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: