Gelöste Aufgaben/JUMP/Car-Body

Scope

In common simulation applications - especially for full-size commercial cars - the pitch-angle of the car is assumed to be small to allow for a linearization of geometry and most parts of the equations of motion. We drop this limitation so we can do more fancy stuff with our model - like climbing steep roads or jumping across ditches.

We employ a spatial x-y coordinate system, x in horizontal, y in vertical, upwards direction. And we’ll briefly employ the z-axis as rotation direction - which is towards you following the “right-hand-rule“ for coordinate systems.

Structure

The driver controls the car's motion via the position of the "gas"-pedal, which is being translated into a torque M_W at the wheel.

This toque will change velocities and thus the position of the car. These state-variables will then create - via info - a feedback to the driver.

We’ll need to invest significant efforts in describing the kinematics of the car-motion and to derive its equations of body-motion since we do not want to limit our study on small pitch-angles of the car. Key accessory will be vectors, which map locations like the center of mass:

{\vec {r}}_{C}(t)

.

This vector has - in 2D - two coordinates $u_{1}(t),u_{2}(t)$ , measured in the inertial x-y frame:

{\vec {r}}_{C}(t)={\vec {\underline {e}}}_{0}\cdot \left({\begin{array}{c}u_{1}\\u_{2}\end{array}}\right)

with

{\vec {\underline {e}}}_{0}=\left({\vec {e}}_{x},{\vec {e}}_{y}\right)

.

representing the unit vectors spanning the x-y space. If we refer to a specific frame, we may drop the vector-notation $({\vec {.}})$ and refer to the column-matrix of coordinates $({\underline {.}})$ only, so

{\underline {r}}_{C}=\left({\begin{array}{c}u_{1}\\u_{2}\end{array}}\right)

.

We’ll also employ coordinate transformations using Euler-rotations.

The car with front-wheel drive consists of the car-body with center of mass “C“, the front wheels “A“ and the rear wheel “B“. Masses of car-body and wheel-sets are M and m respectively.The geometry of the car is described by

a₀: the wheel base;
a₁: longitudinal distance between center of mass and front-wheel-hub;
a₂ = a₀ - a₁ and
a₃: vertical distance between center of mass and front-wheel-hub (relaxed springs).

Model

We use five coordinates to describe the motion of the car:

$u_{1}(t),u_{2}(t),\varphi (t)$ for the location of the center of mass of the car-body and its pitch-angle ,
$u_{3},u_{4}$ for the "vertical" motion of the wheel hubs relative to the car-body - which is synonym to the compression of the springs and

$\psi (t)$ as the rotation angle of the front-wheel - we'll not account for the rotation of the rear-wheel.

So the center of mass of the car-body is

${\vec {r}}_{M}={\vec {\underline {e}}}_{0}\cdot \left({\begin{array}{c}u_{1}(t)\\u_{2}(t)\end{array}}\right)$ ,

the coordinate system of the car is

${\vec {\underline {e}}}_{C}={\vec {\underline {e}}}_{0}\cdot {\underline {\underline {D}}}(\varphi (t))$

where

${\underline {\underline {D}}}(\varphi )=\left({\begin{array}{rr}\cos(\varphi )&-\sin(\varphi )\\+\sin(\varphi )&\cos(\varphi )\\\end{array}}\right)$ .

So the location of the front-wheel “A“ is

${\vec {r}}_{A}={\vec {r}}_{M}+{\vec {r}}_{MA}$

or

${\underline {r}}_{A}={\underline {r}}_{M}+{\underline {\underline {D}}}(\varphi (t))\cdot \left({\begin{array}{c}a_{1}\\-a_{3}+u_{3}(t)\end{array}}\right)$ .

Likewise, the location of the rear-wheel “B“ is

${\underline {r}}_{B}={\underline {r}}_{M}+{\underline {\underline {D}}}(\varphi (t))\cdot \left({\begin{array}{c}-a_{2}\\-a_{3}+u_{4}(t)\end{array}}\right)$