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Single output in Solution for Static Mode (out-solstatic) #453

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harlock974 opened this issue Aug 20, 2024 · 1 comment
Open

Single output in Solution for Static Mode (out-solstatic) #453

harlock974 opened this issue Aug 20, 2024 · 1 comment

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@harlock974
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harlock974 commented Aug 20, 2024

Could you please clarify what is exactly single output in Solution for Static Mode ?

The manual says : "Only one solution for the processing period is output", but it doesn't explain how it is computed.

From Rtklibexplorer blog, there is various explanations :

  • "Out-solstatic=single is meant to be used for static solutions when you only want the final position, rather than a position solution for every epoch."
  • "When "Solution for Static Mode" is set to "Single", RTKLIB tries to save and report the last best solution point in the data set"
  • "The single output in Static mode is the last solution point from the kalman filter which is effectively a best estimate or weighted average of all the previous solution points."

From my own observations with fixed solution results (Q=1), single solution is exactly the same that the last epoch solution when out-solstatic=all. It is different from the median Q=1 values. In the following graph, the blue cross is the last epoch solution or the single solution, and the black one is the median value of all Q=1 observations :
Rplot

So I don't understand if "single" is a solution at a given epoch (the last one) or if there is actually a kind of average to compute it.

In other word, as final coordinates for my measure, is it better to keep the values given by "Single" output, or to average all Q=1 values from "All" output ?

@rtklibexplorer
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I think all three definitions in your list are correct but the last one is the most complete. The single output solution in static mode is the last solution point from the kalman filter. Because the kalman filter is an iterative filter, each solution point is a function of all the previous solution points. However, it is a weighted combination, and in general, more recent points will be weighted more heavily than earlier points, so the solution will not be equal to a simple averaging of the previous solution points.

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