What exactly means universal variable x and z?

Question

I have been studying by myself Orbital Mechanics, and when solving the Lambert Problem, it's common to use the universal variable approach. I understand the algorithm, but I haven't found any book that explains well the physical meaning of the universal anomaly $x$ and the dimensionless variable $z$.

What would be the physical meaning of these two variables?

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de la Torre Sangra & Fantino's Review of Lambert's Problem as well as Izzo's Revisiting Lambert’s problem (also ArXiv) cite Bate 1971 as Fundamentals of Astrodynamics by Roger R. Bate, Donald D. Mueller, Jerry E. White (Dover, 1971) (in google books, pdfs out there as well) which introduces $x$ and $z$:

and later:

Answer 1

Following page of 204 of Fundamentals of astrodynamics:

These are just convenience variables that depend on change of eccentric anomaly from initial to final point of motion analyzed (or predicted).

For elliptical orbit:

$$x = \sqrt{a} ( E - E_0 )$$

or, for negative $a$ (hyperbolic orbit),

$$x = \sqrt{-a} ( F - F_0 )$$

For parabolic orbit,

$$x = D-D_0$$

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For elliptical orbit: $$z = (E - E_0)^2$$

For hyperbolic orbit: $$-z = (F - F_0)^2$$

For parabolic orbit, $z = 0$ (also, when there's no change in eccentric anomaly.)

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where $E$ is the eccentric anomaly (page 183):

$D$ is "parabolic eccentric anomaly" and $F$ - "hyperbolic eccentric anomaly" (always an imaginary value) - counterparts to $E$ for parabolic and hyperbolic trajectories. Subsequent pages explain derivation of these.

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As a side note, I think Eccentric Anomaly deserves a better justification and explanation than what it gets, with extending arbitrary lines to randomly chosen circles for unknown purposes.

As the standard ellipse equation is $({x\over a})^2 + ({y \over b})^2 = 1$ (that's a carthesian coordinate $x$, not the universal variable $x$) the typical parametrization is:

$$ x = a \cos E $$ $$ y = b \sin E $$

Answer 2

@SF.'s answer checks out very nicely! xSF=xOP and zSF = zOP.

I used some equations from this answer.

I am not sure if it answers the OP's question completely, but since it checks out mathematically, I'll award this particular bounty.

As Meatloaf says, everything works if you let it.

def deriv(X, t):
    x, v = X.reshape(2, -1)
    acc  = -x * mu * ((x**2).sum())**-1.5
    return np.hstack((v, acc))

import numpy as np
import matplotlib.pyplot as plt

from scipy.integrate import odeint   as ODEint
from scipy.integrate import cumtrapz as CTrapz

# https://space.stackexchange.com/questions/31032/what-exactly-means-universal-variable-x-and-z

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]

mu     = 1.0
a      = 1.0
peri   = 0.5
apo    = 2.*a - peri
vperi  = np.sqrt(mu*(2./peri - 1./a))
vapo   = np.sqrt(mu*(2./apo  - 1./a))

X0     = np.array([peri, 0, 0, vperi])

time   = np.linspace(0, twopi, 201)

answer, info = ODEint(deriv, X0, time, full_output=True)

x, y, vx, vy = answer.T
theta        = np.arctan2(y, x)
half_theta   = 0.5 * theta


r     = np.sqrt(x**2 + y**2)
xdot  = np.sqrt(mu)/r
xOP   = CTrapz(xdot, time, initial=0)
zOP   = xOP**2/a

# https://space.stackexchange.com/questions/27602/what-is-hyperbolic-eccentric-anomaly-f/27604#27604

ecc       = (apo-peri)/(apo+peri)
term      = np.sqrt((1. - ecc)/(1. + ecc))
tanEover2 = term * np.tan(half_theta)
E         = 2. * np.arctan(tanEover2)
E0        = E[0]
E[E<0]   += twopi  # keep it positive

xSF       = np.sqrt(a)*(E - E0)
zSF       = (E - E0)**2

things    = ( r,   theta,   xOP,   zOP,   xdot,   E,   xSF,   zSF )
names     = ('r', 'theta', 'xOP', 'zOP', 'xdot', 'E', 'xSF', 'zSF')

if True:
    plt.figure()
    for i, (thing, name) in enumerate(zip(things, names)):
        plt.subplot(2, 4, i+1)
        plt.plot(time, thing)
        plt.title(name, fontsize=16)
    plt.show()

if True:
    plt.figure()
    plt.subplot(1, 2, 1)
    plt.title('x and xSF versus time')
    plt.plot(time, xOP,  '-r', linewidth=4)
    plt.plot(time, xSF, '--k', linewidth=2)
    plt.subplot(1, 2, 2)
    plt.title('z and zSF versus time')
    plt.plot(time, zOP,  '-r', linewidth=4)
    plt.plot(time, zSF, '--k', linewidth=2)
    plt.show()

if True:
    x, y, vx, vy = answer.T
    plt.subplot(1, 2, 1)
    plt.title('y versus x')
    plt.plot(x, y)
    plt.plot([0], [0], 'ok')
    plt.subplot(2, 2, 2)
    plt.title('x and y versus time')
    plt.plot(time, x)
    plt.plot(time, y)
    plt.subplot(2, 2, 4)
    plt.title('vx and vy versus time')
    plt.plot(time, vx)
    plt.plot(time, vy)
    plt.show()