[GIS] how to calculate RMS 2D and RMS 1D, using python

accuracygpspython

I would like to calculate:

RMS 2D: standard deviation for the two coordinates of flat (67% confidence level)
and
RMS 1D: standard deviation for height (67% confidence level)
What is the formula and what parameters do I need for dual frquency survey grade receiver?
At the moment I haves something like:

import math
spaceSegment = 0.4 
controlSegment = 3.5 
tropo = 0.2 
iono = 0.4 
multipath = 0.6 
receiver_error = 0.3

hdop = float(raw_input("Enter hdop "))

vdop = float(raw_input("Enter vdop "))

ure = math.sqrt(spaceSegment**2 + controlSegment**2) 
uee = math.sqrt(tropo**2 + iono**2 + multipath**2 + receiver_error**2) 
uere = math.sqrt(ure**2 + uee**2)

horizontal_error = 2. * hdop * uere    #2D horizontal position error
vertical_error = 2. * vdop * uere    ##Vertical Position Error

…but I cannot find gps error budget for survey grade receivers nor formulas
for 67 and 68 percentage calculation.
Is any simple way for my computation?

Above computation is taken from matlab gps toolkit I rewrited in python but it refers to gps budget for single frequency receiver.
I also have two formulas:
enter image description here
and

and their decription:
Compute the rms of the radial errors, i.e. the linear distances between the measure and known (or mean) positions. It can be predicted by using covariance analysis by multiplying the HDOP, a measure of the satellite geometry, by the standard deviation of the observed pseudoranges. This predictability makes it a much more convenient measure in practice.
…so then I am completely confused and any help will be appreciated.

Best Answer

Root mean square (RMS) deviation is similar to standard error. 67% confidence intervals are roughly the average +/- 1 standard error. Consult a Z score table in any entry level statistics book for exact values.

The use of RMS as it relates to confidence is confused by common use of the terms.

I'm not familiar with the formula you are using but it looks like it is estimating an RMS.

This article has a reasonable explanation of RMS and confidence intervals for GPS. http://ohioline.osu.edu/aex-fact/0560.html

This Wikipedia entry discusses the relationship to standard error http://en.m.wikipedia.org/wiki/Root-mean-square_error

Related Solutions

Accepted RMS Error for Topographic Maps – Generally Accepted Root Mean Square (RMS) Error for Rectifying Topographic Maps

No, there is no absolute value for RMS, because it depends on the quality of the map being georeferenced, the quality of the target (base) map, and the purpose of the georeferencing. In particular, any advice that relates RMS to cellsize is misinformed, because cellsize reflects precision in the digital representation of an image whereas the RMS error reflects average accuracy (assuming the basemap is perfectly accurate). Although distinguishing precision and accuracy may seem like aimless pedantry, confusing them is a basic mistake with practical consequences.

All this is rather vague, so let's look at a specific example. Recently I received a series of screenshots of maps showing soil sample locations. To obtain coordinates, I planned to georeference these screenshots to an orthophoto base map and then digitize the points with heads-up digitization. Among the considerations were:

The orthophoto base map has 0.3 m cellsize.
The screenshots have approximately 2 m cellsize.
The soil sample locations were not surveyed; they were located "by eye" on the map when the sampler was in the field. The client estimated the accuracy was about 3 m, but 10 m is more likely.
The screenshots have few sharp details: they are primarily contour lines, with occasional fencelines (which are not clearly visible in the orthophoto). Thus establishing many good links would be time-consuming and error-prone.
There was likely some local distortion in the screenshots, meaning that high accuracy (low RMS) can be achieved only with complex transformations.
It was important to digitize the soil sample locations so that relative distances were fairly accurate for nearby points, but absolute accuracy was unnecessary, because one outcome of the study will be to obtain many more soil samples that refine and make more precise this preliminary survey.

To obtain an RMS of half the larger cellsize would require a high-order polynomial transformation or warping across a grid of points, calling for establishing a network of around 50 - 100 good links between the images: one to several hours of careful work, most likely, given the difficulty of even finding visible links. To obtain an RMS of half the smaller cellsize would require an order of magnitude more effort: days of work. However, for the purposes of the study an RMS of 5 m would be more than sufficient. This was achieved with 7 links and an affine transformation, just a few minutes' work. Note that this RMS is several times greater than the larger of the two cellsizes in the images.

This example illustrates how blindly following bad rules of thumb can be costly. Pay attention first to your data quality objectives; everything else follows from them.

[GIS] Accuracy of GPS and HDOP

DOP relates the User Ranging Error (URE) to position coordinates error (PCE):

PCE = URE * DOP

It seems you already determined both PCE (2 m) and HDOP (1.2). The only unknown left is URE, which is this case equals URE = HPCE / HDOP = 1.67 m.

Now with that information at hand, you can assume the URE will not change, then predict the PCE of interest given DOP predictions, as given in GPS survey planning software.

Best Answer

Related Solutions

Accepted RMS Error for Topographic Maps – Generally Accepted Root Mean Square (RMS) Error for Rectifying Topographic Maps

[GIS] Accuracy of GPS and HDOP

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