STK has two methods for generating TLEs; Trajectory Sampling method and Single Point method.

I have read the differences both of them at http://help.agi.com/stk/index.htm#stk/tools-10.htm but I am still confused about the difference between them.

How might TLEs generated using the two methods differ? Are there guidelines for how to decide which method to use, or which is more accurate?

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    $\begingroup$ I've made some adjustment to the wording of your question to help it fit the Stack Exchange format. Please feel free to edit further or roll back. Welcome to Space! $\endgroup$
    – uhoh
    Sep 9, 2019 at 4:38

1 Answer 1


First, the SGP4 propagator is meant to propagate TLEs. He takes a TLE and a time since the TLE's epoch as inputs and outputs a position and a velocity. By selecting several times, you get a trajectory.

In STK, if you have a trajectory generated by some other method, i.e. the TwoBody propagator or the HPOP propagator, then STK can estimate a TLE that would produce a "similar" trajectory.

And there are two ways of doing so, one is to solve the nonlinear equation $(r,v)=\text{SGP4}(TLE, t_0)$, with $(r,v)$ a single, known point in the trajectory at time $t_0$ (since the TLE's epoch). Thus, you get a TLE that provides you the exact position and velocity you've used as reference at that given epoch. Often, this is good enough, but while this allows you to fit the mean orbital elements in the TLE message, it wouldn't allow you to fit the ballistic coefficient associated with atmospheric drag.

If you need to fit the ballistic coefficient (or the mean motion derivatives if working in deep space), you need more data points. By doing so, you need to provide a trajectory with several points, and STK will sample a some of those point from this trajectory and minimize the quadratic error between the positions and velocities contained in this sample and those that are produced by SGP4. This is what is called "Trajectory Sampling Method". The trajectory provided by SGP4 won't match exactly any point in the original trajectory, but should have a balanced error between different points in the trajectory.

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    $\begingroup$ Great answer! +1 I made a small edit adding a link to Wikipedia, feel free to roll back. Should BSTAR be mentioned as well? Will a non-zero BSTAR come out of the trajectory sampling method? $\endgroup$
    – uhoh
    Sep 9, 2019 at 14:05
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    $\begingroup$ @uhoh : According to [The documentation](If checked, solves for BStar drag and does not use the value in the BStar field.) indicated in the question, the trajectory sampling method may or may not solve for BSTAR depending on the options selected. If the option to solve for it is selected or if the original BSTAR given is non-zero, the result should be non-zero. $\endgroup$
    – Mefitico
    Sep 9, 2019 at 14:10

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