Tech & Wearables

Technical FAQ: Calculating chain tension

How to calculate chain tension, and how it relates to power output and gearing.

Have a question for Lennard? Please email him at veloqna@comcast.net to be included in Technical FAQ.

Dear Lennard,
In this column, you calculate the relative magnitude of friction losses in the chain. For those calculations you use the chain tension ibF, but how can you calculate this?
I have also a question about your solutions: Are those the total friction losses of the drivetrain? Because the solutions you found are really high.
— Robbe

Dear Robbe,
Your question of how to calculate chain tension is a good one, because, even though lots of riders have power meters on their bikes, I think very few of them understand how the tension in their chain relates to their power output and the gear they are in.

Let’s begin by clarifying three things.

The first point is that the calculations in the referenced column were not mine. That handwritten table of calculations that I scanned and posted with the letter was done by the writer of that letter, Brad, whose first letter I had posted in the previous column.

Secondly, “ibF” is actually “lbF”, Pounds of Force: the units that Brad calculated the chain tension in. Brad was calculating the time-averaged value of the force applied by the chain on the rear wheel.

Furthermore, I don’t know what conclusions you seem to be drawing about frictional losses when you say, “Are those the total friction losses of the drivetrain? Because the solutions you found are really high.” There are no frictional losses in the chart because Brad did not calculate any. Frictional losses on a bike are traditionally quoted in watts, though they could also be in other units of power.

Rather than calculating the magnitude of frictional losses, Brad was calculating the percentage of the frictional losses due to chain articulation that were happening at each of the points he labeled A-H (by comparing the number of links bent per revolution at a given point in the drivetrain times the angle of bend in the link at that point times the chain tension at that point). If you’re interested, we measured frictional losses due to the chain for a rider output of 250 watts pedaling at a cadence of 95RPM in a number of different gear ratios.

Now, to answer your question on how to calculate chain tension:

Tension on the upper span of a chain is inversely proportional to chainring size: the smaller the chainring, the higher the chain tension. This is simple physics. Power output on a bicycle can be simplified to torque multiplied by angular speed (in radians per unit time); power = torque x angular speed. To get angular speed from rotational speed (in revolutions per unit time, a/k/a cadence, or RPM), you must multiply RPM by the number of radians in one revolution of the crank circle, which is 2 x π.

So: angular speed = RPM x 2 x π, and power = torque x RPM x 2 x π

Tension on the upper span of a chain is inversely proportional to chainring size. Photo: Lennard Zinn | VeloNews.com

Torque is transferred from the chainring to the cogs by means of the chain tension. Since power output and cadence are both constant in Brad’s calculation, then so is torque, regardless of chainring size. Thus, the chain tension (force) with a smaller chainring must be higher than with a larger chainring, since torque is given by the equation torque = force X radius, and the radius is smaller. Conversely, a larger chainring has a larger radius, and therefore the chain tension—and hence consequent chain friction—must be lower at the same RPM and Power output.

Force, which in this case is chain tension (since that is how the pedal force is transferred to the rear wheel), is torque divided by radius (i.e., force = torque/radius). Since how hard the feet push on the pedals varies as they move around the crank circle, instantaneous chain tension (force), torque, and power are constantly changing as the feet move around the crank circle. However, the rider’s power meter gives an average power, which in this case Brad held constant at 260 watts.

Since power = torque x RPM x 2 x π, then torque = power/(RPM x 2 x pi) (i.e., power divided by angular speed). And force = torque/radius, so force = (power/(2 x π x RPM))/radius, or force = power/(2 x π x RPM x radius).

Power is held constant in this test and is equal to 260 watts, and RPM (i.e., cadence), also constant in this test, is 126 revolutions/minute, or 126 /min; the number of revolutions carries no units.

Wolf Tooth kindly provides the diameters of various sizes of chainrings online, and its table quotes the diameter of a 26-tooth chainring as 111.8mm. The diameter of a 52-tooth chainring should be double that, or 223.6mm (Wolf Tooth’s table actually says it is 216.9mm, which, while close to double 111.8mm, is not precise and makes no sense to me; I’ll just go with 223.6mm). The radius of the 52-tooth chainring is then 223.6/2 = 111.8mm or 0.1118m, and the radius of the 26-tooth chainring is then 111.8/2 = 55.9mm or 0.0559m.

Since force = power/(2 x π x RPM x radius), the chain tension with the 52-tooth chainring is: force(52T) = (260 w)/(2 x π x (126 /min) x 0.1118m)

Watts are defined by 1 watt = 1 joule per second (1w = 1 j/s), where one joule is equal to the work done by a force of one newton acting through one meter (1 J = 1 Nm). There are of course 60 seconds in a minute.

So: force(52T) = (260 Nm/s)/(2 x π x (126 /min) x 1min/60s x 0.1118m) = 176 N

One pound of force (lbf) is the amount of force (called “weight”) exerted by a one-pound mass due to gravity at the equator, where g = 9.8 m/s−2, and one pound of force is equal to 4.44822 newtons. So, the chain tension exerted by a 52-tooth chainring at 260 watts of power output and 126 RPM cadence is (176 N)/(4.45 N/lbf) = 40 lbf. And since 52 = 2 x 26, the chain tension exerted by a 26-tooth chainring at 260 watts of power output and 126 RPM cadence is 40 lbf x 2 = 80 lbf.

Brad instead calculated the chain tension exerted by the 52-tooth chainring to be 85 lbf, and that exerted by the 26-tooth chainring to be 171 lbf. I’m not sure why our results differ by a factor of around 2, but I’m confident in my results.

To check my answers, or to bypass all of this calculation in the future, Jason Smith of Ceramic Speed and founder of Friction Facts turned me on to this calculator for motor power; it is also useful for bicycle calculations.
― Lennard


Lennard Zinn, our longtime technical writer, joined VeloNews in 1987. He is also a custom frame builder (www.zinncycles.com) and purveyor of non-custom huge bikes (bikeclydesdale.com), a former U.S. national team rider, co-author of The Haywire Heart,” and author of many bicycle books including Zinn and the Art of Road Bike Maintenance,”DVD, as well as Zinn and the Art of Triathlon Bikesand Zinn’s Cycling Primer: Maintenance Tips and Skill Building for Cyclists.” He holds a bachelor’s in physics from Colorado College.

Follow @lennardzinn on Twitter.