University of CaliforniaDavis

EME 171 – Lab #3, Part 1BMW S1000RR

Motocross Pitch-Heave ModelGroup # 10

Fall 2010

By:

Manuel Hoog, Konstantin Zamuruyev, Brandon Schroeder, and Michael Schantz

Intro

During this lab we will investigate the response of a motorcycle as it moves over two bumps in the road. The focus of this project is to investigate the heave and pitch motions of the cycle as it travels over such bumps. Furthermore we will investigate the influence of bike rider position on the motorcycle. By changing his position on the bike the center of gravity will be shifted and the dynamics of the systems will change. During this first part we are to construct an initial setup of this problem using camp/g the bond graph modeling method.

System Schematic

g

fvrv

trv

( )fv t( )rv t

tfv

sfksrk fbrb

ftkrt

k

Wheel base

LcgPitch angular velocity p

L

A

Mcr, J

mtrmtf

Vmoto

Schematic model of motorcycle

L

A

Wheel base

Bond graph from GAMP/G

The bond graph corresponding to the motocross Pitch-Heave model.Power flows and velocities are shown in the schematic.

Initial Conditions

We assume that the motorcycle is in equilibrium-that there is no vertical velocity before hitting the bump; therefore, all momentums in vertical directions are zero. Momentum for the mass of the motorcycle are zero and momentum for the front and rear tire are zero.

)( BikeRiderWeight

CG

)(Leg)( Legbase

RF FF

There will be initial displacements (deflections) in the front and rear spring as well as front and rear tire due to the weight of the motorcycle, rider and tires. These initial displacements were found to be:

Doing summation of forces in the vertical direction and moments around rear end and horizontal direction, we found the force acting on the front and rear.

From Hook’s law

Initial displacement of the front spring is Q9 is equal to :

Initial displacement of front tire:

Initial displacement of the rear spring:

Initial displacement of the rear tire:

Initial momentous

P5IN= 0.0 ! Initial momentum of the front tire. P15IN= 0.0 ! Initial angular momentum of the motorcycle and the rider. P17IN= 0.0 ! Initial momentum of the motorcycle and the rider.

Inputs for all efforts and flow sources

We simulate flow source SF1 and SF2 with the help of the step function. The base of the motorcycle is 1.6 meters and the horizontal velocity of the motorbike is 10 meters per second.

Therefore there is a time difference between the front and rear wheel hitting the

bump is found to be:

delay = (base/vmoto) ! m Time delay between the front and rear wheel.

Using the step function for the input for flow:

!.... EXTERNAL INPUTS SE=F(T),SF=F(T) .... SF1 = (hbump*vmoto/(0.5*lbump))*(step(tsf)-2*step(tmf)+step(tef)) SF33 =(hbump/0.5*vmoto/lbump)*(step(tsr)-2*step(tmr)+step(ter))

Effort sources due to gravity are defined as:

SE6 = (mtf*g) SE18 =(mcr*g) SE28 =(mtr*g)

Output equations for the front and rear suspension deflections

For the dynamic deflections of the springs and tires we use the process of integration. We use values for the initial deflection as the starting point of the integral, and integral of flow (relative velocity) that this element has at a particular time. From the bond graph, build and generated in CAMP/G we got the following equations in ACSL.:

Q3= INTEG (dQ3,Q3IN ) ! Dynamic displacement for the front tire. Q9= INTEG (dQ9,Q9IN ) ! Dynamic displacement of the front spring and absorber. Q26= INTEG (dQ26,Q26IN ) ! Dynamic displacement of the rear spring and absorber. Q31= INTEG (dQ31,Q31IN ) ! Dynamic displacement of the rear tire.

Heave Velocity

We know that angular velocity is the same for all points of the bike body, . Knowing the front and rear distances from the center of gravity, velocity of the front and rear can be related to the angular velocity.

From these two equation we can solve for ; or it can be determined with the flows from the bond graph.

Where:P17 is linear momentum of the McrI17 is mass of the motorcycle and biker.

Pitch angular velocity is defined as:

Where:P15 is angular momentum of the McrI15 is J (Moment of Inertia)

System Frequencies

sr sfk k

crMJ

srk sfk trk

mtr

Heave model Pitch modelWheel model

cgL

To find the time control parameters, system frequencies need to found also.Where:

Therefore:

For the Heave Model

For the Pitch model

For the Front wheel model:

For the Rear wheel model:

Period Finish Time [Period*3] Time Steps [Period/1000]Heave Model 1.35 ~4.05 -Pitch Model 0.8452 - -

Wheel Model 0.1405 - ~1.41E-4

Rider Position and dynamics of the system

If the position of the rider changes, the position of the center of gravity will change as well; this in turn will change our frequencies, initial displacements and dynamic displacements. The pitch model frequency is affected the most, and the wheel models will be slightly affected.

ACSL Code

Exporting the bond model form CAMP/G to ACSL environment and modified the script. By defining physical constraints for the system, giving equations for the initial displacements, flow and effort sources and putting constraints on the tire, force the system to not go into tension.Here is our modified ACSL code:

PROGRAM CAMPACSL !...... ACSL INPUT FILE ......

! PHYSICAL PARAMETERS constant vmoto = 10 ! m/s constant mcr = 300 ! kg constant g = 9.81 ! m/s^2 Acceleration of gravity constant rgyr = 0.5 ! m constant base = 1.6 ! m constant sdisp = 0.1 ! m J= (mcr*(rgyr*rgyr))! Moment of angular inertia.

constant ksf = 3000 ! N/m constant bf = 400 ! Ns/m constant mtf = 15 ! kg constant ktf = 30000! N/m constant lcgf = 0.7 ! m constant ksr = 3500 ! N/m constant br = 500 ! Ns/m constant mtr = 20 ! kg constant ktr = 40000! N/m constant lcgr = 0.9 ! m lcgr=(base - lcgf)

constant lbump = 0.5 ! m constant hbump = 0.35 ! m Have to be determined. constant tsf = 0.50 ! s constant tmf = 0.525! s

constant tef = 0.55 ! s delay = (base/vmoto) ! m Time delay betweent the front and rear wheel. tsr =(tsf+delay) tmr =(tmf+delay) ter =(tef+delay)

INITIAL !...... INITIAL CONDITIONS ...... Q3IN= (((mtf+(mcr*lcgr/base))*g)/ktf) ! Initial displacement of the front tire. P5IN= 0.0 ! Initial momentum of the front tire. Q9IN= ((mcr*lcgr*g)/(base*ksf)) ! Initial displacement of the front spring.

P15IN= 0.0 ! Initial angular momentum of the motorcycle and the rider. P17IN= 0.0 ! Initial momentum of the motorcycle and the rider.

Q26IN=((mcr*g*(1-(lcgr/base)))/ksr) ! Initial displacement of the rear spring. P29IN= 0.0 ! Initial momentum of the rear tire. Q31IN=(((mcr*(1-(lcgr/base))+mtr)*g)/ktr) ! Initial displacement of the rear tire. !..... TIME CONTROL (T=TIME)..... CINTERVAL CINT= 0.00003 CONSTANT FINTIM= 3.0 !...... SYSTEM PHYSICAL PARAMETERS ...... C3 = (1/ktf) ! Spring consnstant of the front tire. I5 = mtf ! Mass of the front tire. C9 = (1/ksf) ! Spring constant for the front spring. R10= bf ! Damping constant for the front absorber. T13x14 = lcgf ! Transformer ratio relating Vf to Omega. I15 = J ! Angular momentum. I17 = mcr ! Mass of the motorcycle and the rider. T19x20 = lcgr ! Transformer ratio realating Vr to Omega. R25 = br ! Damping constant for the rear absorber. C26 = (1/ksr) ! Spring constant for the rear spring. I29 = mtr ! Mass of the rear tire. C31 = (1/ktr) ! Spring constant of the rear tire. END $!OF INITIAL DYNAMIC DERIVATIVE !.... EXTERNAL INPUTS SE=F(T),SF=F(T) .... SF1 = (hbump*vmoto/(0.5*lbump))*(step(tsf)-2*step(tmf)+step(tef)) SE6 = (mtf*g) SE18 =(mcr*g) SE28 =(mtr*g) SF33 =(hbump/0.5*vmoto/lbump)*(step(tsr)-2*step(tmr)+step(ter))

!...... SYSTEM EQUATIONS ...... e1=e2 $ f1=SF1 e2=e3 $ f2=f1 e3= bound(0.0,1.0e10,Q3/C3) ! Constraint for the front wheel to be in compression only. $ f3=f2-f4 e4=e3 $ f4=f5 e5=e4-e6-e7 $ f5=P5/I5 e6=SE6 $ f6=f5 e7=e8 $ f7=f5 e8=e9+e10 $ f8=f7-f11 e9=Q9/C9 $ f9=f8 e10=f10*R10 $ f10=f8 e11=e8 $ f11=f12 e12=e11 $ f12=f13+f16

e13=e12 $ f13=f14/T13x14 e14=e13/T13x14 $ f14=f15 e15=e14-e19 $ f15=P15/I15 e16=e12 $ f16=f17 e17=e16-e18+e21 $ f17=P17/I17 e18=SE18 $ f18=f17 e19=e20*T19x20 $ f19=f15 e20=e22 $ f20=f19*T19x20 e21=e22 $ f21=f17 e22=e23 $ f22=-f20+f21 e23=e24 $ f23=f22 e24=e25+e26 $ f24=-f23+f27 e25=f25*R25 $ f25=f24 e26=Q26/C26 $ f26=f24 e27=e24 $ f27=f29 e28=SE28 $ f28=f29 e29=-e27-e28+e30 $ f29=P29/I29 e30=e31 $ f30=f29 e31= bound(0.0, 10e10, Q31/C31) ! Constrain for the rear tire to be in compression only. $ f31=-f30+f32 e32=e31 $ f32=f33 e33=e32 $ f33=SF33 dQ3=f3 $ dP5=e5 dQ9=f9 $ dP15=e15 dP17=e17 $ dQ26=f26 dP29=e29 $ dQ31=f31

!...... STATE VARIABLES ...... Q3= INTEG (dQ3,Q3IN ) ! Dynamic displacement for the front tire. P5= INTEG (dP5,P5IN ) ! Dynamic momentum of the front tire. Q9= INTEG (dQ9,Q9IN ) ! Dynamic displacement of the front spring and absorber. P15= INTEG (dP15,P15IN ) ! Angular momentum of the rider and motorcycle. P17= INTEG (dP17,P17IN ) ! Heave momentum of the motorcycle and the rider. Q26= INTEG (dQ26,Q26IN ) ! Dynamic displacement of the rear spring and absorber. P29= INTEG (dP29,P29IN ) ! Dynamic momentum of the rear tire. Q31= INTEG (dQ31,Q31IN ) ! Dynamic displacement of the rear tire.

bumphf= integ(SF1, 0.0) bumphr= integ(SF33, 0.0)

!TERMINATE CONDITIONS TERMT (T.GE.FINTIM) END $!OF DERIVATIVE END $!OF DYNAMIC

TERMINAL END $!OF TERMINAL END $!OF PROGRAM

Graphs

Figure 1 – Bump height vs. time. Experience by the front and rear wheel. From this graph we can see that height agrees with the input height of 35cm and we also see that impact occurs at the defined times.

(Blue for front wheel and Green for rear wheel)

Figure 2- Deflection in the front and rear tire. From this graph we can see the initial deflection in each tire which corresponds to the distribution of weight and spring constants of the tires.

Figure 3 – Flow sources.

SF1 for front wheel and SF33 for rear wheel. We can see that both flow sources correspond to the time intervals defined in the program.