racer
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Posts posted by racer
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Hi schmi,
That was a wonderful writeup!
Congratulations for confronting your doubts and skepticism while shattering your low expectations with high accomplishments! Mostly, thank you for sharing yourself and your experience. I personally found it very valuable as I am certain many others will.
Cheers,
racer
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Thanks.
Have a safe trip.
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Any comments or questions? Tweek? David? Anyone?
I realize this site is dedicated to riding skills, not math; but, the physical concepts behind the math are fundamental to the riding skills. The "why' behind the "what", so to speak. One can follow a recipe for what to do in a simplified single turn, but, I think that understanding the principles, learning how to apply the fundamentals leads to the ability to adapt to different situations and requires at least some understanding of the "why" ...
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OK...
Tweek, David, et al...
I lack the skill to create a good diagram. If anyone knows some place to copy/paste vector arrows and such, please let me know.
Here are the numbers:
Given:
mass = m = 600 lbs = 272.72 kg
radius = r = 50 ft = 15.38 m
lean angle ( theta θ ) = 50°
Find:
velocity (v) = ?
Formulae:
gravity (g) = 9.8 m/s2
Fgrav = Fvert = m • g = 272.72 kg • 9.8 m/s2 = 2672 Newtons
Fnet = Fhoriz = m • a
And: circular acceleration (a) = v2/r
Hence, substituting for (a): Fhoriz = m • v2/r
Being that we know θ = 50° and Fvert = 2672 N, we can use our trigonometry to find Fhoriz using theta's tangent and substitute the value of Fhoriz in the equation above leaving v as the only variable to solve:
TAN 50° = opposite/adjacent = Fhoriz/Fvert
1.191 = Fhoriz/2672 N
Fhoriz = 1.191(2672) = 3183.2 N
Substituting: Fhoriz = m • v2/r
We get: 3183.2 N = 272.72 kg • v2/15.38 m
v2 = (3183.2/272.72) 15.38 = 179.516
v = 13.398 m/s = 48.234 kph = 29.9 mph
Thank you for your time,
racer
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Now THAT'S more like it!
Thanks for that, Roadman.
Barber labels these turns 14 and 14a.
So... no downshift for 14a as it is labeled by Barber. I can't see if he is dragging the brake at all but he rolls out on the hill well before the entry to 14a. It looks like Skip runs it a bit deeper. Still couldn't see his brake lever. I think if it were me I would be doing some braking while leaned over there and then leaning in more for the tighter part of the turn.
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The "quick flick" is STANDARD OPERATING PROCEDURE for CSS and should be for anyone looking to improve their lap times but.....I can give you one example where there might be opposing theories of approach...
Hey fossilfuel,
Check out a part of Chapter 17 in TOTW II called "Track Positioning" (p. 75 in my copy). I think it might address your question of SOP conflict here.
r
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Hey David, et al,
I apologize I don't have time to create a diagram to upload, so, for now, please refer to the diagram near the middle of the page at link #3:
http://www.glenbrook.k12.il.us/GBSSCI/PHYS...cles/u6l2c.html
Forgive me I must be brief...
In a nutshell, the diagram of component force vectors forms a right triangle allowing us to utilize right triangle trigonometry to find the angle or vice versa to find the component vectors from which we can substitute and solve for velocity using the circular acceleration formula. In this case, the vertical vector over the horizontal vector equals the "opposite" over the "adjacent" sides of an included angle of a right triangle. Those of us who remember our high school trig will recall that opposite over adjacent gives us the tangent, hence, the "inverse tangent" (INVTAN on your calculator) gave us our angle in this example. I originally reversed this formula to solve for the velocity. Unfortunately, because Tweek probably meant 50 degrees from vertical, not horizontal, my answer of ~ 42 kph was, in fact, incorrect.
Now, to correct the situation, we could subtract 50 from 90 and solve the same equation(s) for 40 degrees to find the correct speed... OR we can move theta over the hypotenuse so it refers to the vertical. At that point, David's suggestion of horizontal over vertical becomes the "opposite" over the "adjacent" and the INVTAN function on the calculator will give us the angle, or, allow us to reverse the process to find the velocity according to what Tweek most likely meant.
(We could continue to use the vertical over the horizontal BUT then the function would be adjacent over opposite (co-tangent)... which is more of a pain on my calculator.)
I will try to plug in the numbers before I go to bed tonight. Otherwise, I'll try to get to it in the morning before I leave for work.
Cheers,
racer
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Numbers are right but I think the convension should be higher horizontal force/higher velocity/tighter radius gives a higher lean angle. So it should be Fhoriz/Fvert.
Hi David,
Ah... I think I see what you mean.
Higher velocity and tighter radius requires further lean from vertical. If I follow your point, the issue is that (following the example on the physics site) I applied Tweek's given lean angle (theta) to the horizontal rather than vertical as is typical when describing motorcycle lean angles.
D'oh! My bad.
Thanks for pointing that out, David. I'll re-work it with theta measured from vertical when I get home tonight.
Cheers,
racer
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I am not experienced with this particular tire change, but maybe you misunderstood what Jim Cox meant...
You may need to add a tooth to your rear sprocket to compensate for increased rotating mass of the larger tire, but, if Dunlop tires were SO much heavier than other tires that it required racers to add teeth to the rear sprocket to compensate for using a Dunlop, it would be a distinct disadvantage... and... nobody would use them.
Dontcha think?
Also, more likely, assuming the same aspect ratio (second number that expresses the height of the sidewall as a percentage of the width, 65 % of 195 mm in this case), the wider tire will be taller as well. Hence, you might need to add a tooth (or more) to your rear sprocket to compensate for the larger circumference of the rear tire.
A larger rear tire will cover more ground per revolution. Hence, your speed will be higher for a given RPM. It is effectively the same as making the rear sprocket smaller.
racer
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OK, let's just try it in reverse. Substitute my result for v, solve for theta and see how close to 50° we get.
Given Info:
m = 600 lb = 272.72 kg
v = 42000 m/h = 11.6 m/s
r = 50 ft = 15.38 m
Find:
a = ???
Angle of lean = ???
Fgrav = Fvert = m • g = 272.72 * 9.8 m/s = 2672 N
a = v2/r = (11.6 m/s)2 / 15.38 m = 8.749 m/s2
Fhoriz = Fnet = m • a = 272.72 kg * 8.749 m/s2 = 2386 N
θ = invtan (Fvert/Fhoriz) = invtan (2672/2386) = 48.236°
Not too far off considering how many significant digits I rounded off.
Maybe not quite as simple or straightforward as you thought it might be, Tweek, but... whadya think?
Anybody? Thoughts? Questions?
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OK, so... I am thinking that someone who is really on it in the right gear will definitely be accelerating up the hill and needing to lift up a tad to brake a bit and downshift for 14a. In the meantime, ride the right line and get your gearing sorted. Ask a local racer what they are running. It should really help your drive out of all the corners.
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Oh, and by the way, Barber looks freaking AWESOME!
I am SO adding it to my "to do" list.
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OK, I found this on-bike video "hot lap" of a CSS student at Barber. He's going really, really slow, but, the turns are labeled on screen and I get a better idea of how Turn 14/14a works.
http://video.aol.com/video-detail/superbik...457?icid=acvsv2
This youtube link of the same video lets me go fullscreen:
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The map on the Barber site calls the hill T-14 and the right at the top T-14a. I get the idea you are calling the hill, "the hill", and the right turn at the top T-14.
So, based on that, I take it that you end up to the far left side going up the hill and basically stay there until the top where you find yourself in good position to enter "14" still leaned over where you must at least roll out a bit and downshift before getting back into the gas for 14 (I assume).
Yeah, the whole point of flicking is getting from straight up and down or from a previous lean angle to a new lean angle. If you are already mostly leaned over and in a good position, the only reason to pick it up would be to brake or find a better line for 14 which you already say you don't see the pro's doing. Honestly, you will probably learn more watching them than by me looking at a track map, but, I am thinking that unless you are going pretty slow at the top of the hill, you must need to lean at least a tad more for 14 even with a lot of camber. But, it is hard ot know without seeing the track in real life., Perhaps I will dig up a video online later.
Good luck,
racer
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So, what... no other students of physics out there? Nobody going to check my math for me?
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Hey tweek,
Have you had a chance to study the links I posted above?
Using the ice skater formula from link #3, substituting 50° for θ (angle of lean) and solving for velocity (v), I tentaively came up with ~ 42 kph or ~ 26 mph.
I haven't gone back through to double check my formulas or math; but, did any other "physics students" get anything different?
racer
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The "quick flick" is STANDARD OPERATING PROCEDURE for CSS and should be for anyone looking to improve their lap times but.....I can give you one example where there might be opposing theories of approach, Barber turn 13? and 14? if you come down off the tunnel turn flat out, you are leaned over through the apex of 13 coming into 14 which is the turn at pit out. WHy should I try to pick the bike up for the entry of 14 to "flick" when I could stay at maximum lean angle for both....It seems to me if I am quick flicking in that corner before the last turn for the straight then I am not going fast enough?
Alright, I've never been to Barber, but, I pulled up their website and I am looking at an elevation drawing of the track that appears to include camber. (Man, that looks like a cool track!) So, my comments are based on what I can see in this drawing.
We (you, me, my dog and CSS) have been discussing a fundamental skill applied to single turns in isolation (easiest way to communicate and grasp fundamental concept) ... as opposed to more advanced combinations of turns that require a slightly more complex approach.
So, T13 looks like a dip with some camber as you approach T14 that appears to sweep uphill and due to the elevation change and camber could effectively create a continuous decreasing radius situation. Or perhaps what appears to be DR but effectively is CR or only mildly DR. Not sure what gear you are in or how fast or how much the hill affects acceleration, ie. how much speed do you actually gain up hill toward the crest at the transition to entry for 14a? Do you upshift on the hill? Do you need to brake or downshift for 14a? It looks like there is a lot of camber through there. I don't know if that is exaggerated on the track map or if that is to scale or what.
So... it sounds like you are saying that you personally are able to simply maintain lean angle or lean it over a little more for 14a without picking it up to "reset" like a double apex? Apologies for my lack of track knowledge there. Can you give me more detailed info, ie. gear, rpm's, gear changes, braking, etc?
racer
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I think failure of the traction control system would simply mean that "full power" became available, not that the throttle became stuck wide open. So, it failed to prevent the rear wheel from spinning up, but, it still hooked up when Lorenzo closed the throttle... if TC failure is in fact what happened.
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What do you think?
racer
getting on the gas the soonest vs. getting off the gas the latest. hmmm.
your comments did make me rethink a lot turns that i thought were suitable for trailbraking. they also helped me answer my question that i posted in the other thread. i think i get it now, i need to adjust my lines so that i can turn quicker, reach max lean angle sooner, and throttle on sooner! if im turning in later (keith has mentioned before that riders often turn in too soon) i can shift my braking zone ahead and carry a higher entry speed plus get on the gas quicker, thanks racer.
Check out page 70 of TOTW II. Coming in on the brakes almost forces a rider to turn in early or earlier to compensate for turning in slower and turning in early occurs as a SR for when charging a turn or when a rider feels like they are coming in too hot.
And, honestly... I'm still a novice at thinking through this stuff at this level and trying to verbalize it. So, thanks to you for the opportunity to exercise my noodle!
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My take: Slow in, quick flick, less time in full lean = sooner up out of lean and faster back on the gas, quicker to next corner.
A-HA!
Right. The faster you flick, the sooner you can get back on the gas and begin to accelerate and/or the less lean angle you can use so the more throttle you can apply throughout the corner AND/OR spend less time at max lean and begin the exit phase and hard throttle sooner. Especially important on a bigger bike that has lots of horsepower to put down!
Let me say it again for my own benefit...
Slow in... sacrifice a little speed for a faster flick = earlier throttle and higher corner speed AND/OR less lean and more throttle throughout the corner AND/OR less time at max lean and the sooner you can pick it up and whack it WFO!
Hey, I think I'm starting to get it. All those years on little bikes got my head wired for turn entry and corner speed. Never really turned it around to think about how a quicker flick could affect my exit ...
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I'll bet Lorenzo just loves his new traction control system...
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Setting sag properly willl keep the suspension in the best part (middle third) of the suspension travel and it is the first thing I do. I also reset the sag for trips with luggage or for planned rides with a passenger.
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What about looking at it from the standpoint of the shape of the turn? If you are coming into a combination turn such as a decreasing radius into a right hander. There is no point in getting exited coming into the first turn because it will screw up your line for the right hander? So wouldn't it be better to make sure you come into the d.r. turn slow enough so that you can drive out for the correct line of the r.h. In this case "Slow in - Fast out" ?
Right. Absolutely.
A similar scenario could be a turn that loses camber between entry and exit whereby you'd want to plan your approach to be at max lean at the point of least or negative camber.
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Have some free time so I'll bug you guys some more! Not sure where all this will go, I've not quite planned it.
First, what should be a straight forward math/physics problem/simulation:
Say you have a 200lbs rider with a 400lbs sport bike. The rider and the bike are on a giant flat cement skid pad. The rider is holding the bike in a constant radius turn that makes a circle with a 50' ft radius. How fast does the bike have to be moving in order for the rider to keep it at a 50degree lean angle?
(I really would like to know the equation for this)
So there you have it. Anybody up for some physics calculating?
If we assume infinite traction, define a moment arm between the combined CoG and an arbitrary point at right angles between the contact patches (regardless of f/r balance) at a constant speed to simplify things, I believe the speed at which the component force vectors applied to our moment arm equal each other (and the acceleration of gravity) will be your answer. (Or close enough to it... I think.) I'll have to work on this when I have more time (and sleep).
Edit to add:
Actually, if you study these links, I'll bet anyone who has had high school algebra and trigonometry can solve it for themselves:
1. http://www.glenbrook.k12.il.us/GBSSCI/PHYS...cles/u6l1e.html
2. http://www.glenbrook.k12.il.us/GBSSCI/PHYS...tors/u3l3b.html
3. http://www.glenbrook.k12.il.us/GBSSCI/PHYS...cles/u6l2c.html
racer
Lost My Way...
in Cornering and Techniques
Posted
Well, it is pretty vague.So, sorry I can't offer any specific riding advice. However, my question is this, are your lap times really that bad, does your riding feel "all wrong" or uncomfortable to you or is it a matter of where you are finishing in the races?
Here in the states, the 600 class tends to be the largest and most competitive. Is it possible there is simply more and better competition?
Lastly, if you haven't read Keith's books or taken a school, I would recommend checking them out.
Cheers,
racer