Hi Balata,

could you a little bit more explain about rx,ry and rx please?

It is quite easy. The overall redundancy of your least squares problem - also known as degree of freedom - is the number of unnecessary (reading as redundant) observations. An example: A circle in 2D has u=3 parameters, the radius and the center point represented by the values x0 and y0. To estimate a circle, you need at least n=3 points (which do not lie on a straight line). The degree of freedom is zero: f = n-u = 3-3 = 0. If f is zero, you don't need a least square solver because the derived solution is unique and the residuals v are zero. Thus, a closed solution exists.

A least squares solution is needed, if f > 0 to find the most probable solution with the condition . Whereas u, the number of unknown parameters, is constant, the number of observations (or points) n has to increase to enlarge f.

A circle fit, which was carried out with 4 points, has a degree of freedom of f = n-u = 4-3 = 1. The integer f can be broken down to the influence of each point - the redundancy rx, ry, rz. Whereas a value of r=0 means, the observation completely affected the final solution, denotes a value r=1 an unneeded observation. Therefore, r is an important parameter because it gives an impression of the controlled unit. If r=0 it is impossible to detect blunders, on the other hand r=1 means a full controlled point. A recommended value is 0.4 < r < 0.7 - a balance between cost and benefit.

Do you mean the sum of redundancy of the coordinate components of each point is equal to the degree of freedom of the least squares problem

Yes, because trace(R) = n-u, where R is the so-called redundancy matrix. The values rx, ry and rz are taken from the main diagonal of R.

Regarding the σ{x,y,z} I meant what are represent in the statistic table for each point.

The σ{x,y,z} denote the uncertainties _after_ the adjustment. Because of the imperfection of your instrument, each observation gets a (small) residual during the adjustment. If you have a perfect surface (or a perfect mathematical model), the estimated uncertainties demonstrate the accuracy of your instrument.

Can the software JAG3D able to fit the pyramid and calculate the statistical result of its apex as it does for cone, sphere and others.

A pyramid is currently not supported by the FormFittingToolbox because a pyramid is a composite figure of k planes (with k>2). To get the pyramid-apex, you have to estimate the k planes of each side. Finally, the intersection point of the planes is the apex.

If you separate your cloud for each plane, I try to develop a solution in Octave.

Kind regards
Micha

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applied-geodesy.org - OpenSource Least-Squares Adjustment Software for Geodetic Sciences