COMPONENTES DE UN F1. Opciones

[cibermorcillo]

Bueno, pues a mí me parecía una buena idea. Lástima que los que saben no se hayan animado. Me gustaría que poco a poco este tópic se fuese llenando con muchos datos que se pueden encontrar en otros topics y así "centralizarlo" de alguna manera. Por mi parte voy a hacer una pequeña descripción de 3 componentes de un F1. No es más que una traducción libre de una pagina web pero algo es algo ¿No?.


Deposito de Gasolina:

El combustible está contenido en una recinto muy seguro para prevenir cualquier posible incendio o explosión debido a filtraciones. Se utilizan materiales resistentes al fuego para crear un deposito que no es más que una “bolsa” de caucho especial resistente a posibles pinchazos (y como ya he dicho, al fuego). Esta bolsa se situa entre el motor y el piloto. No hay límite en cuanto a la capacidad de esta bolsa pero normalmente se elige un tamaño que permita una carrera de una sola parada en cada circuito.

Radiador:

Más apropiadamente conocido como intercambiador de calor, se compone de tubos y aletas refrigerantes de un aluminio muy ligero. El monoplaza tiene radiadores de agua y aceite separados que los extraen a ambos del motor, y los enfrían utilizando la corriente de aire entrante debida al movimiento del coche. Esta corriente viene de la toma de aire situada encima del piloto cuya sección de entrada está limitada por la FIA, y también viene de los pontones. Estos fluidos, después de ser enfriados se devuelven de nuevo al motor para controlar su temperatura. Un mal funcionamiento del radiador puede provocar la explosión del motor.

Los Pontones Laterales:

Canalizan el flujo de aire delantero hacia el radiador de una manera suave y controlada. La intención es reducir la cantidad de Drag (o fuerza de arrastre), la cual frena al monoplaza en su desplazamiento. Como cada circuito requiere de diferentes niveles de refrigeración, el flujo que atraviesa los pontones puede ser modificado alterando las salidas traseras (mediante un aumento o reducción de su sección). La parte exterior de los pontones tiene una forma tal que permita llegar al alerón trasero un flujo lo más limpio posible de aire. También podemos ver en ellos unos pequeños alerones que proporcionan un poco más de Downforce (la fuerza que pega al coche al suelo). Estos pequeños alerones se denominan “flick-ups” en ingles, ¿Alguien que me diga como se traducen al español?


Fuente: http://www.renaultf1.com/html/eng/general_...tual_car_3d.htm


Si veo que interesa puedo seguir, aunque mis conocimientos aún no me permiten más que copy-paste .

- visitad www.thehungersite.com -

[Fabio]

This model (adapted from Metz and Maddock, 1986) represents dynamics of the
ground-effects aerodynamics and active suspension of Championship race cars.

As racing cars evolved for speed through streamlining, they became shaped
like wings--air would have farther to travel over the car than under it.
The resulting aerodynamic lift would reduce the weight of the car on the track
thus reducing the friction between the tires and the track surface. As a
result, steering capability was nearly destroyed at high speeds.

In 1969, racing car designer Jim Hall introduced a 'wing' to counteract the
effect. Its success led to the Indianapolis 500 Championship race cars of
today--they incorporate aerodynamic elements like wings near the tires and
venturis along the sides.

The modified aerodynamics increases the tire loads at high speed for better
traction for braking and turning. While the net 'downforce' presses the car to
the road, differences of downforce fore and aft, left and right, can cause the
car to pitch or roll. Also, there is an optimum height which maximizes the
downforce. Below this height, aerodynamic leakage reduces the effect; above it,
the aerodynamic effect decreases.

'[Racing team manager] Larry Curry said, "At Indy, aerodynamics are
critical. A difference in the angle of attack [of the wings] or ride height of
10 to 50 thousandths of an inch is super critical. It can mean the difference
between a lap at 218 mph or 221 mph."'[2]

Active suspension can supply the forces needed to control the heights of
the front and rear ends of the car from the track. Thus it controls the pitch
and ride height to optimize the downforce, ultimately improving the performance
of the car.

Optimal suspension control can also address the problem of road surface and
wind disturbances (gusts), which degrade ride quality and passenger comfort as
speed increases. This would apply not only to racing cars, but passenger road
vehicles and trains as well.

PI Research's Intelligent Information package is an example of how the system
is observed. It connects to the engine Electronic Control Module to collect
and provide access to data on a car's behavior at speed. Available sensors can
monitor engine and chassis information, pitch (angle of attack), ride height,
and vehicle roll.[2]

The tire-wheel-axle assemblies can be modeled as masses on damped springs,
the 'springs' being between the road and the axles. The car chassis is a mass
connected to the axles by the suspension actuators (the spring constant between
these approaches infinity). Forces on the chassis are weight, acting at the
center of gravity, aerodynamic downforce, acting through the center of pressure
and the reaction forces from the suspension. The downforce is modeled as a
negative spring--the farther it is from equilibrium, the greater is the force
away from equilibrium. 'The resulting vehicle model is thus height-unstable.'[1]

For the model, the forward speed of the car is taken to be a standard
200 mph (89.4 m/s). Only motions in the forward-vertical plane are included.

The states are all measured as displacements from equilibrium values:

/ x1 / displacement of front wheel assembly
/ x2 / d(x1)/dt = velocity of front assembly (w.r.t. equilibrium)
/ x3 / vertical displacement of chassis center of gravity (c.g.)
x = / x4 / d(x3)/dt = vertical velocity of chassis c.g.
/ x5 / displacement of rear assembly
/ x6 / d(x5)/dt = velocity of rear assembly
/ x7 / pitch of chassis about c.g.
/ x8 / d(x7)/dt = rate of change of pitch

The inputs are:

u = / Ff / active suspension force on front wheel assembly
/ Fr / active suspension force on rear wheel assembly

The observed outputs are:

y = / y1 / a measure of displacement of the aerodynamic 'spring' f(x3,x7)
/ y2 / pitch (x7)
/ y3 / front end velocity (x2)
/ y4 / rear end velocity (x6)

The state-space representation is (approximately, in MKS units):

/ 0 1 0 0 0 0 0 0 / / 0 0 /
/ -27000 0 0 0 0 0 0 0 / / .044 0 /
/ 0 0 0 1 0 0 0 0 / / 0 0 /
A = / 0 0 83 0 0 0 -15 0 / B = / -.0016 -.0016 /
/ 0 0 0 0 0 1 0 0 / / 0 0 /
/ 0 0 0 0 -27000 0 0 0 / / 0 .044 /
/ 0 0 0 0 0 0 0 1 / / 0 0 /
/ 0 0 -12 0 0 0 2 0 / / -.0019 .0012 /


/ 0 0 1 0 0 0 -.018 0 / / 0 0 /
C = / 0 0 0 0 0 0 1 0 / D = / 0 0 /
/ 0 1 0 0 0 0 0 0 / / 0 0 /
/ 0 0 0 0 0 1 0 0 / / 0 0 /


The equations behind the matrices are:

d(x1)/dt = x2
d(x2)/dt = omegaf * (wf - x1) + Ff / mf
d(x3)/dt = x4
d(x4)/dt = omegaa1 * [x3 - (rhof - lf) * x7] - (Ff + Fr) / m2
d(x5)/dt = x6
d(x6)/dt = omegar * (wr - x5) + Fr / mr
d(x7)/dt = x8
d(x8)/dt = -omegaa2 * (rhof - lf)*[x3 - (rhof - lf) * x7] + [(Fr*lr -Ff*lf)/Im]

where,

kf = 6.13e+5 [N/m] spring constant of front tires
kr = 6.13e+5 [N/m] spring constant of rear tires
ka = 5.25e+4 [N/m] spring constant modeling downforce
mf = 23 [kg] mass of front wheel assembly
mr = 23 [kg] mass of rear wheel assembly
m2 = 635 [kg] chassis mass
Im = 790 [kg m^2] chassis mass moment of inertia in pitch mode about c.g.
rhof = 1.65 [m] distance from center of pressure (c.p.) to front wheels
rhor = 0.78 [m] distance from c.p. to rear wheels
lf = 1.47 [m] distance from c.g. to front wheels
lr = 0.96 [m] distance from c.g. to rear wheels

omegaf = kf/mf; % angular frequency (squared) of front wheel assembly
omegar = kr/mr; % angular frequency (squared) of rear wheel assembly
omegaa1 = ka/m2; % angular frequency (squared) of chassis downforce
omegaa2 = ka/Im; % angular frequency (squared) of chassis pitch motion


(Note: the road-surface disturbance inputs wf and wr are taken to be zero
as if the car were driving on an ideally smooth road. To model a real road
surface, they would be modeled as Gaussian white noise, and there would be a
matrix E in the equation d(x)/dt = A x + B u + E w.)
______

Suppose the car were in a wind tunnel with the air speed equal 200 m.p.h. to
simulate its response at that speed. Since it encounters no bumps or pits,
road surface inputs would be zero.

The motion of the front and rear wheel assemblies are modeled as masses on
springs with external forces supplied by the actuators (for simplicity, the
damping element is left out):
..
mf x1 = - kf x1 + Ff
..
mr x5 = - kr x5 + Fr

The motion of the chassis is modeled as a mass on a 'negative spring' (due to
the aerodynamics) plus the reaction forces from the actuators:
..
m2 x3 = + ka [ x3 - (lp x7) ] - Ff - Fr

where lp = rhof - lf is the distance from c.p. to c.g.

(For the car riding low and level, the displacement of the aerodynamic spring
measured at the c.p. is the same as the displacement of the c.g. If the car
has backward pitch ( x7 < 0 ), for example, the 'spring' located at the c.p. is
stretched an additional - lp sin(x7) > 0 .)

The pitch motion of the chassis around the c.g. is modeled with torques:
..
Im x7 = - ka [ x3 - (lp x7) ] lp + Fr lr - Ff lf

A positive (upward) force at the front end causes the car to tilt backward
because it is applied in front of the c.g.. At the rear end (behind the c.g.), a
positive force causes a positive tilt. Backward tilt from level ( say at
x3 = 0, x7 < 0 ) extends the 'negative spring' and increases the downward force
at the c.p. (behind the c.g.), causing the car to pitch backward farther.
______

For optimization using LQR, the authors chose R to be the identity(2)
matrix and Q to be a scalar qr times the identity(2) matrix. The front
and rear actuators that lead to the R matrix will most likely be of similar
design, they indicate, but the elements of Q would not necessarily be equal.
They chose to minimize

J = Integral( y'Qy + u'Ru )dt

where y is a linear combination of a subset of the states x. Also, the
C matrix they use has only two rows, but the C matrix above has been
augmented with two more rows in order to complement the observability. The
y'Qy term corresponds to x'C'QCx. In this case, C'QC is not diagonal
because there are cross-terms.

{-?- How does using y instead of x affect the results?

The output y is a linear conbination (of a subset) of x states. The cost
associated with the Q term may be affected, and states which are left out
will have no (direct) cost at all.

-?- 'If observability is originally lacking, [the system designer may
change it] by adding additional sensors.' [Brogan p. 376] How does using
[y1 y2 y3 y4] affect the results as compared with just [y1 y2]?

There would be additional elements contributing to the Q cost term. The
relative cost of error would increase with respect to the R (control) cost
for the same qr ratio.

-?- Are the above correct, and are they okay to include?}

Metz and Maddock chose X0 = [ 0 0 0 0 0 0 0.1 0 ] for initial
conditions. These would correspond to the situation when the race car pulls
out from behind another and into 'clear undisturbed air.'

They checked output and corresponding control inputs for values of
qr = 1, 1e3, 1e6, and 1e9. They noted that system performance generally
increases with increasing qr, and indicate that a satisfactory range is
(1e6,1e9).



Four of the eigenvalues remained close to epsilon +/- 164i, where the
epsilons are small numbers O(1e-10), and two stayed near -9.
_________________________

Following is a sample approach to study the racecar system:

In Matlab, type:

Load the racecar.m script >> racecar

(It sets up all the matrices first, so you may exit at the menu if you like.)
(On the Sun workstations, you can run Matlab in one window and an editor
(like emacs) in another. You can copy and paste these commands as follows:
highlight the text from the editor with the left mouse button, move the mouse
into the Matlab window, press the middle button, and press the return key.)

This script contains a state-space model for a Championship racing car with
active suspension. [Metz, Maddock]

x1 = front end displacement from equilibrium x2 = d(x1)/dt
x3 = center of gravity displacement from eq. x4 = d(x3)/dt
x5 = rear end displacement from equilibrium x6 = d(x5)/dt
x7 = pitch of chassis about center of gravity x8 = d(x7)/dt

The control inputs are the additional forces necessary to raise or lower the
front and rear ends of the chassis from their equilibrium positions.

You can transform the A >> [ v, el ] = eig(A)
matrix into canonical form >> vi = inv(v)
with the modal matrix: >> J = vi*A*v

The eigenvalues are >> s = eig(A)
the same for both >> s1 = eig(J)


The eigenvalues determine the stability of the system. [Brogan p. 348]
Continuous time systems are stable when Re[s] < 0. They are stable i.s.L.
(in the sense of Lyapunov) when stable and when Re[s] = 0 for distinct roots.
They are unstable when Re[s] > 0 or when Re[s] = 0 for repeated roots.
______


To check the controllability, >> u = ctrb( A, B )
>> rank(u)

If rank(u) is equal to n, where A is n x n, it is controllable.

You may notice a difference in the eigenvalues for A and J. Also, it may be
the case that the system really is controllable, but this procedure shows it
as uncontrollable. These both are numerical artifacts of the calculation
routines. Try these:

Ideally, this would be the >> Iv=(vi*v)
identity matrix. >> eig(Iv)

Suppose: A and B are: with controllability matrix:

/ 0 10^q 0 / / 0 / / 0 0 10^2q /
A = / 0 0 10^q / B = / 0 / u = [ B, A*B, A*A*B ] = / 0 10^q 10^q /
/ 1 1 1 / / 1 / / 1 1 10^q /

If entries of A are very large in relation to the entries of b, the first column
is negligible in comparison to the remaining columns. This may cause the rank
to be calculated too low. Notice how raising A to higher powers amplifies the
differences. Try this with q = 7 and q = 8.

>> q = 7
(use A1,B1 to avoid overwriting >> A1 = [ 0 10^q 0; 0 0 10^q; 1 1 1]
A,B from the model) >> B1 = [ 0; 0; 1 ]
>> u = [ B1, A1*B1, A1*A1*B1 ]
>> rank(u)

To avoid being led to a possibly incorrect conclusion, use:

normalize A >> u=ctrb(A/norm(A),
>> rank(u)

(You may want to use B/norm( if B is very small.) If rank(u) is still less
than n, then the system is not controllable.

The racecar model has large entries in the A matrix and requires this treatment.

If the system is not controllable, it may still be stabilizable. [Manitius,
Moulden 7.2] If, for all the eigenvalues s such that Re( s[i] ) >= 0,
the rank( [ eye(n)*s[i] - A, B ] ) = n, then the system is stabilizable.
If this is not true for at least one eigenvalue, the system is not stabilizable.

Check the observability >> u = obsv( A/norm(A), C )
the same way: >> rank(u)


By duality, if rank( [eye(n)*s[i] - A', C' ] ) = n for the eigenvalues as above,
then the state corresponding to eigenvalue s[i] is detectable.
_______


To plot the open-loop step response,

set a vector of time instances, >> t = 0 : 0.01 : 1.49;
a scaled step function for first >> u1 = 1.0e-1 * ones(size(t));
and second inputs, >> u2 = u1;
and an initial state. >> X0 = [ 0;0;0;0; 0;0;0.1;0 ]
>> u = [u1; u2]'; % {transposed--time index}
Run the simulation. >> [ y, x ] = lsim( A, B, C, D, u, t, 0*X0 );
Plot the resulting states >> figure,subplot(2,2,1),plot(t,x,'r')
and outputs. >> subplot(2,2,3),plot(t,y,'b')
Show the control used as input. >> subplot(2,2,2),plot(t,u,'w.')

Use title, xlabel, ylabel, grid, gtext, or text to document the plots.
_______


To design the state feedback, start with the original eigenvalues. Change
the unstable roots.

original open-loop eigenvalues >> kap = eig(A)

change sign of the third root >> kap(3) = - kap(3)
set first and second >> kap(1:2) = -[1+1i,1-1i]
find the feedback matrix >> k = place( A, B, kap )

The place command will not accept values repeated more times than there are
inputs--since B is n x 2, you can have a value appear twice (n x m, m times).
Sometimes the eigenvalues cannot be moved from (-?-) or to certain locations.
The eigenvalues tend to move in an ellipse in this model.

-?- Why? because of the properties of the simple harmonic oscillator?

form the closed-loop A matrix >> Ac = A - B * k
You can check >> kap = eig(Ac)
and chart the roots. >> pzmap(Ac,B,C,D)
Choose a non-zero X0. >> X0 =[.01;0;-.002;0; -.01;0;-.008;0]

The meaning of this X0 is that the front end of the car is 1 cm higher than
equilibrium, the rear end 1 cm lower than its equilibrium. Since the distances
from the center of gravity to the front and rear axles respectively are 1.47 m
and .96 m, the center of gravity is .2 cm lower than normal and the pitch is
.008 radians backward.

Run the feedback system. >> [yf,xf]=lsim(Ac,B,C,D,u,t,X0);
Plot the results. >> figure, plot(t,xf,'r'),ylabel('states')

'Zoom in' on a segment >> sr = 1:25;
of the plot. >> figure, plot(t(sr),xf(sr,,'r')

You can also sample the states >> sr = 1:5:length(t);
at regular intervals. >> [t(sr)',xf(sr,1),xf(sr,2)]
>> figure, plot(t(sr),xf(sr,,'r-')

It may be more meaningful to look at a subset of the states--x1, x3, x5, x7
are the states that must be kept small.

>> clf
>> subplot(2,2,1),plot(t,xf(:,1),t,xf(:,3),t,xf(:,5)),ylabel('meters')
>> xlabel('seconds'),title('displacements')
>> subplot(2,2,3),plot(t,xf(:,7)),ylabel('radians'),title('pitch')

Note how the car's front and rear ends oscillate a little and then settle
down. The ride height goes to a steady value at about 1 cm lower than
equilibrium--aerodynamics in action!

This looks pretty good until you look at the pitch--does it make sense that
the ends of the car stay in the range (-.01, +.01) m and the pitch can reach
such a large value? No.

{ -?- How can we explain this?
"the location of the center of pressure is a strong and highly non-linear
function of the pitch angle." Since the c.p. is used in the matrices, because
it is variable, the matrices vary also. That is not accounted for in this
sample solution. }

@@ proofreading marker @@
_______


It is generally better to have the estimator's eigenvalues farther in the
left-half complex plane than the system being followed, so start from the
feedback eigenvalues.
>> lam = 1.5 * kap
The process is similar (under >> l = place( A', C', lam )'
transposition) due to duality. >> Ae = A - l * C

as above, >> [e,z] = lsim(A-L*C,[B L],C,[D D],[u y], t, 0*X0 );

you may plot the step response >> figure
and the estimation both >> subplot(2,2,1),plot(t,x,'r',t,z,'r:')
>> subplot(2,2,3),plot(t,y,'b')
>> hold,plot(t,e,'b:'),hold off
______


To combine the feedback and estimator into one big system,
(following Notes section 6.2)

Form matrix of matrices >> AA = [ A, -B*k; l*C, A-B*k-l*C ]
>> BB = [ B; B ]
{__ check } >> CC = [ C, zeros( size(C,1),size(B,2) ) ]
{__ check } >> DD = zeros( size(C,1),size(B,2) )
composite state vector >> XZ = [X0; 0*X0]
>> [ZZ,XX]=lsim(AA,BB,CC,DD,u,t,XZ);

The composition imposes simultaneity of the feedback and estimation and
produces better estimation results.

>> subplot(2,2,1)
>> plot(t,XX(:,1),'r', t,XX(:,3),'g', t,XX(:,5),'b')
>> title('displacements),xlabel('sec'),ylabel('m'),grid
>> subplot(2,2,3)
>> plot(t,XX(:,7),'g') % pitch
>> title('pitch'),xlabel('sec'),ylabel('rad'),grid
>> disp('here are the observer-states')
>> subplot(2,2,2)
>> plot(t,XX(:, 7),'m', t,XX(:,11),'y', t,XX(:,13),'c')
>> title('est. disp. '),xlabel('sec'),ylabel('m'),grid
>> subplot(2,2,4)
>> plot(t,XX(:,15),'y') % pitch
>> title('est. pitch'),xlabel('sec'),ylabel('rad'),grid
_______


You can also find state-feedback by using the LQR (Linear Quadratic Regulator)
function in Matlab. Since the objective is to minimize

J = Integral( x' Q x + u' R u )dt

the relative magnitudes of Q and R are more important than their absolute
magnitudes. Consider if Q and R were positive scalar multiples of identity
matrices (e.g. Q = q I(n), R = r I(m) ), then

J/r = Integral( x' (q/r) I(n) x + u' I(m) u )dt


This is the starting point for LQR design. Metz and Maddock point out that
this qr ratio 'reflects the compromise between control energy and achievement
of the control objectives.' Equal weight among the elements of the Q matrix
or the R matrix does not distinguish the importance of one element of x or
of u with respect to another element of the same vector--some elements may
dominate, and others may be negligible. Closer examination of physical system
or experimentation may be necessary to assign better relative values.


>> qr = 1e6
>> Q = qr * eye(8)
>> R = eye(2)
>> [K,S,E1]=lqr(A,B,Q,R)
>> k=K
>> Aq = A - B * k
>> [yf,xf] = lsim(Aq,B,C,0*D,u,t,X0);
>> figure,subplot(2,2,1)
>> plot(t,xf(:,1),'r', t,xf(:,3),'g', t,xf(:,5),'b')
>> title('disp. (LQR feedback)'),xlabel('seconds'),ylabel('m'),grid
>> subplot(2,2,3)
>> plot(t,xf(:,7),'g')
>> title('pitch (LQR feedback)'),xlabel('seconds'),ylabel('rad'),grid
>> uf=-k*xf'; % per definition of feedback ' due to time index
>> subplot(2,2,2)
>> plot(t,uf),title('feedback control'),xlabel('seconds'),ylabel('newtons')



[1] Metz, D. and J. Maddock (1986). "Optimal Ride Height and Pitch Control for
Championship Race Cars." Automatica 22 5, pp. 509-520.

[2] Flax, Arthur. "Indy Car Technology." The Men and Machines of Indy Car
Racing, CART 1991-92. N.Y.: Autosport International, Inc., (1992) pp. 138-141.

[3] NOVA "Fast Cars" broadcast 5/27/95

CHUPATE ESA!!!