Landing gear analysis Kinematics and Dynamics -...

Bugatti 100p – Reve Bleu

Landing gear analysis

Kinematics and Dynamics

v0r2 – (preliminary version)

João Paulo PintoEmail: [email protected]

Skype: pintovski

LinkedIN: https://www.linkedin.com/pub/jo%C3%A3o-pinto/19/918/a77

Facebook: https://www.facebook.com/joao.pinto.101

Contents0. Disclaimer................................................................................................................................................................31. Introduction............................................................................................................................................................42. Simulation............................................................................................................................................................... 5

2.2 Kinematics.......................................................................................................................................................52.3 Dynamics....................................................................................................................................................... 10

2.3.1 Acceleration.........................................................................................................................................102.3.2 Forces....................................................................................................................................................11

3. Results...................................................................................................................................................................153.1 Input parameters.........................................................................................................................................153.2 Kinematic results..........................................................................................................................................163.3 Closing cycle with a constant linear actuator velocity of 25mm/s.......................................................183.4 Closing cycle with a constant linear actuator velocity of different velocities....................................20

4 Conclusions............................................................................................................................................................21

Bugatti 100p Landing Gear Analysis Pág. 1 de 21

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0. Disclaimer

The author has performed the analysis voluntarily and the information is available absolutely free of charge. By viewing the content and information of this report, you understand that the information has been compiled to the very best of the author knowledge, but may not be free of errors. The author cannot be held accountable for misrepresentation or inaccurate information presented.

Under no circumstance will the author accept responsibility for any direct or indirect damage that may arise from the use of information presented on this report. If you are not satisfied with the information presented, it is your right and responsibility to stop using it.


1. Introduction

The Bugatti 100p was a remarkable airplane designed to set a world speed record and to compete in the prestigious Coupe Deutsch air race. The airplane was never finished because of World War II and the German invasion of France.

Nowadays, the REVE BLUE team is reconstructing the Bugatti 100p with the intention of flying the airplane. This project is at an advanced stage. One of the tasks at hand is the dimensioning of a linear actuator that retracts/extends the landing gear in flight.

In this report we will present a method for the computation of the kinematics and dynamics of this landing gear retraction system. We will use this method to study the landing gear retraction cycle.

fig. 1-1 – Original drawing of the landing gear retraction system.(Numbers inside brackets denote weight in lbs)

fig. 1-2 – Sketch of the landing gear retraction system.


2. SimulationThe computation of the forces that are present during the retraction cycle requiresthe knowledge of the system geometry. Therefore, the simulation will need to compute the system instantaneous geometry (kinematics) from which it is able to extract the instantaneous velocity and acceleration of each component. This information will then be used to compute each instantaneous force (Dynamics).

KinematicsCompute geometry of system

Compute velocity and accelerationbased on variation in position of each

component

DynamicsCompute Forces and Moments

Solve non-linear system of equations[A] x = b

Solve linear system of equations[A] x = b

Initialization procedureSet initial state, geometry data, ...

Ad

vanc

e in

tim

e: t

n = t

n-1

+ d

to

r ad

vanc

e in

sp

ace:

θn =

θn-

1 +

dθ

fig. 2-1 – Fluxogram identifying the main simulation blocks.

2.2 Kinematics

The kinematics module computes the position and orientation of each system component based on a given parameter. In this simulation we chose to use the variable β , which represents the angle of the main leg of the landing gear with respect to the vertical axis.

Figure 2-2 shows the identification of each component and joint.

Figure 2-3 shows the angles and other dimensions used in the Kinematics formulation.

It is important to note that the angle of the main leg of the landing gear, β , is not the same as θ1as depicted in the following figures. This difference is because point B is offset from the center of the main gear leg by 1in.


Fig. 2-2 – Identification of joints, control points and components.

Fig. 2-3 – Identification of angles and dimensions used in the kinematic analysis.


The following equations compute the position and orientation of each component using the nomenclature presented in figures 2.2 and 2.3.

A x=0mm eq. 1

A y=0mm eq. 2

−Ax+Bx=l1 sinθ1 eq. 3

Ay−B y=l 1cos θ1 eq. 4

D x−C x=l 3cos θ3 eq. 5

−By+C y=l 2 cosθ2 eq. 6

D x=360mm eq. 7

D y=0mm eq. 8

−D x+E x=l 4 cosθ4 eq. 9

−D y+E y=l 4 sinθ4 eq. 10

−Ax+F x=l5 cosθ5 eq. 11

−A y+F y=l5 sinθ5 eq. 12

θ1=variable eq. 13

θ2=sin−1(−B x+C x

l 2 ) eq. 14

θ3=sin−1( D y−C y

l3 ) eq. 15

−θ3+θ4=∢EDC=90º eq. 16

−θ1+θ5=∢ FAB−∢ FAD=25º eq. 17

The Left Hand Side (LHS) of equations 1 to 17 form a linear system of equations that can be solved using conventional mathematical tools. The non-linear terms, that are on the Right Hand Side (RHS) of equations 1 to 17, can be computed iteratively. It was found that this procedure converge in few iterations.

For completeness, we will mention that in the implementation performed, equations 1 to 17 were put in the form

[A] x=b eq. 18


where the matrix [A] is built with the coefficients of the LHS:

[A]=[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

−1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 00 1 0 −1 0 0 0 0 0 0 0 0 0 0 0 0 00 0 0 0 −1 0 1 0 0 0 0 0 0 0 0 0 00 0 0 −1 0 1 0 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 00 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 00 0 0 0 0 0 −1 0 1 0 0 0 0 0 0 0 00 0 0 0 0 0 0 −1 0 1 0 0 0 0 0 0 0

−1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 00 −1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 1 00 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 1

] eq. 19

the x vector contains our solutions:

x=[ Ax A y Bx By C x C y D x D y E x E y F x F y θ1 θ2 θ3 θ4 θ5 ]T

eq. 20

and the b vector contains the non-linear terms of equations 1 to 17 taken from the RHS of each equation.


b=[Ax( parameter)A y( parameter)

l 1sinθ1l1 cosθ1l3 cosθ3l 2 cosθ2

D x ( parameter)D y ( parameter)

l 4 cosθ4l 4 sinθ4l5 cosθ5l 5sinθ5

θ1( parameter)

sin−1(−Bx+C x

l 2)

sin−1(−C y+D y

l 3)

∢ EDC∢FAB−90º

] eq. 21

We used an iterative numerical procedure where the b vector is treated as constant in each iteration so that we only need to solve a linear system of equations. The computed solution with this approximation is used to recompute a new b vector in the following iteration. This procedure is repeated until convergence.


2.3 Dynamics

The dynamics module computes the forces acting on each component and joint based on the system layout computed in the kinematics module and each individual component acceleration.

2.3.1 Acceleration

The linear and angular acceleration of each component is computed based on the second derivative of the part position and rotation, respectively. The following relations are valid for motion in the XY plane and rotation in a single direction perpendicular to the XY plane.

The time interval used in the computation of the acceleration is determined based on the linear actuator velocity (assumed to be an input parameter), and the instantaneous length variation of the line segment that connects its attachment points, EF .

dt=dL linearactuator

∣v linear actuator∣=

√(E xt+dt

−F xt+dt

)2+(E y

t+dt−F y

t+dt)2−√(E x

t−F x

t)2+(E y

t−F y

t)2

∣v linear actuator∣eq. 22

The velocity and acceleration of the center of gravity of each i component/part is computed using equations 23 to 25.

d xCG ,it+dt

= xCG ,it+dt

− xCG , it eq. 23

v CG ,it+dt

=ddtxCG ,it+dt

eq. 24

aCG ,it+dt

=d xCG , i

t+dt−d vCG ,i

t dt12dt 2

eq. 25

The angular velocity and acceleration of the center of gravity of each i component/part is computed using equations 26 to 28.

d θCG ,it+dt

=θCG ,it+dt

−θCG , it eq. 26

ωCG ,it+dt

=ddt

θCG ,it+dt

eq. 27

αCG ,it+dt

=d θCG , i

t+dt−dωCG ,i

t dt12dt2

eq. 28


2.3.2 Forces

The force at each joint is computed using equations 29-41. These equations are constructed based on an analysis of each individual part identified in figure2-2.

Equations for Part 1:

Fig. 2-4 – Identification of the forces used in the dynamics equations for part 1.

∑ F x=F xF+R x

A+F x

B=m1a1, x eq. 29

∑ F y=F yF+Ry

A+F y

B=m1a1, y+P1 eq. 30

∑M zA=−(X y

F−X y

A)F x

F+(X x

F−X x

A)F y

F−(X y

B−X y

A)F x

B+(X x

B−X x

A)F y

B=I 1α1+(X x

CG1−X xA)P1

eq. 31

For clarity and disambiguation, the coordinates of each point are denoted with the letter X . The letter F denotes force and R denotes support reaction force. The subscript denotes Cartesian direction and the superscript denotes the target point/location of interest.




∑ F x=F xC−F x

B=m2a2, x eq. 29

∑ F y=−F yB+F y

C=m2a2, y+P2 eq. 30

∑M zB=(X x

C−X x

B)F y

C−(X y

C−X y

B)F x

C= I 2α2+(X x

CG2−X xB)P2 eq. 31




∑ F x=−F xC+Rx

D+F x

E=m3a3, x eq. 32

∑ F y=−F yC+R y

D+F y

E=m3a3, y+P3 eq. 33

∑M zD=(X y

C−X y

D)F x

C−(X x

C−X x

D)F y

C−(X y

E−X y

D)F x

E+(X x

E−X x

D)F y

E=I 3α3+(X x

CG3−X xD)P3

eq. 34





∑ F x=−F xF−F x

E=m4a4, x eq. 35

∑ F y=−F yF−F y

E=m4a4, y+P4 eq. 36

∑M zE=(X y

F−X y

E)F x

F−(X x

F−X x

E)F y

F=I 4α4+(X x

CG4−X xE)P4 eq. 37


Numerics

The system formed by equations 29 to 37 form a linear system of equations that can be solved using the same procedure used in the kinematics analysis. Since there is are no non-linear terms, there is no need for an iterative procedure.

For completeness, and in a similar fashion with what we did for the kinematics analysis, we will mention that in the implementation performed, equations 29 to 37 were put in the form

[A] x=b eq. 38

where the matrix [A] is built with the coefficients of the LHS of equations 29 to 37:


[A]=[1 0 1 0 0 0 0 0 0 0 1 00 1 0 1 0 0 0 0 0 0 0 10 0 −(X y

B−X y

A) (X x

B−X x

A) 0 0 0 0 0 0 −(X y

F−X y

A) (X x

F−X x

A)

0 0 −1 0 1 0 0 0 0 0 0 00 0 0 −1 0 1 0 0 0 0 0 00 0 0 0 −(X y

C−X y

B) (X x

C−X x

B) 0 0 0 0 0 0

0 0 0 0 −1 0 1 0 1 0 0 00 0 0 0 0 −1 0 1 0 1 0 00 0 0 0 (X y

C−X y

D) −(X x

C−X x

D) 0 0 −(X y

E−X y

D) (X x

E− X x

D) 0 0

0 0 0 0 0 0 0 0 −1 0 −1 00 0 0 0 0 0 0 0 0 −1 0 −10 0 0 0 0 0 0 0 0 0 (X y

F−X y

E) −(X x

F−X x

E)

]eq. 39

the x vector contains each component of the force vectors, which are the required solutions:

x=[ RxA Ry

A F xB F y

B F xC F y

C R xD R y

D F xE F y

E F xF F y

F ]T

eq. 40

and the b vector contains the RHS of equations 29 to 37.

b=[m1a1, x

m1a1, y+P1I 1α1+(X x

CG1−X xA)P1

m2a2, xm2a2, y+P2

I 2α2+(X xCG2−X x

B)P2

m3a3, xm3a3, y+P3


D)P3

m4a4, xm4a4, y+P4


E )P4

] eq. 41


3. ResultsThe procedure presented in chapter 2 was implemented in a FORTRAN program. Please contact the author if you want a copy of this program ([email protected]).

3.1 Input parameters

The geometrical definition of the landing gear that is relevant for the analysis is:

A x=X xA=0

A y=X yA=0

D x=X xD=360mm

D y=X yD=0

l 1=230mml 2=167.64mml 3=284.48mml 4=35mml 5=44mm

eq. 42

The angles are defined as:

θ1=β1−tan−1(25.4mm230mm )

θ4=∢EDC=90º+θ3θ5=25º+θ1

eq. 43

The mass properties are:

mwheel=6.8kgm fork=7.72kgm strut=11.34 kgm1=mwheel+m fork+mstrut

m2=0.816kgm3=1.497 kgm4=0

eq. 44

The center of gravity is located at the center of each part. The only exception is part 1. For this part the CG is estimated considering that each mass (wheel, fork and strut) is located at the longitudinal axis of part 1, at a distance of -970.28, -726.44 and -241.3 mm from point A, respectively.

The moment of inertia is estimated considering that part 1 is rotating at point A, and that the mass is distributed in three cylinders corresponding to the wheel, fork and strut.


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I wheel=112

mwheel (2×457.2mm/1000)2+mwheel×(−907.28mm/1000)2 eq. 45

I fork=112

m fork(2×487.68mm/1000)2+m fork×(−726.44mm /1000)2 eq. 46

I strut=13mstrut(2×−241.3mm/1000)2 eq. 47

I 1=I wheel+ I fork+I strut [ kg.m2] eq. 48

In order to avoid computing the center of rotation of part 2, it is estimated that it is a rod, rotating at its CG.

I 2=112

m2( l2 /1000)2[ kg.m2

] eq. 49

Part 3, is also treated as a rod, and revolves around point D.

I 3=13m3(l 3/1000)

2[kg.m2

] eq. 50

No inertia effects were considered for part 4.

I 4=0 eq. 51

3.2 Kinematic results

The following plots show the trajectory of each point of interest and CG in the opening or closing cycles as well as the orientation of each part as defined in figure 2-2.

Fig. 3.2-1 – Trajectory of each control point and CG location during the landing gear opening/closing cycles.


Fig. 3.2-2 – Closeup view of the trajectory of each control point and CG location during the landing gear opening/closing cycles.

Fig. 3.2-3 – Orientation angle of each component during the landing gear opening/closing cycles.

Fig. 3.2-4 – Closeup of the orientation angle of each component during the landing gear opening/closing cycles.


3.3 Closing cycle with a constant linear actuator velocity of 25mm/s

The following plots show the linear and angular velocities of each part in the retraction cycle, considering that the linear actuator has a constant velocity of 25mm/s.

Fig. 3.3-1 – Velocity Cartesian components at the CG location for each part.

Fig. 3.3-2 – Angular velocity for each part.


Fig. 3.3-3 – Acceleration Cartesian components at the CG location for each part.

Fig. 3.3-4 – Angular acceleration for each part.

Fig. 3.3-5 – Force Cartesian components at each control point.


Fig. 3.3-6 – Linear actuator required force at a constant linear actuator velocity of 25 mm/s. Results from References A and B were taken from the work of Frank van Dalen and Michael C Siegle using an Energy and FBD method.

3.4 Closing cycle with a constant linear actuator velocity of different velocities

The following plot shows the force required by the linear actuator at different working velocities.

Fig. 3.4-1 – Linear actuator required force at different constant linear actuator velocities. Results from References A and B were taken from the work of Frank van Dalen and Michael C Siegle using an Energy and FBD method.

It is interesting to notice that by increasing the linear actuator velocity the force required close to β1=0 is greater. However, the maximum force required in the cycle is slightly lower. By increasing

the linear actuator velocity from close to zero, up to 50 mm/s, a force reduction of 4% can be achieved. The following plot shows the force reduction dependence on linear actuator velocity.


Fig. 3.4-2 – Linear actuator required force reduction dependence on linear actuator velocity.

It should be noted that in the event of the landing gear starts moving from an intermediate position, the required force presented may not be enough to close the landing gear. Even though it can be introduced at a latter version of the model, the model presented does not reproduce the acceleration from a standstill position.

4 ConclusionsIn this document a model for the mathematical and physical representation of the Bugatti 100p landing gear was presented. This model accounts for system inertia due to the motion of each component of the multi-body layout. However, it cannot be used to reproduce the system behavior from a standstill position.

The following topics should be taken into consideration:

1. The linear actuator is being modeled as working at constant velocity, in reality it may not be the case since it can work on constant force or other variable;

2. It is important to note that the position and orientation of the actuator, and force vector, varies with respect to landing gear angle;

3. The impact forces at the end of the cycles due to an abrupt acceleration are not being estimated;

4. The velocity reduces close to the extreme positions, which reduces the stresses at the end of the cycle;

5. The presented analysis is only valid for the retraction cycle;

6. The present model cannot reproduce the system behavior from a standstill position.


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