fixed gear cluetrain: update

Stop the presses. I just read the section in Bicycling Science again, and while the previous post is a generally accurate estimate, math don't like to generalize. As a matter of fact, the page is question is a little vague. Suffice it to say that force = mass times acceleration. Mass is the weight of the rider, duh, and acceleration, on this planet, at least, is 1g, or 9.8 m/sec. However, the author decides that deceleration is about .5g. I don't dispute this, but it may have proven pages earlier. Another point of contention is that Force is calculated according to distribution...that is, the COM is slightly closer to the back wheels, so at a fixed speed 40% of the force (measured in Newtons) would be on the front wheel, and 60% on the back wheel. I certainly don't dispute this, but it gets a little muddy on exactly why this transfers when braking the front wheel. Not only that, but the calculations put 90% of the resulting N on the front wheel...lifting the back wheel ever so slightly and making it very easy to skid the back tire if both brakes are locked up simultaneously. What's unclear is how this 90% figure is derived, and exactly how he achieves the "twice as much downforce" on the front wheel. Apparently, the correlation between stopping time and stopping distance is fairly linear, so "half the time" equals "half the stopping distance" so that's one less thing to clear up. I will transcribe the pages soon (it requires all sorts of annoying markup like f_V,r) and get a certified Mathematical opinion to help sort it out in English.