I do not quite understand your analysis here. Say 600W would be enough for a steady velocity of v and 900W for 1.5v. Then the increase of KE happens during the time 900W is being produced but 1.5v is not quite reached (during this time dv/dt >0). Once the forces are balanced again he moves with constant 1.5v and no acceleration happens.
Math and Electrical Engineering here btw.
Was feeling a bit lazy on Friday afternoon and decided to write that differential equation mentioned before and see what its solution gives, just for kicks. For simplicity, we are going to neglect air and rolling resistance on a steep climb like Muur de Huy. Later, these terms can be included as well if anybody feels like it. Let P_0 be the power right before Pogo+bike system acceleration and let P_max be the total power during the acceleration phase. The general differential equation of the rider+bike motion stating that the total power is equal to the time derivative of the total (potential + kinetic) energy then reads:
mg*sin(a)*v + mv*v'=P,
where sin(a) is the sine of the inclination angle of the given climb, v' is the acceleration, and P is the total power. If the speed is constant, the second term is zero, and we can find the steady speed on the climb given the power. So let us assume 600W constant before the attack and let us use the gradient of 14% -- that is the typical on the slopes on the Muur where the attack took place. Allowing some token amount (say, 50W) for air and rolling resistance and assuming 72kg rider+bike mass, we obtain the steady speed of 5.6 m/s (20.2 km/h) that makes perfect sense.
Now let us consider the acceleration phase. We need to integrate the ODE above from time 0 (when acceleration begins) to the time when it ends. The initial speed is 5.6 m/s, and we assume the final one to be 50% more, i.e. 8.4 m/s or 30.3 km/h. We can use, for example, the variable separation technique to solve that ODE. The solution with the above boundary conditions is as follows.
T=(1/(g*sin(a)))(b*ln((b-v_0)/(b-v))-(v-v_0)),
where T is the total duration of the acceleration phase, v is the final speed, and b =P/(mg*sin(a)), P being the total power.
We can now, assuming v=1.5v_0=8.4 m/s, plug various values of P and calculate the value of T -- how long the acceleration to the final speed lasted.
Recalling that the air resistance is going to grow about three times when the speed increases by 50%, we assume that the extra power required to overcome it at the final speed to be equal to 150W. For the duration of the acceleration phase itself, that power grew from 50 to 150W, so we will assume an average of extra 100W, for simplicity. So let us say Pogo helped himself to 500W worth of assist and added 100W of his own (remember that he looked somewhat labored in the beginning of his attack). Pushing 700W seated is already no small task as we know. So we plug P=700+500-100=1100W for useful total power and obtain T=3.7 s, so pretty quick initial "burst" as we saw. To maintain that final speed, the total power of 550*1.5+150=975W is needed, so Pogo himself would be responsible for 975-500=475W which is not that bad and would further decrease toward the finish as the slope flattens. Thus his nose breathing weak celebration everyone witnessed.
). The science has become very complex and it's very likely that if you have the money you can beat ithe tests (assuming they are adminestered).
Facebook
Twitter
Instagram