Europe Bicycle Touring
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By David May
Theoretical Discussion:

Cycling Speed Math and Ruminations


The author has based this section in part on the French book, "Guide du Vélo en Montagne" published by Altigraph, as well an Internet article by Rainer Pivit, originally published in German in Radfahren magazine, and translated by Damon Rinard.

Please write if you spot errors of fact or emphasis, or have suggestions.

As a cyclist, you have an intuitive understanding of the conditions that affect your riding speed. The following mathematical treatment and discussion, therefore, is given simply as an intellectual exercise. Perhaps you will enjoy knowing the "whys" of riding speed—and therefore distance. Perhaps you will understand better some of the dynamics of road bike racing, and of long distance touring.

Your biking speed depends upon your continuous power output to overcome the forces that slow down your bike: friction, air resistance, and (when going up hill), gravity. In stop and go city riding, and in some racing, your power is also used to accelerate up to speed, but for bicycle touring and long-distance racing, this is a negligible factor, and we shall ignore it here.

This discussion uses mainly metric units, most of which should be familiar. Examples are given in both metric units, in black print, and American units, in red print.

Watts (like horsepower) are a measure of power. In the formulas below "Wrider" means the number of continuous watts of power put out by the efforts of the cyclist.

1000 watts = 1.34 horsepower. "Power" measures "force" times "distance" per unit of "time". Example: Moving an object against a resistive force of 1 pound, for a distance of 10 feet, in 1 second takes the same power as moving it against a force of 10 pounds for a distance of 1 foot in one second, or against a force of 100 pounds for 1/10 of a foot in a second. For a bicycle, it is evident that, using different gear combinations, a cyclist can apply the same power input to either go at high speeds (long distances/second) against low counteracting forces, or at low speeds to overcome high counteracting forces.

Overall Formula:

x means "multiplied by". ^2 means squared, i.e., (V+Vwind) x (V+Vwind) = (V+Vwind)^2. The C's in the formula below are various constant amounts, which vary according to the input units used. Details are discussed below.

Wrider = Cfriction x V x P + Cair x (V + Vwind)^2 x V+ Cslope x P x Slope% x V + acceleration

[Note that V (velocity, i.e., riding speed, i.e., distance ridden per unit of time) can be factored out of all the terms on the right side and left side of the equation; the terms of the equation remaining after this is done are the various forces that the biker must apply and overcome.]

Using the above equation, if we know the total power put out by a rider (Wrider) the coefficient of friction of the bicycle (Cfriction) the weight of the rider (P), the component of the Wind speed acting against (or in favor of) the rider, and the percentage of slope (Slope%), (and if we assume that the rider is not constantly braking and accelerating up to speed) we can calculate the speed (V) that will be attained by the rider, and therefore the distance he will cover in an elapse of time.

Now let us examine the power required on a more detailed basis.

Power of the Rider:

A completely inexperienced rider, for long periods of time, can output 50 or 100 watts of leg power; whereas a Tour de France racer is said to be able to generate 500 watts or more of continuous power—still not up to a horse, but mighty impressive, none the less!

Experience teaches cyclists how much power they can put out on a sustained basis. Some riders may choose to use a heart rate (pulse) monitor as a supplement to their experience: during the course of a ride pulse correlates directly with power output (though not over weeks or years, as aerobic capacity may change).

If there were no forces resisting the cyclist, even the inexperienced rider could accelerate a bike indefinitely—up even to rocket speeds. But there are resisting forces.


At low speeds and on flat surfaces and with no wind, the only resistance that counts comes from friction. That friction goes up proportionally to speed and to total weight. If you have ridden at a gym a stationery bicycle that uses a frictional brake, you know that the total effort required goes up proportionally to speed.

It is less obvious that friction should go up linearly with total weight. In fact, it doesn't exactly. The major components of friction, however, do rise more or less linearly with weight, such as the friction of the wheels on the hubs and the rolling resistance on the road. Compared to these, the resistance caused by pedals, pedal bracket, chain and derailleur are minor. As a first approximation, then, we put down the following formula:

Watts_to_overcome_friction = 0.1 x V x P

When the velocity V is measured in meters/second and the weight P is measured in kilos, for a decent quality bike, with well-inflated racing tires on a smooth road, the metric Cfriction is said to be about 0.1. We can calculate, therefore, that at 5 meters per second, a rider and bike weighing 80 kilos faces a frictional resistance of about 40 watts. Poor quality bicycles, heavy loads, inefficient tires, and especially poor road surfaces can substantially increase the coefficient of 0.1, whereas the highest quality bikes on smooth roads may have coefficients as low as 0.08

One meter per second is 3.6 kilometers per hour or 2.237 miles per hour, and one kilo is about 2.2 pounds. So in American units, roughly:

Watts_to_overcome_friction = .02 x MPH x Pounds.

For our 176 pound rider plus bike going 11.2 miles per hour we calculate 39+ watts of resistance, approximately the same result as above.

Light weight riders are at an advantage over equally strong heavier ones. Similarly, riders with lighter bicycles and lower touring loads have an advantage. This advantage is proportionally more obvious at touring speeds(under 25 km/hr - 15 mph), where air resistance is not much of a factor.

If friction were the only resistance, a typical untrained rider could zoom along at 28 miles per hour, putting out 100 watts to do so. The better the bicycle and the better the road surface, the faster would be his or her speed. But air resistance most definitely comes into play.

Air and Wind Resistance:

Air is a "fluid", so to speak, though a thin one. When you move through a fluid faster, it puts up much more resistance. If you have been swimming, you know you can move your hand very easily in the water if you do so very slowly; but try to move it faster, a huge force is required..

The same is true, as any experienced cyclist knows, in the air. At a few miles per hour, (assuming no wind), you barely feel air resistance, but at 15 miles per hour, it pushes strongly against you. The resistance of fluids—certainly in the case of the wind—goes up with the square of the velocity, and the faster one goes the more air resistance one encounters. Thus a 10% increase in speed requires a 33% increase in power, to overcome air resistance and a 25% increase in speed requires almost a doubling of power.

The formula for the power to overcome air resistance is:

W_to_overcome_air_resistance = Cair x (V + Vwind)^2 x V. If there is no wind, it is simply: Cair x V^3.

For the metric system, Cair ranges from perhaps 0.8 for a city bike without baggage or a hybrid with baggage, to .7 for a mountain bike or loaded touring bike to perhaps 0.45 for a hybrid or upper position on a racing bicycle without baggage, to as low as 0.36 for full racing position on a conventional racing bike. (The numbers for the racing bicycle and the mountain bike, both without baggage, come from a study. Those for the unloaded city bike and the unloaded hybrid are guesses.)

Thus, a rider using a hybrid or the upright position on a racing bike, traveling at 5 meters per second (11.2 mph), will need to expend 56 watts to overcome air resistance. In racing position on a racing bike, the rider need to expend only 45 watts. A rider using a mountain bike may need to expend almost 90 watts to go the same speed.

In American units, Cair ranges from approximately 0.062 for a city bike or fully loaded touring bike to perhaps 0.04 for a hybrid, and 0.032 for racing position on a racing bike.

At 11.2 miles per hour (5 meters per second) on a hybrid bicycle, we calculate (as before) 56 Watts of resistance.

Therefore, at 11.2 miles per hour, approximately slightly more than one-half of all power generated is spent overcoming air resistance—more on a mountain bike, less in racing position on a racing bicycle.

So, suppose you want to go 25% faster? You need to put out 95% more power! Well, at slower speeds, 12 mph, not really. Because half of your original power was used to overcome the force of friction, and only the rest of your power needs only to increase 95%. To go 25% faster (16 mph) you need to increase your power by about 60%. So instead of a novice's 100 watts of power you need to put out 160 watts of power. To ride your hybrid at 20 mph you might need to put out 360 watts of power.

At 20 mph, four-fifths of your total power is already spent overcoming air resistance. To go 25% faster, you need to increase your total power by 83%.

The previous paragraphs assume no wind. If there is a wind, we need to consider it, but not the total wind speed; rather only the portion of that speed that is against or behind the rider. Obviously, if the wind is directly behind a rider at the strong speed of 20 miles per hour, that rider will be able to ride much faster. If that rider could output 90 or 100 Watts of power, and thus ride about 12 mph in no wind, he will now be able to ride at perhaps 24 miles per hour. (At that speed he will have twice the frictional resistance but only a bit of wind resistance.)

On the contrary, if the wind is totally against this rider at 20 miles per hour, a biking speed of about 3 miles per hour for this occasional biker brings the equation into balance.

If a Tour de France racer can continuously put out 5 times the power of a novice biker, then he should be able to ride about twice as fast on the same bike , that is, for (say) a hybrid bike, about 24 miles per hour rather than 12 (about 40 km/hr rather than 20). The racer of course actually goes faster than this: because his bicycle and clothing and position are more aerodynamic; and because, also, he usually rides in a pack that greatly lowers air resistance.

Gravity Climbing Slopes

While climbing steeper hills gravity becomes important, and air resistance becomes unimportant.

It is easy to see why: On the way up slopes, gravity greatly reduces speed, and at low speeds, air resistance is insignificant.

The formula for Gravity is: W_to_overcome_gravity = 9.81 x P x %slope x V.

Where Cslope is a coefficient, P is your weight, and V is your speed on the road.

The faster you go, the more you weigh, and the steeper the slope, the more power is required to take you up the hill.

The "percent of slope" technically measures the altitude gained per horizontal distance. Although this measure may be the one you see in signs, documents and formulas, it is not very useful in everyday bicycle touring. It measures the base of the triangle, not the hypotenuse along which runs the road.

For every day road biking and for use in this formula, we measure the altitude gained per distance of road. Maps or an altimeter show you the elevation gain, and your trip computer shows you the distance ridden. For low percent slopes, the two numbers are very, very close. For higher grades, the two numbers are still quite close: An 18% grade (extremely steep) measured from the road corresponds to an 18.3 % slope measured from the base.

For a 6% slope, in metric units, the formula uses weight in kilos, speed in meters per second, and a constant of 9.81. A rider weighing 80 kilos (with bike), riding at 2 meters per second (7.2 kilometers/hour) must generate 94 Watts of power to overcome the force of gravity.

In American units we use pounds, miles per hour and a constant of 2. Thus the same rider and bike, weighing 176 pounds, and traveling at 4.47 miles per hour expends 94 Watts of power to overcome the force of gravity, as previously calculated.

At this speed (7.2 kilometers/hour - 4.47 miles/hour) our rider only needs to expend 16 Watts to overcome friction. Assuming that there is no wind, air resistance consumes less than 4 Watts. A total of 114 Watts of power is needed to climb the hill at this speed. That is the most this hypothetical novice rider can muster.

If the hill is twice as steep—12%, this rider can only proceed at 3.6 kilometers per hour, or about 2.25 miles per hour. He must use his lowest mountain-bike gears. Maybe he would be better off pushing his bike up the hill!

Most Alpine Cols (passes) in France have average slopes of between 6% and 10%.

Descending Slopes

On curvy mountain roads, total speed will be limited to what is safe. Power and gravitational acceleration make no difference.

In straight line descents without pedaling, on wide roads (ignoring the effects of wind), bikes will accelerate until wind resistance plus frictional resistance equals the acting force of gravity.

We take the general formula above, set the power input to zero, drop out the terms for the wind, and put the coefficient for gravity as a negative number, since gravity will be helping the rider, rather than opposing him.

0 = Cfriction x V x P + Cair x v^2 x v - Cslope x P x% slope x V

Since V appears in every term on the right side of the equation, and since 0 is on the left, V can be dropped out of the equation. The equation now represents forces, rather than power.

In metric units, 0 = Cair x V^2 + 0.1 x P - 9.81 x P x %slope. Transposing and factoring terms, we have:

V^2 = P / Cair x (9.81 x %slope - 0.1)

For the 80 kilo rider (including bike and baggage) on a 7% slope, with a 0.45 aerodynamic coefficient (hybrid) V will be 10.21 meters/sec = 36.8 kilometers/hour = 22 miles per hour. On a 12% slope, the speed obtained without pedaling and no wind would be 52 kilometers/hour or 31 miles/hour.

If the individual rider reduces air resistance by 20%, to a coefficient of 0.36, he will descend slopes of 7% and 12% at speeds that will be respectively 11% and 8% higher; i.e., at 41 and 56 kilometers per hour (25 and 34 miles per hour).

On the 12% descent with a hybrid or touring bicycle, even if a rider wished to add pedal power, at 35 mph his bicycle is unlikely to be equipped with the gears to do so.

Tail winds and head winds can considerably effect the actual speed of a descent.

Concluding remarks

Whereas, on the flat, and out of the pack, the bike racers with the most power, the best aerodynamic position, and the most aerodynamic bicycle will do the best, even if the weight is slightly more, when it comes to going up steep hills, the rider with the best ratio of power to total weight will excel.

(Coming down straight, steep hills, the rider with more weight will gain an advantage, but less so, as his speed does not pick up proportionally.)

For racers, it has always been obvious that it pays to lose fat from body and bike, to improve aerodynamic efficiency, to cut friction, and to increase personal power. Since most of the rest of us are the weight we are, and have the power we have, and own the bicycle we own, the above mathematics are perhaps of theoretical interest only.

We cyclists who do long-distance touring have learned not to exhaust ourselves or strain our muscles by riding too rapidly against the wind or up the steep hill. Experience can tell us better than any calculations how much power and endurance we have available, and, over both short and long intervals, what our optimum power output should be.

For further information on this subject see:

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