The Weight We Wear: Part Three – Cycling shoes are rotational weight!

When looking at the effect of rotational mass of bicycle components, one needs to keep in mind the they are part of a system that includes both an entire bike and a rider. When including the total mass of the bike and rider together, a 100 gram difference at the rim becomes negligible:
https://www.slowtwitch.com/Tech/Why_Wheel_Aerodynamics_Can_Outweigh_Wheel_Weight_and_Inertia_2106.html
and
https://www.cyclingpowerlab.com/AccelerationAndInertia.aspx

This certainly follows for shoes.

Sorry to break the news on you Tyler, but this is complete non-sense. I’m a professional physicist, and can’t help but feel pain when I hear so many conceptual errors and misconceptions, justified as “crazy science” details.

The best way to think about this is in terms of torque. The main torque accelerating a bike is produced by the force exerted on the pedals by the rider. The leg pushing down produces a positive torque, while the leg going up, if it has part of its weight still on the pedal, produces a negative torque. The sum of the two torques is the net torque, which is transmitted to the rear wheel (also as a torque) to fight the torques on the back wheel that slow you down (e.g. friction, wind or gravity) or to accelerate the rider.

The torque produced by the weight of one shoe (and pedal, and sock, and whatever) is always almost exactly equal, but with opposite sign, to the torque from the other shoe. The net torque is zero. That’s why your crank doesn’t start to magically spin by itself if you clip both shoes in the pedals and stare at it.

The only time where shoe mass plays a role is when you change your cadence. Let’s take a pretty extreme example: going from 30 RPM to 90 RPM in just one second. The torque required to do this, for two 100g masses at the ends of the cranks, is about 0.04 N m (Newton – meters; details below). At 90 RPM, that’s an enormous 0.4 W.

0.4W is how much power you need to accelerate a shoe that is 100g heavier than another one from 30 RPM to 90 RPM in one second.

0W is the difference between a heavier and a lighter shoe at constant RPM.

Of course, you have to carry shoes uphill, so they add to your total weight. 100g is the same as 100 ml of water in your bottle; about half a cup.

— details —
Angular velocity at 30 RPM is 3.1 rad/s, 90 RPM is 9.4 rad/s, accelerating constantly between the two gives the angular acceleration of 6.3 rad/s/s.
The torque to produce this acceleration is equal to inertia times angular acceleration; inertia is 2 (because two feet) times cranks length squared, times 0.1 kg. That’s 0.4 Nm for the torque. Power is torque times angular velocity; that’s 0.35 W at 90 RPM.
And of course, linear superposition applies: if you were to really accelerate from 30 RPM to 90 RPM, much of the required torque would go to accelerating your big muscular legs (and you on the bike), not spinning around a few extra grams in your shoes.

Check out https://en.wikipedia.org/wiki/Rotation_around_a_fixed_axis

Indeed, most don’t understand the complex mechanics interacting with aerodynamics. Shoe weight?? Otherwise they would have thought to shield ONLY the upper wheel. We now have proven the fastest conventional road bike, simply by shielding the upper wheel — in downhill coast tests. In this case, a Cervelo P3. Hint Tyler. See you in Reno?

https://youtu.be/tnjEacJybyQ

Yay physics fail Tyler. How about consulting an engineer/physics instructor first before posting things like that in the interwebs?

Waiting for a corrected version of this article, else these types of anecdotes will show up in the cafes how their $500 shoes made them so much more efficient and they can “feel” it.

As already pointed out several times above: this is really hunting small gains.

Here are some numbers to get this in proportion.
To complete a relatively level 10km TT at ~36km/h you spend roughly ~200,000J of energy.

The energy needed to get 100g extra rotating at 90 rpm from 0 rpm (as already pointed out above but with 1kg so divided by 10 here) is 0.13J. And when shifting gears you waste maybe half, say 0.07J. But the largest contributor would be coasting on/off which of course is also 0.13J.

To earn 0.1 second (out of roughly 900s) in my TT example one need to spend ~66J extra. Or the opposite, loose 0.1s by wasting 66J on some uneccessary loss.

So how often do one need to coast to add 0.1s for the TT example by losses due to extra rotational shoe weight?

It turns out to be 500 times, that is every 20 meters of the TT example. (Of couse coasting this often will have other much larger negative implications but that is ignored in this context)

So there might be some tiny gains when riding for a long time in a group/pack with VERY unsteady/erratic pacing. But for a “normal” ride this rotational shoe effect is negligible.

So to the big takeaway from this is that riding at a steady pace (assuming fixed gradient and no wind) is energy efficient!

I have read quite a number of comments posted by claimed “professional physicists/engineers” as well as non-science minded folks. This argument is almost never resolved. So, “anecdotally” I will add: I can attest with firm conviction that “all things” being equal (Bike, rider’s weight, etc) but with the exceptions of the wheel weight, a DRAMATIC difference is immediately noticed in the work required to rotate a heavier wheel set. After decades and thousands of miles of uphill training and racing, you could not pay me enough $ to use a heavy wheel set or heavy shoes. So when I hear arguments using mathematical equations that claim weight savings while peddaling does not matter and that one leg cancels out the other leg, yada yada, stop by a local professional MTB or road race and look at the bike components and rider’s apparel. You won’t see ANY wheel sets in the 1800g range or shoes that are in the 450g range. There is a reason. Lighter wheels, peddals, shoes really do make a person faster by requiring less work.