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Fast Turns Vs. Slow Turns - Where To Make Time?


tzrider
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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:

 

The time advantage is greater if you increase speed from 50mph to 51mph than it is when you increase speed from

100mph to 101mph in a turn of equal length.

 

In a 300ft corner, increasing speed from 50mph to 51mph gains 8/100sec, whereas increasing speed from 100 to 101mph only gains 2/100sec

 

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?

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Adding 1 mph will have a greater impact at slower speed because it represents a larger % of your current speed. At 10 MPH a 1 mph increase will result in a 10% reduction in elapsed time over a distance , at 50 its 2% and 100 its 1% and his math shows this exactly

 

But consider at 100 MPH 300 ft is barely a corner more like a Kink, In TOTW the slow corner example is a 150 ft corner while the fast example is 1280 ft, roughly 8.5x the distance, I think the rider is digging too deep into the math and overlooking what makes a fast corner a fast corner and a slow corner a slow one,

 

Think of it not in terms of your entry speed but in terms of the total distance covered, is it better to go 1.467 FPS (1 MPH) faster for 150 feet , or for 1500 feet ?

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Adding 1 mph will have a greater impact at slower speed because it represents a larger % of your current speed. At 10 MPH a 1 mph increase will result in a 10% reduction in elapsed time over a distance , at 50 its 2% and 100 its 1% and his math shows this exactly

 

But consider at 100 MPH 300 ft is barely a corner more like a Kink, In TOTW the slow corner example is a 150 ft corner while the fast example is 1280 ft, roughly 8.5x the distance, I think the rider is digging too deep into the math and overlooking what makes a fast corner a fast corner and a slow corner a slow 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?

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I think the reason racers generally think this way is because it's so easy to go a little too fast in a slow corner and crash, so the risk is high compared to the potential gain. Around a fast corner, you usually have more options if you're a little to fast since the bike is less reluctant to change attitude at 150 mph than at 30 mph. Also, while you may be able to raise your speed for 30 to 32 mph with great risk for a short corner, you could be raising your speed for 150 to 156 mph around a long bend and not only gain there but for the whole upcoming straight - whereas too high a cornering speed around a hairpin may actually lose you time since you must wait too long before getting on the throttle.

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Having ridden NOLA recently and going to VIR (South Course unfortunately) this weekend, this topic got me looking at track maps to find out just how long corners really are. From what I saw of the VIR maps, I was midly surprised to find most corners were roughly 150 ft to 300 ft in length. Also, the longer corners generally look like slower corners (such as 180 deg turns). Obviously this was but a very small sampling of track layout...

 

Another thing that came to mind, which I think has some relevance, is where racers frequently pass. It seems to me, and I don't race so this is only by spectator observation, most racers are passing in the slower corners. I could well be wrong but I assume this is partly because they can go in just a bit faster (whether through late braking, slipstreaming, etc.) with less effort. I assume trying to outgun another rider at 120+ mph takes a whole lot more power than that needed at 60-ish mph.

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...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?

I've been pondering this question for a while this afternoon... My gut reaction is there is little difference. If 25 mph really is your personal limit in a given corner, then 26 mph might light off all kinds of SRs - likewise going 101 mph in a 100 mph corner. Honestly though, I don't think my sense of speed is sharp enough to recognize a 1 mph difference - I'm think perhaps 5 mph or more before it gets my attention.

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1 mph at 25 I think you would notice. Not that you're going 1 mph faster as such, but it will feel much more than 1 mph if you previously had reached your limit.

 

In Norway, we are required to do some winter driving in order to drive a car on public roads. During summer, they use oil and water to mimic ice. That's when I first learned that there was a massive difference between 25 and 26 mph. Actually, the difference between swerving under control and just going straight with the tyres not managing to find traction to steer the car was less than 1 mph. So while you may not notice if you enter a corner doing 25 or 26 mph, I'm pretty sure the bike's reactions - provided you were close to what it could do - would most certainly make you take notice!

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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.

 

I've been pondering this question for a while this afternoon... My gut reaction is there is little difference. If 25 mph really is your personal limit in a given corner, then 26 mph might light off all kinds of SRs - likewise going 101 mph in a 100 mph corner. Honestly though, I don't think my sense of speed is sharp enough to recognize a 1 mph difference - I'm think perhaps 5 mph or more before it gets my attention.

 

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.

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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 = 150'

Turn length for 90º at this radius = 235.6 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.

 

The time difference for a small increment in speed is independent of the initial speed.

 

Let v = initial speed, A = max. lat acceleration, theta = turn angle (90deg in your example)

 

The radius for the given speed and lat acceleration is r = v^2 / A.

 

The length of the circ segment L = (pi) (theta / 180) r = (pi) (theta / 180) (v^2) / A

 

The time for the segment t = L / v = (pi) (theta / 180) (v) / A

 

Take the derivative, dt / dv = (pi) (theta / 180) / A

 

For a small change in speed dv, the change in time dt = (pi) (theta / 180) dv / A.

 

Example (Turn 1):

 

theta = 90 deg

 

A = 0.8g = 7.848 m/s^2

 

dv = 1 mi/h = 0.447 m/s

 

dt = (3.14159) (90 / 180 ) (0.447 m/s) / ( 7.848 m /s^2 ) = 0.0895 s

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The time difference for a small increment in speed is independent of the initial speed.

 

I'm no mathematician and the rest of your post makes my head spin , but I disagree with this statement. The amount of time difference for a given distance with a small static change in speed totally depends on the initial speed ,

 

Distance traveled : 1 Mile

 

Speed : 1 MPH Travel Time: 60 Min

Speed : 2 MPH Travel Time: 30 min

Time saved: 30 min % Decrease: 50%

 

Speed: 60 MPH Travel Time: 60 seconds

Speed: 61 MPH Travel Time: 59.016 seconds

Time Saved: 0.984 seconds % Decrease: 1.64%

 

Speed 120 MPH Travel Time: 30 seconds

Speed 121 MPH Travel Time: 29.752 seconds

Time Saved: 0.248 Seconds % Decrease: 0.836%

 

the faster you are going the less of a impact 1 MPH will have on your time over a given distance

 

I think that unless you have a garage full of tech's that are tweaking your GPS based fuel maps and reviewing your segment times you can pretty much stick to the old advice of "Go fast in the fast corners"

this also reminds me of a old forum signature that I think applies quite well in this case, and also to the older thread on Friction vs Contact patch

 

"The difference between theory and practice is that in theory there is no difference"

 

 

 

 

 

 

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The time difference for a small increment in speed is independent of the initial speed.

 

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.

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The time difference for a small increment in speed is independent of the initial speed.

 

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.

 

That statement is only true in the context of the situation you presented, a constant radius turn at a defined speed and lat acceleration. These two values determine the radius and length of the turn. Of course, on a straight road the distance is the same for different speeds, thus the difference in elapsed time varies inversely to the speed.

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The time difference for a small increment in speed is independent of the initial speed.

 

Let v = initial speed, A = max. lat acceleration, theta = turn angle (90deg in your example)

 

The radius for the given speed and lat acceleration is r = v^2 / A.

 

The length of the circ segment L = (pi) (theta / 180) r = (pi) (theta / 180) (v^2) / A

 

The time for the segment t = L / v = (pi) (theta / 180) (v) / A

 

Take the derivative, dt / dv = (pi) (theta / 180) / A

 

For a small change in speed dv, the change in time dt = (pi) (theta / 180) dv / A.

 

Example (Turn 1):

 

theta = 90 deg

 

A = 0.8g = 7.848 m/s^2

 

dv = 1 mi/h = 0.447 m/s

 

dt = (3.14159) (90 / 180 ) (0.447 m/s) / ( 7.848 m /s^2 ) = 0.0895 s

:blink:

 

This illustrates why I became a political science major... :P

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the faster you are going the less of a impact 1 MPH will have on your time over a given distance

The corollary then might (should?) be: the slower you are going the more impact 1 mph will have on your time over a given distance.

 

I think that unless you have a garage full of tech's that are tweaking your GPS based fuel maps and reviewing your segment times you can pretty much stick to the old advice of "Go fast in the fast corners"

If the corollary is true (even partially) then "go fast in the fast corners" is incongruent because you stand to gain more from going fast in slower corners.

 

On the other hand, I'm just going give up and agree with Ricky Bobby, "I wanna go fast!" ;)

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If the corollary is true (even partially) then "go fast in the fast corners" is incongruent because you stand to gain more from going fast in slower corners.

 

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.

 

 

Try 5% faster around each bend and see where it gets you ;)

 

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.

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Try 5% faster around each bend and see where it gets you ;)

That's just about the same thought I was having yesterday... I'll be on the track tomorrow and we'll see what happens :)

Finished a track day at VIR (South course) yesterday. It was brutally hot/humid and nearly everyone cut sessions short, skipped a session or two, or just plain quit early. It was my first track day of the year (I'm not counting my two CSS days at NOLA), my first track day on my 2012 Triumph Speed Triple, and my second time on VIR South. I did not want to wreck my new bike so I was not pushing very hard.

 

What I discovered was, for me, the turns with roughly 90 deg or less radius were easier to try to gain some speed without feeling threatened. The longer radius turns (roughly 120 to 180 deg) were harder to gain speed without quickly lighting up SRs. I'm struggling to find the right words... basically: the shorter radius turns were easier to add a little speed, regardless of whether the corner was fast or slow (there essentially were a couple of each); the longer radius turns were harder to add a little speed, and these turns were slow(er).

 

How does this fit with slow vs. fast and where to try for more speed? I'm not sure... With a short radius it didn't seem to make much difference; with the longer radius, the slow turns lit up SRs in a hurry. Maybe the radius of the turn makes the difference? I'd like to know what others have experienced...

 

Two other points: I had to be a bit cautious with my 2d gear roll-ons or I'd immediately wheelie, which made for lots of big grins but didn't help me go faster :) And, trying to hide behind the tiny flyscreen at 120+ mph is comical :)

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Interesting thread... my first thought was that a fast corner will always cover more distance than a slow corner. For instance the fastest corner on a track I regularly ride - just my turn input there takes longer than the whole of the slowest corner on the same track!

 

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?

 

I would like to think I would notice it, but in reality... I don't think I'll notice it in a fast turn as much as a slower turn.

 

The other thing to consider about whether or not the 1mph increase has the same effect on the riders state of mind - would it depend on whether the rider was near their maximum in the fast turn and the slow turn? I am just thinking that because I try not to rush slow corners (because I know how easy it is to mess it up), I will always keep a bit of safety margin there. However in fast corners I am much more likely to push myself, and if I enter at 1mph faster there's a fair chance it will be new territory for me and beyond my comfort level. But I really couldn't say the same for myself in slow corners.

 

But if we are talking about riding an unfamiliar road (or track), yep I'm sure a minor speed increase would seem like much more in a tighter corner than in a wide open corner.

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