- Why talk about torque?
- Torquing nuts and bolts
- Torque in braking
- Torque in drivetrains
- Limits in drivetrains
- Links

As Sheldon says in the glossary on this site, torque is linear force times the radius at which it is applied. For example, a 10-pound force applied two feet from the axis produces the same torque as a 2-pound force applied ten feet from the axis.

The standard units for measuring torque are pound-feet or Newton-meters. The force unit goes first, so as not to be confused with energy/work measurements. A common error is to refer to "foot-pounds" instead of pound-feet of torque. This is not strictly correct, since the foot-pound is a unit of energy/work, not torque.

The tightness of nuts and bolts is measured as torque. Torque is also useful in understanding stresses on the bicycle frame, and the workings of the bicycle drivetrain and brakes.

In connection with bicycles, torque most commonly comes up as a measure of the tightness of threaded parts. Skilled bicycle mechanics have traditionally tightened these by feel. A skilled hand can usually though not always determine when a steel part has reached optimum tightness. The feel at the wrench changes as friction increases and the threading engages tightly.

Softer materials -- aluminum alloy, carbon fiber composites -- have become more common in bicycles, and are easier to overtighten, resulting in cracking, deformation and stripped threads.

Damage from overtorquing can occur when disassembling parts, and corrosion has stuck them together. An ounce of prevention is better than a pound of cure: grease, anti-seize compound or blue threadlock compound (depending on the application) should be used on threads and on surfaces which the threads press together, to prevent corrosion. Aluminum-to-aluminum joints are especially likely to seize up. It is a good idea to disassemble and reassemble parts from time to time while this is still easy, and relubricate them.

A special lubricant called penetrating oil can help to free corroded threading. Special techniques to apply leverage can help with stuck parts such as freewheels, stems and seatposts. Heat or cold to expand or shrink parts slightly also can help. Aluminum expands more when heated than steel, and so an external aluminum part should be heated, an internal one, chilled.

**See for example Jobst Brandt's advice on freeing stuck pedals. **

When a bolt head breaks off, sometimes the remainder of the bolt turns easily, but often it must be drilled out. Drilling can release pressure on the threads, but also there is a special tool called an Easy-Out which threads into the drilled-out shaft of a bolt and gets a grip from inside, with any luck allowing it to be removed.

Besides soft parts, there are a few other ways to fool a skilled hand.

- The wedging action of the expander plug of a quill stem multiplies the tension on the expander bolt. If the bolt is tightened to a torque which feels normal, the expander plug may bulge and permanently deform the fork steerer, possibly leading to failure of the steerer or to insecure clamping of the stem. With an older stem which uses a truncated-cone (cork-shaped) expander plug, the stem cannot be safely tightened enough to make it impossible to turn the handlebar while holding the front wheel from steering right or left. This looseness is unimportant for anything short of technical off-road riding and cn also prevent injury in a crash. A stem with a wedge-shaped expander plug can secure the handlebar more tightly, but also can be overtightened.
- Some threaded parts unscrew clockwise, or appear to. Turning the part counterclockwise in an attempt to loosen it will tighten it and can potentially ruin the part or the wrench. These parts are:
- The left pedal;
- the right side bearing cup or mounting ring of most bottom brackets (exceptions: those with French or Italian threading).
- The left-side locknut and bearing cone of a Thompson/Thun or Ashtabula bottom bracket.
- the outer bearing race of a conventional, threaded freewheel;
- The inner one or two sprockets of some older freewheels, which unscrew toward the inside (though all sprockets unscrew opposite the direction of pedaling).
- Spokes. These are actually threaded normally, but the bolt pulls toward the spoke wrench: As Sheldon puts it, spokes are threaded clockwise but you are looking at the back of the clock -- see his article on wheelbuilding.

- the clamp bolt inside a handlebar-end shifter body. This also is threaded clockwise, but the head is at the far end of the bolt. As with spokes, you are looking at the back of the clock.

- Also, sometimes a part is integral rather than threaded, and no attempt to turn it will remove it. The classic example is the outermost sprocket of the (by now antique) British Cyclo freewheel, which is integral with the freewheel body.

Bicycle and component manufacturers increasingly instruct mechanics to use a torque wrench. To some degree, this reflects a desire not to be held liable for incompetent assembly. We recommend the use of a torque wrench when assembling parts made of softer materials, or whenever unsure about being able to apply appropriate torque by feel. Quoting ace bicycle mechanic Francisco Cornelio, of Harris Cyclery and Cisco's Cycles, "You have to use a torque wrench now, or you're really screwed."

**Sheldon has written about a special torque-measuring tool :-).**

A brake imposes a torque on the wheel, and the road imposes an equal and opposite torque. The torque from the road is the force of deceleration, times the radius of the wheel.

Because of the high center of mass of a conventional bicycle, braking at the front wheel is limited by the possibility of pitchover. The front wheel bears most of the weight and must do most of the braking, The light weight on the rear wheel results in reduced traction. Maximum deceleration which a skillful cyclist can safely achieve is is equal to about half the weight of the bicycle and rider.

Let us assume a 200-pound bicycle and rider decelerating this hard, and using only the front brake. Then the force of deceleration is 100 pounds. We'll also assume a 28-622 tire, with a nominal diameter of 26.76 inches and so a radius of 13.38 inches, or 1.11 feet. Then the torque from braking is 100 pounds times 1.11 feet, or 111 pound-feet. With a braking-surface radius of 1.00 feet, the brake must exert a force of only 111 pounds -- shared equally by the two brake shoes. .

Because a rim brake acts at only a slightly smaller radius than the road surface, the brake need not clamp the rim very tightly, and the force which the brake transfers from the wheel to the bicycle's front fork is only slightly larger than the deceleration force at the road.

The force from the brake and the force from the road act at nearly opposite points on the rim, and both are trying to pull the wheel backward. Adding up the two forces, we can see that the hub must pull the fork backward with a force a bit over 200 pounds. This force is shared equally by the two fork blades.

With a hub brake (disk or drum brake), the torque is imposed on only one fork blade, and at a smaller radius. The brake must clamp much tighter and stress on the fork blade is much higher, so a hub brake requires a stronger fork. A disc brake with the caliper behind the fork blade also tends to pull the hub out of the dropout, and this is not only a theoretical problem.

The bicycle's head angle and fork rake place the front wheel forward of the head tube, and so the headset bearings resist a torque which would pull the front wheel forward, in normal riding. Because front-wheel braking produces a torque which would pull the front wheel rearward, the two torques balance out at a certain level of braking, and the bearings of a loose headset rattle due to road vibration.

Skidding limits rear-wheel braking to approximately 0.3 g -- less if the front brake also is in use. The seatstays, chainstays and seat tube of a conventional bicycle frame resist deceleration in compression and tension -- no torque -- as they form a triangle with one corner at the rear axle. A rear rim brake along the seatstays or chainstays tends to bend them. The reaction arm or caliper of a rear hub brake imposes a greater force, due to the smaller radius at which it acts, usually only on the left seatstay or chainstay. The forces may be calculated the same way as with the front wheel.

In this discussion, I've neglected the slight additional braking needed to reduce the rotational momentum of the bicycle's wheels as a bicycle slows.

If a device is rotating at a steady rate or is stationary, the torques applied to it must add up to zero -- any torque applied at one point must be taken off at another. So, with a bicycle crankset, the torque applied at the pedals is equal and opposite that taken off by the chain, neglecting the small amount lost in friction.

Let's look at a bicycle drivetrain starting with the cyclist's feet. Torque is conveyed from a pedal through the crank -- and from the left crank, then also through the bottom-bracket spindle -- to the chainwheel. Generally, the cyclist's rising leg applies a light torque opposite that of the descending leg. The chain, at the chainwheel, produces a torque equal and opposite the sum of the torques applied at the pedals.

Let's put some numbers to this We assume that the left leg is descending in mid-stroke, applying a force of 100 pounds directly downward on the left pedal, while the rising right leg is applying a reverse force of 10 pounds to the right pedal. Crank length is usually given in millimeters, but we're using English measurement here, as it's more familiar to most English-speaking readers. We'll assume 0.56 foot (170 mm) cranks.

The torque on the bottom-bracket spindle is 56 pound-feet: the 100-pound force at the pedal, times the 0.56 foot length of the crank. The torque at the chainwheel is slightly less, 50.4 pound feet, after we subtract the -5.6 pound-foot torque from the right pedal.

Now, let's assume a 50-tooth chainwheel. This has a radius of about 4 inches, or 0.33 feet. We can now calculate the chain tension:

50.4 pound-feet/0.33 feet = 153 pounds.

We'll assume a 20-tooth sprocket, with a radius of 1.6 inches -- 0.13 foot at the rear wheel. The 153-pound chain tension produces a 20.2 pound-foot torque at this sprocket:

153 pounds * 0.13 foot = 20.2 pound-feet.

(We could also simply note the ratio of tooth counts of the chainwheel and sprocket, 50/20, which would give us the same torque ratio.)

We'll also assume a 28-622 rear tire, with a nominal diameter of 26.76 inches and radius, 13.38 inches, or 1.11 feet.

We can now calculate the drive force where the rubber meets the road:

20.2 pound-feet / 1.11 feet = 18.2 pounds.

The ratio of the force at the pedals to that at the road is 90 : 18.2, or 4.95.

In these calculations, we neglect forces which would not contribute to torque: pedal force not in the direction of rotation, and the weight on the rear wheel. We also neglect friction, which reduces the drive force by a few percent. Numbers are rounded -- close, but not exact.

The ratio of the force at the pedals to drive force at the road is the gain ratio, which can be calculated more simply as the ratio of road speed to pedal speed, like this:

.(1.11 feet /.56 feet) * 50 teeth/20 teeth = 4.95

This calculation is simple and elegant, but our torque calculations give us some additional useful results:

- the torque on the bottom-bracket axle
- the effect of the reverse force from the cyclist's rising leg,
- and the tension on the chain.

Now, let's look at the limits to torque and to drive force.

The limit to drive force is set by front-wheel lifting, which occurs when the drive force is about 1/2 the weight of the cyclist and bicycle.

Let's assume that the weight is 200 pounds. Then the drive force cannot exceed 100 pounds, or the front wheel will lift.

Let's also assume that the cyclist can push down on a pedal with a force of 200 pounds -- somewhat more than the cyclist's own weight, by standing on a pedal and pulling up on the handlebar.

Then at gain ratios below 2, the cyclist's pedaling force is capable of generating over 100 pounds of drive force, and lifting the front wheel. The cyclist must avoid pedaling as hard as possible.

In our example with the 0.56 foot crank length and 1.11 foot wheel radius, this transition occurs almost exactly where the number of chainwheel teeth and of rear sprocket teeth is equal. So, for example, if the chainwheel has 28 teeth and the sprocket, 30 teeth, a hard pedal stroke can lift the front wheel.

Above the transition point, torque at the chainwheel remains the same but torque and drive force at the rear wheel decrease, so the cyclist can pedal with full force, if desired.

and is discussed in a companion article.

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Last Updated: by John Allen