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At what slope will the best bikers (in terms of aerodynamics) on the best bikes roll at a constant speed at 80 km/h on an asphalt surface on a dry day? The reason for this question is because it gives me very accurate information about the force needed to keep the best bikes going at that high a speed. It is also interesting to know the slope at 50 km/h if that is available.

Maybe I could get the answers from watching parts of Tour de France, where they measure speed and slope?

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this belongs on the physics site, if anywhere. – joelmdev Dec 21 '13 at 2:10
@joelmdev: not really, the physics is either trivial or impossible. My answer is the trivial one, perilously close to "try it and see", R.Chung's one gives pointers to some of the more obvious complexities in an analytical answer. – Móż Dec 23 '13 at 1:05
up vote 8 down vote accepted

The Varna, which was one of the best bikes in 2003 has speed + power graphs up from their Battle Mountain runs that year. At 80kph Sam was putting out about 225 watts, but accellerating. There's also a 0.5° slope to consider. This article suggests that the bike uses about 150W at 50mph/80kph.

Sam weighs about 80kg, the bike about 25kg. Let's call it 110kg once he's dressed.

To get 150W out of a 110kg weight we need to be falling at 150/110 = 1.4m/s. 80kph is 80/3.6 = 22m/s, so the slope has to be 1.4:22, or about 1:15.

Combining that to avoid rounding during the calculation: 80/3.6/(150/110) = 1:16.3 or 0.06 (around 3 degrees - you'd know it was sloping, but it wouldn't count as a hill to most people)

So you could expect Sam to coast the Varna down a gentle hill at about 80kph, assuming it was long enough for him to reach that speed.

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As you might expect, the exact down slope you would need for a full answer will depend on how much drag you and your bicycle produce. If you are very aerodynamically efficient and there are few losses through the bearings and tires, the slope can be shallower. If you create a lot of drag either via aerodynamic inefficiency or mechanical and rolling resistance, the slope will need to be steeper. You can calculate the slope yourself if you know the equations for motion on a bicycle and can make reasonable guesses about the parameter values. The equations are given in one of the answers to this bicycles.stackexchange question.

In short, you are looking for the slope so that while coasting at equilibrium speed the power will be zero; that is, at equilibrium speed you are neither accelerating or decelerating (so the kinetic energy is zero) and you are coasting so the power is zero. That is,

Watts = 0 = (Crr + slope) * kg * g * v + 0.5 * rho * CdA * v^3

where Crr is the coefficient of rolling resistance, kg is the total mass of the rider and bike, g is the gravity constant (9.8 meters/sec^2), rho is the air density in kg/m^3, CdA is the "drag area" (Cd is the coefficient of aerodynamic drag and A is frontal area, so their product Cd*A is also known as the drag area), and v is the speed in meters/sec.

Slope for a given speed

The equation above can be re-arranged to solve for slope:

slope = (0.5*rho*CdA*v^2)/(kg*g) - Crr

In your question, 80 km/h is 22.2 m/s and 50 km/h is 13.9 m/s.

Air density, rho, at sea level is typically near 1.2 kg/m^3

Crr for a high quality tire on an asphalt road is typically around .0045.

CdA for a rider on a conventional bike in the drops is typically in the range of 0.24 to 0.3 m^2; for a rider on a TT bike in the aerobars, CdA is typically in the range of 0.2 to 0.28 m^2; for a rider in a streamlined fully-faired recumbent bicycle the CdA has been measured at under .02 m^2.

Let's say you and your bike together weigh a total of 80 kg, rho=1.2, Crr=.0045, and CdA = .24. What is the slope so that you would reach an equilibrium speed of 50 km/h (13.9 m/s) while coasting? That would be

slope = (0.5*rho*CdA*v^2)/(kg*g) - Crr or slope = (0.5*1.2*0.24*13.9^2)/(80*9.8) - .0045

Thus, the downslope is -0.04, or -4%.

Suppose you wanted to know how steep the slope would need to be to attain an equilibrium speed of 80 km/h with the same mass, rho, and CdA? That would be -0.095, or -9.5%.

Suppose you wanted to know how steep the slope would need to be to attain an equilibrium speed of 80 km/h in the streamlined Varna Diablo? If one had access to the power and speed data for a speed run, and one also knew the road profile, the air density on the day, the wind speed, and the total mass of vehicle and rider, one can estimate the Crr and CdA given the method described in the linked bicycles.stackexchange answer. I have done that: Crr and CdA appear to be in the range of .0045 and .0175, as noted above. Battle Mountain, NV, where the world human powered vehicle speed runs are attempted, is at an elevation of around 1375 meters (4500 ft) above sea level, but if you could make the run at sea level with a total mass of 90 kg, you could attain an equilibrium coasting speed of 80 km/hr at a down slope of -.011, or -1.1%.

Terminal Speed

A related question is, given a slope, what would the equilibrium speed be, assuming the slope is long enough to reach maximum speed? This is also known as the "terminal speed." In this case, we can re-arrange the equations of motion to determine the speed in meters per second, v. That equation is

v (in m/s) = sqrt( -2*(Crr+slope)*mass*g/(rho*CdA))

For example, if Crr = .0045, slope = -3% = -.03, mass=80, g=9.8, rho=1.2, and CdA = 0.24, then v = 11.78 m/s. To convert to km/h, multiply v by 3.6; v = 11.78 m/s is equivalent to 42.4 km/h or 26.4 mph.

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Why does this site not support equation, as in Mathematics.SE? I tired beautifying your formulas, but the site is not parsing them. – Vorac Dec 20 '13 at 16:23
@Vorac Consider going to Bicycles Meta. – Bleeding Fingers Dec 21 '13 at 19:35

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