pulling up is more efficient mechanically



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
 
K

Ken Roberts

Guest
[ start of this post is the same as previous, but then lots more details
below -- way too much and way too technical. ]

In the midst of the "pedaling tricks" thread 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 limits of knee joint.
(details further below.)

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 are 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.
That's a difference 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 get 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.)

Ken
_______________________________________________
Data from a scientific study:
The often-cited Kautz Coyle 1991 study of pedaling technique showed that
elite racers used much more upward-pull work when pedaling at a high
workload (around 90% of VO2max power) than for pedaling at low workload
(under 70% of VO2max power) -- which fits with C + D being more significant
for high-power performances -- and fits with a strategy of (unconsciously?)
"sparing" the upward-pull muscles during lower-power periods so they'll more
available for any possible high-power (or high-force) requirement which
might arise later.

Details on sources of inefficiency

(A) + (B) transmission loss of power thru ankle + knee joints.
"Translational" forces are transmitted from bones above the joint, thru the
joint, to the bones below. As force load comes onto the joint, the soft
stuff in the joint compresses. Then as the load comes off the joint, the
stuff expands. Although the starting and finishing configuration are the
same, there's friction and turbulence during the change, which result in
some of the work being lost as heat. I think this effect is well-known to
engineers + physicists for power transmission in this kind of situation --
the main question for me about it is:
*** How much power is lost?

"Rotational" torques are transmitted thru the tendons + muscles associated
with the "extension" articulation of each joint. The knee-extension muscles
contract to apply the force and relax and extend to prepare for the next
force -- but there's some stiffness and resistance in the "extend" phase,
and the resistance is a little higher if the force and/or speed of
contraction are higher -- which are negative drags on work. Ankle-extension
muscles have somewhat different power losses. The knee and ankle tendons
stretch a little as the torque load is applied, and the "snap back" a little
when the torque load is released. Although the starting and finishing length
of the tendon are the same, there's strain and vibration during the change,
which result in some of the work being lost as heat. I think these effects
are well-known to engineers + physicists for power transmission in this kind
of situation -- the main question for me about it is:
*** How much power is lost?

Seems to me that the loss of Work in A + B has a component that's roughly
the same for each sequence move pair of loading and unloading the joints and
tendons and muscles, and another component that is higher or lower depending
on whether the magnitude of the force + torque load is higher or lower. But
I don't know whether the fixed or variable part is larger. There have been
studies which tried to measure the internal muscular losses in pedaling, but
there are several kinds of losses, so it's hard to know what portion is due
to A + B. I'd guess at least one other kind of internal muscular loss is
larger.
______________________________
C + D overview:
These other two kinds of power loss are tricky, so first a "big picture" of
what's going on, then some detail on the assumptions, then detail on each
kind of loss.

basic idea:
When you're trying to operate a mechanical system with two or more
real-world active power sources and a "linkage" with two rotational
joints -- trying to achieve its maximum sustainable power -- you're up
against several constraints at once. If you succeed in modifying one or more
of the power sources to deliver a little more power, this higher power level
in the system pushes the contraints tighter, so you have less freedom in how
to satisfy all of them. Less freedom to optimize will not usually lead to
as high a power delivery as more freedom. Thus even if the increase in
down-push muscle power is 1 Watt, usually somewhat less than that gets
delivered to the pedals and wheel. A key reason is that a normal way to deal
with these contraints is to increase cadence frequency a little, but this
results in a higher rate of loss inside each muscle in its pair of
contracting / extending moves once in each stroke-cycle (as in the Neptune
Herzog 1999 article). On the other hand, an improvement in Upward-pull
power
capacity does not have to satisfy any complicated contraints, so there is
full freedom to transmit it directly and fully to the pedal.

*** Warning *** : The following more detailed explanation gets truly gory in
the details of mechanics. Unless you're techy beyond hard-core, just stop
reading here (if not earlier)

Because C + D are about "high power output" performance, each transmission
joint has been pushed to some sort of limit. Adding 1 Watt of power to the
output of the upward-pull muscles makes no change for these joints near
their limit, because it goes directly to the pedal. But trying to add 1
power to the output of the down-push muscles puts more force + torque load
thru the ankle + knee joints in the downstroke. Since each joint it near
some limit, it cannot simply accept that additional load, but must change
its configuration. For the ankle joint, such a change to its configuration
results in less Work per stroke-cycle -- so the only way to attain the
desired 1 Watt increase in Power rate is to increase the _cadence_ frequency
of pedaling (more stroke-cycles per minute). For the knee joint, the change
in configuration moves it to a different point on its maximum Force / Speed
/ Angle-range function, which requires a higher Speed, which means a higher
_cadence_ frequency of pedaling.

Cadence is significant because there are losses of Work (measured in Joules)
in the contracting and extending of the pushing muscles which are known to
be higher at higher cadence. The leg muscles do not fully relax immediately
after making a strong contraction to push or pull. Instead the muscle is
"stiff" for a while, so it resists getting extended back to stretched out
long enough to be ready to make its next contraction, and it requires real
work from some other muscle to overcome this resistance. The Neptune Herzog
1999 study found that the level of these Work losses rises non-linearly with
cadence above 90rpm. That work could have been used to drive the pedal, but
instead it's cancelled against stiffness + resistance. The more times per
minute this sequence of contracting and extending occurs, the more Work is
lost per minute -- so the higher the cadence, the larger the difference
between the Power generated by the muscles and the Power delivered to the
pedals and the wheel. (Likely it's easier to demonstrate this result with a
well-designed realistic computer simulation of the muscles + limbs + joints
+ pedals + cranks, but we don't have that, so first we're trying to
demostrate the result conceptually, or at least try to "motivate" belief in
the likelihood of the result.)

R. R. Neptune + W. Herzog, "The association between negative muscular work
and pedaling rate." Journal of Biomechanics 32 (1999) 1021-1026.

Assumptions shared for C + D:
Scenario: The situation is that the cyclist is trying to maximize Power to
the pedals and wheel sustained for a specific Time period, say like 45
minutes. We assume that this hypothetical cyclist's control system has found
the optimal pedaling technique to deliver the maximum sustainable power rate
given the current capabilities of the leg muscles. We then improve the
capacity of either the down-push or the upward-pull muscles to be able to
produce 1 Watt more of power, then the hypothetical cyclist's control system
finds a new modified optimal pedaling technique, and we find how much higher
its power to the pedals is than before.

We assume that the resistance on the wheel and pedals can be "fine-tuned" to
match the power transmitted from the muscles (i.e. as by slightly changing
the effective grade of the hill by a "tacking" strategy, or slightly
changing the gear ratio), so that the pedal speed and cadence can remain
unchanged even though the Power transmitted from the muscles has increased.
(This assumption just means that fewer other things must be re-calculated to
adjust for an incremental increase in power capacity of one muscle. It
should be easy to relax this assumption if we had a sophisticated computer
model simulation.)

(D) ankle in downstroke -- problem with complicated contraints on high-power
performance due to limit on quasi-static torque thru ankle joint.

The ankle joint in the downstroke operates in quasi-static or
quasi-isometric mode -- its angle is not fixed, and the ankle-extension
muscles have no positive-work motion, only some slight negative-work motion.
The maximum down-push with the ankle muscles and tendon can transmit
sustainably for that period Time depends on the configuration of the ankle
relative to the pedal axis (which corresponds roughly to the "pedal angle").
Basically, if the ankle has an overall higher relative position thru the
down-push, then it can handle a higher push force from the upper leg muscles
without suddenly "collapsing" in the midst of the push. If the
ankle-relative-to-pedal position is overall lower, then the amount of push
force that can be sustainably transmitted is lower -- because the effective
radius of the force to the pedal is larger, so the torque on the ankle joint
and its tendons and muscles is larger -- and there is a limit to how much
torque the ankle joint can sustain repeatably thru the whole performance
Time. The assumption is that the ankle position is the lowest which can
handle the overall Force and Work level required to deliver the Power from
the muscles at the desired cadence.

If (according the Assumptions shared Scenario above) we increase Upward-pull
muscular power capacity by 1 Watt without changing cadence, then there is no
change to the conditions on the downstroke Ankle joint, so there is no loss
of any portion of this increase to the downstroke ankle joint.

If (according the Assumptions shared Scenario above) we try to increase
Down-push muscular power without changing cadence, then the down-push Force
must increase. To transmit this force without collapsing, the ankle position
must be lower relative to the pedal. If the knee position is roughly
unchanged, then the pedal must start the down-push lower. Therefore the
Distance is smaller than before, so the increase in Work per cycle is lower
then expected, so the full additional Power is not utilized. We could try to
get the full additional power by increasing cadence -- but this would result
in higher power losses as discussed further above and in the Neptune Herzog
1999 article.

If we try to avoid this problem by keeping the starting pedal position
unchanged, then we must start with a higher ankle position relative to crank
center and seat, which also requires a higher knee position. But a higher
knee position is a different "range of motion" segment -- a starting
position associated with a lower sustainable Force (knee starting closer
to chest is more strenuous) -- but that it against our idea of adding 1
Watt to down-push power capacity. The compensating change is to
significantly increase cadence -- but this would result in higher power
losses as discussed further above and in the Neptune Herzog 1999 article.

Therefore there's no way to increase Down-push power by increasing Force
only while holding cadence unchanged. Some increase in Cadence is required,
but this increases power losses -- not a problem for increasing Upward-push
power with respect to the Ankle joint. Therefore with respect to the
constraint of force / torque transmission thru the downstroke ankle,
increasing Upward-push power is more efficient mechanically.

(D) knee in downstroke -- problem with complicated contraints on high-power
performance due to dynamic force / speed / range-of-motion limits of knee
joint.

The knee joint in the downstroke operates in active movement mode -- its
angle changes rapidly, and it does a substantial amount of positive work.
For a given performance Time period and the specific functional muscle
group's "aerobic capacity" level, the pushing capabilities it can sustain
depend on: (a) the average Force / Torque magnitude thru its main push; (b)
the Speed of its push; (c) the Range of Motion starting angle; and (d)
finishing angle. The Power the knee-extension muscles can generate is (a)
multiplied by (b), but only certain combinations of values of {(a), (b),
(c), (d)} are permitted. (The set of permitted combinations forms a region
in a 4-dimensional space, and the subset of likely candidate combinations
for maximum power forms a 3-dimensional "surface" of this region in
4-space). Once the cyclist's control system (or our computer simulation) has
found the permitted combination which produces the optimal maximum power,
any change to one of the four input parameters must result in a decrease in
Power.

A key trade-off in this constraint function is between Force versus Speed.
The highest sustainable Force is at a muscle speed of zero: isometric
contraction -- but no additional power is delivered with this Force, so this
isometric mode is only for transmitting power from other muscles. There is a
maximum speed above which the muscle cannot sustainably deliver Force (this
is a critical constraint for human _running_ on flat terrain). Somewhere in
between these two extremes is a Force / Speed combination which has maximum
Power.

If (according the Assumptions shared Scenario above) we increase Upward-pull
muscular power capacity by 1 Watt without changing cadence, then there is no
change to the conditions on the downstroke Knee joint, so there is no loss
of any portion of this increase to the downstroke ankle joint.

If (according the Assumptions shared Scenario above) we try to increase
Down-push muscular power without changing cadence, then the down-push Force
must increase. In seated pedaling, the leg's linkage of hip + knee + ankle
joints is fixed above to the seat position, and constrained below by
connection thru the pedal to the crank circle. Because of this geometrical
constraint, some of the increased force must be coming from increased Hip
joint torque -- because for much of the downstroke range, the torque thru
the Knee is "aimed" mostly radially away from the crank center, rather than
tangent along the crank circle, more aligned with the direction of the pedal
motion. The "aiming" of the hip-extension motion tends to "track" the
direction of the pedal velocity thru a larger portion of the downstroke.

The problem is that all of the increased active torque through the Hip joint
must be transmitted thru the Knee joint, because the Knee joint is between
the hip and the pedal. But the knee joint received only a portion of the
increase in power capacity of this "Assumptions shared Scenario". Therefore
in order to handle the higher total torque / force, its Speed must be a
little lower than before the attempted increase in power capacity -- lower
than expected for this capacity increase -- so the full increase is not
utilized.

So the only remaining way to try to draw the full additional power is by
increasing cadence -- but this would result in higher power losses as
discussed further above and in the Neptune Herzog 1999 article. Which is not
a problem for increasing Upward-push power with respect to the Knee joint.
Therefore with respect to the constraint of Force versus Speed trade-off in
the downstroke Knee, increasing Upward-push power is more efficient
mechanically.
___________________________________________________
 
G

G.T.

Guest
Ken Roberts wrote:
> 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


KISS. Just ride your bike and the efficiencies will follow.

Greg
--
"All my time I spent in heaven
Revelries of dance and wine
Waking to the sound of laughter
Up I'd rise and kiss the sky" - The Mekons
 
K

Ken Roberts

Guest
Car Fogel wrote
> What fools we've been to push down on the pedals!


Just say No.

Ken
 
T

Tim McNamara

Guest
In article
<[email protected]>,
"Ken Roberts" <[email protected]> wrote:

> [ start of this post is the same as previous, but then lots more
> details below -- way too much and way too technical. ]


And way too over thought. You've talked yourself into nonsense. A trip
to the gym will settle the question for you: can you raise more weight
by extending or contracting your leg?

Human evolution has resulted in legs that are stronger pushing against
the ground in a normal gravitational field, and weaker lifting up.
Pushing down works better, whether walking, climbing or pedaling.
 
C

cel

Guest
You don't want to continue to push down when the pedal is on the upstroke,
but actively pulling up is generally worst than having the other leg push it
up during the down stroke. The limiting factor is usually your aerobic
capacity. Now I have been know to pull-up when I have a leg that is cramping
up and I have to make it up the hill! Relax on the down stroke, pull up on
the upstroke.

cel


"G.T." <[email protected]> wrote in message
news:[email protected]
> Ken Roberts wrote:
> > 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

>
> KISS. Just ride your bike and the efficiencies will follow.
>
> Greg
> --
> "All my time I spent in heaven
> Revelries of dance and wine
> Waking to the sound of laughter
> Up I'd rise and kiss the sky" - The Mekons
 
W

Werehatrack

Guest
On Mon, 30 Oct 2006 18:40:25 GMT, "Ken Roberts"
<[email protected]> wrote:

>In the midst of the "pedaling tricks" discussion it hit me that pulling up
>has an advantage over pushing down.


Really? How so? I can use gravity to assist with the downward push,
but all of the upward pull must come from my legs...and frankly, I can
push with them a whole heck of a lot harder than I can pull.

>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 do not believe this is true. Where do you draw this conclusion
from? It is directly contrary to my own experience.

>I'm not saying that pulling up on the _pedal_ is good idea.


Then what the *********** *are* you trying to imply? There's exactly
one foot/bike interface at which propulsion effort can be applied, and
if you're not proposing to pull up on the pedals, what exactly is the
point?

>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.


How?

>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.


A specious assertion unless the statement can be supported by
something demonstrating a path for power transmission that does not
involve the knees and ankles; it should also be noted that those
joints are designed to transmit large and/or continuous load ONLY
under compression, NOT under tension; the physiological effects of
trying to do so are probably nontrivial.

>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.


I'm wondering if your source (and I'm implying here that you're not
the one who actually came up with this, but are merely posting it for
an nth opinion) understood the losses involved, either.

>How much? I'm guessing


Aha!

>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%


Absent hard data, all of this is WAG. (Not even SWAG, which is still
not good enough to draw conclusions from.

>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.)


Gosh, it would be just wonderful if the vascular system had
voluntarily manipulable points of flow control in it somewhere, but
the funny thing about that is that it doesn't.

>But I'd be surprised if the gain in total power from
>improved mechanical efficiency is larger then 1%.


I'd be even more surprised if there was one at all.

>And I doubt many skilled
>riders use _no_ upward-pull work in a time-trial performance,


Acording to their direct statements to me, your doubts are baseless.

>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.)


This is the assertion that drives the followers of "spinning", but yet
they mysteriously seem to fare no better than mash-alone cyclists when
it comes to getting power to the ground.

>* 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)


For any given Miracle Trick, there will be adherents. Once in a
while, one of them will win something somewhere; inevitably, the
miracle trick will be given credit, and a group of credulous
competitors will immediately rush to adopt the technique. In rare
instances such as the Fosbury Flop (high jumping) a new method of
doing an old task is shown to produce better results, but I do not
believe this will be the case for pulling up when pedalling in part
because it has been tried, and hasn't produced the results expected.

>The magnitude of the mechanical efficiency gain from shifting training
>proportions is not likely to be substantial for a well-trained cyclist.


Experience would seem to predict that it would be statistically
indistinguishable from "none" at best.

>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.)


After reading all of that, I am convinced that somebody spent way too
much time extrapolating from a poor assumption. Pretty much the only
part with which I can agree is the last three parenthetical words.


--
Typoes are a feature, not a bug.
Some gardening required to reply via email.
Words processed in a facility that contains nuts.
 
E

Ernst Blofeld

Guest
Tim McNamara wrote:
> And way too over thought. You've talked yourself into nonsense. A trip
> to the gym will settle the question for you: can you raise more weight
> by extending or contracting your leg?


Yes. But I suspect that the spin will be a little smoother if the
cyclist doesn't push down quite as much on the pedal upstroke.

Back in the elder days after I spent some time on a fixed gear
or on rollers and my pedal technique seemed halfway decent,
it seemed that on the upstroke the force down on the pedal
during the upstroke was small, if not quite zero. It seemed
biomechanically better. Or at least it looked more stylin'.

Going too far on this and actually pulling up might send you
over the edge back to wobbly pedal technique.
 
K

Ken Roberts

Guest
Tim McNamara wrote
> Human evolution has resulted in legs that are stronger pushing
> against the ground in a normal gravitational field, and weaker lifting up.


I agree. The leg-extension muscles in most humans are bigger, stronger, and
have much more aerobic capacity than the leg-flexion muscles.

In the numerical examples I gave in my post, I had the leg-extension muscles
generating about 83% of the muscular power, and the leg-flexion muscles only
17%. Seems like pretty close agreement with the Tim's point about "stronger"
and "weaker" above.

But "weaker" is not the same thing as "nothing". We use the hip-flexion and
knee-flexion muscles effectively in human running, so why not use them some
to help with pedaling? The Kautz Coyle 1991 study showed that the 16 elite
racers in the study used the upward-pull muscles a lot -- but of course
nowhere near is much as the downward-push muscles.

Then just because the hip-extension and knee-extension muscles are much
bigger and stronger does not say anything about how mechanically efficient
is their connection to the pedal.

Ken
 
K

Ken Roberts

Guest
Werehatrack wrote
> There's exactly one foot/bike interface at which propulsion
> effort can be applied, and if you're not proposing to pull up
> on the pedals, what exactly is the point?


To use the leg's own muscles to pull up on a portion of the weight of the
leg, which will otherwise be pressing _down_ on the upstroke pedal. The
point is to take some of the weight of the leg off from resting on the
pedal, but not (usually) take all the weight off -- which is what would
be required to actually pull up on the pedal.

Another force to pull against is the resistance of the leg's own down-push
"extension" muscles. When making a high-power effort, these muscles push so
hard on the down-push that they tend to continue exerting significant
down-force on the pedal during part of the upstroke. So to some extent the
leg's own muscles often work against each other during part of the upstroke
in high-power pedaling. Sounds strange, but it's worth it, to extract the
maximum power out of the big down-push muscles.

So using upward pull muscles can help, even if they're not pulling directly
on the "foot/bike interface" -- because they pull on other things which do
connect directly with the foot/bike interface. (It's just an accident that
that help is mechanically more efficient than the down-push.)

Ken
 
K

Ken Roberts

Guest
cel wrote
> Now I have been know to pull-up when I have a leg that
> is cramping up and I have to make it up the hill! Relax on
> the down stroke, pull up on the upstroke.


Yes, even some people who are intellectually opposed to using upward-pull
muscles in normal pedaling recognize that using upward-pull muscles for
climbing up a steep hill is a helpful idea.

> The limiting factor is usually your aerobic capacity.


Not "usually" in my riding. I like to do long single-day tours, and my
central cardio-vascular system can easily deliver enough oxygen pressure to
keep my legs pushing.

I think my limiting factor riding in the afternoon of a long day is fuel.
How much glycogen and fatty acids is remaining stored locally in my leg
muscles; and At what rate can I digest the food that I eat and transport it
to the muscles that need it?

Lots of smart people think that the limiting factor for 60-minute time trial
is not the maximum total "aerobic capacity" of the body, but the sustainable
power at "Lactate Threshold" - LT. (Not that lactate is itself the limiter,
it's a useful fuel. But the presence of higher lactate concentrations in the
blood is often a useful _indicator_ of getting near a limit on 30-60 minute
performance.)

Anyway, in performance situations where the limiting factor really is
maximum aerobic capacity, then the highest riding speed will be achieved by
_prioritizing_ the limited oxygen, sending it first to the muscles with
highest mechanical efficiency -- which for seated pedaling is the
upward-pull muscles.

Ken
 
Ken,

you're talking rubbish.

This kind of rubbish has been tested scientifically and and is about as
real as Iraqi Weapons of Mass Destruction or alternative medicine or
psychics.

Just go out and pedal! ;-)
 
K

Ken Roberts

Guest
[email protected] wrote
> This kind of rubbish has been tested scientifically


Please give me a hint: _What_ stuff has been tested scientifically?

I might agree with you if I only knew what you were specifically talking
about.

Ken
 
D

dvt

Guest
Ken Roberts wrote:
> 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.


If this is true, then it should be more efficient to put the pedal
spindle at the heel. Then the ankle joint could bear the load without
flexing. Why do we put the pedal spindles forward of the heel?

I don't know the answer to the above question. Today as I was riding
with street shoes, I moved my foot around on the pedal. Having the
spindle somewhere near the ball of the foot is far more comfortable than
any other position. I didn't adjust anything else on the bike, so it
wasn't really a fair test. But I doubt I'd be able to get comfortable
with the spindle at the heel, no matter how I adjusted the rest of the bike.

--
Dave
dvt at psu dot edu

Everyone confesses that exertion which brings out all the powers of body
and mind is the best thing for us; but most people do all they can to
get rid of it, and as a general rule nobody does much more than
circumstances drive them to do. -Harriet Beecher Stowe, abolitionist and
novelist (1811-1896)
 
On Oct 31, 3:23 pm, "Ken Roberts" <[email protected]>
wrote:
> [email protected] wrote
>
> > This kind of rubbish has been tested scientificallyPlease give me a hint: _What_ stuff has been tested scientifically?

>
> I might agree with you if I only knew what you were specifically talking
> about
>
> Ken





Ken, can you explain the difference between unweighting and pulling up
?
 
On Tue, 31 Oct 2006 14:06:04 -0500, dvt <[email protected]> wrote:

>Ken Roberts wrote:
>> 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.

>
>If this is true, then it should be more efficient to put the pedal
>spindle at the heel. Then the ankle joint could bear the load without
>flexing. Why do we put the pedal spindles forward of the heel?
>
>I don't know the answer to the above question. Today as I was riding
>with street shoes, I moved my foot around on the pedal. Having the
>spindle somewhere near the ball of the foot is far more comfortable than
>any other position. I didn't adjust anything else on the bike, so it
>wasn't really a fair test. But I doubt I'd be able to get comfortable
>with the spindle at the heel, no matter how I adjusted the rest of the bike.


Dear Dave,

Thud, thud, thud . . .

You'll also be uncomfortable just walking on your heels, even on level
ground.

Now try climbing stairs on your heels. Things are even worse, though
the height involved is less than the pedal cycle.

The size of the calf muscle indicates that the rest of the foot beyond
the heel performs some useful function.

You can easily raise your body weight on tip-toe by pushing down.
Trying to pull a weight up with your toes is pretty much a waste of
time. In normal use, the foot dangles on the upstroke, whether it's
walking or running--there's nothing to be gained by straining the toes
toward the knee.

Cheers,

Carl Fogel
 
Tim McNamara wrote:

> And way too over thought. You've talked yourself into nonsense. A trip
> to the gym will settle the question for you: can you raise more weight
> by extending or contracting your leg?


Underthought.

> Human evolution has resulted in legs that are stronger pushing against
> the ground in a normal gravitational field, and weaker lifting up.


Human evolution has until recently required legs to pull themselves
forward to take the next step. If they are not used similarly on a
bicycle, then they are not contributing what they could be.

> Pushing down works better, whether walking, climbing or pedaling.


As Jobst likes to say, you failed to append the word "period" at the
end of your claim to give it absolute credibility.

dkl
 
K

Ken Roberts

Guest
[email protected] wrote
> explain the difference between unweighting and pulling up


You know maybe "unweighting" is a more helpful word for what I'm talking
about here -- so thanks for the suggestion, Noel. And for raising the
concern that perhaps my terminology is getting in the way of understanding.
Another word I like is "lifting". Or maybe we could try "lightening"; as in
"lightening" the load on the upstroke pedal.

What I'm talking about is taking some of the leg's weight off the upstroke
pedal, but (usually) not all of it. So literally it is "unweighting", though
some might interpret that as meaning to take _all_ the weight off the pedal,
which is not my point. "Lightening" the weight gets further from the concept
of _all_ the weight.

"Pulling up":
Actually succeeding in pulling up on the _pedal_ requires 100%
unweighting, or we might say "extreme lightening". I've been saying "pulling
up" on the weight of the leg, because in the objective biomechanics, the
hip-flexion and knee-flexion muscles do contract and pull, and the tendons
do "pull" on their attachment points on the upper and lower leg, and then
the masses of the upper and lower leg to go "up". But just because that
usage of "pulling up" is objectively justifiable doesn't mean it's helpful.
So I'll stop using it that way, and instead reserve it for pulling up on the
pedal.

Ken
 
K

Ken Roberts

Guest
Dave dvt wrote
> If this is true, then it should be more efficient to put the pedal spindle
> at the heel. Then the ankle joint could bear the load without flexing.


Yes, moving the position of the pedal relative to the ankle could reduce the
amount of energy lost in transmission torque thru the ankle. I would not
think that literally under the _heel_ would be the best place, more like
roughly intersecting the line of the bones of the lower leg thru the ankle
joint, assuming the angle of the ankle is set so that the tendons need to do
very little work to keep the ankle stable while pushing. I'd guess that
point would be in front of the heel. Note that would reduce mainly the
energy lost thru tightening and loosening the tendon, and not change much
the amount of energy lost thru compression and re-expansion of the "soft
stuff" in between the bones above the ankle and bones below. (And it doesn't
do anything about energy loss from transmission of force / torque from the
big hip-extension muscles thru the _knee_ joint.)

> I doubt I'd be able to get comfortable with the spindle at the heel


I'm not surprised.

> If this is true ...


From the perspective of biomechanics or engineering or physics, there's no
"if" about energy losses in transmission of force / torque thru human
joints. The only question is "how much". And that's one of the key questions
I'm asking.

> Why do we put the pedal spindles forward of the heel?


Let me rephrase as, "Why don't we put the pedal spindle a little further
back where it would minimize one kind of energy loss thru the ankle joint
and foot?"

First, it's clear from pedal angle / force measurements that elite racers
do make use of the relaxed freedom of their ankle joints (in a style which
could be called "reverse ankling" -- because it's the opposite of what you'd
do to try to get propulsive work from the muscles that act thru the ankle
joint). In the upstroke they lift their knee higher than necessary, which
gives them a larger range of motion at the start of the down-push, and in
the down-stroke they extend the down-push range-of-motion further so it
"collapses" the ankle joint a little.

The result is that the the big down-push muscles get to push thru a longer
Distance, so they are able to produce more Work (in Joules) in each
stroke-cycle. Both the "knee higher" and "ankle collapse" moves require
freedom in the ankle joint. If the pedal spindle were closer to the ankle
joint, the elite racers would have much less freedom to do those moves.

(Second, maybe the energy losses thru the ankle tendon aren't that large,
so moving the pedal back isn't worth it. That's why I repeat the question
"How much?")

But couldn't the elite racers achieve the same longer range-of-motion
Distance for the big down-push muscles just by using longer cranks? I think
maybe ... but also maybe longer cranks would have other side effects I don't
understand.

Here's my try at justifying why retaining freedom in the ankles is better
than a longer fixed crank:
Because sometimes a skilled rider wants to _vary_ the range-of-motion
Distance in the down-push.

My experience doing strength training in the gym on a "leg press" machine is
that my leg-extension muscles can exert higher force if the range-of-motion
is smaller. So perhaps in high-force situations, like getting thru a short
steep climb standing in a not-low-enough gear, and pedaling at low cadence,
a skilled rider (unconsciously) shortens the range of motion Distance a
little, so the big leg muscles can generate a higher Force without hurting
them for the rest of the ride.

Ken
 
P

Phil Holman

Guest
"Ken Roberts" <[email protected]> wrote in message
news:[email protected]
> Dave dvt wrote
>> If this is true, then it should be more efficient to put the pedal
>> spindle at the heel. Then the ankle joint could bear the load without
>> flexing.

>
> Yes, moving the position of the pedal relative to the ankle could
> reduce the amount of energy lost in transmission torque thru the
> ankle. I would not think that literally under the _heel_ would be the
> best place, more like roughly intersecting the line of the bones of
> the lower leg thru the ankle joint, assuming the angle of the ankle is
> set so that the tendons need to do very little work to keep the ankle
> stable while pushing. I'd guess that point would be in front of the
> heel. Note that would reduce mainly the energy lost thru tightening
> and loosening the tendon, and not change much the amount of energy
> lost thru compression and re-expansion of the "soft stuff" in between
> the bones above the ankle and bones below. (And it doesn't do anything
> about energy loss from transmission of force / torque from the big
> hip-extension muscles thru the _knee_ joint.)
>
>> I doubt I'd be able to get comfortable with the spindle at the heel

>
> I'm not surprised.
>
>> If this is true ...

>
> From the perspective of biomechanics or engineering or physics,
> there's no "if" about energy losses in transmission of force / torque
> thru human joints. The only question is "how much". And that's one of
> the key questions I'm asking.
>
>> Why do we put the pedal spindles forward of the heel?

>
> Let me rephrase as, "Why don't we put the pedal spindle a little
> further back where it would minimize one kind of energy loss thru the
> ankle joint and foot?"
>
> First, it's clear from pedal angle / force measurements that elite
> racers do make use of the relaxed freedom of their ankle joints (in a
> style which could be called "reverse ankling" -- because it's the
> opposite of what you'd do to try to get propulsive work from the
> muscles that act thru the ankle joint). In the upstroke they lift
> their knee higher than necessary, which gives them a larger range of
> motion at the start of the down-push, and in the down-stroke they
> extend the down-push range-of-motion further so it "collapses" the
> ankle joint a little.
>
> The result is that the the big down-push muscles get to push thru a
> longer Distance, so they are able to produce more Work (in Joules) in
> each stroke-cycle. Both the "knee higher" and "ankle collapse" moves
> require freedom in the ankle joint. If the pedal spindle were closer
> to the ankle joint, the elite racers would have much less freedom to
> do those moves.
>
> (Second, maybe the energy losses thru the ankle tendon aren't that
> large, so moving the pedal back isn't worth it. That's why I repeat
> the question "How much?")
>


Having the foot move forward of the pedal spindle puts it in the way of
the front wheel when slow turning. This can be overcome with some frame
geometry changes but I'm not sure this would be an overall improvement.

Phil H