Gelöste Aufgaben/JUMP/Car-Body

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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:

${\overrightarrow{r}}_C(t)$.

This vector has - in 2D - two coordinates $u_1(t),u_2(t)$, measured in the inertial x-y frame:

${\overrightarrow{r}}_C(t)={\overrightarrow{\underset{\_}{e}}}_0\cdot \left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)$ with ${\overrightarrow{\underset{\_}{e}}}_0=({\overrightarrow{e}}_x,{\overrightarrow{e}}_y)$.

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

${\underset{\_}{r}}_C=\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)$.

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:

• $u_1(t),u_2(t),\phi (t)$ 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

${\overrightarrow{r}}_M={\overrightarrow{\underset{\_}{e}}}_0\cdot \left(\begin{array}{c}{u}_1(t)\\ u_2(t)\end{array}\right)$,

the coordinate system of the car is

${\overrightarrow{\underset{\_}{e}}}_C={\overrightarrow{\underset{\_}{e}}}_0\cdot \underset{\_}{\underset{\_}{D}}(\phi (t))$

where

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

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

${\overrightarrow{r}}_A={\overrightarrow{r}}_M+{\overrightarrow{r}}_{MA}$

or

${\underset{\_}{r}}_A={\underset{\_}{r}}_M+\underset{\_}{\underset{\_}{D}}(\phi (t))\cdot \left(\begin{array}{c}{a}_1\\ -a_3+u_3(t)\end{array}\right)$.

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

${\underset{\_}{r}}_B={\underset{\_}{r}}_M+\underset{\_}{\underset{\_}{D}}(\phi (t))\cdot \left(\begin{array}{c}-a_2\\ -a_3+u_4(t)\end{array}\right)$

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

$\left(\begin{array}{c}{U}_3\\ U_4\end{array}\right)=:{\underset{\_}{r}}_A$

and

$\left(\begin{array}{c}{U}_5\\ U_6\end{array}\right)=:{\underset{\_}{r}}_B$.

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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

$\begin{array}{ll}{t}_i(x)& =y_i\cdot {\tilde{x}}_i+y_{i-1}\cdot (1-{\tilde{x}}_i)\text{ for }{\tilde{x}}_i=\frac{x-x_{i-1}}{x_i-x_{i-1}}\\ & =T_{i,1}\cdot x+T_{i,0}\end{array}$.

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

$p_i(x)=\sum _{j=0}^2{P}_{i,j}\cdot x^j$

which connect two neighboring tracks.

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

$\begin{array}{ll}{t}_i(x_i-\mathrm{\Delta }{x}_i)& =p_i(x_i-\mathrm{\Delta }{x}_i),\\ {\frac{d{t}_i}{d{x}_i}|}_{x_i-\mathrm{\Delta }{x}_i}& ={\frac{d{p}_i}{d{x}_i}|}_{x_i-\mathrm{\Delta }{x}_i},\\ t_{i+1}(x_i+\mathrm{\Delta }{x}_i)& =p_i(x_i+\mathrm{\Delta }{x}_i),\\ {\frac{d{t}_{i+1}}{d{x}_{i+1}}|}_{x_i+\mathrm{\Delta }{x}_i}& ={\frac{d{p}_i}{d{x}_i}|}_{x_i+\mathrm{\Delta }{x}_i}.\end{array}$

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.

$\begin{array}{ll}{b}_3& =0,\\ b_2& =\frac{a_{2,1}-a_{1,1}}{4\mathrm{\Delta }{{x}}_{{i}}},\\ b_1& =\frac{(a_{2,1}+a_{1,1})\mathrm{\Delta }{{x}}_{{i}}-x_i{a}_{2,1}+x_i{a}_{1,1}}{2\mathrm{\Delta }{{x}}_{{i}}},\\ b_0& =\frac{(a_{2,1}-a_{1,1}){\mathrm{\Delta }{{x}}_{{i}}}^2+(-2x_i{a}_{2,1}-2x_1{a}_{1,1}+4y_i)\mathrm{\Delta }{{x}}_{{i}}+x_i^2{a}_{2,1}-x_i^2{a}_{1,1}}{4\mathrm{\Delta }{{x}}_{{i}}}\end{array}$

and

$\begin{array}{ll}{a}_{1,0}& =y_i-x_i{a}_{1,1},\\ a_{1,1}& =\frac{y_{i-1}-y_i}{x_{i-1}-x_i},\\ a_{2,0}& =y_i-x_i{a}_{2,1},\\ a_{2,1}& =\frac{y_i-y_{i+1}}{x_i-x_{i+1}}\end{array}$

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

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1+1=2

tmp

Track

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1+1=2

tmp

Track

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1+1=2

tmp

Track

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1+1=2

tmp

Track

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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 →

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References

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