Study on the rigorous accuracy of any point on a linear element

Abstract

In order to obtain the rigorous accuracy of any point on the straight line element, based on existing research, this article considers the error of the straight line element and the error of any point to its end point of the straight line element, attempts to discuss the accuracy calculation of any point on the straight line element in three different situations, and through case analysis, compares the position covariance matrix of any point on the straight line element in three situations and draws its error ellipse using MATLAB programming. Finally, different accuracy sizes and different error ellipses are obtained. This research has certain guiding significance and role for error calculation in precision engineering measurement, such as large bridges or large through projects.

Keywords

Rigorous accuracy linear element covariance matrix error ellipses MATLAB

1. Introduction

In practical measurements involving spatial phenomena like roads and railways, a common simplification is representing them as straight lines. The position of such a line is typically determined by the coordinates of its two endpoints. Any positional errors in these endpoints can introduce inaccuracies in both the position and direction of the line [1, 2]. In geographic data processing, a linear element holds significance as a geographic information entity, making the accurate description of its positional errors crucial. Consequently, a thorough investigation into the positional errors of linear elements is of paramount importance.

In general, determining the positional error of a linear element involves assessing the errors at its starting and ending points. Numerous scholars have delved into this subject, with Sun, for instance, assuming independence between errors at the two endpoints. This assumption led to the derivation of error curve equations and corresponding error ellipse equations for any point along the line. The calculation also yielded the conjugate diameter of the error ellipse [3]. Cai et al. [4] incorporated modeling errors in GIS line objects into their analysis, utilizing the error propagation law to estimate a two-dimensional polyline uncertainty model. They scrutinized the characteristics of its error ellipse. Other scholars, such as Wei, Li, and Liu, explored the propagation of positional errors in linear elements, providing insights into error curves and error ellipses [5, 6, 7] based on $\sigma_{x},\sigma_{y},\sigma_{xy}$ . Further research by Shi, Liu, Tang, Li, and Huang et al. [8, 9, 10, 11, 12] involved error calculations for any point on a line using the error transmission law. They depicted error ellipses for these points and determined error bands of various shapes like “g” bands and “e” bands based on the error band’s shape.

This paper considers the error in the length of the linear element and the error from any point on the linear element to its endpoint, attempts to avoid discussing the accuracy calculation of any point on the linear element in different situations, and draws an error ellipse using MATLAB on this basis.

2. Propsoed methods

The following discusses the accuracy calculation of any point on the straight line element in three cases.

2.1 Error at any point on the linear element

If errors exist in the positions of the two endpoints $A$ and $B$ of the line, then the digital coordinates of the two endpoints are set as $A(x_{1},y_{1})$ , $B(x_{2},y_{2})$ . The covariance matrix is:

$\displaystyle D=\left[\begin{array}[]{llll}\sigma_{x_{1}}^{2}&\sigma_{x_{1}y_{% 1}}&\sigma_{x_{1}x_{2}}&\sigma_{x_{1}y_{2}}\\ \sigma_{x_{1}y_{1}}&\sigma_{y_{1}}^{2}&\sigma_{y_{1}x_{2}}&\sigma_{y_{1}y_{2}}% \\ \sigma_{x_{1}x_{2}}&\sigma_{y_{1}x_{2}}&\sigma_{x_{2}}^{2}&\sigma_{x_{2}y_{2}}% \\ \sigma_{x_{1}y_{2}}&\sigma_{y_{1}y_{2}}&\sigma_{x_{2}y_{2}}&\sigma_{y_{2}}^{2}% \end{array}\right]$ (1)

Try to find the $P$ coordinates $(x,y)$ of $\textit{AP}=S_{1}$ on line AB and its covariance matrix.

Figure 1.

Linear element and arbitrary point P.

It can be seen from Fig. 1 that:

$\displaystyle\left\{\begin{array}[]{l}x=x_{1}+\Delta x_{\textit{AP}}=x_{1}+% \frac{S_{1}}{S}(x_{2}-x_{1})=(1-r)x_{1}+rx_{2}\\ y=y_{1}+\Delta y_{\textit{AP}}=y_{1}+\frac{S_{1}}{S}(y_{2}-y_{1})=(1-r)y_{1}+% ry_{2}\\ \end{array}\right.$ (2)

where the scale number $r=\frac{S_{1}}{S}$ is considered error-free.

Then, the variance of P coordinates is

$\displaystyle\left\{{\begin{array}[]{l}\sigma_{x}^{2}=(1-r)^{2}\sigma_{x_{1}}^% {2}+r^{2}\sigma_{x_{2}}^{2}+2(1-r)r\sigma_{x_{1}x_{2}}\\ \sigma_{y}^{2}=(1-r)^{2}\sigma_{y_{1}}^{2}+r^{2}\sigma_{y_{2}}^{2}+2(1-r)r% \sigma_{y_{1}y_{2}}\\ \sigma_{xy}=(1-r)^{2}\sigma_{x_{1}y_{1}}+(1-r)r\sigma_{x_{1}y_{2}}+(1-r)r% \sigma_{x_{2}y_{1}}+r^{2}\sigma_{x_{2}y_{2}}\\ \end{array}}\right.$ (3)

The formula provided above represents the general method for calculating the coordinates of any point on the line AB, along with its variance and covariance. Currently, this formula is widely adopted in general teaching materials [1, 2].

2.2 Formula considering S error

The determination of S is determined by the two endpoints:

$\displaystyle S=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ (4)

The total differentiation of Eq. (4) is obtained:

$\displaystyle{d}S=\frac{(x_{2}-x_{1})}{S}({d}x_{2}-{d}x_{1})+\frac{(y_{2}-y_{1% })}{S}({d}y_{2}-{d}y_{1})$ (5)

Total differentiation of Eq. (2) yields:

$\displaystyle\left\{{\begin{array}[]{l}{d}x={d}x_{1}+r({dx}_{2}-{dx}_{1})-% \frac{r}{S}(x_{2}-x_{1}){dS}\\ {d}y={d}y_{1}+r({dy}_{2}-{dy}_{1})-\frac{r}{S}(y_{2}-y_{1}){dS}\\ \end{array}}\right.$ (6)

Substitute Eq. (5) into Eq. (6) to get:

$\displaystyle\left\{{\begin{array}[]{l}{d}x={dx}_{1}+r({dx}_{2}-{dx}_{1})-% \frac{r}{S}(x_{2}-x_{1})\left[{\frac{(x_{2}-x_{1})}{S}({dx}_{2}-{dx}_{1})+% \frac{(y_{2}-y_{1})}{S}({dy}_{2}-{dy}_{1})}\right]\\ {d}y={dy}_{1}+r({dy}_{2}-{dy}_{1})-\frac{r}{S}(y_{2}-y_{1})\left[{\frac{(x_{2}% -x_{1})}{S}({dx}_{2}-{dx}_{1})+\frac{(y_{2}-y_{1})}{S}({dy}_{2}-{dy}_{1})}% \right]\\ \end{array}}\right.$ (7)

It can be obtained by sorting out Eq. (7) that

$\displaystyle\left\{{\begin{array}[]{ll}{d}x=\left[1-r+\frac{r}{S^{2}}(x_{2}-x% _{1})^{2}\right]{d}x_{1}+\left[r-\frac{r}{S^{2}}(x_{2}-x_{1})^{2}\right]{d}x_{% 2}\\ \qquad+\frac{r}{S^{2}}(x_{2}-x_{1})(y_{2}-y_{1}){d}y_{1}-\frac{r}{S^{2}}(x_{2}% -x_{1})(y_{2}-y_{1}){d}y_{2}\\ {d}y=\left[1-r+\frac{r}{S^{2}}(y_{2}-y_{1})^{2}\right]{d}y_{1}+\left[r-\frac{r% }{S^{2}}(y_{2}-y_{1})^{2}\right]{d}y_{2}\\ \qquad+\frac{r}{S^{2}}(x_{2}-x_{1})(y_{2}-y_{1}){d}x_{1}-\frac{r}{S^{2}}(x_{2}% -x_{1})(y_{2}-y_{1}){d}x_{2}\\ \end{array}}\right.$ (8)

Let

$\displaystyle\Delta x=x_{2}-x_{1},\Delta y=y_{2}-y_{1}$ (9)

Then Eq. (8) can be simplified as

$\displaystyle\left\{\begin{array}[]{l}dx=\left[1-r+\frac{r}{S^{2}}\Delta^{2}x% \right]dx_{1}+\frac{r}{S^{2}}\Delta x\Delta ydy_{1}+\left[r-\frac{r}{S^{2}}% \Delta^{2}x\right]dx_{2}-\frac{r}{S^{2}}\Delta x\Delta y{d}y_{2}\\ dy=\frac{r}{S^{2}}\Delta x\Delta y{d}x_{1}+\left[1-r+\frac{r}{S^{2}}\Delta^{2}% y\right]{d}y_{1}-\frac{r}{S^{2}}\Delta x\Delta y{d}x_{2}+\left[r-\frac{r}{S^{2% }}\Delta^{2}y\right]{d}y_{2}\\ \end{array}\right.$ (10)

Equation (10) can be obtained by writing it in matrix form

$\displaystyle\left[{\begin{array}[]{l}dx\\ dy\\ \end{array}}\right]=\left[{{\begin{array}[]{llll}{1-r+\frac{r}{S^{2}}\Delta^{2% }x}&{\frac{r}{S^{2}}\Delta x\Delta y}&{r-\frac{r}{S^{2}}\Delta^{2}x}&{-\frac{r% }{S^{2}}\Delta x\Delta y}\\ {\frac{r}{S^{2}}\Delta x\Delta y}&{1-r+\frac{r}{S^{2}}\Delta^{2}y}&{-\frac{r}{% S^{2}}\Delta x\Delta y}&{r-\frac{r}{S^{2}}\Delta^{2}y}\\ \end{array}}}\right]\left[{{\begin{array}[]{c}{{d}x_{1}}\\ {{dy}_{1}}\\ {{d}x_{2}}\\ {{dy}_{2}}\\ \end{array}}}\right]$ (11)

According to the covariance propagation law

$\displaystyle\left[\begin{array}[]{cc}\sigma_{x}^{2}&\sigma_{x}y\\ \sigma_{x}y&\sigma_{y}^{2}\\ \end{array}\right]=\left[\begin{array}[]{llll}{1-}r+\frac{r}{S^{2}}\Delta^{2}x% &\frac{r}{S^{2}}\Delta x\Delta y&r-\frac{r}{S^{2}}\Delta^{2}x&-\frac{r}{S^{2}}% \Delta x\Delta y\\ \frac{r}{S^{2}}\Delta x\Delta y&1-r+\frac{r}{S^{2}}\Delta^{2}y&-\frac{r}{S^{2}% }\Delta x\Delta y&r-\frac{r}{S^{2}}\Delta^{2}y\\ \end{array}\right]\times\left[\begin{array}[]{cccc}\sigma_{x_{1}}^{2}&\sigma_{% x_{1}y_{1}}&\sigma_{x_{1}x_{2}}&\sigma_{xy_{2}}\\ \sigma_{x_{1,y_{1}}}&\sigma_{y_{1}}^{2}&\sigma_{yy_{2}x_{2}}&\sigma_{yy_{2}}\\ \sigma_{x_{1}x_{2}}&\sigma_{y_{1}x_{2}}&\sigma_{x_{2}}^{2}&\sigma_{x_{2}y_{2}}% \\ \sigma_{x_{1}y_{2}}&\sigma_{yy_{2}y_{2}}&\sigma_{x_{2}y_{2}}&\sigma_{y_{2}}^{2% }\end{array}\right]\left[\begin{array}[]{cc}1-r+\frac{r}{S^{2}}\Delta^{2}x&% \frac{r}{S^{2}}\Delta x\Delta y\\ \frac{r}{S^{2}}\Delta x\Delta y&1-r+\frac{r}{S^{2}}\Delta^{2}y\\ r-\frac{r}{S^{2}}\Delta^{2}x&-\frac{r}{S^{2}}\Delta x\Delta y\\ -\frac{r}{S^{2}}\Delta x\Delta y&r-\frac{r}{S^{2}}\Delta^{2}y\end{array}\right]$ (12)

It can be obtained by calculation of Eq. (12) that

$\displaystyle\sigma_{x}^{2}=(1-r+\frac{r}{S^{2}}\Delta^{2}x)^{2}\sigma_{x_{1}}% ^{2}+2\sigma_{x_{1}y_{1}}\left[(1-r+\frac{r}{S^{2}}\Delta^{2}x)\frac{r}{S^{2}}% \Delta x\Delta y\right]+2\sigma_{x_{1}x_{2}}\left[(1-r+\frac{r}{S^{2}}\Delta^{% 2}x)(r-\frac{r}{S^{2}}\Delta^{2}x)\right]-2\sigma_{x_{1}y_{2}}\left[(1-r+\frac% {r}{S^{2}}\Delta^{2}x)\frac{r}{S^{2}}\Delta x\Delta y\right]+\frac{r^{2}}{S^{4% }}\Delta^{2}x\Delta^{2}y\sigma_{y_{1}}^{2}+2\sigma_{x_{2}y_{1}}\left[(r-\frac{% r}{S^{2}}\Delta^{2}x)\frac{r}{S^{2}}\Delta x\Delta y\right]-2\frac{r^{2}}{S^{4% }}\Delta^{2}x\Delta^{2}y\sigma_{y_{1}y_{2}}+(r-\frac{r}{S^{2}}\Delta^{2}x)^{2}% \sigma_{x_{2}}^{2}-2\sigma_{x_{2}y_{2}}\left[(r-\frac{r}{S^{2}}\Delta^{2}x)% \frac{r}{S^{2}}\Delta x\Delta y\right]+\frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^{2% }y\sigma_{y_{2}}^{2}$ (13)

$\displaystyle\sigma_{xy}=\left(1-r+\frac{r}{S^{2}}\Delta^{2}x\right)\frac{r}{S% ^{2}}\Delta x\Delta y\sigma_{x_{1}}^{2}+\sigma_{x_{1}y_{1}}\left[\left(1-r+% \frac{r}{S^{2}}\Delta^{2}x\right)\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right)+% \right.\left.\frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^{2}y\right]+\sigma_{x_{1}x_{% 2}}\left[\left(2r-2\frac{r}{S^{2}}\Delta^{2}x-1\right)\frac{r}{S^{2}}\Delta x% \Delta y\right]+\sigma_{x_{1}y_{2}}\left[\left(1-r+\frac{r}{S^{2}}\Delta^{2}x% \right)\right.\left.\left(r-\frac{r}{S^{2}}\Delta^{2}y\right)-\frac{r^{2}}{S^{% 4}}\Delta^{2}x\Delta^{2}y\right]+\sigma_{x_{2}y_{1}}\!\!\left[\left(1-r+\frac{% r}{S^{2}}\Delta^{2}y\right)\!\!\left(r-\frac{r}{S^{2}}\Delta^{2}x\right)-\frac% {r^{2}}{S^{4}}\Delta^{2}x\Delta^{2}y\right]+\sigma_{y_{1}y_{2}}\left[\left(2r-% 2\frac{r}{S^{2}}\Delta^{2}y-1\right)\frac{r}{S^{2}}\Delta x\Delta y\right]{-}% \left(r-\frac{r}{S^{2}}\Delta^{2}x\right)\frac{r}{S^{2}}\Delta x\Delta y\sigma% _{x_{2}}^{2}+\sigma_{x_{2}y_{2}}\left[\left(r-\frac{r}{S^{2}}\Delta^{2}x\right% )\left(r-\frac{r}{S^{2}}\Delta^{2}y\right)+\frac{r^{2}}{S^{4}}\Delta^{2}x% \Delta^{2}y\right]+\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right)\frac{r}{S^{2}}% \Delta x\Delta y\sigma_{y_{1}}^{2}-\left(r-\frac{r}{S^{2}}\Delta^{2}y\right)% \frac{r}{S^{2}}\Delta x\Delta y\sigma_{y_{2}}^{2}$ (14)

$\displaystyle\sigma_{y}^{2}=\frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^{2}y\sigma_{x% _{1}}^{2}+2\sigma_{x_{1}y_{1}}\left[\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right% )\frac{r}{S^{2}}\Delta x\Delta y\right]-2\frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^% {2}y\sigma_{x_{1}x_{2}}+2\sigma_{x_{1}y_{2}}\left[\left(r-\frac{r}{S^{2}}% \Delta^{2}y\right)\frac{r}{S^{2}}\Delta x\Delta y\right]+2\sigma_{y_{1}y_{2}}% \left[\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right)\left(r-\frac{r}{S^{2}}\Delta% ^{2}y\right)\right]+\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right)^{2}\sigma_{y_{% 1}}^{2}{-}2\sigma_{x_{2}y_{1}}\left[\left({1-}r+\frac{r}{S^{2}}\Delta^{2}y% \right)\frac{r}{S^{2}}\Delta x\Delta y\right]+\frac{r^{2}}{S^{4}}\Delta^{2}x% \Delta^{2}y\sigma_{x_{2}}^{2}-2\sigma_{x_{2}y_{2}}\left[\left(r-\frac{r}{S^{2}% }\Delta^{2}y\right)\frac{r}{S^{2}}\Delta x\Delta y\right]+\left(r-\frac{r}{S^{% 2}}\Delta^{2}y\right)^{2}\sigma_{y_{2}}^{2}$ (15)

2.3 Consider the formula for S and S₁

The total differentiation of Eq. (2) is obtained

$\displaystyle\left\{{\begin{array}[]{l}{d}x={d}x_{1}+r({dx}_{2}-{dx}_{1})+% \frac{(x_{2}-x_{1})}{S}{d}S_{1}-\frac{r}{S}(x_{2}-x_{1}){dS}\\ {d}y={d}y_{1}+r({dy}_{2}-{dy}_{1})+\frac{(y_{2}-y_{1})}{S}{d}S_{1}-\frac{r}{S}% (y_{2}-y_{1}){dS}\\ \end{array}}\right.$ (16)

Substitute Eq. (9) into Eq. (16) to get

$\displaystyle\left\{{\begin{array}[]{l}{d}x={d}x_{1}+r({dx}_{2}-{dx}_{1})+% \frac{\Delta x}{S}{d}S_{1}-\frac{r}{S}\Delta{xdS}\\ {d}y={d}y_{1}+r({dy}_{2}-{dy}_{1})+\frac{\Delta y}{S}{d}S_{1}-\frac{r}{S}% \Delta{ydS}\\ \end{array}}\right.$ (17)

Substitute Eq. (5) into Eq. (17) to derive

$\displaystyle\left\{\begin{array}[]{l}{d}x=\left[{1-}r+\frac{r}{S^{2}}\Delta^{% 2}x\right]dx_{1}+\frac{r}{S^{2}}\Delta{x}\Delta{ydy}_{1}+\left[r-\frac{r}{S^{2% }}\Delta^{2}{x}\right]{dx}_{2}-\frac{r}{S^{2}}\Delta{x}\Delta{ydy}_{2}+\frac{% \Delta x}{S}{d}S_{1}\\ {dy}=\frac{r}{S^{2}}\Delta x\Delta y{dx}_{1}+\left[{1-}r+\frac{r}{S^{2}}\Delta% ^{2}y\right]dy_{1}-\frac{r}{S^{2}}\Delta x\Delta y{dx}_{2}+\left[r-\frac{r}{S^% {2}}\Delta^{2}{y}\right]{dy}_{2}+\frac{\Delta y}{S}{d}S_{1}\\ \end{array}\right.$ (18)

Write Eq. (18) in matrix form as

$\displaystyle\left[{\begin{array}[]{l}dx\\ dy\\ \end{array}}\right]=\left[{{\begin{array}[]{ccccc}{1-r+\frac{r}{S^{2}}\Delta^{% 2}x}&{\frac{r}{S^{2}}\Delta x\Delta y}&{r-\frac{r}{S^{2}}\Delta^{2}x}&-\frac{r% }{S^{2}}\Delta x\Delta y&{\frac{\Delta x}{S}}\\ {\frac{r}{S^{2}}\Delta x\Delta y}&{1-r+\frac{r}{S^{2}}\Delta^{2}y}&{-\frac{r}{% S^{2}}\Delta x\Delta y}&r-\frac{r}{S^{2}}\Delta^{2}y&{\frac{\Delta y}{S}}\\ \end{array}}}\right]\left[{{\begin{array}[]{c}{{d}x_{1}}\\ {{dy}_{1}}\\ {{d}x_{2}}\\ {dy}_{2}\\ {d}S_{1}\\ \end{array}}}\right]$ (19)

According to the covariance propagation law

$\displaystyle\left[{{\begin{array}[]{cc}{\sigma_{x}^{2}}&{\sigma_{xy}}\\ {\sigma_{xy}}&{\sigma_{y}^{2}}\\ \end{array}}}\right]=\left[{{\begin{array}[]{ccccc}{{1-}r+\frac{r}{S^{2}}% \Delta^{2}x}&{\frac{r}{S^{2}}\Delta x\Delta y}&{r-\frac{r}{S^{2}}\Delta^{2}x}&% {-\frac{r}{S^{2}}\Delta x\Delta y}&{\frac{\Delta x}{S}}\\ {\frac{r}{S^{2}}\Delta x\Delta y}&{1-r+\frac{r}{S^{2}}\Delta^{2}y}&{-\frac{r}{% S^{2}}\Delta x\Delta y}&{r-\frac{r}{S^{2}}\Delta^{2}y}&{\frac{\Delta y}{S}}\\ \end{array}}}\right]\times\left[\begin{array}[]{ccccc}\sigma_{x_{1}}^{2}&% \sigma_{x_{1}y_{1}}&\sigma_{x_{1}x_{2}}&\sigma_{x_{1}y_{2}}&0\\ \sigma_{x_{1}y_{1}}&\sigma_{y_{1}}^{2}&\sigma_{x_{2}y_{1}}&\sigma_{y_{1}y_{2}}% &0\\ \sigma_{x_{1}x_{2}}&\sigma_{x_{2}y_{1}}&\sigma_{x_{2}}^{2}&\sigma_{x_{2}y_{2}}% &0\\ \sigma_{x_{1}y_{2}}&\sigma_{y_{1}y_{2}}&\sigma_{x_{2}y_{2}}&\sigma_{y_{2}}^{2}% &0\\ 0&0&0&0&\sigma_{S}^{2}\end{array}\right]\left[\begin{array}[]{cc}1-r+\frac{r}{% S^{2}}\Delta^{2}x&\frac{r}{S^{2}}\Delta x\Delta y\\ \frac{r}{S^{2}}\Delta x\Delta y&1-r+\frac{r}{S^{2}}\Delta^{2}y\\ r-\frac{r}{S^{2}}\Delta^{2}x&-\frac{r}{S^{2}}\Delta x\Delta y\\ -\frac{r}{S^{2}}\Delta x\Delta y&r-\frac{r}{S^{2}}\Delta^{2}y\\ \frac{\Delta x}{S}&\frac{\Delta y}{S}\end{array}\right]$ (20)

It can be obtained by calculation of Eq. (20) that

$\displaystyle\sigma_{x}^{2}=\left(1-r+\frac{r}{S^{2}}\Delta^{2}x\right)^{2}% \sigma_{x_{1}}^{2}+2\sigma_{x_{1}y_{1}}\left[\left(1-r+\frac{r}{S^{2}}\Delta^{% 2}x\right)\frac{r}{S^{2}}\Delta x\Delta y\right]+2\sigma_{x_{1}x_{2}}\left[% \left(1-r+\frac{r}{S^{2}}\Delta^{2}x)(r-\frac{r}{S^{2}}\Delta^{2}x\right)% \right]-2\sigma_{x_{1}y_{2}}\left[\left(1-r+\frac{r}{S^{2}}\Delta^{2}x\right)% \frac{r}{S^{2}}\Delta x\Delta y\right]+\frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^{2% }y\sigma_{y_{1}}^{2}+2\sigma_{x_{2}y_{1}}\left[\left(r-\frac{r}{S^{2}}\Delta^{% 2}x\right)\frac{r}{S^{2}}\Delta x\Delta y\right]-2\frac{r^{2}}{S^{4}}\Delta^{2% }x\Delta^{2}y\sigma_{y_{1}y_{2}}+\left(r-\frac{r}{S^{2}}\Delta^{2}x\right)^{2}% \sigma_{x_{2}}^{2}-2\sigma_{x_{2}y_{2}}\left[\left(r-\frac{r}{S^{2}}\Delta^{2}% x\right)\frac{r}{S^{2}}\Delta x\Delta y\right]+\frac{r^{2}}{S^{4}}\Delta^{2}x% \Delta^{2}y\sigma_{y_{2}}^{2}+\frac{\Delta^{2}x}{S^{2}}\sigma_{s}^{2}$ (21)

$\displaystyle\sigma_{xy}=\left(1-r+\frac{r}{S^{2}}\Delta^{2}x\right)\frac{r}{S% ^{2}}\Delta x\Delta y\sigma_{x_{1}}^{2}+\sigma_{x_{1}y_{1}}\left[\left(1-r+% \frac{r}{S^{2}}\Delta^{2}x\right)\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right)+% \right.\left.\frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^{2}y\right]+\sigma_{x_{1}x_{% 2}}\left[\left(2r-2\frac{r}{S^{2}}\Delta^{2}x-1\right)\frac{r}{S^{2}}\Delta x% \Delta y\right]+\sigma_{x_{1}y_{2}}\left[\left(1-r+\frac{r}{S^{2}}\Delta^{2}x% \right)\right.\left.\left(r-\frac{r}{S^{2}}\Delta^{2}y\right)-\frac{r^{2}}{S^{% 4}}\Delta^{2}x\Delta^{2}y\right]\!+\sigma_{x_{2}y_{1}}\!\!\left[\left(1-r+% \frac{r}{S^{2}}\Delta^{2}y\right)\left(r-\frac{r}{S^{2}}\Delta^{2}x\right)-% \frac{r^{2}}{S^{4}}\Delta^{2}x\Delta^{2}y\right]+\sigma_{y_{1}y_{2}}\left[% \left(2r-2\frac{r}{S^{2}}\Delta^{2}y-1\right)\frac{r}{S^{2}}\Delta x\Delta y% \right]{-}\left(r-\frac{r}{S^{2}}\Delta^{2}x\right)\frac{r}{S^{2}}\Delta x% \Delta y\sigma_{x_{2}}^{2}+\sigma_{x_{2}y_{2}}\left[\left(r-\frac{r}{S^{2}}% \Delta^{2}x\right)\left(r-\frac{r}{S^{2}}\Delta^{2}y\right)+\frac{r^{2}}{S^{4}% }\Delta^{2}x\Delta^{2}y\right]+\left(1-r+\frac{r}{S^{2}}\Delta^{2}y\right)% \frac{r}{S^{2}}\Delta x\Delta y\sigma_{y_{1}}^{2}-\left(r-\frac{r}{S^{2}}% \Delta^{2}y\right)\frac{r}{S^{2}}\Delta x\Delta y\sigma_{y_{2}}^{2}+\frac{% \Delta x\Delta y}{S^{2}}\sigma_{s}^{2}$ (22)

3. Case analysis

Within a wire network, there exists a wire segment defined by endpoints $A$ (100 m, 100 m), $B$ (200 m, 200 m), as illustrated in Fig. 1.

Calculating by coordinates $S=\sqrt{(x_{B}-x_{A})^{2}+(y_{B}-y_{A})^{2}}=$ 141.4 m, the covariance matrix of $A$ and $B$ is:

$\displaystyle D(X)=\left[{{\begin{array}[]{cccc}{3.50}&{0.15}&{0.05}&{0.05}\\ {0.15}&{2.50}&{0.20}&{0.20}\\ {0.05}&{0.20}&{2.00}&{0.10}\\ {0.05}&{0.20}&{0.10}&{3.00}\\ \end{array}}}\right](\text{cm}^{2})$ (24)

Determine the coordinates $(x,y)$ and the covariance matrix at the intersection point, $P$ , on line $S_{1}=$ 50 m, where the length of the line segment is being measured. The measurement accuracy is denoted by $\sigma_{s}^{2}=$ 25 mm², and the analysis is conducted along the line $A B$ .

Solution: $r=\frac{S_{1}}{S}=\frac{\sqrt{2}}{4}$ calculated from Eq. (2):

$\displaystyle\left\{{\begin{array}[]{l}x=100+\frac{\sqrt{2}}{4}\times 100=135.% 35∼{}\text{m}\\ y=100+\frac{\sqrt{2}}{4}\times 100=135.35∼{}\text{m}\\ \end{array}}\right.$

Three cases are discussed to obtain the coordinates and coordinate differences.

(1) Without considering the influence of $S$ and $S_{1}$ , the covariance matrix of point P coordinates can be obtained by calculating according to Eq. (3):

$\displaystyle\left[{{\begin{array}[]{cc}{\sigma_{x}^{2}}&{\sigma_{xy}}\\ {\sigma_{xy}}&{\sigma_{y}^{2}}\\ \end{array}}}\right]=\left[{{\begin{array}[]{cc}{1.7355}&{0.1323}\\ {0.1323}&{1.5112}\\ \end{array}}}\right](\text{cm}^{2})$ $\displaystyle\left\{{\begin{array}[]{l}\sigma_{x}^{2}=(1-r)^{2}\sigma_{x_{1}}^% {2}+r^{2}\sigma_{x_{2}}^{2}+2(1-r)r\sigma_{x_{1}x_{2}}=1.7355∼{}\text{cm}^{2}% \\ \sigma_{y}^{2}=(1-r)^{2}\sigma_{y_{1}}^{2}+r^{2}\sigma_{y_{2}}^{2}+2(1-r)r% \sigma_{y_{1}y_{2}}=1.5112∼{}\text{cm}^{2}\\ \sigma_{xy}=(1-r)^{2}\sigma_{x_{1}y_{1}}+(1-r)r\sigma_{x_{1}y_{2}}+(1-r)r% \sigma_{x_{2}y_{1}}+r^{2}\sigma_{x_{2}y_{2}}=0.1323∼{}\text{cm}^{2}\\ \end{array}}\right.$ (25)

(2) Considering the influence of S, the covariance matrix of point P coordinates can be calculated according to Eq. (10)

$\displaystyle\left[{\begin{array}[]{l}dx\\ dy\\ \end{array}}\right]=\left[{{\begin{array}[]{cccc}{0.8232}&{0.1768}&{0.1768}&{-% 0.1768}\\ {0.1768}&{0.8232}&{-0.1768}&{0.1768}\\ \end{array}}}\right]\left[{{\begin{array}[]{c}{{d}x_{1}}\\ {{dy}_{1}}\\ {{d}x_{2}}\\ {{dy}_{2}}\\ \end{array}}}\right]$ (26)

Equation (26) can be obtained by calculating according to Eq. (12)

$\displaystyle\left[{{\begin{array}[]{cc}{\sigma_{x}^{2}}&{\sigma_{xy}}\\ {\sigma_{xy}}&{\sigma_{y}^{2}}\\ \end{array}}}\right]=\left[{{\begin{array}[]{cc}{2.6437}&{0.8295}\\ {0.8295}&{1.9973}\\ \end{array}}}\right](\text{cm}^{2})$ $\displaystyle\left\{{\begin{array}[]{l}\sigma_{x}^{2}=2.6437∼{}\text{cm}^{2}\\ \sigma_{y}^{2}=1.9973∼{}\text{cm}^{2}\\ \sigma_{xy}=0.8295∼{}\text{cm}^{2}\\ \end{array}}\right.$ (27)

Considering the influence of S and S₁, calculate the covariance matrix of the coordinates of point P, and calculate according to Eq. (3)

$\displaystyle\left[{\begin{array}[]{l}dx\\ dy\\ \end{array}}\right]=\left[{{\begin{array}[]{ccccc}{0.8232}&{0.1768}&{0.1768}&{% -0.1768}&{0.7071}\\ {0.1768}&{0.8232}&{-0.1768}&{0.1768}&{0.7071}\\ \end{array}}}\right]\left[{{\begin{array}[]{c}{{d}x_{1}}\\ {{dy}_{1}}\\ {{d}x_{2}}\\ {\begin{array}[]{l}{dy}_{2}\\ {d}S_{1}\\ \end{array}}\\ \end{array}}}\right]$ (28)

Let

$\displaystyle\bar{X}=[{d}x_{1}\ {dy}_{1}\ {dx}_{t}{2}\ {dy}_{2}\ {dS}_{1}]^{T}$

There is

$\displaystyle D(\bar{X})=\left[{{\begin{array}[]{ccccc}{3.50}&{0.15}&{0.05}&{0% .05}&0\\ {0.15}&{2.50}&{0.20}&{0.20}&0\\ {0.05}&{0.20}&{2.00}&{0.10}&0\\ {0.05}&{0.20}&{0.10}&{3.00}&0\\ 0&0&0&0&{0.25}\\ \end{array}}}\right](\text{cm}^{2})$ (29)

According to Eqs (28) and (29), it can be calculated according to Eq. (20)

$\displaystyle\left[{{\begin{array}[]{cc}{\sigma_{x}^{2}}&{\sigma_{xy}}\\ {\sigma_{xy}}&{\sigma_{y}^{2}}\\ \end{array}}}\right]=\left[{{\begin{array}[]{cc}{2.7687}&{0.9545}\\ {0.9545}&{2.1223}\\ \end{array}}}\right](\text{cm}^{2})$ $\displaystyle\left\{{\begin{array}[]{l}\sigma_{x}^{2}=2.7687∼{}\text{cm}^{2}\\ \sigma_{y}^{2}=2.1223∼{}\text{cm}^{2}\\ \sigma_{xy}=0.9545∼{}\text{cm}^{2}\\ \end{array}}\right.$ (30)

Draw the error ellipses of this point in three cases by using MATLAB programming, as shown in Fig. 2.

Figure 2.

Point error ellipse in different cases.

It can be seen from Fig. 2 that the accuracy of any point obtained is not the same in the three cases. The error ellipses of the three can be seen that the parameters of the error ellipses are different. This research has certain guiding significance and role for error calculation in precision engineering measurement, such as large bridges or large through projects.

4. Conclusions

This paper studies the accuracy of any point on a straight line element in three cases, and obtains the accuracy calculation process and corresponding error ellipse for each case. It is evident that there are differences between the three cases. By plotting different accuracy sizes and different error ellipses, it has certain guiding significance and role for error calculation in precision engineering measurements such as large bridges or large-scale through projects.

References

Xiao

. Error Theory and Measurement Adjustment. Beijing: Science Press, 2019.

Zhu

Zuo

Song

. Error Theory and Basis of Measurement Adjustment. Beijing: Surveying and Mapping Science Press, 2013.

Sun

. Point position plane accuracy determined by two uncorrelated position differences and its application. Journal of Surveying and Mapping.2020; (3): 221-224.

Cai

Zhu

. Unified uncertainty model of line position based on spatiotemporal scale. Science of Surveying and Mapping.2011; (4): 27-30.

Wei

Cui

. Propagation model of linear element position uncertainty in GIS. Journal of Henan Polytechnic University (Natural Science Edition).2005; (2): 152-154.

Lin

Zhang

. Direct derivation of the formula for calculating the extreme direction of point position difference. Mining Survey.2017; (5): 68-70.

Liu

. New thinking on error ellipse teaching content design. Bulletin of Surveying and Mapping.2017; (5): 143-146+151.

Shi

Liu

. Error circle band of linear element position Uncertainty in GIS. Surveying and Mapping Information and Engineering.1998; (2): 6-8.

Gong

Xie

, et al. Error entropy model of surface elements in GIS. Journal of Surveying and Mapping.2003; (1): 31-35.

10.

Tang

. GIS vector linear entity equal probability density error model. Wuhan: Wuhan University, 2004.

11.

. Research on Location uncertainty model of spatial data based on information entropy. Wuhan: Wuhan University, 2003.

12.

Huang

Liu

et al. Error analysis and foundation of GIS spatial data. Wuhan: China University of Geosciences Press, 1995.