Gelöste Aufgaben/JUMP/Car-Body

← Back to Start

Scope

Car-Body

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.

Block Diagram.

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).

tmp

Body

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

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

and in the following, we’ll be using the abbreviations

$\left({\begin{array}{c}U_{3}\\U_{4}\end{array}}\right)=:{\underline {r}}_{A}$

and

$\left({\begin{array}{c}U_{5}\\U_{6}\end{array}}\right)=:{\underline {r}}_{B}$.

tmp

Track

Tack specification

We describe the road by road-sections, which we call tracks. We start by defining each track as a line-segment, i.e. a first-order polynomial. The road is thus defined by pairs of track-endpoints, i.e. (x_i, y_i).

The track-polynomials are

${\begin{array}{ll}t_{i}(x)&=y_{i}\cdot {\tilde {x}}_{i}+y_{i-1}\cdot \left(1-{\tilde {x}}_{i}\right){\text{ for }}{\tilde {x}}_{i}={\frac {\displaystyle x-x_{i-1}}{\displaystyle x_{i}-x_{i-1}}}\\&=T_{i,1}\cdot x+T_{i,0}\end{array}}$.

During the numerical integration of the initial-value-problem - when the car runs along the track - the solver needs to cope with the transition between two straight lines - which is numerically more efficient if we convert the road in a continuously differentiable function.

Rounding the edges.

For this purpose, we define transition polynomials p_i of second degree - parabolas - as

$p_{i}(x)=\displaystyle \sum _{j=0}^{2}P_{i,j}\cdot x^{j}$

which connect two neighboring tracks.

The boundary conditions for the parabolas (red) and the neighboring lines (blue) yield

${\begin{array}{ll}\color {blue}t_{i}(x_{i}-\Delta x_{i})&=\color {red}p_{i}(x_{i}-\Delta x_{i}),\\\color {blue}\left.{\frac {\displaystyle dt_{i}}{\displaystyle dx_{i}}}\right\vert _{x_{i}-\Delta x_{i}}&=\color {red}\left.{\frac {\displaystyle dp_{i}}{\displaystyle dx_{i}}}\right\vert _{x_{i}-\Delta x_{i}},\\\color {blue}t_{i+1}(x_{i}+\Delta x_{i})&=\color {red}p_{i}(x_{i}+\Delta x_{i}),\\\color {blue}\left.{\frac {\displaystyle dt_{i+1}}{\displaystyle dx_{i+1}}}\right\vert _{x_{i}+\Delta x_{i}}&=\color {red}\left.{\frac {\displaystyle dp_{i}}{\displaystyle dx_{i}}}\right\vert _{x_{i}+\Delta x_{i}}.\end{array}}$

And though we have four boundary conditions, they allow for a solution with three unknown polynomial coefficients P_i if Δx_i is identical left and right from x_i - which we have already implied.

${\begin{array}{ll}{b_{3}}&=0,\\{b_{2}}&={\frac {\displaystyle {a_{2,1}}-{a_{1,1}}}{\displaystyle 4{\mathit {\Delta x_{i}}}}},\\{b_{1}}&={\frac {\displaystyle \left({a_{2,1}}+{a_{1,1}}\right)\,{\mathit {\Delta x_{i}}}-{x_{i}}\,{a_{2,1}}+{x_{i}}\,{a_{1,1}}}{\displaystyle 2{\mathit {\Delta x_{i}}}}},\\{b_{0}}&={\frac {\displaystyle \left({a_{2,1}}-{a_{1,1}}\right)\,{{\mathit {\Delta x_{i}}}^{2}}+\left(-2{x_{i}}\,{a_{2,1}}-2{x_{1}}\,{a_{1,1}}+4{y_{i}}\right)\,{\mathit {\Delta x_{i}}}+{{x}_{i}^{2}}\,{a_{2,1}}-{{x}_{i}^{2}}\,{a_{1,1}}}{\displaystyle 4{\mathit {\Delta x_{i}}}}}\end{array}}$

and

${\begin{array}{ll}{a_{1,0}}&={y_{i}}-{x_{i}}\,{a_{1,1}},\\{a_{1,1}}&={\frac {\displaystyle {y_{i-1}}-{y_{i}}}{\displaystyle {x_{i-1}}-{x_{i}}}}\operatorname {,} \\{a_{2,0}}&={y_{i}}-{x_{i}}\,{a_{2,1}},\\{a_{2,1}}&={\frac {\displaystyle {y_{i}}-{y_{i+1}}}{\displaystyle {x_{i}}-{x_{i+1}}}}\end{array}}$

But instead of employing the above, we solve the linear systems of equations for the coefficients numerically.

Now each polynomial creates new endpoints x_i - Δx_i and x_i + Δx_i which substitute for the initial x_i-point. The tracks are represented by a succession of lines and parabolas.

Next we need to find the contact point between wheel and road!

1+1=2

tmp

Contact-Point and -Normal

Text

1+1=2

tmp

Contact Forces

Text

1+1=2

tmp

Track

Text

1+1=2

tmp

Track

Text

1+1=2

tmp

Track

Text

1+1=2

tmp

Track

Text

1+1=2

Parameter

x

y

z

a

b

Contact conditions.

c

Contact angles

Wheel kinematics

Contact forces F, N

Characteristic for contact force.

Friction characteristic

next workpackage: driver-controls →

---

References

• ...