Maximum torque on the crank?

Crononauta wrote:
Petacchi is worth to produce a max of 130 kg during his fast sprint at
Milano-Sanremo.

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

-Ron

On 7 Aug 2005 04:36:30 -0700, Ron Ruff wrote:
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". Considered that Pinarello would demonstrate strength of its
frames, I suppose it was the peak force.

How they measured it, it wasn't explained. It could be both comparing
developped speed (74 kmh - 46 mph during last 100 metres) with some
measure made with telemetry during tests; or Petacchi had that device
to measure power and effort, applied to crankset, like that:

http://biketechreview.com/archive/po...s/image021.jpg

I know that some team used it during Giro d'Italia to monitor
performance of cyclers, but I don't know if Petacchi had it.

--
Massimo Bacilieri AKA Crononauta
Ravenna, Italy

Per Earls61:
Could anyone tell me the maximum torque that is applied to the bottom
bracket spindle of a road bicycle? Assume worst case scenarios, i.e.
Very strong rider, very steep hill, etc. I would prefer a value in
foot-pounds.

What's the intended use of the info?

ReasonForAsking: Although I don't pretend to understand the Newtonian
physics/engineering stuff that greater minds will post, if you're looking for
what's needed to deal with your body weight; use may be a factor.

I'm about 220#, have broken one crankset - and attribute that break to extended
periods of trying to learn to bunnyhop. Like I said, I don't understand the
theoreticals - but there's definately a diff between somebody just lunging up a
hill and the same person repeatedly trying to jump the bike - even though the
weight would seem tb spread out over both cranks in the latter situation.
--
PeteCresswell

Werehatrack wrote:
On 6 Aug 2005 19:25:10 -0700, "Earls61"
wrote:

Could anyone tell me the maximum torque that is applied to the bottom
bracket spindle of a road bicycle? Assume worst case scenarios, i.e.
Very strong rider, very steep hill, etc. I would prefer a value in
foot-pounds.

Max-torque on the spindle would probably be achieved when doing a
"trackstand" with the cranks nearly horizontal, in which the rider's
weight is applied about equally to both pedals. Given a crank of
175mm in length and a rider weight of 350 lbs (not common, but
certainly within the realm of what exists), I get 200 ft/lbs of torque
applied to the spindle.

the force and moment arm are multiplied so the unit is ft-lb, not
ft/lb.

Clearly this is wrong. With equal opposing torque on the ends of
the spindle, it carries only 100 ft-lb applied at each pedal.

Any "weight" applied to the left crank (put it forward) that had been
applied to the right in your example would increase the spindle torque.


I doubt that this would be exceeded when pedalling in most cases,
though it might on hard climbs if the rider is both standing and
applying additional force by pulling up on the bars. The difference
is probably small, however...and with the exception of one of the more
heroically-sized denizens of this group, I would not expect to find a
350lb rider trying that tactic very often.

Whenever standing, when the left pedal is forward the 350 pounder
would get about 200ft-lb.

There is a small reduction due to the downward acceleration of the
foot & leg minimized by a steep gear. Pulling up on the bars or
lifting the right (rearward) pedal increases the number attainable
without accelerating the body upward.


The OP said maximum so I'll get ridiculous:
A big guy jumps off his seat onto a forward pedal
on the front of a tandem when there's somebody really
big on the back seat so the tire doesn't slip.

If nothing breaks, the spindle torque is limited by
the force which the jumper can produce in his
"landing" leg. 1000 lb two legged squats are possible
so over 500 lb from one leg ought to be doable
briefly. Lets put it on a 200mm crank instead of
wimpy 170s like I use. about 350 ft-lb!

On 6 Aug 2005 19:25:10 -0700, "Earls61"
wrote:

Could anyone tell me the maximum torque that is applied to the bottom
bracket spindle of a road bicycle? Assume worst case scenarios, i.e.
Very strong rider, very steep hill, etc. I would prefer a value in
foot-pounds.

Thanks,

Bob

Dear Bob,

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. If the rider exerts more force than his body
weight on the pedal with one leg, his body rises.

To see what would really happen, try this abnormal maneuver:

While already standing up to pedal, start coasting and rise
up further by putting the pedals horizontal and
straightening both legs.

This weird and awkward posture does indeed raise your center
of mass (your hips are higher), so it allows you to drop
your weight down on the pedal, but only once and with no
more force than you exerted when you raised yourself up.

When we speak of "throwing" our weight on the pedal, it's
pretty much wishful thinking. We really just lean from side
to side to move our center of mass over the pedal and push
down. Unless you see a rider's hips rising and falling, he's
just shifting his weight back and forth as he pushes down
with alternate legs, not throwing his weight up or down.

During normal riding (meaning something sustainable for more
than a single downstroke), the hips stay at about the same
distance from the bottom bracket, neither rising nor
falling, with one leg straightening and extending, while the
other leg bends and retracts. Our center of mass doesn't
"jump" up and down even as much as it does when we walk or
run.

Carl Fogel

Alfred Ryder writes:

Could anyone tell me the maximum torque that is applied to the
bottom bracket spindle of a road bicycle? Assume worst case
scenarios, i.e. Very strong rider, very steep hill, etc. I would
prefer a value in foot-pounds.

That is so simple that I might take a shot at it.

The left pedal is the only one that puts torque on the spindle. The
right pedal cannot put torque on the spindle. The crank arm is
usually right at 7 inches long. Very rarely is a rider going to put
more than 200 pounds of force on a single pedal. However, if the
rider weighs, say, 300 pounds and then bounces on the pedal while
the crank arm is parallel to the ground, he might get 350 pounds on
it. Thus 350*7/12 is about 200 foot pounds. And the more reasonable
rider does 200*7/12=117 foot pounds. It probably does not matter
much how strong the rider is or how steep the hill is.

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. Not a lot of
that came from pulling up on the other pedal, this being done with
worn metal cleats and foot straps. I also id this in the rain,
showing that traction on wet concrete is not a problem although the
roughness of the hand towelled surface is.

Further, if the chain is in a relatively small chain ring, say one
with a radius of 2 1/2 inches, the force on the chain from the above
heavy rider is 200*12/2.5 which is about 960 pounds.

Of course gradients and pedal force are all related by the chain ratio
and wheel diameter but that was not the question here.

Jobst Brandt

wrote:

When we speak of "throwing" our weight on the pedal, it's
pretty much wishful thinking. We really just lean from side
to side to move our center of mass over the pedal and push
down. Unless you see a rider's hips rising and falling, he's
just shifting his weight back and forth as he pushes down
with alternate legs, not throwing his weight up or down.

I agree, if we are just looking at what a person "normally" does when
they ride out of the saddle at a relatively easy pace... but I think
the OP is interested in the worst case scenerio.

If you perform an uphill sprint; a "violent" all-out effort for a few
seconds... particularly if you start in a fairly tall gear... it's easy
to get a force much greater than your weight during the power stroke.
The hips don't need to rise and fall very much... even if you
completely unweight your forward leg, and "punch" down with all your
might.

Using Jobst's example of climbing Filbert, he had to have an *average*
tangential force on the cranks of 326 lbs just to keep going forward!
Seems obvious that in some portions of the stroke, the force was much
higher than this... and this was probably not even a true sprint; I
don't know how long the effort was.

-Ron

Crononauta wrote:
On 7 Aug 2005 04:36:30 -0700, Ron Ruff wrote:
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".

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?

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).

-Ron

On 7 Aug 2005 14:05:24 -0700, "Ron Ruff"
wrote:

wrote:

When we speak of "throwing" our weight on the pedal, it's
pretty much wishful thinking. We really just lean from side
to side to move our center of mass over the pedal and push
down. Unless you see a rider's hips rising and falling, he's
just shifting his weight back and forth as he pushes down
with alternate legs, not throwing his weight up or down.

I agree, if we are just looking at what a person "normally" does when
they ride out of the saddle at a relatively easy pace... but I think
the OP is interested in the worst case scenerio.

If you perform an uphill sprint; a "violent" all-out effort for a few
seconds... particularly if you start in a fairly tall gear... it's easy
to get a force much greater than your weight during the power stroke.
The hips don't need to rise and fall very much... even if you
completely unweight your forward leg, and "punch" down with all your
might.

[snip]

Dear Ron,

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?)

Carl Fogel

On Sun, 07 Aug 2005 17:01:47 -0700, "(PeteCresswell)"
wrote:

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.

Dear Pete,

I like your kangaroo-like point, even though I was thinking
of steadily repeatable pedal motion.

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

There's no torque on the spindle on landing if the crank is
vertical.

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.

Carl Fogel