K
Ken Roberts
Guest
In the midst of the "pedaling tricks" discussion it hit me that pulling up
has an advantage over pushing down. So now I'm wondering "Which parts of
that idea are correct?" and "How much is it worth?" . . . so I'll start with
this
Claim:
Using the leg's own muscles to pull up the weight of the leg during the
upstroke transmits a higher percentage of its power (measured in Watts) to
the pedal and wheel -- a higher percentage than pushing down in the
downstroke (during seated pedaling in most situations).
I'm not saying that pulling up on the _pedal_ is good idea. I'm convinced
that for me trying to pull up on the _pedal_ is generally a bad idea which
will tend to reduce my net total power to the pedals. Trying to use the
leg's own muscles (hip-flexion and knee-flexion) to pull up against some
portion the leg's own weight (and against some resistance from the leg's
own down-push muscles) is less radical and somewhat different from trying to
pull up on the pedal.
Upward-pull work has greater mechanical efficiency because its force and
work is applied directly to the pedal and crank (by cancelling a portion
of negative forces on the pedal) -- while Down-push work must first be
transmitted through the ankle-joint (some also thru the knee-joint) in
order to reach the pedal and crank.
There are four kinds of power losses with down-push power:
(A) transmission loss of both knee-extension + hip-extension power thru
ankle joint.
(B) transmission loss of hip-extension power thru knee joint.
(C) problem with complicated contraints on high-power performance due to
limit on quasi-static torque thru ankle joint.
(D) problem with complicated contraints on high-power performance due to
dynamic force / speed / range-of-motion constraints of knee joint.
(more detail on these in another post.)
A + B are fairly straightforward to understand, but C + D are kind of
tricky, so
*** I'm wondering if I've understood these losses correctly.
How much? I'm guessing that the simple transmission losses are around one
half percent per functional muscle group per transmission joint -- thus (A)
1% for the ankle joint; (B) 0.5% for knee joint. And I'll guess that the
other two more complicated losses cost: (C) 0.5% for ankle; (D) 0.25% for
knee. So the additional mechanical inefficiency losses for push-down power
are around 2.25%.
*** I'd love to get some clues for better estimates for each part of those
power losses. (The reason I'm giving these guesses is to prod somebody to
try to offer something more accurate.)
Suppose a cyclist in a 40km time trial at 90rpm produces 350 Watts all in
down-push muscles, of which 8 Watts is lost related to downstroke knee and
ankle joints by A + B + C + D, and say 60 Watts is lost in negative work by
down-push muscles during the upstroke, so the power to the pedals is 282
Watts. It the cyclist instead shunts some of the oxygen to upward-pull
muscles, then perhaps the Down-push muscles produce 290 Watts, Upward-pull
muscles 60 Watts, the A + B + C + D losses are reduced to 6.5 Watts, the
resistance during upstroke reduced to 50 Watts, but now there is some
additional resistance by upward-pull muscles during the downstroke of say 10
Watts. So out of the 350 Watts produced by the muscles, the power delivered
to the pedals would now be 283.5 Watts -- for a gain of 0.5%
Implications:
* In high-power performance, slightly greater power to the pedals and wheel
should be achieved if oxygen is shunted somewhat disproprortionately to the
upward-push muscles. (It's possible that some experienced cyclists do this
unconsciously.) But I'd be surprised if the gain in total power from
improved mechanical efficiency is larger then 1%. And I doubt many skilled
riders use _no_ upward-pull work in a time-trial performance, so lots of us
have already received a significant portion of the mechanical efficiency
gain that's available. (Even apart from efficiency, the upward-pull muscle
fibers are attached to the leg anyway, so it makes sense to use them for
pedaling.)
* Training strategy: Disproportionately larger adaptation of upward-pull
muscles over the long term should result in higher total power to the wheel
than adaptation proportional to current power distribution between
upward-pull and down-push muscles. (There are reports that some pro racers
are pursuing this strategy)
The magnitude of the mechanical efficiency gain from shifting training
proportions is not likely to be substantial for a well-trained cyclist.
Suppose a racer currently can produce 290 Watts in 40km time trial
performance from down-push muscles and 60 Watts from upward-pull muscles --
out of which 283.5 Watts gets delivered to the pedals. Then suppose this
racer can achieve a 5% increase in muscular power sustainable for a 40km TT.
If this increase is distributed proportionally according to current muscle
capacities, then the power to the pedals rises to 297.5 Watts. If the
increase were made _completely_ thru adaptation of upward-pull muscles, with
no adaptation of down-push (a very unlikely occurrence), the power to the
pedals rises to 280 Watts, due to reducing A + B + C + D losses by 0.5 Watt.
Gain of less than 0.2%.
So either the down-push power losses are much larger than I guesses, or
there would need to be some other special benefit to persuade me to put a
major focus on training upward-pull muscles. (The radical truly long-term
strategy would be to first allow all my leg muscles atrophy, then completely
rebuild them to a different proportion of muscles skewed toward upward-pull.
Not for me.)
Lots more detail (way too much) in following post.
Ken
has an advantage over pushing down. So now I'm wondering "Which parts of
that idea are correct?" and "How much is it worth?" . . . so I'll start with
this
Claim:
Using the leg's own muscles to pull up the weight of the leg during the
upstroke transmits a higher percentage of its power (measured in Watts) to
the pedal and wheel -- a higher percentage than pushing down in the
downstroke (during seated pedaling in most situations).
I'm not saying that pulling up on the _pedal_ is good idea. I'm convinced
that for me trying to pull up on the _pedal_ is generally a bad idea which
will tend to reduce my net total power to the pedals. Trying to use the
leg's own muscles (hip-flexion and knee-flexion) to pull up against some
portion the leg's own weight (and against some resistance from the leg's
own down-push muscles) is less radical and somewhat different from trying to
pull up on the pedal.
Upward-pull work has greater mechanical efficiency because its force and
work is applied directly to the pedal and crank (by cancelling a portion
of negative forces on the pedal) -- while Down-push work must first be
transmitted through the ankle-joint (some also thru the knee-joint) in
order to reach the pedal and crank.
There are four kinds of power losses with down-push power:
(A) transmission loss of both knee-extension + hip-extension power thru
ankle joint.
(B) transmission loss of hip-extension power thru knee joint.
(C) problem with complicated contraints on high-power performance due to
limit on quasi-static torque thru ankle joint.
(D) problem with complicated contraints on high-power performance due to
dynamic force / speed / range-of-motion constraints of knee joint.
(more detail on these in another post.)
A + B are fairly straightforward to understand, but C + D are kind of
tricky, so
*** I'm wondering if I've understood these losses correctly.
How much? I'm guessing that the simple transmission losses are around one
half percent per functional muscle group per transmission joint -- thus (A)
1% for the ankle joint; (B) 0.5% for knee joint. And I'll guess that the
other two more complicated losses cost: (C) 0.5% for ankle; (D) 0.25% for
knee. So the additional mechanical inefficiency losses for push-down power
are around 2.25%.
*** I'd love to get some clues for better estimates for each part of those
power losses. (The reason I'm giving these guesses is to prod somebody to
try to offer something more accurate.)
Suppose a cyclist in a 40km time trial at 90rpm produces 350 Watts all in
down-push muscles, of which 8 Watts is lost related to downstroke knee and
ankle joints by A + B + C + D, and say 60 Watts is lost in negative work by
down-push muscles during the upstroke, so the power to the pedals is 282
Watts. It the cyclist instead shunts some of the oxygen to upward-pull
muscles, then perhaps the Down-push muscles produce 290 Watts, Upward-pull
muscles 60 Watts, the A + B + C + D losses are reduced to 6.5 Watts, the
resistance during upstroke reduced to 50 Watts, but now there is some
additional resistance by upward-pull muscles during the downstroke of say 10
Watts. So out of the 350 Watts produced by the muscles, the power delivered
to the pedals would now be 283.5 Watts -- for a gain of 0.5%
Implications:
* In high-power performance, slightly greater power to the pedals and wheel
should be achieved if oxygen is shunted somewhat disproprortionately to the
upward-push muscles. (It's possible that some experienced cyclists do this
unconsciously.) But I'd be surprised if the gain in total power from
improved mechanical efficiency is larger then 1%. And I doubt many skilled
riders use _no_ upward-pull work in a time-trial performance, so lots of us
have already received a significant portion of the mechanical efficiency
gain that's available. (Even apart from efficiency, the upward-pull muscle
fibers are attached to the leg anyway, so it makes sense to use them for
pedaling.)
* Training strategy: Disproportionately larger adaptation of upward-pull
muscles over the long term should result in higher total power to the wheel
than adaptation proportional to current power distribution between
upward-pull and down-push muscles. (There are reports that some pro racers
are pursuing this strategy)
The magnitude of the mechanical efficiency gain from shifting training
proportions is not likely to be substantial for a well-trained cyclist.
Suppose a racer currently can produce 290 Watts in 40km time trial
performance from down-push muscles and 60 Watts from upward-pull muscles --
out of which 283.5 Watts gets delivered to the pedals. Then suppose this
racer can achieve a 5% increase in muscular power sustainable for a 40km TT.
If this increase is distributed proportionally according to current muscle
capacities, then the power to the pedals rises to 297.5 Watts. If the
increase were made _completely_ thru adaptation of upward-pull muscles, with
no adaptation of down-push (a very unlikely occurrence), the power to the
pedals rises to 280 Watts, due to reducing A + B + C + D losses by 0.5 Watt.
Gain of less than 0.2%.
So either the down-push power losses are much larger than I guesses, or
there would need to be some other special benefit to persuade me to put a
major focus on training upward-pull muscles. (The radical truly long-term
strategy would be to first allow all my leg muscles atrophy, then completely
rebuild them to a different proportion of muscles skewed toward upward-pull.
Not for me.)
Lots more detail (way too much) in following post.
Ken