Maximum torque on the crank?

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

"Earls61" wrote in message
ups.com...
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


That is so simple that I maight 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 reasoonable 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.

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.

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.

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

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,

If a 200-lb rider stands on a horizontal 175mm crank and
doesn't go anywhere because the hill is too steep for the
gearing, then he's applying . . .

Let's see, 175mm / (25.4 mm/inch) is about 6.9 inches.

So you'd have 200 lbs on a lever 6.9 inches long, and we
want to convert it to pounds on a longer 12-inch lever . . .

200 lbs * (6.9 inches / 12 inches/foot )

200 * 0.575 = 115 lb-feet

So it looks as if the motionless 200-lb rider balancing on
one horizontal pedal is applying the equivalent of 115 lbs
on a 12-inch lever, or 115 ft-lbs of torque.

If he pulls down hard on the handlebars, he might raise his
200-lb effect on the pedal to say 300 pounds--which gives
172.5 ft-lbs of torque.

Heavier or stronger riders could raise the figure.

Basically, stand on one foot on a bathroom scale, then pull
up on a stout railing with both hands--the weight that you
reach is then applied to a 6.9 inch lever, so multiply it by
0.575 to get foot-lbs.

Carl Fogel

Alfred Ryder wrote:
Very rarely is a rider going to put more than 200 pounds of
force on a single pedal.

It is actually easy to put in excess of 200 lbs of force onto the
pedals when sprinting... particularly if it is a steep uphill. In this
case you are using the momentum of your weight and the power of your
leg to produce a force that is much higher than what you weigh... and
you don't even need to be pulling up on the bars to do it.

If you have a bathroom scale, try a little experiment. Stand on it with
one leg. Then let your body drop down a little and push on it hard like
you would when sprinting. When you let your body drop the scale goes
close to zero, then it maxes out when you push down.

Or you could think about jumping up and down. Your *average* weight
pushing down on the floor will be whatever you weigh... but obviously
there is no weight pushing down when you are in the air, so it must be
a lot more than your weight in some point in the cycle.

I frankly don't know what the peak torque would be, but I'd guess it
make sense to look at a force of something around 2-4 times body weight
in a sprint. Then there is the issue of landing from jumps...

I wonder, does the OP want to know the total stresses in the bottom
bracket spindle or the torque only? Torque is the greatest factor, but
there are other forces besides torque.

-Ron

Your track stand analogy fits in with Carl Fogel's "stationary rider"
analogy somewhat. I do them all the time, however, and I know I don't
put any where near 350 ft.-lbs. of torque on the spindle. I think this
is because this is one of the few things I am actually good at!

Alfred wrote:

"The left pedal is the only one that puts torque on the spindle. The right
pedal cannot put torque on the spindle"

That is a good point. I didn't think about that. the pedal force is
applied directly to the chanring through the chainring bolts.

Thanks for the clear explanation, Carl. I would think a stationary
rider would put more torque on the spindle as you say, rather than Ron
Ruff's sprinting rider mentioned in his post. But for some reason, I
was thinking that the torque would be larger 170-180ft-lbs.

Basically, stand on one foot on a bathroom scale, then pull
up on a stout railing with both hands--the weight that you
reach is then applied to a 6.9 inch lever, so multiply it by
0.575 to get foot-lbs.

Later this morning, I am going to take the bathroom scale out to the
garage and do exactly that. The workbench out there is bolted securely
to the floor, so I can use that to pull on.

Bob

Earls61 wrote:
Thanks for the clear explanation, Carl. I would think a stationary
rider would put more torque on the spindle as you say, rather than Ron
Ruff's sprinting rider mentioned in his post.

You'd certainly get the most torque at lower revs, that's basic
physics. Sprinting riders aren't using their maximum strength because
they're pedalling too fast, but riders overgeared on a steep hill will
be much closer.

However, a stationary rider isn't putting in the same *peak* force as a
moving one. Look at it like this: you've been cutting down a tree in
the garden and now you have a branch you want to break. To put the
most force into it, do you stand on it gently rocking back and forth,
or do you jump up in the air and stamp down on it?

So what you want for your worst-case scenario is:
a big heavy rider (250lb)
with big long cranks (180mm)
grinding away at very low speed (so he puts maximal force into each
pedal stroke)
flinging his considerable weight into each pedal stroke (which
multiplies his effective weight by some unknown factor)
and pulling around on the bars (ditto).

Unfortunately we've got two unknowns in there so you can't really get a
good answer, but it's easy to calculate the torque of a stationary
rider and then you've just got to guesstimate how much gets added by
throwing weight around and hauling on bars.

Just out of interest, why do you want to know?

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.

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

With a crank of 175mm it means an impressive torque of 130*0.175 = 22.75
kgm.

Assuming 1 kg = 2.20 pounds and 1 m = 3.2 feet, this should result in:
22.75 * 2.20 * 3.2 = 160.16 pounds*feet.

I suppose this is the maximum power a man could produce on a bottom
brack!

--
Massimo Bacilieri AKA Crononauta
Ravenna, Italy.