What variables are needed to calculate simple horseshoe orbit times? EDIT This was NOT a duplicate of Horseshoe orbit cycle times.

But that other question has been deleted, regardless. My original question asked for the answer to the equation(s), and provides many of the variables.  But since no one was apparently willing/able to answer it, this new question asks only for the form of the equation(s) itself and a list of variables involved, so that I can either provide more information on a future question, or try and solve the equations myself (though I doubt my math ability to do that, hence the original question, if I know what equations and variables are involved I can at least make an attempt myself).

What variables need to be known in order to calculate a horseshoe orbit cycle time?



In other words: What is the equation, and what do the variables in that equation represent (speed? mass? semi major axes? etc.), to calculate the time it takes for an object to go from Point A on that image, through Points B, C, D, and E and back to A, along the light blue line in the image above (image also available here)?

For example, Earth and Cruithne complete their cycle in 770 years, while Janus and Epimetheus complete theirs in 8 years.  I'm confident that semi-major axes are key factors in the equation, but I don't know what other variables are included, or how the variables relate to each other in the calculation.  

Also, I understand that these orbits are not stable, in the long term, and that my example of Cruithne is especially complex, as it will actually change orbit types periodically, and will likely be ejected from the solar system entirely, or impact the Sun or Jupiter, at some point.  But those complexities are for another time.  For this question, I just want to know the equation in its simplest form.
That depends on the accuracy you want to work with. To zeroth order, as outlined in Murray &amp; Dermott, "Solar System dynamics", Chapt 3., you can do the following:


The zero-velocity contours that are plotted in your image will not be coinciding with particle orbits to infinite precision, but they're a good zeroth order approximation for objects with low eccentricity w.r.t the star ($v_{r}/v_{\theta} \ll 1$)
A particle on those orbits is on a regular Keplerian orbit of radius $r_{\rm H}$, out of reach of the gravitational influence of the planet at $r_{\rm P}$. Thus, to get the largest "chunk" of traveltime of one part of the horseshoe orbit, either inner or outer part, you can work out with relative velocities and the assumption of keplerian velocities.
Caution is required as to which orbital time you're interested in: If $a$ is Earth's distance to the sun, and the horseshoe is $d$ away from a perfectly circular orbit, thus alternating between distances $a\pm d$, then the orbital time in the rest-frame of the sun will be $v_{-}=\sqrt{\frac{GM}{a-d}}$, for a large number of orbits until close encounter, and after that $v_{+}=\sqrt{\frac{GM}{a+d}}$.
Therefore, the simplified case that the satellite is on its own Keplerian orbit is true for most of the time. Armed with this knowledge, we can approximate the Planetocentric recurrence time as $t_{\rm rec}=\frac{2\pi a}{v_{K} - v_{-}}$ as a simple catch-up time between objects on different orbits.
The relative velocity $v_{\rm rel} = v_{K} - v_{-} = \sqrt{\frac{GM}{a}}-\sqrt{\frac{GM}{a-d}}$ can be expanded in the limit of $d/a\ll1$ into $v_{\rm rel} \approx \frac{1}{2} \sqrt\frac{GM}{a} \frac{d}{a}$ and thus I derive from this $t_{rec}\approx \frac{4\pi}{\sqrt{GM}} \frac{a^{5/2}}{d}$. As expected, the cycle time diverges for $d=0$, as in the co-orbiting case with Earth this time must be infinite. An upper limit on $d$ cannot be derived from this, one would need to turn for the full solution for that.


Just out of curiosity, I've plugged in a few values into this formula and written something quick in python:

import numpy as np

#Basic physics quantities
G      = 6.678e-8 #cgs units
pi     = 3.141592
navo   = 6e23 # particles per mole
sigma  = 5.67e-5   #erg cm-2 s-1 K-4
kb     = 1.38e-16  #erg/K
km     = 1e5 #kilometers in cm
mearth = 5.98e27  #g
msun   = 2.0e33   #g
au     = 1.49e13  #cm
yr   = 365*24*3600
rearth   = 6370e5
rjupiter = 74000*km

#
# Returns the approximate horseshoe-cycle time in the reduced 3body problem
# Masses of bodies: m0&gt;&gt;m1&gt;&gt;m2
# Semimajor axis distance is from m0 to m1, radial distance is a(m0-&gt;m1)-a(m0-&gt;m2)
#
def hs_cycle(mcentral, semimajor_axis, radial_distance):
    return 4*pi/np.sqrt(G*mcentral)*semimajor_axis**(5./2.)/radial_distance/yr

#
# https://en.wikipedia.org/wiki/(419624)_2010_SO16 around the Sun
#
# Quoted cycle time ~350 years, with d=0.004 AU
#
print("Predicted 2010_SO16 cycletime [years] = ", hs_cycle(msun, 1*au,0.004*au), " predicted = 350 yrs")

#
# Janus/Epimetheus around Saturn
#
# a = 151410 km, d = 25 km, as stated in https://en.wikipedia.org/wiki/Epimetheus_(moon)
# Quoted cycle time = 8 years (from comments)
#
print("Predicted Janus/Epimetheus cycletime [years] = ", hs_cycle(95*mearth, 151410*km,50*km), " predicted = 4 yrs")


#
# 3753 Cruithne
#
# a = 1 AU and semimajor axis difference from https://en.wikipedia.org/wiki/3753_Cruithne
# Quoted cycle time = 770 years
#
print("Predicted 3753 Cruithne cycletime [years] = ",hs_cycle(msun, 1*au, (1.0-0.99774)*au), " predicted = 770 yrs")


and the results I get are

Predicted 2010_SO16 cycletime [years] =  495.7747141830971  predicted = 350 yrs
Predicted Janus/Epimetheus cycletime [years] =  11.542076781209305  predicted = 8 yrs
Predicted 3753 Cruithne cycletime [years] =  877.4773702355546  predicted = 770 yrs


So the formula may be off by up to a factor of ~2. This is for sure simply because reality is more complex than a simple circular-orbit approximation, but also due to the quality of the values used.
Wikipedia isn't well known for researching particular values well. I took those that I found there. For SO16 that was particularly confusing to select, so I took those two that were mentioned in the same line of text, hoping they would come from the same source.

Anyone finding more consistent values, is free to comment.