 second mark, there is a head-on shot that shows the rider steer the ski to rider's right, which snaps the bike over to rider's left. You may have to advance this frame by frame to see it. As an aside, I want one of these things! What do you think?

Tim, Cobie was making the point that to catch a fall to the right, you steer the front wheel farther to the right to make the bike stand up. That is definitely an example of countersteering. If nobody has said it yet, countersteering is used merely to cause a lean angle change. Point the front wheel to the left, the bike changes lean angle to the right. That can be true whether the bike was vertical and you want to lean into a right hander or if the bike was leaned left and you want to bring it upright. It's countersteering either way.