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Directional stability is stability of a moving body or vehicle about an axis which is perpendicular to its direction of motion. Stability of a vehicle concerns itself with the tendency of a vehicle to return to its original direction in relation to the oncoming medium (water, air, road surface, etc.) when disturbed (rotated) away from that original direction. If a vehicle is directionally stable, a restoring moment is produced which is in a direction opposite to the rotational disturbance. This "pushes" the vehicle (in rotation) so as to return it to the original orientation, thus tending to keep the vehicle oriented in the original direction.

Directional stability is frequently called "weather vaning" because a directionally stable vehicle free to rotate about its center of mass is similar to a weather vane rotating about its (vertical) pivot.

With the exception of spacecraft, vehicles generally have a recognisable front and rear and are designed so that the front points more or less in the direction of motion. Without this stability, they may tumble end over end, spin or orient themselves at a high angle of attack, even broadside on to the direction of motion. At high angles of attack, drag forces may become excessive, the vehicle may be impossible to control, or may even experience structural failure. In general, land, sea, air and underwater vehicles are designed to have a natural tendency to point in the direction of motion.

Example: road vehicle

Arrows, darts, rockets, and airships have tail surfaces (fins or feathers) to achieve directional stability; an airplane uses its vertical stabilizer for the same purpose. A road vehicle does not have elements specifically designed to maintain stability, but relies primarily on the distribution of mass.

Introduction

These points are best illustrated with an example. The first stage of studying the stability of a road vehicle is the derivation of a reasonable approximation to the equations of motion.

The diagram illustrates a four-wheel vehicle, in which the front axle is located a distance ${\displaystyle a}$ ahead of the centre of gravity and the rear axle is a distance ${\displaystyle b}$ aft of the cg. The body of the car is pointing in a direction ${\displaystyle \theta }$ (theta) whilst it is travelling in a direction ${\displaystyle \psi }$ (psi). In general, these are not the same. The tyre treads at the region of contact point in the direction of travel, but the hubs are aligned with the vehicle body, with the steering held central. The tyres distort as they rotate to accommodate this mis-alignment, and generate side forces as a consequence.

The net side force Y on the vehicle is the centripetal force causing the vehicle to change the direction it is traveling:

${\displaystyle MV{\frac {d\psi }{dt}}=Y\,\cos(\theta -\psi )}$

where M is the vehicle mass and V the speed. The angles are all assumed small, so the lateral force equation is:

${\displaystyle MV{\frac {d\psi }{dt}}=Y}$

The rotation of the body subjected to a yawing moment N is governed by:

${\displaystyle I{\frac {d^{2}\theta }{dt^{2}}}=N}$

where I is the moment of inertia in yaw. The forces and moments of interest arise from the distortion of the tyres. The angle between the direction the tread is rolling and the hub is called the slip angle. This is a bit of a misnomer, because the tyre as a whole does not actually slip, part of the region in contact with the road adheres, and part of the region slips. We assume that the tyre force is directly proportional to the slip angle (${\displaystyle \phi }$). This is made up of the slip of the vehicle as a whole modified by the angular velocity of the body. For the front axle:

${\displaystyle \phi (front)=\theta -\psi -{\frac {a}{V}}{\frac {d\theta }{dt}}}$

whilst for the rear axle:

${\displaystyle \phi (rear)=\theta -\psi +{\frac {b}{V}}{\frac {d\theta }{dt}}}$

Let the constant of proportionality be k. The sideforce is, therefore:

${\displaystyle Y=2k(\phi (front)+\phi (rear))=4k(\theta -\psi )+2k{\frac {(b-a)}{V}}{\frac {d\theta }{dt}}}$

The moment is:

${\displaystyle N=2k(a\phi (front)-b\phi (rear))=2k(a-b)(\theta -\psi )-2k{\frac {(a^{2}+b^{2})}{V}}{\frac {d\theta }{dt}}}$

Denoting the angular velocity ${\displaystyle \omega }$, the equations of motion are:

${\displaystyle {\frac {d\omega }{dt}}=2k{\frac {(a-b)}{I}}(\theta -\psi )-2k{\frac {(a^{2}+b^{2})}{VI}}\omega }$

${\displaystyle {\frac {d\theta }{dt}}=\omega }$

${\displaystyle {\frac {d\psi }{dt}}={\frac {4k}{MV}}(\theta -\psi )+2k{\frac {(b-a)}{MV^{2}}}\omega }$

Let ${\displaystyle \theta -\psi =\beta }$ (beta), the slip angle for the vehicle as a whole:

${\displaystyle {\frac {d\omega }{dt}}=2k{\frac {(a-b)}{I}}\beta -2k{\frac {(a^{2}+b^{2})}{VI}}\omega }$

${\displaystyle {\frac {d\beta }{dt}}=-{\frac {4k}{MV}}\beta +(1-2k{\frac {(b-a)}{MV^{2}}})\omega }$

Eliminating ${\displaystyle \omega }$ yields the following equation in ${\displaystyle \beta }$:

${\displaystyle }$Zdroj:https://en.wikipedia.org?pojem=Directional_stability
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