Bicycle Mechanics - Installing pedals - how much torque?

Hi all. I've been riding for years, but here's a "noobie" style question: just bought new Look Keo pedals which install with an 8mm hex -- instructions recommend 40nm of torque. I have never installed my own pedals and do not own a torque wrench, an am worried about stripping out the crank threads. So here's the question: should I just go ahead and put these on myself without a torque wrench, and if so, how far should I torque them -- in other words, how critical is it, and how easily will I strip out the crank threads (Ultegra 9-sp cranks)? Thanks in advance for your advice.

Don't worry about the torque wrench. Just tighten the pedals nice and tight - tight enough so you're using your muscles, but not so tight that you're straining them. Be sure to apply grease to the threads. Your feet will do the rest, over time, as you pedal.

The tighter you put them on, the more likely you will have blood involved when you are taking them off.

So, pedals will not fall off if they are under-torque'd?

My method is once the pedal is tight to the crank, I torque it one-half of a turn.

snug.

snug.
+1 I use antiseize compound rather than grease. It is only a few $ at the automotive store and will last for years.

So, pedals will not fall off if they are under-torque'd?

doubtful. You'd have to REALLY undertorque them, and pedal backwards a lot.

40nm is firm enough to tense your arm, but your jaw should still be relaxed.

So, pedals will not fall off if they are under-torque'd?The reason pedals are threaded left and right the way they are is so they don't loosen. Don't obsess over torque. Bicycle builders got by without torque wrenches for centuries. In my opinion, on bikes they're only required by people with absolutely no mechanical empathy.

Bicycle builders got by without torque wrenches for centuries.
Now that's a well researched statement.

Hi all. I've been riding for years, but here's a "noobie" style question: just bought new Look Keo pedals which install with an 8mm hex -- instructions recommend 40nm of torque. I have never installed my own pedals and do not own a torque wrench, an am worried about stripping out the crank threads. So here's the question: should I just go ahead and put these on myself without a torque wrench, and if so, how far should I torque them -- in other words, how critical is it, and how easily will I strip out the crank threads (Ultegra 9-sp cranks)? Thanks in advance for your advice.
Worry about having them too loose, not too tight.

Stripping the threads happens when they're too loose, so they move around in the threads.

If you're using a typical length 8 mm Allen wrench, make them as tight as you can get them.

Normally pedals are installed with a wrench over a foot long.

Sheldon "ELDI" Brown

[QUOTE=Eatadonut]doubtful. You'd have to REALLY undertorque them, and pedal backwards a lot./QUOTE]

Don't bet the rent money on that one. If pedals aren't put on pretty snugly they can definitely come off.

I put them in snug enough while not going crazy on it. I've struggled with too many stuck pedals to make the mistake of over tightening. I count on the fact they'll be kept tight by the forward pedal motion

The tighter you put them on, the more likely you will have blood involved when you are taking them off.


Amen

The tighter you put them on, the more likely you will have blood involved when you are taking them off.

since keo's dont have flats on the pedal spindle for a pedal wrench and only an 8mm allen, I make it tight enough so it doesn't just spin off, but loose enough that I can easily take off the pedal w/o taking off all the skin on my knuckles.

The reason pedals are threaded left and right the way they are is so they don't loosen. Don't obsess over torque. Bicycle builders got by without torque wrenches for centuries. In my opinion, on bikes they're only required by people with absolutely no mechanical empathy.
or on CF components

Really tight, with grease = you can take them off again.

since keo's dont have flats on the pedal spindle for a pedal wrench and only an 8mm allen,

Wow, that seems like a pretty boneheaded design decision.

Originally Posted by nitropowered:
since keo's dont have flats on the pedal spindle for a pedal wrench and only an 8mm allen,


"Wow, that seems like a pretty boneheaded design decision."


Is it gonna be murder to eventually remove those pedals?

Originally Posted by nitropowered:
since keo's dont have flats on the pedal spindle for a pedal wrench and only an 8mm allen,


"Wow, that seems like a pretty boneheaded design decision."


Is it gonna be murder to eventually remove those pedals?

I'd really like to see what happens when someone uses an 8mm and strips out the the hole, now what?

I'd really like to see what happens when someone uses an 8mm and strips out the the hole, now what?


http://www.matadorcoupe.com/seebee65/18v_driver2.jpg

I rarely use my pedal wrench anymore and usually just go with the hex key. I've never had any trouble, although I do have my pedals on and off fairly often switching my favourites between bikes. Don't forget those pedals were put on with a key too, so they won't be as tight as normal anyhow.


If someone did strip the hex you could always go the ez-out route (well, on the rh pedal at least.) Same argument if you round off the flats on a standard pedal anyhow...

Originally Posted by nitropowered:
since keo's dont have flats on the pedal spindle for a pedal wrench and only an 8mm allen,


"Wow, that seems like a pretty boneheaded design decision."


Is it gonna be murder to eventually remove those pedals?

Same on my Time pedals. it means everything can be more snugly over against the cranks as there doesn't need to be any room for a spanner [wrench] to grab.

Now really bizarrely, each side of my pedals uses a different size hex wrench. No one has been able to give me a good reason why that would be.

40NM is about 28lb-ft of torque. Sheldon is right of course (big surprise); it will be close to impossible to overtorque these with the typical 8mm Allen wrench which is about 5" long.You would have to be capable of putting out 80 lbs of force with one hand, on a short, thin allen wrench. Superman-maybe-the rest of us, no way. Now, if you have allen bits for a socket wrench(-7"-8" long) you would still have to put out close to 60 lbs of force-a lot for one hand.
Put them on with antiseize or grease.Clean the threads on the crankarms very well first.
Luck,
Charlie

40NM is about 28lb-ft of torque.
Wow, somebody other than me who remembers the correct English units for torque! :-)

I've given up correcting folks who ignorantly use "foot pounds" when they should be referring to lb-ft.

The foot pound is a unit of work or energy. The pound foot is the unit of torque. At least that's what I learned in high-school physics.

Sheldon "Units" Brown

+----------------------------------------------------------------+
| Anyone who cannot cope with mathematics is not fully human. |
| At best he is a tolerable subhuman who has learned to wear |
| shoes, bathe, and not make messes in the house. |
| --Robert A. Heinlein |
+----------------------------------------------------------------+

ft.lb is the how torque is normally expressed in imperial measurement. I have couple of torque wrenches, Stahlwille (great) and a Warren & Brown, both have ft.lb scales. Multiplication is commutative so ft.lb = lb.ft.

Don't worry about it. I have a ****ty rusted 8mm allen wrench or whatever it is that my pedals use, so I only install loosely so that I don't damage the allen socket. I did have it come loose once, but nothing bad happened. It just kept going loose tight as I pedaled, making an odd sound. Irritating, but big deal. So I tightened it up slightly more and all's good. For reference, when I put that pedal in it was pretty damn loose, like maybe 10 lbs of force at 5". So you can get away with it, but I would still go with Sheldon's advice and put it on tight as you can.

Wow, somebody other than me who remembers the correct English units for torque! :-)

I've given up correcting folks who ignorantly use "foot pounds" when they should be referring to lb-ft.

The foot pound is a unit of work or energy. The pound foot is the unit of torque. At least that's what I learned in high-school physics.

I beg to differ.

The most general form of the torque equation is the vector equation:

T=r X F

Where r is the moment arm in units of length and F is the force applied. "X" is the cross product of two vectors, not multiplication. The cross product operation is not commutative. In other words, r X F does not equal F X r. Therefore, T in ft-lbf is correct.

The most general form of the equation for Work (not energy, that comes later!) is the vector equation:

W=F • d

where F is the force vector, and d is the distance vector. "•" is the dot product of two vectors, not multiplication. Again, this operation is not commutitive, so lbf-ft would be correct.

Energy is:

W/t

Not a vector equation! "W" is the work applied over "t", time. Those units are in lbf-ft/sec, or Hp, Watts, whatever.

I have heard people contend that "lbf-ft" is correct for torque, but I have yet have anyone explain why this is so, mathematically. That's why I presented the equations, which can be found in any Engineering Statics book.

So I may be ignorant, but do I know the difference between torque, work, and energy! :D

In the end, it really doesn't make much difference, as long as you keep track of the units.

Wow! Scholarly cyclists. Suddenly, I'm feeling kinda igorant. Anybody wanna discuss centripital and centrifugal forces, to make me feel more igorant?

The foot pound is a unit of work or energy. The pound foot is the unit of torque. At least that's what I learned in high-school physics.

I'm amazed that there's anyone on BF this time of day that can remember, or ever knew, any HS physics.
(excepting SB of course)

I beg to differ.

The most general form of the torque equation is the vector equation:

snip...

It's not about the mathematics, it's about convention.

The official convention is to put the force first for torque, and the length first for work/energy.

This is sort of like how in bicycle tires, 26 x 1.75 is not equal to 26 x 1 3/4.

Sheldon "High School Physics" Brown

+---------------------------------------------+
| The nice thing about standards is that |
| there are so many of them to choose from. |
| --Andrew S. Tanenbaum |
+---------------------------------------------+

I beg to differ.

The most general form of the torque equation is the vector equation:

T=r X F

Where r is the moment arm in units of length and F is the force applied. "X" is the cross product of two vectors, not multiplication. The cross product operation is not commutative. In other words, r X F does not equal F X r. Therefore, T in ft-lbf is correct.

The most general form of the equation for Work (not energy, that comes later!) is the vector equation:

W=F • d

where F is the force vector, and d is the distance vector. "•" is the dot product of two vectors, not multiplication. Again, this operation is not commutitive, so lbf-ft would be correct.

Energy is:

W/t

Not a vector equation! "W" is the work applied over "t", time. Those units are in lbf-ft/sec, or Hp, Watts, whatever.

I have heard people contend that "lbf-ft" is correct for torque, but I have yet have anyone explain why this is so, mathematically. That's why I presented the equations, which can be found in any Engineering Statics book.

So I may be ignorant, but do I know the difference between torque, work, and energy! :D

In the end, it really doesn't make much difference, as long as you keep track of the units.
See, this is what happens when you go to school for far too long! I think you could reduce this whole statement to be "In the end, it really doesn't make much difference, as long as it's tight"

(I do see you're member status as "philosopher" so that explains a lot!) :p

30 ft-lbs. bk

I stand by my original post (#2).

See, this is what happens when you go to school for far too long! I think you could reduce this whole statement to be "In the end, it really doesn't make much difference, as long as it's tight"

(I do see you're member status as "philosopher" so that explains a lot!) :p

Yeah, you're right...far too many years of skool. :D

But if the screw is under an alternating load, torque becomes extremely important...oh, forget it...


It's not about the mathematics, it's about convention.

The official convention is to put the force first for torque, and the length first for work/energy.

Well, OK, but I've still not had anyone explain lbf-ft to my satisfaction. If it's convention, it doesn't match the math. I'll stick to ft-lbf.

I'm all about mathematics.

Oh, and before I got all off topic...

Just snug the pedals...they will keep themselves tight. :D

Tighten till stripped then back off 1/2 turn. :D

My method is once the pedal is tight to the crank, I torque it one-half of a turn.
That's an awful lot of torque, if by "tight to the crank" you mean what I mean. I'd feel really good about 1/4 turn (or even less; I go by feel on pedals rather than use the torque wrench, but next time I'll try the wrench to see how 40 nm equates to turns).

so i hate to be "that guy" but..


I beg to differ.

The most general form of the torque equation is the vector equation:

T=r X F

Where r is the moment arm in units of length and F is the force applied. "X" is the cross product of two vectors, not multiplication. The cross product operation is not commutative. In other words, r X F does not equal F X r. Therefore, T in ft-lbf is correct.

the cross product is not commutative, but the units "pound" and "feet" are. energy and torque both have units of pounds times feet or feet times pounds but the convention is to refer to "foot pounds" when you're talking about energy or "pound force feet" when referring to torque so as not to confuse the units of energy (a scalar) and torque (a pseudovector (http://en.wikipedia.org/wiki/Pseudovector))



The most general form of the equation for Work (not energy, that comes later!) is the vector equation:

W=F • d

where F is the force vector, and d is the distance vector. "•" is the dot product of two vectors, not multiplication. Again, this operation is not commutitive, so lbf-ft would be correct.

the dot product is commutative. d • F = F • d



Energy is:

W/t

Not a vector equation! "W" is the work applied over "t", time. Those units are in lbf-ft/sec, or Hp, Watts, whatever.

I have heard people contend that "lbf-ft" is correct for torque, but I have yet have anyone explain why this is so, mathematically. That's why I presented the equations, which can be found in any Engineering Statics book.

So I may be ignorant, but do I know the difference between torque, work, and energy! :D

In the end, it really doesn't make much difference, as long as you keep track of the units.

the units of work and energy are the same, the british call them joules.

work = the change in energy = delta E

i believe you're thinking of power, which for a constant rate of energy transfer is equal to W/t.

Wikipedia uses the same term, "foot-pounds force (ft·lbf)" both for torque and for work. The people have spoken! (And so it would seem, as confused as ever. http://www.thesmilies.com/smilies/winking0023.gif)

But! Torque is defined in this wise: ""rotational force" or "angular force" which causes a change in rotational motion." That's work, not just force. So the confusion is one of usage.

"Pound-foot", btw, is nowhere defined as anything other than a synonym for "foot-pound".

the units of work and energy are the same, the british call them joules.
Actually, everybody but the British calls them "joules", unless you mean "Brits with metric hats on".

"One joule is the work done or energy required to exert a force of one newton for a distance of one metre."

However, it's immediately followed by this, which serves to light the fire again:

"Thus, the same quantity may be referred to as a newton metre or newton-metre with the symbol N·m. However, since the newton metre usually refers to torque, which is a quite different measure, the joule is a less ambiguous unit. In elementary units:" J = kg * m² / sec²

Wikipedia uses the same term, "foot-pounds force (ft·lbf)" both for torque and for work. The people have spoken! (And so it would seem, as confused as ever. http://www.thesmilies.com/smilies/winking0023.gif)

But! Torque is defined in this wise: ""rotational force" or "angular force" which causes a change in rotational motion." That's work, not just force. So the confusion is one of usage.

"Pound-foot", btw, is nowhere defined as anything other than a synonym for "foot-pound".

hate to disagree with the people, but again i feel the distinction in nomenclature stems from the difference between a vector and a scalar quantity.

work = force dot distance = a scalar

torque = force cross radius = a vector

they are not the same thing, and can not be added. to avoid writing "foot pound force the scalar" and "foot pound force the vector," physicists just call it "foot pounds" for energy and "pound force feet" for torque.

Actually, everybody but the British calls them "joules", unless you mean "Brits with metric hats on".

yep, stupid me lumping the brits in with the rest of europe.

I-and most folks-use LB-FT for torque to avoid confusion with FT-LBS which is usually energy. I don't know how to put a period between, and in the middle, of LB-FT so I use a dash.
Thanks,
Charlie

But! Torque is defined in this wise: ""rotational force" or "angular force" which causes a change in rotational motion." That's work, not just force. So the confusion is one of usage.

As often, Wikipedia (which is, after all the internet) is wrong.

The "which causes a change in rotational motion" part is bull. Since we're talking bikes, hold that rear brake real tight, step on your pedal, and put all your weight on it. Nothing's moving. You mean that there is no torque on the crank, nor reaction torque on the frame, nor torque on the rear wheel from the chain, nor opposite torque from the rear brake, nor a reaction torque on the rear brake mounting?

If so, I am eagerly awaiting your proposal for a solution of the resultant force on the rear brake pads, knowing the force exerted on the pedal, an all the geometry data, but having to assume all torques are zero, since there is no "change in rotational motion" caused anywhere.

And to settle the lb-ft/ft-lb thing, go metric: it's Nm, no debate there.

so i hate to be "that guy" but..
i believe you're thinking of power, which for a constant rate of energy transfer is equal to W/t.

You are correct...thanks for calling me out.

I messed up, and I know the difference. I often have to size motors at work.

Sorry about that.

As often, Wikipedia (which is, after all the internet) is wrong.

The "which causes a change in rotational motion" part is bull. Since we're talking bikes, hold that rear brake real tight, step on your pedal, and put all your weight on it. Nothing's moving. You mean that there is no torque on the crank, nor reaction torque on the frame, nor torque on the rear wheel from the chain, nor opposite torque from the rear brake, nor a reaction torque on the rear brake mounting?

If so, I am eagerly awaiting your proposal for a solution of the resultant force on the rear brake pads, knowing the force exerted on the pedal, an all the geometry data, but having to assume all torques are zero, since there is no "change in rotational motion" caused anywhere.

And to settle the lb-ft/ft-lb thing, go metric: it's Nm, no debate there.
My post was made with tongue firmly implanted in cheek. The thing Wikipedia (along with a few posters here) doesn't do is make a distinction between torque as a practical measure and torque as a physical concept. I doubt seriously that anyone is going to mistake one context for the other, so terminology such as that suggested is pointless.

Btw, all those pretty mathematical formulas work fine until you start factoring in things like friction and material deformation. :eek: