$$|F_d| = \frac\ll1$ makes sense: the drag is due to shearing of the air near the surface, rather than pushing air aside the way that a parachute does, and air doesn't have any shear strength. Each of the wheels was a five spoke, 17´´ aluminum rim (Figure 3).It seems textbooks on classical mechanics covers linear drag very well but frequently leaves out angular drag. For given air conditions, shape, and inclination of the object, we must determine a value for Cd to determine drag. Rims were tested in Volvo aerodynamic wind tunnel. The drag equation states that drag D is equal to the drag coefficient Cd times the density r times half of the velocity V squared times the reference area A. In order to measure ventilation resistance in the wind tunnel there were used wheels with different rim designs (4).
Air drag free#
Where Fvtrac is the ventilation resistance force that correspondent to ventilation resistance force Fvent, CD(vent) is ventilation resistance coefficient, analogous to aerodynamic resistance drag coefficient, ρ is air density, V∞ is a free stream velocity and A is reference area.ĬD(vent) ventilation resistance coefficient can be expressed as: The ventilation resistance force Fvtrac can be expressed as (3):
The moment Mvent is called ventilation moment and it occurs due to the wheel rotation in the airflow.
There is possible to see in Figure 2 moment Mvent that acting on the rotating wheel. Because the underbody and wheelhouses are responsible for a larger part of the aerodynamic drag force, reducing the aerodynamic drag of these parts would reduce consumption of propulsion energy and thus extend the range of vehicles. Wheels and wheelhouses account for approximately 25% of passenger car aerodynamic drag (Figure1) (2). From experiments in both research (Cogotti 1983) and development (Mercker et al 1991) it can be concluded that wheels and wheel wells contribute approximately half the drag of a low-drag car (1).
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