In the efficiency focused automotive world, we all know that mass matters, but how much thought do we give to where that mass is located? Vehicle dynamics, of course, keeps a close eye on the issues of centre of gravity position, and unsprung mass is generally considered less desirable than sprung mass, but there is another aspect of the mass location within the vehicle that receives rather less attention.
Of course, the automotive industry is not unique in its interest in engineering efficiency, the world of cycling has long pursued the most minute of gains, with aluminium technology now being widely replaced by carbon fibre not just for the professional racers of the Tour du France and Olympics, but increasingly filtering into the volume production market aimed at the enthusiastic amateur and club racer.
This drive for lightweight has led to a heated debate about just where that weight should most effectively be shaved off in order to maximise performance. Central to this debate are the wheel and frame manufacturers. With both components now making extensive use of high tech and high cost materials, the answer to the question of where best to spend considerable sums of money to achieve the greatest performance gain is key to securing sales for the respective manufacturers.
Looking impartially from outside the cycle industry (though many may remember the Lotus’ bike which famously carried Chris Boardman to a world record and gold medal in the 1992 Olympics) at some of the questions the debate has raised, an examination of the science might put the claims and counter claims into context.
Firstly, why does the weight of a wheel play a more significant factor in performance than the weight of other parts of a vehicle?
The weight (or mass) of a wheel isn’t more significant, but the inertia of a wheel is. If the weight of the wheel changes, then typically the inertia changes too. Inertia is the result of a mass at a distance from the axis about which it is being rotated, in the case of a wheel it is the distance of the mass from the spindle axis.
How is the force required to make something rotate different from the force that makes something move generally?
The force required to move (translate) a mass faster (i.e. to accelerate it) is defined by Newton’s Second Law, from this can be derived calculations (see bottom of page) for the individual forces required to accelerate the rotating mass of a bicycle wheel, the non-rotating mass of the frame and their contributions to the total effort required to accelerate bike and rider.
Using typical data for a lightweight road cycle, the real world example gives a clear indication of the potential gains from weight reduction of the wheel and frame.
We can see that reducing the wheel inertia does have a beneficial effect over reducing non-rotating mass. In reality, weight saved from the wheels is likely to be up to 75% more beneficial than an equal weight saving from other parts of the bike.
However, the beneficial effect over reducing weight in the wheels versus the rest of the bike as a percentage is reduced further when the extra force required to maintain constant speed is calculated.
When a vehicle stops accelerating, is force still required to make a wheel rotate?
Yes, because we have energy losses due to friction, aerodynamic drag and tyre rolling resistance to overcome.
Is the force of rotation made more significant when climbing a gradient? If so will the distribution of weight through the wheel have an effect on this?
When climbing a gradient the force required to maintain constant speed is increased.
If the rider cannot provide the required force to maintain constant speed, the bike will slow down. The rate at which it slows down is governed by the equation for force required to accelerate the bike (in physics, speeding up and slowing down are both a change in speed and hence an ‘acceleration’, but one is positive and one is negative). Since mass reduces the level of acceleration, the bike slows down less quickly the greater the mass.
This effect is well known as momentum.
But slowing down is reduced more if wheel mass increases, due to the inertia effect. So high wheel mass and inertia is desirable for maintaining speed up a gradient.
However, if the rider wishes to increase his speed up a gradient, then the effect of inertia is the same as on the flat where low inertia is desirable.
So where should you try to conserve the most amount of weight from a wheel? Is a wheel which is lighter at its furthest extremities (its circumference) more effective than one which is heavier?
It is advantageous to concentrate the mass at the hub, for minimum inertia, however this has to be balanced against the need for sufficient strength and stiffness.
When a wheel rotates it develops a certain gyroscopic, stabilising force, is that right?
Yes. When a rotating inertia is rotated about a second axis, a gyroscopic torque is produced which resists the rotation. This means that if a rotating bike wheel starts to fall over, a gyroscopic force will act to slow down the rate at which it falls.
If so, will a heavier wheel be a more powerful gyroscope than a lighter one? Would this be desirable?
A higher inertia wheel would have a more powerful gyroscopic effect. A bike is kept upright by the riders balance and use of counter steer. By slowing down the rate at which the bike can fall over, the gyroscopic effect of the rotating wheel is helping the rider balance the bike by giving more time to make balance and counter steer corrections.
How might this affect cornering?
This is more complicated. In order to corner, the bike must be leaned into the turn. This is achieved by a combination of counter steer (so small that most riders don’t realise they are doing it) and movement of the riders weight towards the direction of turn, to make the bike start to fall in this direction. As the bike falls, so the tyres will generate a sideways force towards the direction of turn. This sideways force produces a centripetal acceleration which acts sideways at the centre of gravity of the bike and rider and tries to stand the bike up again. Hopefully these two effects balance, and the bike negotiates the desired turn at a balanced lean angle.
We have already said that gyroscopic effects reduce the rate of lean of the bike, however there is a second gyroscopic effect. This is a result of the rate at which the rider turns the handlebars. The faster the handle bars are steered, the greater the gyroscopic resistance to being turned. As for the case of leaning the bike, the faster the wheels are rotating about the spindle axis and the higher the inertia, the greater the gyroscopic resistance torque. So high inertia makes it more difficult to turn the handlebars – good for stability, but harder to initiate a turn and lean the bike. However, it also requires less quick reactions to maintain a stable lean angle during a turn.
What effect does the weight of wheels have in general stability?
As with many things related to human control of dynamic machines, there is a sweet spot for wheel inertia and the resulting gyroscopic effects. Too little and the bike will be difficult to keep stable in a straight line and through turns, too much and the bike will lose agility. In theory, very high inertia and the resulting gyroscopic effects could slow the response rate to such a degree that it becomes unstable again, as the rider is forced to make exaggerated inputs to try and get response.
In summary, in the world of competition cycling, reduced wheel mass and inertia gains a distinct advantage over non-rotating mass reduction on the bike.
Back in the automotive world, the same science is at play, affecting the efficiency of our vehicles and the amount of energy they consume. Four wheeled vehicles may not rely on gyroscopic effects to balance roll and prevent overturning, but they still have a noticeable influence on vehicle cornering agility.
Fortunately, they still succumb to the application of physics, enabling us to understand where we should target investment between rotating and non-rotating mass reduction in order to achieve the best balance for whole vehicle efficiency.
Writer: Steve Williams⎢Manager Vehicle Dynamics CAE
The maths behind the science
Newton’s Second Law:
- Force (F) = mass (M) x acceleration (a) (translational motion)
- Torque = Inertia x rotational acceleration (rotational motion)
- Torque = F x tyre radius (r)
- Inertia (I) = M x [distance from axis of rotation (k)]2
- Rotational acceleration (α) = translational acceleration (a) / r
The total force required to accelerate a bicycle might be summarised as:
Faccel = (Mbike + Mwheels + Mrider) x a + (I x α) / r where, replacing I and α; Faccel = (Mbike + Mwheels + Mrider) x a + (Mwheel x k2 x a / r) / r
This simplifies to:
Faccel = (Mbike + Mwheels + Mrider + Mwheel x k2 / r2) x a
So, the extra force to accelerate a the rotating wheel is
Mwheel x k2 / r2 x a,
And this is added to the force required to maintain constant speed on a flat surface:
Fconstant speed = rolling resistance (Frr) + friction (Ffriction) + aerodynamic drag (Fdrag)
But when climbing a gradient the equation becomes:
Fconstant speed = Frr + Ffriction + Fdrag + M x gravity x gradient
Where gradient is expressed as a ratio i.e. 1/10 for a 10% gradient. If the rider cannot provide this force, the rate at which the bike will slow down can be found by rearranging the equation to:
Faccel = (Mbike + Mwheels + Mrider + Mwheel x k2 / r2) x a
a = Frider – Frr + Ffriction + Fdrag + M x gravity x gradient
(Mbike + Mwheels + Mrider + Mwheel x k2 / r2)
A Simple Example:
- Total bike mass (including wheels) = 5 kg
- Wheel mass = 0.9 kg
- Rider mass = 75 kg
k = 0.35 m (~27 inches) – assumes wheel mass is all concentrated at rim
r = 0.40 m (tyre outer radius)
Faccel = (75 + 5 + 2 x 0.9 x 0.352 / 0.42) x a
Faccel = 81.378 Ns2/m
The force to accelerate the cycle above is 81.378 Ns2/m lightening the bike frame by 0.6 kg reduces this to 80.778 Ns2/m, so 0.74% less force is required.
Faccel = (75 + (5 – 2 x 0.3) + 2 x 0.9 x 0.352 / 0.42) x a
Faccel = 80.778 Ns2/m
However, by lightening each of the wheels by 0.3 kg (0.6 kg off the cycle in total) the force required to accelerate is reduced by more than lightening 0.6 kg from the frame, so 1.3% less force is required.
Faccel = (75 + (5 – 2 x 0.3) + 2 x (0.9-0.3) x 0.352 /0.42) x a
Faccel = 80.319 Ns2/m