Let us then draw (in our heads) the graph of the deduced Pogo+bike system produced power as a function of time. Up to the moment the acceleration starts, we have a constant 600W (horizontal line on our graph). Then we have an instantaneous in our model (very quick in reality) jump to 600+540=1140W. At the moment acceleration begins, the speed is still 6m/s, so first term in that sum (power going into potential energy increase) is still 600W. Then the speed starts growing, and the first term grows as well (say, linearly) from 600W to 900W. The acceleration is still in place during that whole second period and the power producing it (the part of it going into kinetic energy increase) is still (in our model) a constant 540W. Thus the power grows from the initial value of 1140W to the final one of 900+540=1440W. Now the acceleration is finally done, and the power drops instantaneously (in our simplified model) to the new constant value of 600+300=900W where it continues until the finish or close to it. So we indeed have a peak power of over 1400W which had to be maintained for at least a couple of pedal strokes. The latter would require, according to our earlier calculation,
a force of 2 (two!) Pogo's body weights. And he still exerted that seated?!!! Are you going to give me that break or do I have to take it from you?
Remember also that we've totally neglected the supralinear dependence of the constant velocity component of the required power on the velocity itself. That power consists -- when going up a steep hill -- of the "gravity fighting" component and that producing work against air and rolling resistance (plus a bit of internal friction as well). Now, the first component, as we noted, is the dominant one on a hill like Huy, and it exhibits linear dependence. But the air resistance at 9m/s is already quite significant exceeding that at 6m/s by a factor of approximately 1.5^3=3.3 (rounding down to your benefit). So, in our model, that constant component of power, to which Pogo+bike system goes to after the acceleration phase is done, would be more than 900W.
Now that we have done -- at your prompt -- these simple approximate calculations, I actually start thinking that the power of Pogo's bike assist that he seemingly activated by an under-hood button press was more than even 500W (easy-peasy for a brushless+lipo system -- I know that from experience). In fact, the likely reason he did not stand on the pedals and looked somewhat "disjointed" during that acceleration phase was that the assist -- "dialed' by the old ghoul Mauro -- was so strong that he was literally afraid to lose his balance by standing on the pedals. That also explains his unhappy look at the finish.
P.S, You can easily feel the power dependence we discussed here by doing, for example, a standing time trial start. In the beginning you really mash the pedals (standing, of course) and can literally feel the power flowing through you (it is still easy at this point as you are totally fresh). Then you reach your intended cruising speed (say, 45 km/h) and, in spite of your speed being the highest so far, can get back on the saddle as the power demand has just dropped.
A little extension of that analysis while you are refreshing the laws of classical dynamics in your memory. What we described was -- as was explicitly stated -- a much simplified model good for obtaining ballpark figures. In the Muur de Huy reality, the acceleration phase probably lasted a bit longer and the acceleration itself was not constant, tapering off towards the end of the acceleration phase. What was (approximately) constant is the e-assist power likely initiated by that under-brake-hood button press. That power likely stayed approximately constants almost to the finish, possibly tapering off somewhat due to the controller's heat management function (but, given cool and rainy weather, that effect probably was not very pronounced). So the real Pogo+bike power output profile likely looked something like the following.
Before the acceleration phase, we have around 600W (possibly a bit less since overall that was not a fastest Muur climb). Then assist goes on and pushes the total power to something like 1100-1200W (remember that initial 1140W in the simplified calculation) where it stays throughout the whole acceleration phase. Pogo himself possibly adds an extra 100 almost instinctively (this is the "go" moment, after all). As we know, the result of this almost constant system power output is the acceleration behaving as a decreasing function of time (we could even write a differential equation to obtain that dependence, if we felt like it). Qualitatively, the reason for that decrease is that the increasing speed makes more of the total power go into work production against gravity (and air resistance) or, equivalently, into potential energy increase. So less is left to contribute to the kinetic energy gain (aka acceleration).
After the acceleration phase is over, the Pogo+bike system settles into constant 900W (or maybe even a bit less) power output mode, simply to make sure that the speed up the Muur does not go into some comical 40+ km/h territory. But, with power assist activated, it likely keeps pumping that 500-600W into the road (or, rather, into potential energy). So Pogo himself can ease into a pleasant 300W. That readily explains his easy looking nose breathing finish noticed by everyone.
That's some solid background. Much better than philosophy.
You should have no difficulties following these calculations and their meaning.
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.
First of all I would like to thank you for the nice discussion. Second, I am sorry to all the folks who are not that much into physics, my post will contain some numbers as well and this whole back and forth with
@Casual cyclist may be more suited for a physics forum, but maybe some not so technically inclined people will find these posts interesting as well.
I finally had some time today to think about what you wrote and do some analysis on my own. Below is the segment I want to analyze. This is from when the attack took place until 20 seconds after. I chose this segment because I can see from the broadcast that Healy needs 24s to ride this segment. FYI on the Discovery broadcast that lasts 2:25:24 hours the attack happens at 1:58:31 (basically when the director cuts to the frontal view of Pog) and it lasts until 1:58:51 when Pog is at point A. Healy passes point A at 1:58:55.
Now, I did some modeling and I believe I got the constants pretty spot on.
It is a shame I cannot use LaTeX, but here is my DE:
P(t) = dE/dt = m*g*sin(\theta)*v + m*v*v' + k*v + q*v^3 (for these speeds I assume that the rolling resistance (RR) is constant)
m = 72kg (Pog+bike)
g = 9.81m/s^2
k = 3.6 kg*m/s^2
q = 0.25 kg/m
\theta = 0.2 rad (we see from the Strava file that the average slope of the segment of interest is 20% and the angle is small enough so sin(\theta) = 0.2). I have to add that the model could be improved here by making the angle be a function of time, but I believe even this constant value is good enough to illustrate my ideas.
Rewritten for v (the speed) it is:
v' = P(t)/m/v - g*sin(\theta) - k/m - q/m*v^2
Let us now turn to modeling the input power P(t). I suggest we use the following model:
P(t) = IP+EP*(1-exp(-t*10)),
where the Initial Power (IP) is the steady state power before the attack and the Extra Power (EP) is the increase of power during the attack. This makes it so that the maximum power is reached after about 0.5s after the attack.
Let us try to infer what the IP should be. According to Strava, in the 14s before the attack the slope is 10.4% and his average speed is 5.36 m/s and that would require about 450W (I am including the RR and Aerodynamic Resistance (AR) at this speed), so IP = 450W. Now what should the EP be in this model so that he covers a little more than 98m in 20s? The answer is about 320W.
Let us now see what does this model predict graphically.
Let us do a sanity check. As argued before, the KE increases as the power increases but eventually the increase in KE goes to 0. The increase per unit time in PE is higher for higher total power input. The RR and AR are small but non-negligible as you correctly argued. Notice the interesting point where the AR=RR (just put of curiosity, nothing to do with the analysis). You might argue that your analysis assumed 6m/s at the beginning of the acceleration and 9 m/s at the end at a constant slope much smaller than what I have here. OK, but the reality is that Pog attacked at the steepest point and when the gradient changed so there is not such a big change of the KE of 1500J as in your model but a mere 572J in mine (I computed the integral under the d(KE)/dt curve).
Edit: You predicted accurately that it would take about 3s to reach max velocity.
If we assume a more modest EP = 180W for Healy the (equivalent) model for Healy would be something like this:
Note that it would take 4 more seconds to cover the 98m, so we are on the right track.
Now, I argue, Pog's power output of 450W + 320W = 770 W is doable whilst seated. Easily doable I would say since I managed to do about 600W seated for about 10s. Therefore, I do not see any evidence for motor doping by Pog during the FW race.
Let us now turn to your analysis. First, you argue that for some reason it is not enough to just increase the power from 450W to 770W, but you need an additional power to accelerate (I still do not understand why). Please look at the curves above and verify that everything is kosher so to speak. In any case, just to indulge you and for my understanding as well, I tried to model your assumptions as well. Here is what I get:
The parameters for this model are as follows:
P(t) = (exp(-t/4)/2+0.7).*(IP+EP*(1-exp(-t*10))),
where IP = 450W and EP = 520W.
Note that some of the KE was converted to PE near the end (lower speed), so the lower total energy at the input. However, this does not seem to be a good model (maybe I did not do justice to all of your thoughts, assumptions and considerations).
To conclude, I believe there is no evidence for motor doping stemming from Pog's performance on FW's final climb.
P.S. If you find the model interesting and want to play around with it, I can share the MATLAB code with you.
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