pedal force

bicycle_disciple said:

For an average, what would the pedal force exerted by the a foot of a
rider be? For experiments, how is this data obtained?

This can be done, and has been done, with a strain gauge.

Strain gauges on both pedals quickly reveal how much force is applied at
various angle of rotation, as well as demonstrating differences between
the two legs.

A rear hub such as the PowerTap uses a strain gauge in order to
calculate how much power the rider is generating. A useful training
tool.

--
Ted Bennett

"bicycle_disciple" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

For an average, what would the pedal force exerted by the a foot of a
rider be? For experiments, how is this data obtained?

For a rider who outputs 300w (say a time trial where he/she travels at
approx. 26mph), the average pedal force with 175mm cranks at a cadence
of 90rpm would be:

300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)

Because of the cyclic nature of force application to the pedal, maximum
force could be as much as 37 kgf and minimum close to zero for this
example.

Phil H

Phil Holman said:

"bicycle_disciple" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

For an average, what would the pedal force exerted by the a foot of a
rider be? For experiments, how is this data obtained?

For a rider who outputs 300w (say a time trial where he/she travels at
approx. 26mph), the average pedal force with 175mm cranks at a cadence
of 90rpm would be:

300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)

Because of the cyclic nature of force application to the pedal, maximum
force could be as much as 37 kgf and minimum close to zero for this
example.

Phil H

Nice formula Phil. Thanks all. -B.D

Phil Holman said:

"bicycle_disciple" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

For an average, what would the pedal force exerted by the a foot of a
rider be? For experiments, how is this data obtained?

For a rider who outputs 300w (say a time trial where he/she travels at
approx. 26mph), the average pedal force with 175mm cranks at a cadence
of 90rpm would be:

300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)

Because of the cyclic nature of force application to the pedal, maximum
force could be as much as 37 kgf and minimum close to zero for this
example.

Phil H

Actually, to deliver the most power the rider should be clipped in, and
pulling up for about half of the cycle. This means that the force would
go from a positive 18.5 Kg (on the downstroke) to some negative value
(on the upstroke).

HTH,
EJ in NJ

"Ernie Willson" <[email hidden]> wrote in message
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Quoted message said:

Phil Holman said:

"bicycle_disciple" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

For an average, what would the pedal force exerted by the a foot of a
rider be? For experiments, how is this data obtained?

For a rider who outputs 300w (say a time trial where he/she travels
at approx. 26mph), the average pedal force with 175mm cranks at a
cadence of 90rpm would be:

300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)

Because of the cyclic nature of force application to the pedal,
maximum force could be as much as 37 kgf and minimum close to zero
for this example.

Phil H

Actually, to deliver the most power the rider should be clipped in,
and pulling up for about half of the cycle. This means that the force
would go from a positive 18.5 Kg (on the downstroke) to some negative
value (on the upstroke).

Unless specifically trained to pull up (Powercranks), most riders will
not pull up any more than lifting the weight of their leg (most
exceptions occur with mtb riders). I'll stick by my original numbers
having seen the pedal force data on numerous elite road cyclists.

Phil H

Phil Holman said:

"Ernie Willson" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Phil Holman said:

"bicycle_disciple" <[email hidden]> wrote in message
news:[email hidden]...

>For an average, what would the pedal force exerted by the a foot of a
>rider be? For experiments, how is this data obtained?
>

For a rider who outputs 300w (say a time trial where he/she travels
at approx. 26mph), the average pedal force with 175mm cranks at a
cadence of 90rpm would be:

300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)

Because of the cyclic nature of force application to the pedal,
maximum force could be as much as 37 kgf and minimum close to zero
for this example.

Phil H

Actually, to deliver the most power the rider should be clipped in,
and pulling up for about half of the cycle. This means that the force
would go from a positive 18.5 Kg (on the downstroke) to some negative
value (on the upstroke).

Unless specifically trained to pull up (Powercranks), most riders will
not pull up any more than lifting the weight of their leg (most
exceptions occur with mtb riders). I'll stick by my original numbers
having seen the pedal force data on numerous elite road cyclists.

Phil H

You are no doubt correct. It is highly doubtful that anyone will pull up
with a greater force than the down force. My only point was to make the
original poster aware that there can be a significant reversal of forces
in the crank, because I don't know what he intends to use the info for.

HTH,
EJ in NJ

Ernie Willson said:

Phil Holman said:

"Ernie Willson" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Phil Holman wrote:

>"bicycle_disciple" <[email hidden]> wrote in message
>news:[email hidden]...
>
>
>>For an average, what would the pedal force exerted by the a foot of a
>>rider be? For experiments, how is this data obtained?
>>
>
>
>For a rider who outputs 300w (say a time trial where he/she travels
>at approx. 26mph), the average pedal force with 175mm cranks at a
>cadence of 90rpm would be:
>
>300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)
>
>Because of the cyclic nature of force application to the pedal,
>maximum force could be as much as 37 kgf and minimum close to zero
>for this example.
>
>Phil H

Actually, to deliver the most power the rider should be clipped in,
and pulling up for about half of the cycle. This means that the force
would go from a positive 18.5 Kg (on the downstroke) to some negative
value (on the upstroke).

Unless specifically trained to pull up (Powercranks), most riders will
not pull up any more than lifting the weight of their leg (most
exceptions occur with mtb riders). I'll stick by my original numbers
having seen the pedal force data on numerous elite road cyclists.

Phil H

You are no doubt correct. It is highly doubtful that anyone will pull up
with a greater force than the down force. My only point was to make the
original poster aware that there can be a significant reversal of forces
in the crank, because I don't know what he intends to use the info for.

HTH,
EJ in NJ

Dear Ernie,

I'm not sure, but there may a misunderstanding here.

Tests indicate that riders normally exert almost no pulling-up force
on the back-stroke of the pedal cycle at typical cadences, despite the
common belief that they need to be clipped in so that they can tug the
pedal upward.

What may feel like pulling up to the rider appears to be little more
than unloading the pedal.

Here's a nice graph from Robert Chung's site, showing the torque
measured on an actual rider at 90.3 rpm:

http://anonymous.coward.free.fr/rbr/kautz.png

The basically flat line on the right side shows the trivial pulling-up
force.

Cheers,

Carl Fogel

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Quoted message said:

Ernie Willson said:

Phil Holman said:

"Ernie Willson" <[email hidden]> wrote in message
news:[email hidden]...

>Phil Holman wrote:
>
>>"bicycle_disciple" <[email hidden]> wrote in message
>>news:[email hidden]...
>>
>>
>>>For an average, what would the pedal force exerted by the a foot
>>>of a
>>>rider be? For experiments, how is this data obtained?
>>>
>>
>>
>>For a rider who outputs 300w (say a time trial where he/she travels
>>at approx. 26mph), the average pedal force with 175mm cranks at a
>>cadence of 90rpm would be:
>>
>>300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)
>>
>>Because of the cyclic nature of force application to the pedal,
>>maximum force could be as much as 37 kgf and minimum close to zero
>>for this example.
>>
>>Phil H
>
>Actually, to deliver the most power the rider should be clipped in,
>and pulling up for about half of the cycle. This means that the
>force
>would go from a positive 18.5 Kg (on the downstroke) to some
>negative
>value (on the upstroke).
>

Unless specifically trained to pull up (Powercranks), most riders
will
not pull up any more than lifting the weight of their leg (most
exceptions occur with mtb riders). I'll stick by my original numbers
having seen the pedal force data on numerous elite road cyclists.

Phil H

You are no doubt correct. It is highly doubtful that anyone will pull
up
with a greater force than the down force. My only point was to make
the
original poster aware that there can be a significant reversal of
forces
in the crank, because I don't know what he intends to use the info
for.

HTH,
EJ in NJ

Dear Ernie,

I'm not sure, but there may a misunderstanding here.

Tests indicate that riders normally exert almost no pulling-up force
on the back-stroke of the pedal cycle at typical cadences, despite the
common belief that they need to be clipped in so that they can tug the
pedal upward.

What may feel like pulling up to the rider appears to be little more
than unloading the pedal.

Here's a nice graph from Robert Chung's site, showing the torque
measured on an actual rider at 90.3 rpm:

http://anonymous.coward.free.fr/rbr/kautz.png

The basically flat line on the right side shows the trivial pulling-up
force.

Cheers,

Carl Fogel

This was probably derived from Steve Kautz's data.........
http://isbweb.org/data/kautz/

Click on the average for 14 cyclists which is for a power output of 385
watts. Its easy to compute the pedal force from the instantaneous torque
values. Notice also the pedal forces are bi-directional.
Individual data is also available on this link and IIRC, one or two of
the individuals did pull up, allbeit minimally.

Phil H

Phil Holman said:

<[email hidden]> wrote in message

Quoted message said:

Here's a nice graph from Robert Chung's site, showing the torque
measured on an actual rider at 90.3 rpm:

http://anonymous.coward.free.fr/rbr/kautz.png

The basically flat line on the right side shows the trivial pulling-up
force.

This was probably derived from Steve Kautz's data.........
http://isbweb.org/data/kautz/

Click on the average for 14 cyclists which is for a power output of 385
watts. Its easy to compute the pedal force from the instantaneous torque
values. Notice also the pedal forces are bi-directional.
Individual data is also available on this link and IIRC, one or two of
the individuals did pull up, allbeit minimally.

Can't find Kautz's paper online. Were the data taken in real
riding situations or on stationary trainers? Were the subjects
cognizant that their technique was under scrutiny or were they
just told go ride hard? Were any data taken with the recommendation
that the cyclist try to smooth his stroke? Were there data for
seated, steady riding, accelerating to a sprint, and standing either
on a flat or a climb?

--Blair

"Blair P. Houghton" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

Phil Holman said:

<[email hidden]> wrote in message

Quoted message said:

Here's a nice graph from Robert Chung's site, showing the torque
measured on an actual rider at 90.3 rpm:

http://anonymous.coward.free.fr/rbr/kautz.png

The basically flat line on the right side shows the trivial
pulling-up
force.

This was probably derived from Steve Kautz's data.........
http://isbweb.org/data/kautz/

Click on the average for 14 cyclists which is for a power output of
385
watts. Its easy to compute the pedal force from the instantaneous
torque
values. Notice also the pedal forces are bi-directional.
Individual data is also available on this link and IIRC, one or two
of
the individuals did pull up, allbeit minimally.

Can't find Kautz's paper online. Were the data taken in real
riding situations or on stationary trainers? Were the subjects
cognizant that their technique was under scrutiny or were they
just told go ride hard? Were any data taken with the recommendation
that the cyclist try to smooth his stroke? Were there data for
seated, steady riding, accelerating to a sprint, and standing either
on a flat or a climb?

It is my understanding this was on a stationary trainer but I haven't
found the full article either. My interest was purely in the shape of
the force plot which is very close to sinusoidal.

Phil H

Carl Fogel wrote:

|> Tests indicate that riders normally exert almost no pulling-up force
|> on the back-stroke of the pedal cycle at typical cadences, despite the
|> common belief that they need to be clipped in so that they can tug the
|> pedal upward.
|>
|> What may feel like pulling up to the rider appears to be little more
|> than unloading the pedal.

What about pushing forward before it's time to pedal downward? Does
this have any effect?

I ride with sandals on and so am not properly "clipped in" nevertheless
clips make a big difference for me.

--
ciao,
Bruce

drift wave turbulence: http://www.rzg.mpg.de/~bds/

On Mon, 4 Sep 2006 13:31:05 +0200 (MEST), Bruce Scott TOK

[email protected] said:

Carl Fogel wrote:

|> Tests indicate that riders normally exert almost no pulling-up force
|> on the back-stroke of the pedal cycle at typical cadences, despite the
|> common belief that they need to be clipped in so that they can tug the
|> pedal upward.
|>
|> What may feel like pulling up to the rider appears to be little more
|> than unloading the pedal.

What about pushing forward before it's time to pedal downward? Does
this have any effect?

I ride with sandals on and so am not properly "clipped in" nevertheless
clips make a big difference for me.

Dear Bruce,

If anyone knows of tests showing significant power changes for normal
riders at normal cadences using clips, they're keeping the data to
themselves.

All the tests that I've seen show a power graph like this: /\_

The flat part on the backstroke may have a small, irregular wiggle,
but it's insignificant.

I've never seen any tests for sprinting or climbing style effort where
the riders try to yank the pedal up at low cadences, but I've wondered
how much actual useful force they add, given the huge difference
between the strength of our legs extending and flexing.

We can easily lift our own weight by extending two legs, but very few
of us can hang by our feet and do the equivalent of a pull-up. Compare
the weights involved in leg presses versus leg curls at a gym.

Some posters have mentioned pulling out of their shoes in sprints, but
it takes surprisingly little force to do so, as children, soldiers,
and observed trials riders sometimes learn in mud sections. Take a
moment, hold the heel of your riding shoe with one hand, and see how
much your heel rises out of the shoe if you try to pull up.

If a rider does indeed pull up with significant force very often,
there should be some consequences, but I've never heard of anyone
complaining about them.

At worst, there should be foot pain, since pulling up will cause the
sides of the shoe to squeeze the foot in an unnatural handshake
fashion.

At best, there should be blistering and abrasion. No one, as far as I
know, designs shoes for pulling up--the sides and top of the foot will
be doing things with enough force to cause problems, even with socks.
Anyone who's blistered an hand or finger in just a few minutes of what
turned out to be foolish without a glove knows how easy it is to do.

As for the forward, near-top-of-pedal problem that you mention, any
normal shoe/pedal interface is pretty much solid throughout the front
180 degrees of the cycle.

Your sandals, however, may be so loose that they cause problems. That
is, most people are reluctant to try to run or even just trot in
sandals, but they're perfectly happy to do so in $10 sneakers.

I occasionally look the soles of ordinary shoes used by other
bicyclists. So far, I've seen no signs of wear that show any slipping,
just faint grooves where the shoe presses against the lightly toothed
pedal rims.

Cheers,

Carl Fogel

Ernie Willson said:

Phil Holman said:

"bicycle_disciple" <[email hidden]> wrote in message
news:[email hidden]...

Quoted message said:

For an average, what would the pedal force exerted by the a foot of a
rider be? For experiments, how is this data obtained?

For a rider who outputs 300w (say a time trial where he/she travels at
approx. 26mph), the average pedal force with 175mm cranks at a cadence
of 90rpm would be:

300/(.175*2Pi*90/60) = 182 Newtons (18.5 kgf) (41lbs)

Because of the cyclic nature of force application to the pedal, maximum
force could be as much as 37 kgf and minimum close to zero for this
example.

Phil H

Actually, to deliver the most power the rider should be clipped in, and
pulling up for about half of the cycle. This means that the force would
go from a positive 18.5 Kg (on the downstroke) to some negative value
(on the upstroke).

Does anyone actually pull up for half the cycle? Maybe at very low rpm
they do, but I doubt it at normal cadence.

JT

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John Forrest Tomlinson said:

Does anyone actually pull up for half the cycle? Maybe at very low rpm
they do, but I doubt it at normal cadence.

All the time, at high (->125) and low (->30) cadence. I didn't realise
I did until I borrowed a friend's mtb with flat pedals. I kept lifting
my foot off the pedal. At a high cadence I'm just lifting my leg, sort
of completing the circle and at a low cadence I'm actualy pulling on
the pedal.

Laters,

Marz