In Theory

Wheel diameter and performance: part 1

One of my guilty pleasures is reading internet forums. A bad habit, but one that I find particularly hard to break. Though much of the content you can find on such platforms is of questionable quality, sometimes you stumble across a hidden gem.

Some time ago I read this post on the Bicycles Stackexchange site. The original poster was interested in the influence wheel diameter has on various aspects of ride quality. This seemingly innocuous question turns out to be a real rabbit, taking us deep into the realm of vehicle dynamics.

Since vehicle – especially those moving on two wheels – dynamics is a fascinating subject, I decided to give this question the respect it deserves by writing a series of posts dedicated to answering it. Each one will be a self-contained analysis of a particular aspect of the problem. And although each will contain some mathematical formulas, feel free to skip more technical passages if you so wish. They (hopefully) won’t be necessary to understand the bigger picture.

Scenario: riding over obstacles

One of the most fundamental aspects of bicycle performance is its ability to ride over obstacles. And it’s also quite obvious why. Unless you spend most of your time riding on a silky smooth, Siberian spruce-decked velodrome, you’ll need to ride over some sort of an obstacle rather sooner than later. A crack in the pavement, a fallen branch on the trail, or a curb in the city, all fall into the same category.

Poor obstacle-crossing performance has short and long-term effects on both the rider and the bike. The most immediate consequence is losing control of the bike and crashing, potentially shattering your ego as collateral. Chronic exposure to violent jolts, however, can lead to conditions such as hand-arm vibration syndrome or pudendal nerve entrapment syndrome. It’s thus in the rider’s best interest to ride a bike that provides the most comfortable ride possible. That’s also true for competitive riders, as a smoother ride causes less fatigue and allows for higher power output.

You also should keep in mind that harsher rides are more taxing for the gear and negatively impact service life. You’ll thus need to replace parts more often, and those usually aren’t cheap either. But while spending more money is bad enough, having a gear failure in the middle of nowhere and arranging for a ride back home is even worse – and I’m speaking from experience.

Rigid wheel model

Let’s start with a simple mathematical model: a rigid wheel going over a rigid, rectangular bump. By ‘rigid’ we mean not deforming at all, regardless of the forces acting on it. At this point, mathematicians rejoice, but physicists and engineers might start to shift uneasily in their seats. It’s easy enough to swallow a rectangular obstacle, as curbs come very close to it. But rigid? In reality, everything deforms to some extent, especially pneumatic wheels. After all, the entire circumference is literally wrapped in a rubber balloon!

A close-up shot of a granite kerb.
Common European curb (Curbus vulgaris). They come as close to rigid rectangular steps as humanely possible.

That is, without a doubt, absolutely true. But starting with a simple model allows us to gain some insight into the problem without overly complicating it at first. It also models a real-world situation of riding with underinflated tires, so that they bottom out on impact. Finally, we’ll use some of the results obtained here in more complicated scenarios.

The maths

Because this model is so simple, we can readily derive the relation for the vertical displacement of the wheel as a function of the distance traveled since the initial point of contact with the obstacle. Using the Pythagoras’ theorem only, we arrive at the following equation:

    \[y(x)=h - r +\sqrt{r^2 -( \sqrt{h(2r-h)}-x)^2}\]

where y is the vertical displacement, x is the horizontal distance, h is the obstacle height, and r is the wheel radius. We silently assumed that the wheel stays in contact with the obstacle at all times, i.e. it does not bounce up. We’ll get to the validity of it soon, but for now, let’s assume that it’s true.

The graphs

We can plot the path followed by the wheel for different wheel sizes and obstacle heights:

The first thing we notice is that larger wheels follow gentler, longer paths. On the other hand, a wheel is only able to cross an obstacle smaller than its radius, in which case it must come to a complete stop before climbing vertically. That’s far from ideal. We can thus conclude that larger wheels have better overall climbing abilities.

Confirmation of this theory comes from plots of vertical acceleration. Below you see how acceleration changes with the distance traveled from the initial point of contact with the obstacle for different obstacle heights and wheel radii. The speed is the same as before (i.e. 36 km/h).

Larger wheels are indeed subject to smaller vertical accelerations and, thus, forces. Interestingly, horizontal forces are smaller as a result. That’s because the force on the wheel acts mainly in the radial direction, from the point of contact with the obstacle to the hub. This radius lies more horizontally for small wheels, yet the vertical force component is bigger than in their larger counterparts. As a result, the force magnitude and its horizontal component must be bigger, too.

One thing to be said about the acceleration graphs is that the values are humongously huge. Even for obstacles as small as 3 cm and fairly large wheels, we still get almost 40 g for the largest wheel, going well over 100 for smaller ones. 40 g is actually enough to trigger a blackout, and about 100 g is what you can expect in a car crash. Something must be wrong, right? After all, nobody dies riding a scooter or rollerblades on rough pavement.

That’s true, and that’s just an artifact of our model. The key is that acceleration values are actually negative, as the vertical speed decreases as the wheel climbs. We assumed that the wheel stays attached to the ground at all times, but in reality, it would get airborne as soon as it reaches negative 1 g, which is how much Earth’s gravity alone can provide. Finally, the graphs do not show the infinite acceleration at x=0, which exists only in the distribution sense.

We will fix those problems by including the tire in our model.

Influence of the tire

As we saw in the last section, a rigid wheel would be impossible to ride on anything but the smoothest of surfaces. Let’s thus improve our model by accounting for the influence of the tire. In this way, we get closer to reality for a large category of bikes. Indeed, virtually all road bikes have no suspension, and so do many city bikes. For such bikes, most of the cushioning from the road unevenness is provided by the tire alone. The same applies to the rear wheels of hardtail mountain bikes.

So how do we model the tire? The standard way is to think of the bike and the rider as a rigid body and replace the tires by springs, with one end following the road profile. We will go one step further and focus on one wheel only, by considering a mass equal to front wheel loading. Under certain assumptions, this simplification is exact, but even if not, the results we’re interested in will not be far off. Finally, the road profile is just the path followed by the wheel in the rigid wheel model.

A bike without suspension can be modelled as a simple mass m on a spring with stiffness k and damping c.

The maths

To get the path followed by the wheel, we first need to write an appropriate equation of motion. If we denote by y(t) the vertical wheel displacement at time t, and by f(t) the corresponding displacement of the rigid wheel model at the same time, then the Newton equation of motion is

    \[m \frac{d^2}{dt^2}y(t)=-k \left( y(t)-f(t) \cdot \sin{\theta(t)}\right) -c \frac{d}{dt}\left( y(t)-f(t) \cdot \sin{\theta(t)}\right)\]

where m is the mass loading of the wheel, k is the tire radial stiffness, and c is the tire damping coefficient.

The graphs

Let’s plot the first 0.2 seconds of the solution for two road bike wheel standards, i.e. 700c (29 inches) and 650b (27.5 inches). Tire radial stiffness corresponds to a 25 mm tire on a 17 mm rim pumped to 6 bar as found by Silca, while the damping coefficient chosen corresponds to critical damping. The speed is 36 km/h, while the obstacle height is 1 cm, which corresponds to a rough road patch.

Path followed by a wheel with a deformable tire going over a 1 cm bump. Note how smoother it is in comparison to the rigid wheel model.

If you don’t see any blue in this graph, don’t worry, neither do I. Those two paths are so close, that the difference between them is smaller than the line width. But the paths followed are much smoother than that without the tire.

And what about the acceleration? Unsurprisingly, the difference is also very small:

The tire decreases peak acceleration substantially, but not enough to prevent the wheel from getting air.

Peak acceleration differs by less than half g, partly due to small obstacle height, and partly due to the cushioning effect of the tire. However, it might make the difference between winning and losing a race, so, it may still be worth it. Also, note that the acceleration the handlebar is subject to would be pretty much the same as that of the wheel. That’s because the handlebar sits almost directly above the wheel, and the connection between the two is rigid.

Moving over to mountain bikes case, the difference is more conspicuous, as obstacle heights are bigger, and so is the difference between 29″ and 26″ wheels. For instance, take a look at accelerations for 26″ and 29″ wheels and a 5 cm obstacle at 20 km/h:

Peak acceleration of 26″ and 29″ MTB wheels going over a 5 cm bump at 20 km/h differs by almost 7%.

Here, the peak acceleration difference is about 2 g, or 6.7%. Note, however, that the model predicts negative acceleration below -1 g which is just an artifact of the simplification used – in reality, the wheel would lose contact with the ground at that point.

Influence of the suspension

But what about bikes with suspension? After all almost every mountain bike on sale right now has at least a front shock, not to mention all motorcycles.

To account for suspension, we can modify our previous model and consider two, instead of just one, masses, connected with a spring. The lower mass is the so-called unsuspended weight, i.e. the wheel and the lower part of the fork. The upper, or suspended mass, is equal to the wheel loading. As before, we’ll assume a road profile to be the same as the path followed by the wheel in the rigid wheel model.

Simple model of a bike

The maths

Denoting by y_1 and y_2 the displacements of the wheel and the handlebar, respectively, the equations of motion have the following form:

    \[\left\{\begin{array}{@{}l@{}}     m_2 \frac{d^2}{dt^2}y_1 = -k_2(y_2-y_1) - c_1 \frac{d}{dt}(y_2 - y_1) \\     m_1 \frac{d^2}{dt^2}y_2 =  k_2(y_2-y_1) + c_1 \frac{d}{dt}(y_2 - y_1) - k_1(y_1-f \cdot \sin{\theta}) - c_1 \frac{d}{dt} (y_1-f \cdot \sin{\theta}) \\  \end{array}\]

where we used the symbols from the figure above.

The graphs

Those equations might seem complicated, but we don’t need to find a general solution to them. What’s much more interesting is how the wheel radius affects performance. Below you can see graphs of vertical acceleration for both the wheel and the handlebar for 26 and 29-inch wheels hitting a 5 cm obstacle at 36 km/h:

We clearly see that peak handlebar acceleration is about 0.25 g smaller with the larger wheel. That might not sound as much, but it amounts to a whopping 6%. The same can be said about the wheel itself, with peak acceleration being about 5% smaller for the larger wheel.

To be continued

It is clear that bigger wheels provide better performance when it comes to riding over obstacles. The ride is more comfortable and planted, and the components are subject to much less stress.

It the next part, we’ll take a closer look at how good they are at smoothing road irregularities.

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    […] the last post, we saw how bigger wheels offer better performance when riding over obstacles. In the next […]

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