Gelöste Aufgaben/JUMP

Problem Statement

This example from Mechatronics combines mathematical models of mechanical and electrical components to simulate an RC-car-model. We've taken some reference-parameters from commercial vendors and adopted them as appropriate.

Jump!

Trajectories

Create a mathematical Model for an RC-Model Car.

The model shall account for the basic car-functions in a 2D-driving environment including vehicle dynamics, drive-train, battery and driver.

A 2D-test-parkour shall be provided to assess basic climbing and jumping capabilities of the car.

Overall Scope

We are tasked to provide a mathematical performance model for an RC-car model with a given reference-configuration. The mathematical model shall account for different operational envelopes and test for e.g. climbing capabilities, typical loads on springs and car-agility.

We follow a generic approach in modelling and simulation which consists of three overall steps:

• Structuring
• Modelling
• Simulating

Structure

The function of “structuring” is to break down this complex task - modelling and simulating the performance of our RC-car - into work-packages. We aim at reducing the complexity of the model until the remaining entities appears to be rather trivial to manage.

The structure we are aiming at is a pre-requisite for step 2: building the mathematical model. This model is the projection of the physical processes into mathematical equations. We’ll be implementing our model as an Initial-Value-Problem (IVP) , written in mathematical terms as

${\dot {\underline {Q}}}={\underline {f}}({\underline {Q}},t);{\dot {(.)}}={\frac {\displaystyle d}{\displaystyle dt}}(.)$

The left-hand side of the equation, q, are the state variables of our system. They are the minimum-set of values to describe the physical state of our model at a certain time t. The state variables stem from the car's sub-systems. We choose four workpackages to contribute to the state-variables:

• Battery:the power-source for the car
• Car Body: consisting of the car's chassis and its wheels
• Driver-Controls: processes that control acceleration and deceleration of the car
• E-Motor and Drive-Train: the mechanisms that transfer battery power into mechanical energy.

State Variables

Each workpackage contributes state variables to the column vector q:

The IVP establishes mathematical equations describing how these state variables change over time, e.g. how a driver actuates the "gas"-pedal, which increases the motor-voltage, thus increasing the motor-torque and hence accelerating the car.

Block Diagram.

Our first approach is to depict each workpackage as graphical block and connect information / property exchange between blocks as in a diagramming tool:

The driver chooses a "gas"-pedal state "p" depending on information form the car's location, speed and battery temperature "T_B".

At the E-motor, "p" sets the nominal voltage "U_M" of the E-motor, this implies a motor current "I_M" and wheel torque "M_W".

The motor current "I_M" then drives both the battery temperature and the available battery voltage "U_B".

And the wheel torque propels the car, delivering new locations and velocities.

These are the submodels to be developed in each workpackage. Please follow the links:

1. Battery
2. Car-Body
3. Driver Controls
4. E-Motor and Drive-Train

Model

The mathematical model consists of XX differential equations with coordinates for each subsystem:

${\underline {q}}_{B}(t):=\left({\begin{array}{c}\color {orange}Q_{B}(t)\\\color {orange}U_{B2}(t)\\\color {orange}T_{B}(t)\\\end{array}}\right)$ for the battery,

${\underline {q}}_{C}(t):=\left({\begin{array}{c}\color {blue}u_{1}(t)\\\color {blue}u_{2}(t)\\\color {blue}u_{3}(t)\\\color {blue}u_{4}(t)\\\color {blue}\varphi (t)\\\color {blue}\psi (t)\\\end{array}}\right)$ for the car-body

${\underline {q}}_{D}(t):=\left({\begin{array}{c}\color {green}E(t)\\\end{array}}\right)$ for the driver and

${\underline {q}}_{E}(t):=\left({\begin{array}{c}\color {red}I_{M}(t)\\\color {red}\psi _{M}(t)\end{array}}\right)$ for the E-motor.

However, we have no elasticity accounted for between the E-motor and the wheel, thus

${\dot {\tilde {\psi }}}(t)=n_{G}\cdot {\dot {\tilde {\psi }}}_{M}(t)$

and so the remaining coordinates are

${\underline {q}}(t):=\left({\begin{array}{c}\color {orange}Q_{B}(t)\\\color {orange}U_{B2}(t)\\\color {orange}T_{B}(t)\\\color {blue}u_{1}(t)\\\color {blue}u_{2}(t)\\\color {blue}u_{3}(t)\\\color {blue}u_{4}(t)\\\color {blue}\varphi (t)\\\color {blue}\psi (t)\\\color {green}E(t)\\\color {red}I_{M}(t)\\\end{array}}\right)$.

We have to deal with first and second order time derivatives of coordinates from q. All mechanical coordinates (e.g. u_i, φ) are represented with their second-order time-derivatives that come from their D'Alembert's Inertia Forces. In order to transfer the equations of motion into a uniform set of differential equation of first order, we employ e.g.

$\color {lightblue}{\frac {\displaystyle du_{1}(t)}{\displaystyle dt}}\color {black}=\color {blue}v_{1}(t)\,,\,\color {lightblue}{\frac {\displaystyle d\varphi (t)}{\displaystyle dt}}\color {black}=\color {blue}\Omega (t){\text{ and }}\color {lightblue}{\frac {\displaystyle d\psi (t)}{\displaystyle dt}}\color {black}=\color {blue}\omega (t)$

so that the column matrix of state variables of the system is

${\underline {Q}}(t):=\left({\begin{array}{c}\color {orange}Q_{B}(t)\\\color {orange}U_{B2}(t)\\\color {orange}T_{B}(t)\\\color {blue}v_{1}(t)\\\color {blue}v_{2}(t)\\\color {blue}v_{3}(t)\\\color {blue}v_{4}(t)\\\color {blue}\Omega (t)\\\color {blue}\omega (t)\\\color {lightblue}u_{1}(t)\\\color {lightblue}u_{2}(t)\\\color {lightblue}u_{3}(t)\\\color {lightblue}u_{4}(t)\\\color {lightblue}\varphi (t)\\\color {blue}\psi (t)\color {black}\!\!\!\!\!\!\!\!\!\!\!--\\\color {green}E(t)\\\color {red}I_{M}(t)\\\end{array}}\right)$.

But as indicated above the rotation angle of the front wheel ψ_M never appears in our equation - we remove it from the list of state variables.

Now in matrix-notation, the resultant equations of motion are

${\underline {\underline {M}}}\cdot {\dot {\underline {Q}}}={\underline {f}}({\underline {Q}})$

with

${\underline {\underline {M}}}=\left({\begin{array}{cccccccccccccccc}\color {orange}1&&&\\&\color {orange}1&&\\&&\color {orange}1&\\&&&\color {blue}2{m_{2}}+{m_{1}}&\color {blue}0&\color {blue}0&\color {blue}0&\color {blue}2{m_{2}}\,{a_{3}}&\color {blue}0\\&&&\color {blue}0&\color {blue}2{m_{2}}+{m_{1}}&\color {blue}{m_{2}}&\color {blue}{m_{2}}&\color {blue}{a_{1}}\,{m_{2}}-{a_{2}}\,{m_{2}}&\color {blue}0\\&&&\color {blue}0&\color {blue}{m_{2}}&\color {blue}{m_{2}}&\color {blue}0&\color {blue}{a_{1}}\,{m_{2}}&\color {blue}0\\&&&\color {blue}0&\color {blue}{m_{2}}&\color {blue}0&\color {blue}{m_{2}}&\color {blue}-{a_{2}}\,{m_{2}}&\color {blue}0\\&&&\color {blue}2{m_{2}}\,{a_{3}}&\color {blue}{a_{1}}\,{m_{2}}-{a_{2}}\,{m_{2}}&\color {blue}{a_{1}}\,{m_{2}}&\color {blue}-{a_{2}}\,{m_{2}}&\color {blue}{m_{2}}\left(2{{a}_{3}^{2}}+{{a}_{2}^{2}}+{{a}_{1}^{2}}\right)+{J_{1}}\color {black}+\color {red}\left(n_{G}^{2}+2\,n_{G}+1\right)\,J_{3}&\color {blue}0\color {red}-\left(n_{G}^{2}+n_{G}\right)\,J_{3}\\&&&\color {blue}0&\color {blue}0&\color {blue}0&\color {blue}0&\color {blue}0\color {red}-\left(n_{G}^{2}+n_{G}\right)\,J_{3}&\color {blue}{J_{2}}\color {black}+\color {red}n_{G}^{2}\,J_{3}\\&&&&&&&&&\color {lightblue}1&&&&\\&&&&&&&&&&\color {lightblue}1&&&\\&&&&&&&&&&&\color {lightblue}1&&\\&&&&&&&&&&&&\color {lightblue}1&\\&&&&&&&&&&&&&\color {lightblue}1\\&&&&&&&&&&&&&&\color {green}1\\&&&&&&&&&&&&&&&\color {red}1\end{array}}\right)$

and

${\underline {f}}({\underline {Q}})=\left({\begin{array}{l}\color {orange}-\alpha (I_{B})\cdot \beta (T_{B})\cdot I_{B}\\\color {orange}{\frac {\displaystyle I_{C}}{\displaystyle C}},\\\color {orange}{\frac {\displaystyle 1}{\displaystyle m_{B}\;c_{B}}}\cdot \left({\dot {Q}}_{H}-{\dot {Q}}_{D}\right)\\\color {blue}\cos(\gamma _{A})\,F_{T}-\sin(\gamma _{2})\,F_{N2}-\sin(\gamma _{A})\,F_{N1}\\\color {blue}\left(-2\,m_{2}-m_{1}\right)g+\sin(\gamma _{A})\,F_{T}+\cos(\gamma _{2})\,F_{N2}+\cos(\gamma _{A})\,F_{N1}\\\color {blue}-F_{T}\sin \left(\varphi (t)-\gamma _{A}\right)+F_{N1}\cos \left(\varphi (t)-\gamma _{A}\right)-m_{2}g\cos \left(\varphi (t)\right)-b_{1}\,{\dot {u}}_{3}-k_{1}\,u_{3}(t)\\\color {blue}\sin(\gamma _{2})\,F_{N2}\sin \left(\varphi (t)\right)+\left(\cos(\gamma _{2})\,F_{N2}-m_{2}g\right)\cos \left(\varphi (t)\right)-b_{2}\,{\dot {u}}_{4}-k_{2}\,u_{4}(t)\\\color {blue}\left(a_{3}\,F_{N1}-a_{1}\,F_{T}\right)\sin \left(\varphi (t)-\gamma _{A}\right)+\left(a_{3}\,F_{T}+a_{1}\,F_{N1}\right)\cos \left(\varphi (t)-\gamma _{A}\right)+\sin \left(\varphi (t)\right)+\left(\sin(\gamma _{2}(t))\,F_{N2}u_{4}(t)+\left(\sin(\gamma _{1}(t))\,F_{N1}-\cos(\gamma _{2}(t))\,F_{T}\right)u_{3}(t)+\left(a_{2}-a_{1}\right)\,m_{2}g+\left(-a_{3}\sin(\gamma _{2}(t))-a_{2}\cos(\gamma _{2}(t))\right)\,F_{N2}\right)\cos(\varphi (t))+\left(\sin(\gamma _{1}(t))\,R\,F_{T}+\cos(\gamma _{2}(t))F_{N1}\,R\right)\sin(\gamma _{1}(t))+\left(\cos(\gamma _{2}(t))\,R\,F_{T}-\sin(\gamma _{1}(t))\,F_{N1}\,R\right)\cos(\gamma _{2}(t))+B\,\left(\omega (t)+\Omega (t)\right)+M_{W}\\\color {blue}\left(-\sin(\gamma _{A})\,R\,F_{T}-\cos(\gamma _{A})\,F_{N1}\,R\right)\sin(\gamma _{A}(t)+\left(\sin(\gamma _{A})\,F_{N1}\,R-\cos(\gamma _{A})\,R\,F_{T}\right)\cos {\left({{\gamma }_{A}}(t)\right)}+B\,\left(\omega (t)+\Omega (t)\right)+M_{W}\\\color {lightblue}v_{1}\\\color {lightblue}v_{2}\\\color {lightblue}v_{3}\\\color {lightblue}v_{4}\\\color {lightblue}\Omega \\\color {green}e\\\color {red}{\frac {\displaystyle 1}{\displaystyle L}}\left(U_{B}\cdot p(t)-R\,I_{M}(t)-e\right)\\\end{array}}\right)$

Simulation

We implement the system of differential equations

${\underline {\underline {M}}}\cdot {\underline {\dot {Q}}}={\underline {f}}({\underline {Q}})$

in Matlab^Ⓡ - the nonlinear characteristics and the algorithm that computes the contact points between car and track are too demanding for Maxima.

Implementation in Matlab^®

We follow [[Werkzeuge/Software/Matlab/Grundstruktur eines Matlab^®-Programms|our basic process schema]] for implementing this initial-value-problem in Matlab^®.

The JUMP-script has these functions:

• pre-processing,
• solving and
• post-processing.

Key to the solution process is a call to Matlab^®'s own solver

ode15s

for stiff differential equations. Though the equations of motion are not intrinsically stiff, this solver provided the best performance if the car encounters e.g. a rough touch-down after a jump-phase. The call to ode15s is

[t,y] = ode15s(@(t,y)jumpdydt(t,y,parC,parN,b,c,d,e,f),tspan,y0);

with parC, parN being maps of parameters.

The objects b, c, d, e, f are classes that hold parameters and functions related to each subsystem of the problem - from 'battery' to contact-'formulations'.

Matlab^©-files

Folder Structure

The folder-structure shows the main script JUMP.m. Classes and Functions are located in the respective dolfers. The Excel-file holds all system parameters.

To download the full Matlab^®-project download the zip-file from the link to your right.

archive

download compressed archive →

Systamparameter

Tabs of the spreadsheet-file

We collect all systems parameters in a spreadsheet file. Each system component has its own tab - e.g. the "car-body". We have two additional tabs, one for constants (gravity) and one for numerics (simulation end-time).

data

download spreadsheet →

Running the program will deliver these results:

target speed: 10 m/s. (click to enlarge)

Take your time to study the individual diagrams. With such a high velocity prescribed, the integrator part of the controller is simply being filled up, the diagram to the lower right (row 3, column 3) shows that for "E".

We'll investigate a few scenarios to the acquainted to the system, we start by reducing the target cruising speed to 2 m/s:

target speed: 2 m/s

Now our controller struggles to hold the prescribed speed downhill - remember, we have no breaks installed - and again on the steep climb upwards. But it works.

Let's see what happens, if we beef-up the motor (higher M_St and ω_Nl) and switch-off the slip-control. We get a very different behaviour which shows our model-car speeding with screetching tires for 9 s < t < 15 s.

target speed: 2 m/s

As the highlight of this model, we want to let our model-car jump over a gap of - say L=3 meters. We could try to simulate several combinations of initial values and system parameters to succeed - but that's too tiresome. We need to understand, what the "lift-off" velocity v₀on the ramp should be to make it to the other side.

So let us define a ramp with an inclination of 1:10 thus a ramp angle of γ=atan(1/10), then a gap of three meters and then a ramp downwards - and ask for an analytical solution to the trajectory.

Assuming - during the jump-phase - a motion of

${\begin{array}{ll}{\underline {r}}_{C}(t)&=\left({\begin{array}{c}u_{1}\\u_{2}\end{array}}\right)\\&=\left({\begin{array}{c}v_{0,x}+u_{1,0}\;t\\-{\frac {1}{2}}g\;t^{2}+v_{0,y}\;t+u_{2,0}\end{array}}\right)\end{array}}$,

which fulfills

${\begin{array}{ll}{\underline {\ddot {r}}}_{C}&=\left({\begin{array}{c}0\\-g\end{array}}\right)\end{array}}$,

we find the boundary-conditions for the jump to be

${\begin{array}{ll}{\underline {r}}_{C}(0)&=\left({\begin{array}{c}0\\0\end{array}}\right)\end{array}}$,

${\begin{array}{ll}{\underline {\dot {r}}}_{C}(0)&=\left({\begin{array}{c}v_{0}\;\cos(\gamma )\\v_{0}\;\sin(\gamma )\end{array}}\right)\end{array}}$ and for the landing point ${\begin{array}{ll}{\underline {r}}_{C}(t^{*})&=\left({\begin{array}{c}L\\0\end{array}}\right)\end{array}}$.

We compute that

$t^{*}=0.2\;{\frac {\displaystyle s^{2}}{\displaystyle m}}\;v_{0}$ and thus $v_{0}=12.3{\frac {\displaystyle m}{\displaystyle s}}$,

so we need to plan for a long run-up!

Simulation Runs

We show a few examples of what can go wrong when doing this kind of simulations. I did not include the one where the car drives right through the ramp - because the stiff solver chose step-sizes which passed out on this object ....

First "jump" attempt: a test-run

We select a speed-setpoint of V_set = 1 m/s - and test the drive-range carefully.

Jump-attempt 1: easy test-run.

Of course we did not jump - with such a low speed. We we know, the simulation performs as expected.

Second "jump" attempt: ups - nearly made it

No we give full throttle:

Jump-attempt 2: crashed

We did not just have the minimum speed to make the jump. Instead the car-model crashes into the ramp with its rear wheels and tips over. In our simulation it continues driving upside down - but with the car-body-center below the wheels.

Third "jump" attempt: YES!

No we use a little more run-up and ...

Jump-attempt 3: YES!

... yes - we made it!

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Associated pages of the submodels:

• Battery
• Car-Body
• Driver Controls
• E-Motor and Drive-Train

Links

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Literature

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