Maximum torque on the crank?

Ron Ruff writes:

Thanks, could be some good info. Do you know how this force was
determined, and was it average or peak force?

Well, I don't know how scientific it is. It's what declared by
Pinarello just to demonstrate strength and stiffness of its frames.
Description was: "Petacchi applied to pedals a force of 130 kg each
rev".

A road sprint at the close of a stage is endurance, not peak force.
This is not a reasonable place to look for that. Besides, the quote
doesn't clarify whether this force had a sinusoidal characteristic or
something else. It certainly was not continuous.

If it was an average force, we'd have 130kg x 9.81m/s^2 x 3.14 x 2 x
.175m x 120?rpm x 1/60 min/sec = 2,803W... which I think is more
power than he could sustain in a sprint... so it probably is a peak
value. The cadence I'm not sure of; it would likely be higher than
120 unless it was uphill. What can a good sprinter put out these
days? Isn't it around 1,500W?

Forget about cadence, climbing an exceptionally steep grade is a high
force maximum exertion effort that is accomplished anaerobically.
That is, one doesn't begin breathing hard till it's over. That is why
the term sprint applies. Pedaling rate for this is probably around
20rpm or less. Guys on MTB's do this more often. I get the
impression that you haven't tried this. If you watch the San
Francisco Grand Prix of Cycling bicycle race up Fillmore street, it
has a lot in common except that it is longer and not as steep. There
riders are in an aerobic mode because they were so before staring the
hill.

That part of this is anaerobic should be apparent in that if the hill
were longer, many of the riders would get off and walk as they did in
some stages of the Giro d'Italia on the Muro di Sormano for instance.

Just for fun, lets put him on a steep hill with a starting cadence
of 60rpm instead of 120rpm. Would he be able to produce the same
power? The technique would be a little different, but I'd wager
that he could get close... and the ratio of peak force to average
force would be at least as great. To do that he'd have to double
the peak pedal force to 260kg (572lbs).

Who's fun. The whole idea of young bucks attempting such feats is for
fun. Try it... oops that was twenty years ago for you so I guess
unless you know Oscar Boom and the time machine that slipped by. How
do these things get so lost in side tracks. What was the max force on
the pedal again???

Jobst Brandt

Per :
Various posts in this thread have mentioned "flinging" and
"jumping" and similar notions in connection with the rider's
weight.

The only time that this occurs normally is when the rider
stands up.

Or bunny hops or lands from a drop.
--
PeteCresswell

wrote:

And there's no net torque on the spindle on landing if the
crank isn't vertical, assuming that the rider's weight is
distributed evenly on the two pedals.

If cranks are horizontal with equal force on each pedal, then the
torque on the spindle is the same as it would be if we were applying
that force to the left crank while riding. I'm not sure what you mean
by "net torque"... obviously the torque is balanced in this case, so
there wouldn't be any rotation induced... but that isn't relevant to
determining the torque on the spindle.

-Ron

On Sun, 07 Aug 2005 19:42:33 GMT,
wrote:

[snip]

I have ridden up the 31.5% grade of Filbert St in SF in a 47-21 ratio.
At a combined weight of rider and bicycle of a little over 200lbs that
gives about 326lbs to hold the bicycle at a stand still.

[snip]

Dear Jobst,

Could you show us your calculations and the details for the
stand-still forces?

Thanks,

Carl Fogel

wrote:

Sorry, but I'm not following you.

When you say "completely unweight your forward leg" . . .

Where is the pedal? (Let's say noon is straight up, 3
o'clock is toward the rear axle, 6 o'clock is straight down,
and 9 o'clock is toward the front axle.)

What is the leg's bend? (Let's try straight, half-bent,
full-bent.)

Where did your weight shift to? (Other leg? Hurled upward
into the air?)

Sorry... I meant "completely unweight the forward *moving* leg" meaning
the top leg before it begins the power stroke. Lets put the clockface
where the chainring is, which I think is easier to visualize since the
cranks would then be turning clockwise. When I'm sprinting on a steep
hill it feels like I'm pushing down pretty hard from 2 to 5 or so, and
by the time my other leg gets to 11 or so it becomes completely
unweighted as it transitions from pulling up earlier in the stroke (8
to 10). This coincides with the CG rising and falling slightly, as you
mentioned.

I guess an analogy would be a running stride. There is a period where
both feet are off the ground, and the forces at initial contact and
when pushing off are greater than the runner's weight.

-Ron

wrote:
A road sprint at the close of a stage is endurance, not peak force.
This is not a reasonable place to look for that.

I agree that it isn't the best place to find peak force... mostly
because of the high cadence.

Forget about cadence, climbing an exceptionally steep grade is a high
force maximum exertion effort that is accomplished anaerobically.
That is, one doesn't begin breathing hard till it's over. That is why
the term sprint applies.

I assumed that it was an all-out effort, but if it was more than a few
seconds long I think you were not even producing your absolute maximum
force. For instance, couldn't you have produced a higher force if you
were only trying to complete 3 pedal strokes as quickly as possible,
instead of trying to make it to the top? The amount of power we can
produce falls off pretty quickly with the duration of the effort.

Who's fun. The whole idea of young bucks attempting such feats is for
fun. Try it... oops that was twenty years ago for you

Yes, but I usually like to pretend I'm 25... until the aching back and
creaking knees remind me that I'm not.

-Ron

writes:


I have ridden up the 31.5% grade of Filbert St in SF in a 47-21 ratio.
At a combined weight of rider and bicycle of a little over 200lbs that
gives about 326lbs to hold the bicycle at a stand still.

Could you show us your calculations and the details for the
stand-still forces?

First a correction. When I did this I was still using a Regina
5-speed 13-15-17-20-25t so the ratio was 47-20. Now that I review it,
I can't seem to find where I got the number that I recalled from
earlier discussions. Here is what I get now and cannot readily
explain where the numbers went.

Maybe you can find what I left out. I might have used a greater rider
weight. I weighed 190 lbs in those days.

31.5% grade = 17.484 deg
vertical bicycle and rider = 200 lbs
horizontal force 200*31.5% = 63 lbs
wheel radius = 333,59mm
crank length = 180mm
wheel/crank ratio = 333.59/180 = 1,853
chain ratio = 47t/20t = 2.35
pedal force = 63 * 1.856 * 2.35 = 274.8 lbs static

It's late so maybe you can make more of this. I have a good side view
picture of this but I'm sure you can imagine it. The bicycle is
lunged forward to get over TDC.

Moments like this make fatigue failure obvious to me because if
failures were caused by excess rider strength or too rough a road,
components would break immediately, not several thousand miles later.

Jobst Brandt

Ron Ruff writes:

A road sprint at the close of a stage is endurance, not peak force.
This is not a reasonable place to look for that.

I agree that it isn't the best place to find peak force... mostly
because of the high cadence.

Forget about cadence, climbing an exceptionally steep grade is a
high force maximum exertion effort that is accomplished
anaerobically. That is, one doesn't begin breathing hard till it's
over. That is why the term sprint applies.

I assumed that it was an all-out effort, but if it was more than a
few seconds long I think you were not even producing your absolute
maximum force. For instance, couldn't you have produced a higher
force if you were only trying to complete 3 pedal strokes as quickly
as possible, instead of trying to make it to the top? The amount of
power we can produce falls off pretty quickly with the duration of
the effort.

This is all done at stall speeds, it is not a matter of cadence but
more like a bench press.

Who's fun. The whole idea of young bucks attempting such feats is
for fun. Try it... oops that was twenty years ago for you

Yes, but I usually like to pretend I'm 25... until the aching back
and creaking knees remind me that I'm not.

Jobst Brandt

Per :
But if we think of the original poster's question about
maxium torque on the crank spindle . . .

I acknowledge the drift. But when my users/clients come to me with such
specific questions I'm always a little anxious that the real issue may require
quite a different answer. The trick being to smoke out the real issue without
hurting anybody's feelings...
--
PeteCresswell

wrote in message
...
On Sun, 07 Aug 2005 19:42:33 GMT,
wrote:

[snip]

I have ridden up the 31.5% grade of Filbert St in SF in a 47-21 ratio.
At a combined weight of rider and bicycle of a little over 200lbs that
gives about 326lbs to hold the bicycle at a stand still.

[snip]

Dear Jobst,

Could you show us your calculations and the details for the
stand-still forces?

Thanks,

Carl Fogel

I ran the numbers and came out with a requirement of 259 pounds which is a
little less than the 326 that Jobst gave. But I tend to make a lot of math
errors. But if it is only 259 pounds, it still is impressive.

A 31.5% slope is one where the rise is 31.5 percent of the horizontal
distance. So 31.5% is 17.5 degrees since the tangent of 17.5 degrees is
0.315. Since the total weight is about 200 pounds, the down-hill force is
200*sin 17.5 = 60.1 pounds. Thus the rear wheel sees a torque of about 67.6
foot pounds. Using the ratio 47/21, this means that the torque on the crank
has to be about 151 foot pounds. Using a crank arm length of 7 inches, this
requires a force on one pedal of 259 pounds. Thus if you weigh 180 pounds
and you are standing on one pedal and pulling up on the handlebars with, say
30 pounds, you would have to be pulling up on the other pedal with 49 pounds
of force. This assumes the crank arms are horizontal and you are not
accelerating. If you are actually moving up the hill, I guess you have to
apply quite a bit more force than this while the crank arms are horizontal
since, when they are vertical, you would not be able to put anywhere near
this force on them.

I run a "girly man" triple with a ratio of 30/27 instead of a more manly
47/21. Thus on the same hill, I would have to exert a torque of only 75.1
foot pounds and a force of 129 pounds, which sounds a little more
reasonable. Still not sure I could climb the Filbert Street hill though.