Powertrain and Vehicle Dynamics, Dr Con Doolan 1

MECH ENG 3036 POWERTRAIN AND VEHICLE DYNAMICS

MSC.ADAMS/Car Assignment

Aim To numerically investigate the role of tyre properties on the static stability and control of a road vehicle. To compare the 2DOF bicycle model with a complex numerical model. To obtain a good knowledge of how to analyse vehicles using MSC.ADAMS/Car.

What is ADAMS?

ADAMS: Automatic Dynamic Analysis of Mechanical Systems.

Engineers use ADAMS to simulate the dynamics of mechanical components and systems.

■ Technology was implemented about 25 years ago. ■ Mechanical Dynamics Incorporated (MDI) formed by researchers who developed the base ADAMS code at the University of Michigan, Ann Arbor, MI, USA. MDI has been part of MSC.Software Corporation since 2002. ■ Large displacement code. ■ Systems-based analysis. ■ Original product was ADAMS/Solver, an application that solves nonlinear numerical equations. You build models in text format and then submit them to ADAMS/Solver. ■ In the early 90’s, ADAMS/View was released, which allowed users to build, simulate, and examine results in a single environment. ■ Today, industry-specific products are being produced, such as ADAMS/Car, ADAMS/Rail, and ADAMS/Engine.

We only use ADAMS/Car in this course.

ADAMS Training Excellent materials are available for you to learn how to use ADAMS/Car. These have been provided for you on MyUni. Download the document “Getting Started Using ADAMS/Car” (primer.pdf). Complete the “Suspension Analysis Tutorial” and the “Full-Vehicle Analysis Tutorial” in this document. This will give you the skills needed to complete the assignment. A tutorial has been arranged in the CAT Suite to help you with any difficulties in using ADAMS/Car.

Powertrain and Vehicle Dynamics, Dr Con Doolan 2

An additional document ‘ADAMS/Car Training Guide’ has also been provided for students who wish to obtain an in-depth knowledge of ADAMS/Car.

Background Please review the 2DOF bicycle model, steady-state response and stability factor before attempting this assignment. Appendix A gives background theory to aid your review.

Assignment Set-up Start ADAMS/Car (2005) (Standard Interface). Open the Assembly MDI_Demo_Vehicle.asy from the <acar_shared> directory. Change the tyre properties for ALL TYRES using the file provided on MyUni:

Fiala_ADAMS_ASSIGNMENT.tir

This replaces the tyre model from the Pacejka ‘Magic-Tyre’ model to the Fiala model. The Fiala model is based on the elastic foundation model as described in lectures, so should be easy to understand. The important parts of the tyre file are listed below, with comments showing you what they mean: PROPERTY_FILE_FORMAT = 'FIALA' FUNCTION_NAME = 'TYR902' USE_MODE = 2.0 $----------------------------------------------------------------------dimension [DIMENSION] UNLOADED_RADIUS = 326.0 Radius of tyre WIDTH = 245.0 Width of tyre

ASPECT_RATIO = 0.35 Aspect Ratio of tyre $----------------------------------------------------------------------parameter [PARAMETER] VERTICAL_STIFFNESS = 310.0 Vertical Stiffness (For Ride) VERTICAL_DAMPING = 3.1 Vertical Damping (For Ride) ROLLING_RESISTANCE = 0.0 Ignore CSLIP = 1000.0 Longitudinal stiffness (ignore)

CALPHA = 800.0 LATERAL STIFFNESS (N/deg)

CGAMMA = 0.0 Camber stiffness (ignore) UMIN = 0.9 Min friction coefficient UMAX = 1.0 Max friction coefficient

Highlighted is the LATERAL stiffness of the tyre. You will be changing this parameter for this assignment. SEE APPENDIX C ABOUT THIS VALUE.

Powertrain and Vehicle Dynamics, Dr Con Doolan 3

Assignment Tasks (1) Perform a Quasi-Static Constant Radius Cornering Test in ADAMS/Car using the supplied FIALA tyre model for ALL tyres. Assume a maximum lateral acceleration of 0.9 and a radius of 50 m. Don’t worry if ADAMS gives you an error; check the animation to see if the simulation worked. EXPORT REQUIRED DATA FROM THE POSTPROCESSING WINDOW (SEE APPENDIX D). [10 Marks] (2) Using the original lateral tyre stiffness (SEE NOTE IN APPENDIX C); plot the steady-state curvature control response of the MDI_Demo_Vehicle for longitudinal velocities 0-80 km/hr using the 2DOF bicycle model. (Suggestion: convert to radians before doing calculations). Calculate all stability and control derivatives. [10 Marks] (3) Calculate and plot the steady-state curvature control response of the MDI_Demo_Vehicle for longitudinal velocities 0-80 km/hr using the ADAMS data. Plot this on the same graph as your 2DOF bicycle data. [10 Marks] (4) Calculate the stability factor K from your data. [10 Marks] (5) Repeat steps (1-4) for a 30% increase in the front tyre stiffness. Use a text editor to change the stiffness. [15 Marks] (6) Repeat steps (1-4) for a 30% decrease in the front tyre stiffness. Use a text editor to change the stiffness. [15 Marks] (7) Do ADAMS and the bicycle model agree? Explain your answer. [10 Marks] TOTAL [80 Marks]

Powertrain and Vehicle Dynamics, Dr Con Doolan 4

APPENDIX A – BACKGROUND THEORY Equations of Motion for the Bicycle Model Here we will describe the equations of motion for our elementary road vehicle. This is a linear model with 2 degrees of freedom (2DOF) that will enable the calculation of the motion of the vehicle as a function of the forces and moments applying to the vehicle. Figure 1 is shown below and our 2DOF model will be based upon information in this figure.

b

a

l+N +r

x & y axes are horizontal

+Y

Turning centre

CG

R

RF

FF

+β

+x

+y+v

u

+δ

V

Figure 1. The 2-DOF “bicycle” model. Figure adapted from Milliken and Milliken,

Race Car Vehicle Dynamics, SAE, 1995.

The motion variables of interest are forward velocity, u, lateral velocity, v and the yaw rate, r. The path velocity, V, (the vector sum of u and v) is perpendicular to the turn radius R. The side or lateral force is Y and the yawing moment about the CG is

N. The small angle assumption is used, therefore 1cos ≈β and uV ≈ . Hence if V is

specified as the independent variable, there are only two dependent variables r and v, therefore the model is 2DOF. The reference axis system has its origin at the vehicle centre of gravity (accelerated reference frame). The perturbations in the dependent variables are measured relative to inertial space and the moments and products of inertia are constant. If the actual path of the vehicle relative to the ground is required, another set of Earth-fixed reference axes are needed. We can now write Newton’s 2nd Law for our bicycle model:

Powertrain and Vehicle Dynamics, Dr Con Doolan 5

CG)on (Forces

CG)about (Moments

y

ZZ

maY

rIdt

drIN

=

== &

Equation 1

Where N and Y are the resultant yawing moment and lateral force that the tyres apply to the vehicle. Note that aerodynamic forces are not considered in our 2DOF vehicle model. The moment of inertia about the z-axis is Iz in our model. It is important to realise that if we did not use vehicle fixed axes for our 2DOF model, this moment of inertia term would become very complicated and would lose physical meaning (it would become a complicated tensor matrix and a real pain to analyse! – Actually, this is why ADAMS is used to analyse complicated machines.) The y-axis acceleration term (ay) is made up of a centrifugal term and a direct lateral acceleration term:

( )ββ&

&

&

+=

=+=

+=

rV

VVVr

vVray

)constant(

Equation 2

-br + v

-br +v

R +Vv-brα =

+V

+V+v

+β =

b

+r

CG+v

+V

Figure 2. Velocities at the rear tyre. Figure adapted from Milliken and Milliken,

Race Car Vehicle Dynamics, SAE, 1995.

Figure 2 shows the velocities at the rear tyre. From the figure, it can be seen that if the vehicle experiences a positive yaw velocity, r, about the z-axis, the rear tyre will experience an additional lateral velocity –br. Therefore the rear slip angle can be determined from the vector sum of the velocities at the rear tyre:

Powertrain and Vehicle Dynamics, Dr Con Doolan 6

V

br

V

br

V

v

V

brv

R

R

−=

−=−

=

βα

α

Equation 3

The front tyre is analysed in a similar manner. Figure 3 shows the velocities acting on the front tyre. This time, the steering angle is included in the representation of the front tyre slip angle:

δβα

δδα

−+=

−+=−+

=

V

ar

V

ar

V

v

V

arv

F

F

Equation 4

V

V

+v

+ar

F+α

+v

a

+r

CG

+δ

Figure 3. Velocities at the front tyre. Figure adapted from Milliken and Milliken,

Race Car Vehicle Dynamics, SAE, 1995.

Recalling that we are using the small angle assumption and that the lateral tyre forces vary linearly with angle for the bicycle model, the forces at the front and rear track are:

−=

−=

−

+=

−+=

V

brCC

V

brCY

CV

arCC

V

arCY

SRSRSRR

SFSFSFSFF

ββ

δβδβ

Equation 5

Therefore the total lateral force becomes:

Powertrain and Vehicle Dynamics, Dr Con Doolan 7

( ) ( ) δβ

βδβ

SFSRSFSRSF

SRSRSFSFSFRF

CrbCaCV

CCY

V

brCCC

V

arCCYYY

−−++=

−+−

+=+=

1

Equation 6

The total yawing moment can be determined in a similar fashion:

( ) ( ) δβ

βδβ

SFSRSFSRSF

SRSRSFSFSF

RFRF

aCrCbCaV

bCaCN

V

rbCbCaC

V

raCaC

bYaYNNN

−++−=

+−−

+=

−=+=

22

22

1

Equation 7

Derivative Notation We can use a derivative form to represent the effects of perturbations in the motion

variables β, r and δ. In other words, the derivatives or slopes of the force/moment curves versus the motion variables can be used to determine the overall forces and moments on our vehicle. The simplified model we are deriving here is linear. As such, we can use the principle

of superposition and the components of force and moment due to β, r and δ are additive. Therefore, Y and N can be written:

( )

( )

δβ

δδ

ββ

δβ

δβ

δδ

ββ

δβ

δβ

δβ

NrNNN

Nr

r

NNrfN

YrYYY

Yr

r

YYrfY

r

r

++=

∂∂

+

∂∂

+

∂∂

==

++=

∂∂

+

∂∂

+

∂∂

==

,,

,,

Equation 8

The constant partial derivatives used in Equation 8 are known as the stability and control derivatives. They are usually written in a shorthand notation. For example:

δ

β

δ

β

∂∂

=

∂∂

=

NN

YY

Equation 9

Comparing Equations 6, 7 and 8, the stability and control derivatives can be deduced:

Powertrain and Vehicle Dynamics, Dr Con Doolan 8

( )

( )

SF

SRSFr

SRSF

SF

SRSFr

SRSF

aCN

CbCaV

N

bCaCN

CY

bCaCV

Y

CCY

−=

+

=

−=

−=

−

=

+=

δ

β

δ

β

221

1

Equations 10

Equations of Motion in Derivative Notation Now we can write the equations of motion in derivative notation using the above information:

( ) δββ

δβ

δβ

δβ

YrYYrmV

NrNNrI

r

rZ

++=+

++=

&

&

Equations 11

Physical Significance It is convenient to think of vehicle motion in terms of the derivatives derived above. The text ‘Race Car Vehicle Dynamics’ by Milliken and Milliken (SAE, 1995) provides an excellent interpretation of the derivatives. An edited (so as to maintain our symbol and English conventions) portion of this interpretation is given below: Yawing Moment Derivatives:

Nδ This derivative is the proportionality factor between the yawing moment and the steering angle. It is the control moment derivative. It increases with the cornering stiffness of the front tyres and their distance from the CG. It is always positive.

Nr This derivative is the proportionality factor between the yawing moment on

the vehicle (produced by the tyres) and the yawing velocity of the vehicle. It is the yaw damping derivative. It is always negative, i.e., damping, in the linear range of the tyres. It is exactly analogous to an angular viscous damper, always trying to reduce the yawing velocity. It is a function of the front and rear cornering stiffnesses, the distances from the front and rear wheels and the CG squared and is inversely proportional to forward velocity. Nr is theoretically infinite at zero velocity since a small yawing velocity, r, will

produce an infinite change in αF = ar/V and αR = br/V. Nr is zero at infinite speed. For a simple vehicle with CSF and CSR independent of load, it has its smallest magnitude when the CG is at the midpoint. The fall off in steady-state directional stability with speed is frequently attributed to changes in understeer but is more properly attributed to the reduction in yaw damping.

Powertrain and Vehicle Dynamics, Dr Con Doolan 9

Nβ This is the “static” directional stability slope or under/oversteer derivative. On this simple vehicle it is the difference between the moment (about the CG)

produced per unit β, by the rear and front wheels, i.e., -bCSR and aCSF. It is a pure “weathercock” effect. If the rear wheels (stabilising) produce a greater moment slope than the front wheels (destabilising), the vehicle is stable and

wants to reduce β. Nβ is basically independent of speed. With a positive Nβ, the vehicle is always trying to align itself with its relative velocity vector, thus US.

Lateral Force Derivatives:

Yδ This derivative is the proportionality factor between the side force from the front wheels due to steering and the steer angle; in short, the side (lateral) force due to steering – the control force derivative. It is always positive.

Yβ This is the side (lateral) force slope. It is similar to the slope of the lateral force vs. slip angle curve for a single tyre, but it is the slope of the lateral force curve for the entire vehicle. As such, it is a measure of the rate that lateral

force is developed as the vehicle is sideslipped. Since the slope Yβ is always negative in the linear range, i.e., a negative lateral force for a positive sideslip velocity, it corresponds to the rate in a linear damper and is called the damping-in-sideslip.

Yr This derivative is the side force due to yawing velocity and arises from the

difference in the lateral tyre forces that comprise Nr. The sign of Yr follows

that of Nβ. Yr is inversely proportional to V. In general, Yr is small. It is called the lateral force/yaw coupling derivative.

In summary, there are six derivatives, two associated with control (Nδ, Yδ), two with

damping (Nr, Yβ) and two cross-coupling between the two degrees of freedom (Nβ, Yr). Steady-State Responses

For our 2DOF vehicle in a steady turn, r&=0 and β& = 0 and hence the equations of motion become:

δβ

δβ

δβ

δβ

NrNN

YrYYmVr

r

r

++=

++=

0

Equation 12

In a steady turn, r = V/R. Therefore:

( )

+=−

−+=−

RVNNN

RmVVYYY

r

r

1

12

βδ

βδ

βδ

βδ

Equation 13

Powertrain and Vehicle Dynamics, Dr Con Doolan 10

Control Response

The linear steady-state control response characteristics (i.e. response of vehicle motion to driver control input at the steering wheel) are given as a set of ratios:

Curvature response defined as: (1/R)/δ Yawing velocity response: r/δ Lateral acceleration response: (V2/R)/δ Sideslip angle response: β/δ

Now, solve for β in the second equation of Eq. 13 and substitute into the first:

( )

( )

( ) ( )[ ]

( )

( )rr

rr

rr

rr

rr

r

NYmVNYNV

YNNY

NVYmVVYN

YNNYR

RNVYmVVYNYNNY

N

RN

N

YVmVVYYN

N

Y

RmVVY

RN

N

YVN

N

YY

RN

NV

N

N

βββ

δβδβ

ββ

δβδβ

ββδβδβ

β

β

βδδ

β

β

β

βδ

β

βδ

ββ

δ

δ

δ

δ

δδ

δβ

−−

−=

−−

−=

−−=−

−−=

−

−+

−

−=−

−

−=

2

2

2

2

/1

:is response curvature thetherefore

1

:by hrough multiply t

1

:rearrange

11

:equationfirst theinto substitute

1

Equation 14

The curvature response can be simplified further:

VQ

YNNYR δβββ

δ

−=

/1

Equation 15

where

rr NYmVNYNQ βββ −−=

Equation 16

The yawing velocity response can be determined by:

Powertrain and Vehicle Dynamics, Dr Con Doolan 11

Q

YNNYRV

r δβββ

δδ

−==

/1

Equation 17

The lateral acceleration response is obtained by multiplying Equation 17 by V2:

( )Q

YNNYVRV δβββ

δ

−=

/2

Equation 18

Sideslip angle response can be obtained by eliminating r from Equations 12 in a

similar manner to the β elimination for curvature response:

( )Q

mVYNNY rr −−= δδ

δβ

Equation 19

Side Force Response

We now determine the steady-state responses of the car to an applied side force at the

CG. If this side force is Fy and δ = 0, and the steady-state turning conditions ( r&=0 and β = 0 ) apply, then Equations 12 become:

( )

+=

−+=−

RVNN

RmVVYYF

r

ry

10

12

β

β

β

β

Equation 20

The side force control characteristics are:

Q

N

F

Q

VN

F

RV

Q

N

F

r

VQ

N

F

R

r

y

y

y

y

+=

−=

−=

−=

β

β

β

β

/

/1

2

Equation 21

Powertrain and Vehicle Dynamics, Dr Con Doolan 12

Yaw Moment Response

Again, we apply the same procedure to find the response to a steady, externally applied yawing moment, N:

( )

+=−

−+=

RVNNN

RmVVYY

r

r

1

10 2

β

β

β

β

Equation 22

The yaw response control characteristics are:

( )Q

mVY

N

Q

VY

N

RV

Q

Y

N

r

VQ

Y

N

R

r −−=

=

=

=

β

β

β

β

/

/1

2

Equation 23

Superposition

The response to simultaneous application of side force and yawing moment can be found by using the principle of superposition. That is, because the system is linear, the response to each input can be added. For example, the response in sideslip to an applied side force (such as an aerodynamic force due to a wind gust) and moment is:

( )

( )mVNNYFNQ

NQ

mVYF

Q

N

ryr

ry

r

+−=

−−=

1

β

Equation 24

Neutral Steer Response We have shown that a neutral steer (NS) car follows the geometric path or Ackermann steering angle. We also showed that the car would respond to a steady-state side load (the tilted road exercise) in a way so that there was uniform slip angles at the front and rear tyres. We can enhance these descriptions and say that for the NS car, the

static directional stability is zero, i.e., Nβ = 0. What this says is that no moment arises

on the vehicle when a sideslip (β) exists (with δ = r = 0).

Powertrain and Vehicle Dynamics, Dr Con Doolan 13

If Nβ = 0, Q = -YβNr, the product of the two damping derivatives. The response characteristics are then: Response to Control – Neutral Steer

rN

N

V

R δ

δ1/1

−=

Equation 25

rN

Nr δ

δ−=

Equation 26

rN

NV

RV δ

δ−=

/2

Equation 27

( )mVYN

N

YY

Yr

r

−+−= δ

ββ

δ

δβ 1

Equation 28

Response to Applied Side Force – Neutral Steer

0/1

=yF

R

Equation 29

0=yF

r

Equation 30

0/2

=yF

RV

Equation 31

β

βYFy

1−=

Equation 32

Response to Applied Yawing Moment – Neutral Steer

rVNN

R 1/1−=

Equation 33

rNN

r 1−=

Equation 34

rN

V

N

RV−=

/2

Equation 35

Powertrain and Vehicle Dynamics, Dr Con Doolan 14

( )r

r

NY

mVY

N β

β −+=

Equation 36

From Equation 34:

rrNN −=δδ

Equation 37

This states that in a steady turn, the control moment just balances out the yaw

damping moment, for a NS vehicle. Please remember:

The NS vehicle has zero response to an applied side load at the CG except for the slip

angle β. This slip angle is given by Equation 32:

β

βY

Fy−=

Equation 38

Recall that Yβ is the total tyre stiffness:

SRSF CCY +=β

Equation 39

Therefore maximising the tyre stiffness will reduce the side slip response to a side load. For a yawing moment applied to the vehicle, from say a side gust with the centre of pressure forward of the CG, the side slip response is from Equation 36:

( )r

r

NY

mVYN

β

β−

+=

Equation 40

Hence increasing the stiffness of the tyres will again reduce yaw response. But the other term in the denominator is a function of tyre stiffness, geometry and the inverse of vehicle speed:

( )SRSFr CbCaV

N 221+

=

Equation 41

Therefore, at high speed, the effectiveness of the tyres to damp the β response to yaw is reduced. It is interesting to note that controlling vehicle side slip, there is a trade-off between vehicle speed, geometry and tyre properties, giving the engineer a few options for vehicle design.

Powertrain and Vehicle Dynamics, Dr Con Doolan 15

Stability Factor: K Let’s go to work on Equation 15, the general curvature response to a steer angle:

( )

( )δβδβ

β

δβδβ

ββ

βββ

δβδβ

δβδβ

δ

YNNY

mVN

YNNY

NYYNV

NYmVNYNV

YNNY

VQ

YNNYR

rr

rr

−−

−

−=

−−

−=

−=

2

1

/1

Equation 42

Now, from our earlier definitions:

( ) wheelbase the,lba

YNNY

NYYNV rr =+=−

−

δβδβ

ββ

Equation 43

Then:

2

2

2

2

1

/1

11

/1

1/1

KV

l

VNYYN

mVN

l

l

YNNY

mVNl

R

+=

−

+

=

−−

=

δβδβ

β

δβδβ

βδ

Equation 44

Where:

( )

( )( )

−

−=

−=

δββδ

ββ

δβδβ

β

YYNN

YN

l

mK

NYYNl

mNK

/

/

or

Equation 45

Where K is the stability factor. Yawing velocity response is therefore:

Powertrain and Vehicle Dynamics, Dr Con Doolan 16

21

/

KV

lVr

+=

δ

Equation 46

The stability factor, K, depends on vehicle mass, wheelbase length and the stability and control derivatives. The derivatives are independent of speed in our simple, linear 2DOF model. However, in situations where aerodynamic downforce is significant, such as in racing, the derivatives may become coupled with speed and become more complicated. NS, US, OS and K Equation 44 can be used to describe the steering characteristics of a vehicle. Rearranging, the path radius of a vehicle in a steady-state turn is:

( )21 KVl

R +=δ

Equation 47

Recall that for a Neutral Steer (NS) vehicle, as lateral acceleration is applied, it neither “oversteers” or “understeers” – the vehicle follows the intended geometric path described by the Ackermann steering angle. In Equation 47, the Ackermann

relationship is described by having K=0 so that R = l/δ.

Therefore, for an NS vehicle, K=0.

Also recall that as lateral acceleration is increased, the US vehicle “understeers” the geometric path described by the Ackermann steering angle. This is described in

Equation 47 by having K > 0, so that R > l/δ.

Therefore, for a US vehicle, K is positive.

Finally remember that the OS vehicle will “oversteer” the intended geometrical path as described by the Ackermann steering angle. This is described in Equation 47 by

having K < 0, so that R < l/δ.

Therefore, for an OS vehicle, K is negative.

Powertrain and Vehicle Dynamics, Dr Con Doolan 17

APPENDIX B FERRARI VEHICLE STATISTICS Mass = 1528.68 kg Wheelbase = 2560 mm Steering Gear Ratio = 23.9 (Big, I know) a = 1482 mm b = 1078 mm Mass Inertia Tensor: IXX : 5.8339295752E+008 kg-mm^2 IYY : 6.1291213262E+009 kg-mm^2 IZZ : 6.0223575551E+009 kg-mm^2 IXY : -1.907091268E+006 kg-mm^2 IZX : 1.1607572609E+009 kg-mm^2 IYZ : -1.3046453885E+006 kg-mm^2

Powertrain and Vehicle Dynamics, Dr Con Doolan 18

APPENDIX C ADDITIONAL REQUIRED

INFORMATION The steer angle output from ADAMS is negative for positive lateral acceleration – opposite to the SAE vehicle axis system used in this course. This results in some unusual sign conventions. In order to do the assignment, please recognise that the negative steer angle will require that the following form of equations be used:

( )VQ

YNNYR δβδβ

δ

−−=

/1

( ) ( )211

2

11

KVKV

lR lR

+−

=⇒+=δδ

ALSO, PLEASE USE A POSITIVE LATERAL STIFFNESS IN YOUR

CALCULATIONS TO COMPENSATE FOR THE ADAMS VEHICLE AXIS

SYSTEM

REMEMBER THAT LATERAL STIFFNESS IS USUALLY NEGATIVE IN

VEHICLE DYNAMICS CALCULATIONS – THIS IS THE CONVENTION

USED FOR THE REST OF THIS COURSE

THE FIALA TYRE MODEL LATERAL STIFFNESS IS

2.67 TIMES THE 2DOF LATERAL STIFFNESS – YOU

MUST COMPENSATE FOR THIS IN YOUR ANALYSIS, SO IF THE FIALA MODEL LATERAL STIFFNESS IS 800 N/deg,

USE 300 N/deg FOR YOUR 2DOF MODEL.

Powertrain and Vehicle Dynamics, Dr Con Doolan 19

APPENDIX D EXPORTING DATA FROM ADAMS/Car

(1) After your analysis is complete, press F8 for the post-processing window.

(2) Then click on the menu: File ⇒ Export ⇒ Spreadsheet to bring up the Export window.

(3) In the Export window, right click on the File Name box and select the directory where you want to store your results. Type in your file name.

(4) Right click on the Result Set Name in the Export window and choose Result

Set ⇒ Browse, this will bring up the Database Navigator in a separate window

(5) Select your Analysis Results by double clicking on the analysis name (sometimes it is already opened).

(6) Select the results you want by Ctrl-clicking on each of them (e.g. we need chassis_velocities and jms_steering_wheel_angle_data), click OK

(7) Click OK in the export window – you have now saved your selected results to a tab limited text file.

(8) Import into Excel by opening your file in any text editor like Notepad. Then select the data, copy and paste into Excel. You may think of an easier way of doing this, or even use Matlab.