On Wednesday, May 27, 2020 at 6:37:32 AM UTC-7, Axel Reichert wrote:
jbeattie writes:

On Tuesday, May 26, 2020 at 2:54:50 PM UTC-7, wrote:
This combination makes out and back TT's in the wind, better to
expend as much energy as possible into the wind and then as much as
you have left downwind. It's a math problem and actually works.

Slaughter yourself on the climbs and recover on the descents. You
can do it with headwinds, too, assuming you don't over-spend on the
way out.

I think we have to clearly separate two things he

1. External (wind and climbs are the prime examples, but different surface,
e.g., sand versus ultra-smooth asphalt, would also be possible)

2. Internal (the effort you are spending at a particular point of
the trip, also known as pacing).

Now with regards to 1, if you use

http://kreuzotter.de/english/espeed.htm

go to the "drops" position, enter 400 W, 0 wind, 0 slope and 20 km
(default data elsewhere), you will get 27 min 39 sec for the time. Do
it again with 400 W, 30 km/h wind, 0 slope and 10 km, you will get 22
min 18 sec for the outbound time. Do it again with 400 W, -30 km/h
wind, 0 slope and 10 km, you will get 9 min 22 sec for the inbound time.
In total this 31 min 40 sec for the round trip, so considerably longer
than without wind. The same effect happens if you use +5 and -5 for
slope out-/inbound, respectively. This is left as an exercise to the
reader.

This is the point I was trying to make. External, additional forces WILL
slow you down.

Now with regards to 2, let us assume that you can sustain 400 W only for
10 min, and after that you bonk and have to limp home running on the
emergency power system with 50 W.

First approach is to spend your energy against the wind. With 400 W, 30
km/h wind, 0 slope and 4.483 km your "strong" 10 min are spent. With 50
W, 30 km/h wind, 0 slope and 5.517 km still to go, you will need 45 min
21 sec for the remaining outbound distance. Then it is 50 W, -30 km/h
wind, 0 slope and 10 km go inbound, which need 15 min, 47 sec. In total
that is 71 min 8 sec.

Second approach is to spend your energy with the wind. We know from
above that the inbound leg takes only 9 min 22 sec with 400 W. So you
have 38 sec of 400 W left on the outbound leg, which bring a mere 0.284
km. So 9.716 km to go with 50 W, 30 km/h wind, 0 slope, needing a
whopping 79 min 52 sec. In total that is 89 min 52 sec.

This is the point you were trying to make. Spend your energy there where
you need it most (against the wind or uphill).

In the hilly/windy time trials that were mentioned, these two effects
superimpose. Hills and wind will slow you down, but smart pacing will
give you an advantage over the less smart competition. If, however, you
record the power as a function of time and apply this to a course
without hills or without wind, you will always see that you are able to
achieve a higher average speed: For the above power profile (400 W for
10 min, then 50 W) with 0 km/h wind and 0 slope your "strong" 10 min
last for 7.233 km. 12.767 km to go with 50 W, which take 39 min 5 sec,
for a total of 49 min 5 sec. Much faster than both bad and good pacing.

So in summary, we were both right, but things were a little bit
muddled. I hope this clarifies things.

Axel

I wasn't really advocating any position, but that is an interesting explanation.

I train based on FT threshold. I was climbing with a friend on my lunchtime ride through the west hills yesterday, and he sprints to the top of a maybe two mile climb. I'm just grinding along but decide to give it a big effort, knowing that when we hit the top, we'll go right or left on a rolling ridge road. So, I empty the tank, hit the top -- and he goes straight ahead on a route we never take up a little 20% goat road and on to a dirt trail. The guy just wants to punish me this year. I yell "f*** that!"

That's my FT threshold. If I exceed my FT threshold, all models fail, my output reduces to near zero, and swearing ensues. They need to factor FT threshold into those on-line calculators.

-- Jay Beattie.