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Kalkat

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About Kalkat

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    Cornering Apprentice
  1. Keith, I got into an argument with a racer friend over whether load transfers front or back when you accelerate. I was VERY positive that load transfers back when you accelerate, and forward when you brake/decelerate. My friend thought it was the other way around. and he said he read so in TOTW, so i decided to go home and check myself. Wat i found was funny, and confusing. At numeroous places, Keith talked about loading the back by accelerating ( one such instance was chapter 10, page 81, in TOTW). but then, in TOTW II, chapter 3, i found a mention where Keith says tha
  2. yeaaaah, now we r talking. yes, higher COG will make the bike stand up(wat i called outward force in my eariler post), but u want to keep it down, thats why u wld like to keep as low a COG as possible. however, at some point, traction, surface conditions, and other factors wld come in, and therefore, u want to look for that "optimum" COG, and not necessarily the lowest COG, as somebody posted above. now next few lines, only those inclined towards theory shld read.:-) as i posted once earlier, a simplified equation for balance in corner is m*g*x=m*v^2*d/r now its simple, lower COG is
  3. I dont know why we always end up going tangentially into theory all the time. probably too much knowledge and too little application. i dont know much, and i dont care for much. All I care for is speed. watever gives me higher speed thru the corner is the RIGHT answer. and answer in this case is LOW. i dont know how to explain it well, but they dont make a formula1 car 100m high, do they?point here is that u want to MINIMIZE the force that pulls u outwards, bcos other things constant, u will get a chance to have higher speed then!
  4. Jonathan Livingston Seagull, Siddhartha, Zen, being "there" while simultaneously being "here"...this thread... just flow!!!
  5. Thanks Andy, for jolting me back out of the "geek" mode...lol. Racer, thanks.. u have a gr8 time with ur vacation here in Penn and then NZ!
  6. Your point is taken Sir. I will leave mine to myself. About riding, if u would be in Texas, I would gladly come out and ride with u and learn a few good things I am sure.
  7. gracias for the comments senor.. but it aint my theory... I will try to find the link to it...but I agree..and I did mention... that it does not take into account MANY factors...so u r right about that. As u might have seen on other posts..Keith himself says that u can do 2 things on a bike 1.change direction 2. change speed (though he was trying to keep it simple).. Now if u would diagnose these ACTIONS... and see the FACTORS which lead to these changes, u would c that m, g,x, v, d, r ALONG WITH MANY other factors like suspension, road condition, traction , tyre, temp, etc etc arewat ena
  8. I think u will stay "on" the bike even if u were totally in air.. forget having just outside foot unlocked.. thats simple physics..newton's law..aint it? If u r weighing the outside peg, then obviously u have a grasp becuase u r putting that small downward pressure on it..isnt it(this is more of a ques, I m confused)
  9. Ok, lets take out braking. but you would still decelerate till the apex..or just stay on steady RPMs..right? But deceleration, in a way, is also braking!!! Constant RPMs too are deceleration(one of the posts below talks about lesser tyre circumference, hence lesser distance covered, hence lesser speed, implying deceleration). so anyway, u r braking till apex! now to answer Andy, yes u r right, this was a generic, simplified and theoretical statement. in practice, each turn has to be tackled differently. but if u will analyze, then the principle would still remain same. even if it were a
  10. A non-technical answer would be: brake till your apex,and accelerate all the way after apex. Now I do not mean to say that that is easy to do, but that is the right thing to do if you can. To give a technical answer: The basic(and a little simplified) logic of cornering is: m*g*x = m*square(v)*d/r where m=mass, g= gravity, v=velocity, r=radius of turn at that istant, d= height of centre of mass and x= distance of centre of mass from rear contact point(hence roughly factoring in ur lean angle) where mgx is the force that makes u fall into corner and m*square(v)*d/r is the force tha
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