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Fluid Dynamics Drag

Initial Thought

Have any of you cycled against 40mph headwind? Is it even possible? If 90% of resistance to moving forward at this speed, is wind resistance, then you would have to be able to average ~36mph in still air, to be able to make forward progress. I have cycled through a tropical storm, where I was moving very, very slowly, and there was water lapping the bottom bracket. What would it take to not be able to make forward progress at all? The following youtube video is an example of just that:

The Dilemma

Equations for drag and power, where (v) is the speed of the object relative to the fluid. Notice the “relative” bit. That means, that land speed is irrelevant for this equation. Even if you stood still, bulk fluid flow is being converted into a form less capable of doing work, increasing the amount of entropy. See For simplification, let's say that all of the work produced by the cyclist is fully converted into entropy of the fluid (as a mind exercise), then:
Power from cyclist = Rate of production of Entropy
Power from cyclist = Drag Force x (v) = a_constant x (v)^3

Or you can say that you have a Perfect Fluid, where:
Power from cyclist = Rate of decreasing momentum of the fluid

Aside from spokes spinning into the wind, I don't see a difference in drag between a 20mph direct headwind, and rider moving through still air at 20mph. However, the equations from disagrees with this belief. His equation is below in this article.

The equation I am challenging, used drag force multiplied by the velocity of the rider relative to the ground, to calculate power production. After all, F x V gives the units for power. The power produced by the cyclist can also be expressed (and measured), as torque multiplied by angular velocity, or P = T ?. However, if the cyclist comes to a stop, where the forward force from the cyclist is equal to the drag force, then according to this equation, the cyclist is no longer doing work! Which is true, because that's what I was taught. I just don't find it very useful for this particular situation. The fluid is undergoing a rate of change of momentum, despite the complete lack of forward progress from the cyclist. In this case, how do you set up an equation, for the conservation of energy? I believe the correct way to do this, is to calculate the power required to overcome drag as the one above, rather than saying drag power is relative to the forward progress of the cyclist.

Speed & Power Calculations from another site, that I am challenging

The following equations take into account all of the relevant resistance components: Rolling friction including the dynamic (speed-dependent) rolling friction, air drag including the influence of wind speed, mechanical losses, and uphill/downhill forces.

PRider's power
VVelocity of the bicycle
WWind speed
HnnHeight above sea level (influences air density)
TAir temperature, in ° Kelvin (influences air density)
gradeInclination (grade) of road, in percent
ß(“beta”) Inclination angle, = arctan(grade/100)
mbikeMass of the bicycle (influences rolling friction, slope pulling force, and normal force)
mriderMass of the rider (influences rolling friction, slope pulling force, and the rider's frontal area via body volume)
CdAir drag coefficient
ATotal frontal area (bicycle + rider)
CrRolling resistance coefficient
CrVCoefficient for velocity-dependent dynamic rolling resistance, here approximated with 0.1
CrVnCoefficient for the dynamic rolling resistance, normalized to road inclination; CrVn = CrV*cos(ß)
CmCoefficient for power transmission losses and losses due to tire slippage (the latter can be heard while pedaling powerfully at low speeds)
?(“rho”) Air density
?0Air density on sea level at 0° Celsius (32°F)
p0Air pressure on sea level at 0° Celsius (32°F)
gGravitational acceleration
FrgRolling friction (normalized on inclined plane) plus slope pulling force on inclined plane

Air density via barometric formula:
Air density via barometric formula

Rolling friction plus slope pulling force:
Equation for rolling friction force and slope pulling force

Equation for the required human power

In order to solve this Power equation for Velocity V, we write it in the implicit form
Equation for the required human power

so we can use the cardanic formulae to obtain the solutions:

If a2 + b3 = 0:
Velocity equation
If a2 + b3 < 0 (casus irreducibilis; in case of sufficient downhill slope or tailwind speed):
Velocity equation (casus irreducibilis)

Expression "a" of the velocity equation
Expression "b" of the velocity equation


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physics/fluid-dynamics-drag.txt · Last modified: 2023/12/21 04:33 by