

Thread Tools  Display Modes 
#21




Braking while turning
All these numbers are quite unnecessary. Look at it this way:
In a race in any vehicle:  at the apex of the corner, you want to be using up 99% (or a bit more) of your available traction from turning. If you are not, you could be going faster.  Before entering a turn, you want to be using up 99% of your available traction due to braking. If you are not, then you could have held your speed longer. What you want to do, therefore, is match the two so that they add up to almost 100%. You brake as hard as you can before a turn, and you back off the brakes as cornering force increases. Doug Christopher Brian Colohan wrote in message ... Joe Riel writes: ... Do you corner anywhere near that lean angle? Few do. Ok, so you first establish that on flat pavement with no surface gunk friction is very high. And you argue that few folks ride anywhere near this limit on corners. So for most people fc is not that high. Second, because the total force on the contact patch is the vector sum of the cornering force and the braking force, which are applied perpendicular to each other, it is possible to apply considerable braking force while barely changing the total force. For example, let the braking force (fb) be 20% of the cornering force (fc). The total force is then You then argue that 20% of fc (which we established is not that high) is a "significant braking force". For some reason, I disagree. :) ftot = sqrt(fc^2 + fb^2) = fc*sqrt(1+(fb/fc)^2) ~ fc*(1+(fb/fc^2)/2) for fb fc = fc*(1+(2/10)^2/2) = 1.01*fc Here is a much easier way of solving (and understanding) this, with the bonus of getting a more accurate answer: ftot = sqrt(fc^2 + fb^2) Assume that fc = 1unit. Assume that fb = .2*fc = .2units ftot = sqrt(1^2 + .2^2) = 1.02 Now, if you assume that the rider is cornering somewhat conservatively, because they fear that there might be a patch of sand or tar on the corner somewhere, then you can assume that fc is not too high. What happens if they are also going down a steep hill, and want to brake to maintain their speed? If fc=fb, then this changes to: ftot = sqrt(1^2 + 1^2) = 1.41 You have just lost 40% of the margin of safety you planned to have in the corner. I do agree that the best way to learn this is not through math, but through practice. I found that a great way to learn how to deal with low traction conditions in corners is to go out and ride just after a fresh snowfall (before the plows come by). Very low traction, loads of fun, and if you happen to fall then the snow offers some padding... Chris 
Ads 
#22




Braking while turning
wrote in message
... One thing I have noticed is how differently bicycles handle when the rear brake is applied compared to the front brake. While braking with the front wheel and turning, the rear wheel naturally swings around follows through the turn. The steering feels the same as if the brake wasn't applied. If I brake while turning, the front suspension compresses and the entire steering geometry changes drastically. The first ride after getting the RS Pilot XC's put on, I damned near came unstuck on the first corner that I applied even light braking on the front. I now try to avoid the front brake whilst in turns. Trentus 
#23




Braking while turning
"Douglas Landau" wrote in message
om... All these numbers are quite unnecessary. Look at it this way: In a race in any vehicle:  at the apex of the corner, you want to be using up 99% (or a bit more) of your available traction from turning. If you are not, you could be going faster.  Before entering a turn, you want to be using up 99% of your available traction due to braking. If you are not, then you could have held your speed longer. What you want to do, therefore, is match the two so that they add up to almost 100%. You brake as hard as you can before a turn, and you back off the brakes as cornering force increases. From the FAQ: "Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G centrifugal acceleration, he can still apply 0.1 G braking and hardly increase the load on the tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake substantially near maximum cornering. The centrifugal acceleration changes as the square of the speed, so braking rapidly reduces the required lean angle and allows increased braking. Being aware of this relationship should leave no doubt why racers are nearly always applying brakes at the apex of max speed turns." 
#24




Braking while turning
"Peter Cole" writes:
"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G centrifugal acceleration, he can still apply 0.1 G braking and hardly increase the load on the tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. A bit of a tangent: does this mean there are bike tires with a coefficient of friction larger than 1? If so, how do they do itglue? super sticky rubber? Just curious... Chris  Chris Colohan Email: PGP: finger Web: www.colohan.com Phone: (412)2684751 
#25




Braking while turning
"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G centrifugal acceleration, he can still apply 0.1 G braking and hardly increase the load on the tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake substantially near maximum cornering. The centrifugal acceleration changes as the square of the speed, so braking rapidly reduces the required lean angle and allows increased braking. Being aware of this relationship should leave no doubt why racers are nearly always applying brakes at the apex of max speed turns." Have you guys actually DONE this? or are you just repeating things? Mike 
#26




Braking while turning
"Mike S." mikeshaw2@coxDOTnet wrote in message
news:7KyTa.14678$Bp2.738@fed1read07... "Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G centrifugal acceleration, he can still apply 0.1 G braking and hardly increase the load on the tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake substantially near maximum cornering. The centrifugal acceleration changes as the square of the speed, so braking rapidly reduces the required lean angle and allows increased braking. Being aware of this relationship should leave no doubt why racers are nearly always applying brakes at the apex of max speed turns." Have you guys actually DONE this? or are you just repeating things? Since it was (stated as) a quote from the FAQ, I certainly was "repeating things". But it is only simple vector math, something bicycles (and everything else in the universe) must comply with. 
#27




Braking while turning
"Christopher Brian Colohan" wrote in message
.. . "Peter Cole" writes: "Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G centrifugal acceleration, he can still apply 0.1 G braking and hardly increase the load on the tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. A bit of a tangent: does this mean there are bike tires with a coefficient of friction larger than 1? If so, how do they do itglue? super sticky rubber? It's an example. Plug in whatever numbers you want. Maximum centrifugal acceleration may be limited by other factors than the coefficient of friction of the rubber tread. Read the FAQ for a more complete description. 
#28




Braking while turning
Have you guys actually DONE this? or are you just repeating things? Since it was (stated as) a quote from the FAQ, I certainly was "repeating things". But it is only simple vector math, something bicycles (and everything else in the universe) must comply with. The math's great, but have you actually DONE what you're preaching? I've done it both ways. Braking going straight, then turning seems to work better for me. Seems that I can corner harder when I'm not braking in the middle of the turn. Then again YMMV... Mike 
#29




Braking while turning
"Mike S." mikeshaw2@coxDOTnet wrote in message news:EzzTa.14689$Bp2.7758@fed1read07... Have you guys actually DONE this? or are you just repeating things? Since it was (stated as) a quote from the FAQ, I certainly was "repeating things". But it is only simple vector math, something bicycles (and everything else in the universe) must comply with. The math's great, but have you actually DONE what you're preaching? I've done it both ways. Braking going straight, then turning seems to work better for me. Seems that I can corner harder when I'm not braking in the middle of the turn. Then again YMMV... Most of us do not get off the bike before a turn, measure the friction coefficient of the road surface, the camber and the angle of the turn and then perform a simple mathematical equation to determine how hard we can squeeze the brakes  although I think I saw Beloki with his sliderule out just before the big crash. Most of us have an experiencedbased sense of when an how to brake. Those lacking that sense  or those riding on a descent with obscured sight lines  are best served with a general rule that braking should be done before the turn. This avoids panic braking during the turn, which, depending on the angle and pitch of the turn, can be the kiss of death. I don't think anyone could reasonably argue that you should wait until you are in the middle of a steep, blind or offcamber corner to brake.  Jay Beattie. 
#30




Braking while turning
From the FAQ:
"Take for example a rider cornering on good traction, leaning at 45 degrees. With this 1 G centrifugal acceleration, he can still apply 0.1 G braking and hardly increase the load on the tires, which is given by the square root(1^2+0.1^2)=1.005 or 1/2%. In other words, you can brake substantially near maximum cornering. The centrifugal acceleration changes as the square of the speed, so braking rapidly reduces the required lean angle and allows increased braking. Being aware of this relationship should leave no doubt why racers are nearly always applying brakes at the apex of max speed turns." From the FAQ or not, this paragraph falls apart in multiple ways halfway through. The first part is correct  up through the statement "you can brake substantially near maximum cornering". However, second half of the next sentence is incorrect. 1. Reduced speed, not braking itself, reduces required lean angle. 2. Even this is not a result of the fact that centrifugal acceleration changes as the square of the speed, so the word "so" in the middle of that sentence is incorrect. 3. Reduced lean angle does not permit increased braking, excepting for subtleties arising from, for example, lessthanperfectlyrigid wheels. 4. Increased braking is not what you want at the apex of a turn. There is a proper place for braking up until very close to the apex. 
Thread Tools  
Display Modes  


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Thoughts on braking  John Appleby  General  76  August 11th 03 10:30 AM 
Braking Technique  asqui  Racing  55  July 25th 03 04:16 PM 
brake pads are wider than braking surface  Michael  Techniques  2  July 10th 03 05:45 PM 