More bike bits here.
Rumor has it "a gram on the wheels is like two on the frame".
The typical argument is accellerating a gram at the rim or tire takes energy both to get it moving foward and to get it spinning.
No, on both counts. Now let's see why:
The energy of a linear-motion mass m moving at velocity v is
|Elinear = 0.5 m v2||(1)|
The energy of a rotating but not translating mass is
|Erotating = 0.5 I ω2||(2)|
where the mass m is at the rim or tire, at radius r.
I is called the moment of inertia and is
|I = 0.5 m r2||(3)|
ω is the rate of rotation, expressed is radians/second:
|ω = 2 π rev/sec||(4)|
|rev/sec = v / circumference = v / (2 π r)||(5)|
|ω = 2 π (v / (2 π r)) = v / r||(6)|
|Erotating = 0.5 (0.5 m r2) (v2/r2)||(7)|
|= 0.25 m v2||(8)|
That is, the rotating energy is 1/2 the linear (translational) energy.
So the first answer is at most "not quite": when accellerating, a gram on the wheels is like 1.5 grams on the frame.
Put another way, for a gram on the wheel, 2/3 of the acceleration energy is simply being mass anywhere on the bike. 1/3 comes from being rotating mass.
A lighter bicycle is easier to lift — both lifting in the sense of "climbing a hill" and lifting in the sense of "hopping over an obstacle", such as a pothole, tree branch, and so on.
From the standpoint of lifting, a gram on the wheel is equal a gram on the frame or elsewhere: it is no easier or harder to lift a bicycle when the wheels are spinning than when they are stationary.
In this case, a gram on the wheels is the same as a gram on the frame.
The total energy to get from A to B depends on rolling and aerodynamic drag. Suppose a rider goes one city block (0.1 km), starting from a standstill, proceeding at 20 kph for the block, and coming to a rest at the end of the block. Further suppose riding at 20 kph takes 100 W to overcome rolling and aerodynamic drag.
The energy to accellerate a 100 kg rider+bicycle to 20 kph (which is 5.56 m/s) with 2 kg of rims plus tires (but zero-mass spokes and nipples) is:
|Elinear = 0.5 (100 kg) (5.56 m/s)2 = 1540 J||(9)|
|Erotating = 0.25 (2 kg) (5.56 m/s)2 = 15.5 J||(10)|
|Eacceleration = Elinear + Erotating = 1540 J + 15.5 J = 1556 J||(11)|
Let us assume the rider goes the whole bock at 20 km/h — that is, instantaly gets to full speed and stops instantly at the end of the block. That is not true, but it simplifies the calculation,* although it overstates the contribution of drag and understates the contribution of inertia. If we assume that, the rider's energy in to drag is:
* A reasonable "fix" is to assume the rider is at full speed for a fraction of the block, and has no rolling or aerodynamic drag for the rest of the block. For example, the first and last 10% could be lossless and the middle 80% at full speed. Some percentage approximates reality, and a smaller percentage understates the energy required to overcome moving drag, and thus show accelleration energy as a larger percentage of the total. For most of the calculations below, a simple linear interpolation is correct. For example, if a calculation shows a rolling drag energy cost is 10x the rotating mass energy, then assuming losses for (say) 60% of the block means the rolling drag energy cost is only 6x the rotating mass energy. Alternatively, the numbers can be construed as the energy over a longer distance, with 1 km having drag motion, and the start/stop effects taking before and after the 1 km. For example, the numbers could be construed as 1.2 km, of which 1 km is at speed, and the start and stop 0.1 km have no rolling or aerodynamic drag.
|Edrag = (100 W) time||(12)|
|time = 0.1 km / (20 km/h) = 0.05 h = 180 s||(13)|
|Edrag = (100 W) (180 sec) = 18,000 J||(14)|
The total energy is:
|Etotal = 18,000 J + 1540 J + 15.5 J = 19,600 J||(15)|
Assuming all kinetic energy is thrown away in braking at the end, we can compute the portion of energy spent on accelleration vs. the portion spent on drag as
|Eaccelleration/Etotal = 1556 J / 19,600 J = 0.079 = 7.9%||(16)|
or about 8% of the total energy for the block.
From (10), the rotating energy is 15.5 J, so:
|Erotating/Etotal = 15.5 J / 19,600 J = 0.00079 = 0.079%||(17)|
or about 8 parts in 10,000.
That is quite small.
If the rider goes faster, say 40 km/h, both accellerating and steady-state energy rise equally, since both rise as v2.
Air drag rises as the square of velocity and power as the cube. So (12)/(13)/(14) at 40 km/h are:
|Edrag,40 = (800 W) time||(18)|
|time = 0.1 km / (40 km/h) = 0.025 h = 90 s||(19)|
|Edrag,40 = (800 W) (90 sec) = 72,000 J||(20)|
Similarly, (9)/(10)/(11) for 40 km/h:
|Elinear,40 = 0.5 (100 kg) (11.1 m/s)2 = 6,160 J||(21)|
|Erotating,40 = 0.25 (2 kg) (11.16 m/s)2 = 62 J||(22)|
|Eacceleration,40 = Elinear,40 + Erotating,40 = 6,160 J + 62 J = 6,280 J||(23)|
The energy to accellerate the same rider and bicycle is, like (17):
|Eaccelleration,40/Etotal,40 = 6,280 J / (72,000 J + 6,280 J) = 0.080 = 8.0%||(24)|
The rotating mass portion, Erotating,40 is thus 62/(72,000+6280) = 0.00079, or still 0.079%.
Thus, the inertia effect of the rotating weight is the same at 40 km/h as it is at 20 km/h, or at any other speed. The inertia effect in this example is 0.079% or somewhat less than one part in 1,000.
Note also that 40 km/h and 20 kmk/h are the same relative effect of rotating mass: in both caases, 2/3 of the inertial effect of rotating mass is from linear energy, and thus the same as frame weight; and in both cases 1/3 is in rotating energy.
In flat riding, weight affects accelleration. In hilly riding, weight makes the cyclist slower up hills and faster down hills. Slower up and faster down results in a net speed loss for at least two reasons. First, climbing speed is typically limited by rider power output, while descending speed is often limited by road conditions. Second, air drag energy rises as the square of rider speed, so while increased weight directly slows climbing, it provides much less benefit descending.
However, for steady-speed climbing and descending, the effect of mass is the same whether it is on the frame or on the wheels. A gram saved is a gram saved, but it does not matter whether it is saved on the wheels or frame.
There is a transition from slow climbing to fast descending, and during this brief accelleration wheel mass does play a small role — as in the city block described above. However, it is worth noting that the transition from climbing to descending is often a few seconds of the total time, but climbing often goes on for many minutes. Thus, the total transitional energy is again a small part of the total energy.
Competition cyclists in timed events have repeatedly demonstrated lighter wheels make them faster, and ordinary cyclists often report they can feel the effect of lighter wheels. There is no doubt that saving wheel weight makes a difference. However, there is more going on here than the old saying "saving a gram on the wheel is like saving two from the frame".
Lighter wheels usually means lighter tubes and tires, which usually have lower rolling drag. At 20 km/h, rolling drag may be as much as 50% of total power, and tire rolling drag (as opposed to bearing drag) may be 80% of that — or 40% of the total steady-speed power.
Suppose 1 kg lighter tires and tubes also have a 1% reduction in rolling drag. For the 20 km/h city block example above, that means:
|Erolling,20 = (18,000 J) (50%) (80%) (1%) = 72 J||(25)|
From (10), removing 1 kg of rim/tire mass saves 7.7 J in rotating energy. Thus, a 1% reduction in rolling drag with no reduction in mass gives nearly 10x the energy savings of 1 kg less mass with no reduction in rolling drag. Reduced mass is important when accellerating, but reduced rolling drag helps in all situations.
At 40 kph, both aerodynamic and rolling drag have increased substantially, and rolling drag is more like 20% of the total. From (20), the drag energy is 72,000 J, so a 1% reduction is
|Erolling,40 = (72,000 J) (20%) (80%) (1%) = 115 J||(26)|
From (22), the effect of removing 1 kg of mass is about 31 J, so a 1% reduction in rolling drag has about 4x the energy savings of losing 1 kg of mass.
Thus, the effect of lighter tires may be easy to feel, but the major effect is reduced rolling drag, not reduced rotating mass.
In addition, cyclists often run lighter tires at higher inflation pressure, and often maintain pressure more consistently, both of which can further reduce rolling drag.
Lighter tires are often narrower, which reduces aerodynamic drag. Light wheels often have fewer spokes, which also reduces aerodynamic drag. The benefit is harder to estimate simply, but note that at steady-state 20km/h aerodynamic drag is about 50% of the total power and at steady-state 40 km/h, aerodynamic drag is about 80% of the total power (doubling the speed roughly doubles rolling drag but increases aerodynamic power by roughly 8x). The rider is the biggest part of total air drag, but wheel drag is typically reported as significant for racing cyclists.
Let us consider the effect of reduced wheel air drag using some made-up numbers. The numerical result will be wrong, but an estimate that is "in the ballpark" will at least tell us approximately how air drag compares to the effects of reduced mass.
Suppose at steady 20 km/h, aerodynamic drag is 50% of the rider's total power output; that wheel air drag is 3% of the total, and that the effect of narrower tires and fewer spokes is to reduce the wheel's aerodynamic drag by 6% (that is, 0.18% of total air drag):
|Eaero-savings,20 = (18,000 J) (50%) (3%) (6%) = 16.2 J||(27)|
So for the 20 km/h example, reducing wheel aerodynamic drag by 6% leads to a savings of 16 J. By (10), the total energy of 2 kg rotating mass is 15.5 J. If narrower tires and fewer spokes saves 1 kg wheel mass, the comparison is aerodynamic savings of 16 J compared to a mass savings of 7.8 J, or about 2x aerodynamic savings compared to mass savings.
As a second example, suppose at steady 40 km/h aerodynamic drag is 80% of the rider's total power output; and that as above, wheel drag is 3% of the total bike+rider drag, and the effect of narrower tires and fewer spokes is to reduce wheel's aerodynamic drag by 6%.
|Eaero-savings,40 = (72,000 J) (80%) (3%) (6%) = 104 J||(28)|
For the 40 km/h example, reducing wheel aerodynamic drag by 6% saves 104 J. By (22), the total rotaitng energy is 62 J, and saving 1 kg of rotating mass would save half that, or 31 J. If a 1 kg lighter tire, etc., also saved 6% aerodynamic drag, the savings in energy from air drag would be about 3x the savings from reduced mass.
It is worth reiterating the air drag numbers used in this calculation are made-up, and the model is quite a bit simpler than reality. But at first approximation it appears the savings from air drag are much larger than from rotating weight; and also air drag savings are much more than the difference in savings from wheel vs. frame weight.
It is also worth noting that the effects of reduced rolling drag and reduced aerodynamic drag help all the time the bicycle is moving, whereas the effects of wheel vs. frame mass are an issue only when accellerating. Reduced rolling drag helps slightly on hills, and reduced aerodynamic drag on flats or going down hill, whether or not the cyclist is accellerating.
A lighter wheel often gives a "snappier" steering feel. Even if that has no mechanical advantage, it still can help the rider go faster: If a bicycle "feels" snappier, the rider will go faster.
A U.S. Olympic coach famously told the team they had a "secret weapon", which was helium in the tires instead of air. "Helium is lighter". If you do the PV=nRT calculation you find the weight advantage is smaller than plucking out a few eyebrow hairs, but the riders, convinced they had a signficant advantage, rode faster.
Riders with "race" and "training" wheels often ride the race wheels freshly-pumped, where the training wheels often run at reduced pressure — tires lose significant pressure over a few days, but many riders do not pump up their training tires daily; also, lower pressures often give a more comfortable ride, which can be an advantage for training.
Higher pressure can reduce rolling drag signficantly, roughly linearly. Tires vary, but as an example, one test reports dropping pressure about 13% from 8 kg/cm^2 (7.85 bar or 114 PSI) to 7 kg/cm^2 (6.86 bar or 100 PSI) on one brand/model of tire under 50 kg load increased rolling drag from 266 grams to about to 282 grams, or about 5.7%.
As noted in (25) and (26), a 1% change in rolling drag can be a bigger effect than 1 kg of weight loss.
Air pressure is not always at play in wheel comparisons, but where it is, the effect can be much larger than the difference in weight.
Reducing weight is useful, whether removed from the wheels or frame.
One important question is whether a gram from the wheels is like two grams removed the frame. The answer is "no", it is at most like 1.5 grams removed from the frame. But for things like climbing and hopping, a gram removed from the wheels is no different than a gram removed from the frame.
A second question is whether reduced rotating mass effects are significant. Again, the short answer is "no". The big energy savings of lighter tires, fewer spokes, and so on is lower rolling and aerodynamic drag, not inertia. Further, reduced drag helps always, reduced mass anywhere on the bicycle helps during climbing and accelleration, but reduced rotating mass helps only during accelleration.
This analysis is useful in that it can help riders make smarter buying decisions. If the same price saves either 1g of rotating mass or 2g of non-rotating ("frame") mass, the 2g frame mass avings is almost always the better deal. Indeed, in most cases, whatever is the cheapest way to save a gram gets the most benefit for a given price: by going with the best value, a fixed spending removes the most grams and thus gets the biggest advantage.
This analysis is also useful in that it shows improving aerodynamics without losing mass can also be a good deal. Such changes are not quantified here, but it shows the effect of improved aerodynamics is often quite large, and thus can be a good value for the money.