求解轨道力学二体意义下的Lambert方程（兰伯特方程）的Matlab程序

轨道力学中二体问题下求解兰伯特方程需要解决一个迭代问题。

这是一个老外写的，有很多注释，相信大家应该能看懂，经实际检测，切实可用

function [v1,v2]=solve_lambert(r1,r2,t,GM,lw,N,branch)

%This routine implements a new algorithm that solves Lambert‘s problem. The

%algorithm has two major characteristics that makes it favorable to other

%existing ones.

%

%   1) It describes the generic orbit solution of the boundary condition

%   problem through the variable X=log(1+cos(alpha/2)). By doing so the

%   graphs of the time of flight become defined in the entire real axis and

%   resembles a straight line. Convergence is granted within few iterations

%   for all the possible geometries (except, of course, when the transfer

%   angle is zero). When multiple revolutions are considered the variable is

%   X=tan(cos(alpha/2)*pi/2).

%

%   2) Once the orbit has been determined in the plane, this routine

%   evaluates the velocity vectors at the two points in a way that is not

%   singular for the transfer angle approaching to pi (Lagrange coefficient

%   based methods are numerically not well suited for this purpose).

%

%   As a result Lambert‘s problem is solved (with multiple revolutions

%   being accounted for) with the same computational effort for all

%   possible geometries. The case of near 180 transfers is also solved

%   efficiently.

%

%   We note here that even when the transfer angle is exactly equal to pi

%   the algorithm does solve the problem in the plane (it finds X), but it

%   is not able to evaluate the plane in which the orbit lies. A solution

%   to this would be to provide the direction of the plane containing the

%   transfer orbit from outside. This has not been implemented in this

%   routine since such a direction would depend on which application the

%   transfer is going to be used in.

%

%Usage: [v1,v2,a,p,theta,iter]=solve_lambert(r1,r2,t,GM,lw,N,branch)

%

%Inputs:

%           r1=Position vector at departure (column,km)

%           r2=Position vector at arrival (column, same units as r1,km)

%           t=Transfer time (scalar,s)

%           GM=gravitational parameter (scalar, units have to be

%           consistent with r1,t units,km^3/s^2)

%           lw=1 if long way is chosen

%           branch=1 if the left branch is selected in a problem where N

%           is not 0 (multirevolution)

%           注:天体运行是右分支.所以branch一般选择0

%           N=number of revolutions

%

%           说明:当N~=0时,旋转方向不光用lw来控制还要先用branch来控制.

%Outputs:

%           v1=Velocity at departure        (consistent units)(km/s)

%           v2=Velocity at arrival              (km/s)

%           iter=number of iteration made by the newton solver (usually 6)

%

%    当需要时可以加上.

%补充说明:

%                [v1,v2,a,p,theta,iter]=lambertI(r1,r2,t,GM,lw,N,branch);
%branch=1;%here 1 is represent left

%Preliminary control on the function call
if t<=0

    disp(‘Negative time as input‘);

    v1=NaN;

    v2=NaN;

    return

end
tol=1e-11;  %Increasing the tolerance does not bring any advantage as the

%precision is usually greater anyway (due to the rectification of the tof

%graph) except near particular cases such as parabolas in which cases a

%lower precision allow for usual convergence.
%Non dimensional units

R=sqrt(r1‘*r1);

V=sqrt(GM/R);

T=R/V;
%working with non-dimensional radii and time-of-flight

r1=r1/R;

r2=r2/R;

t=t/T;
%Evaluation of the relevant geometry parameters in non dimensional units

r2mod=sqrt(r2‘*r2);

theta=real(acos((r1‘*r2)/r2mod)); %the real command is useful when theta is very

%close to pi and the acos function could return complex numbers
%计算夹角,并确定是大弧还是小弧.

if lw

    theta=2*pi-theta;

end

c=sqrt(1+r2mod^2-2*r2mod*cos(theta)); %non dimensional chord

s=(1+r2mod+c)/2;                      %non dimensional semi-perimeter

am=s/2;                               %minimum energy ellipse semi major axis

lambda=sqrt(r2mod)*cos(theta/2)/s;    %lambda parameter defined in BATTIN‘s book
%We start finding the log(x+1) value of the solution conic:

%%NO MULTI REV --> (1 SOL)

if N==0

    inn1=-.5233;    %first guess point

    inn2=.5233;     %second guess point

    x1=log(1+inn1);

    x2=log(1+inn2);

    y1=log(x2tof(inn1,s,c,lw,N))-log(t);

    y2=log(x2tof(inn2,s,c,lw,N))-log(t);
%Newton iterations

    err=1;

    i=0;

    while (err>tol) && (y1~=y2)

        i=i+1;

        xnew=(x1*y2-y1*x2)/(y2-y1);

        ynew=log(x2tof(exp(xnew)-1,s,c,lw,N))-log(t);

        x1=x2;

        y1=y2;

        x2=xnew;

        y2=ynew;

        err=abs(x1-xnew);

    end

    x=exp(xnew)-1;
%%MULTI REV --> (2 SOL) SEPARATING RIGHT AND LEFT BRANCH

else

    if branch==1

        inn1=-.5234;

        inn2=-.2234;

    else

        inn1=0.2;

        inn2=.5234;

    end

    x1=tan(inn1*pi/2);

    x2=tan(inn2*pi/2);

    y1=x2tof(inn1,s,c,lw,N)-t;
y2=x2tof(inn2,s,c,lw,N)-t;

    err=1;

    i=0;
%Newton Iteration

    while ((err>tol) && (i<90) && (y1~=y2))

        i=i+1;

        xnew=(x1*y2-y1*x2)/(y2-y1);

        ynew=x2tof(atan(xnew)*2/pi,s,c,lw,N)-t;

        x1=x2;

        y1=y2;

        x2=xnew;

        y2=ynew;

        err=abs(x1-xnew);

    end

    x=atan(xnew)*2/pi;

end
%The solution has been evaluated in terms of log(x+1) or tan(x*pi/2), we

%now need the conic. As for transfer angles near to pi the lagrange

%coefficient technique goes singular (dg approaches a zero/zero that is

%numerically bad) we here use a different technique for those cases. When

%the transfer angle is exactly equal to pi, then the ih unit vector is not

%determined. The remaining equations, though, are still valid.
a=am/(1-x^2);                       %solution semimajor axis

%calcolo psi

if x<1 %ellisse

    beta=2*asin(sqrt((s-c)/2/a));

    if lw

        beta=-beta;

    end

    alfa=2*acos(x);

    psi=(alfa-beta)/2;

    eta2=2*a*sin(psi)^2/s;

    eta=sqrt(eta2);

else %iperbole

    beta=2*asinh(sqrt((c-s)/2/a));

    if lw

        beta=-beta;

    end

    alfa=2*acosh(x);

    psi=(alfa-beta)/2;

    eta2=-2*a*sinh(psi)^2/s;

    eta=sqrt(eta2);

end

p=r2mod/am/eta2*sin(theta/2)^2;     %parameter of the solution

sigma1=1/eta/sqrt(am)*(2*lambda*am-(lambda+x*eta));

ih=cross(r1,r2)/norm(cross(r1,r2));

if lw

    ih=-ih;

end
vr1 = sigma1;

vt1 = sqrt(p);

v1  = vr1 * r1   +   vt1 * cross(ih,r1);
vt2=vt1/r2mod;

vr2=-vr1+(vt1-vt2)/tan(theta/2);

v2=vr2*r2/r2mod+vt2*cross(ih,r2/r2mod);
v1=v1*V;

v2=v2*V;

if err>tol

    v1=[100 100 100]‘;

    v2=[100 100 100]‘;

end
function t=x2tof(x,s,c,lw,N)

%Subfunction that evaluates the time of flight as a function of x

am=s/2;

a=am/(1-x^2);

if x<1 %ELLISSE

    beta=2*asin(sqrt((s-c)/2/a));

    if lw

        beta=-beta;

    end

    alfa=2*acos(x);

else   %IPERBOLE

    alfa=2*acosh(x);

    beta=2*asinh(sqrt((s-c)/(-2*a)));

    if lw

        beta=-beta;

    end

end

t=tofabn(a,alfa,beta,N);
function t=tofabn(sigma,alfa,beta,N)

%subfunction that evaluates the time of flight via Lagrange expression

if sigma>0

    t=sigma*sqrt(sigma)*((alfa-sin(alfa))-(beta-sin(beta))+N*2*pi);

else

    t=-sigma*sqrt(-sigma)*((sinh(alfa)-alfa)-(sinh(beta)-beta));

end

求解轨道力学二体意义下的Lambert方程（兰伯特方程）的Matlab程序,布布扣,bubuko.com