Example 6


This example has two goals:

  1. To extend the earlier Example 2 in order to demonstrate how to linearize the basic functional model of the distance between two points; and
  2. To demonstrate good practice in documenting your solution to such a problem, i.e. to demonstrate what I am looking for in your written responses to questions of this type.


We first looked at the simple case of measuring the distance between two points in Example 2.

There, we were asked to imagine that we wanted to know the coordinates of two unknown points A and B in a planar 2D coordinate system, and that we had gone out and to further imagine that we had the ability to measure the distance between them.

Then you were asked to write out the functional model and identify what form it took.

Here I want to start with the same situation and I want you to further imagine that you are given some initial approximate coordinates for the two unknown points: E_A^0 = 201.0\text{ m}N_A^0 = 111.6\text{ m}E_B^0 = 525.1\text{ m}, and N_B^0 = 612.5\text{ m}.

And that we went out and measured the distance between them once to be 596.8\text{ m}.


Now I want you to do the following:

  1. Practice deriving the functional model and put it into one of the three main forms
  2. Linearize it

Sample solution

When responding to a problem like this I would encourage you to have the following sections.

1. Sketch

The situation was sketched for us in Example 2, so I’d simply repeat that here:

2. Given

Next you need a statement of what you’ve been given in terms and using variables from the course.

We know that the desired parameter vector \mathbf{x} is made up by the coordinates of the stations A and B. In other words, E_AN_A, E_B, and N_B will become part of \mathbf{x}. The \mathbf{x} vector could be written out in full if you wish.

And the measurements l_i of the distance d_{AB} will become part of the measurement vector \mathbf{l}_{measured}.

We’ve also been given that:

    \begin{equation*} \mathbf{x}^0 =\begin{bmatrix} E_A^0  \\ N_A^0  \\ E_B^0  \\ N_B^0 \end{bmatrix} =\begin{bmatrix} 201.0  \\ 111.6  \\ 525.1  \\ 612.5 \end{bmatrix} \text{ m} \end{equation*}


    \begin{equation*} \mathbf{l}_{measured} = 596.8\text{ m} \end{equation*}

We also know that:

    \begin{equation*} u = \text{ number of unknowns} = 4 \end{equation*}

    \begin{equation*} n = \text{ number of measurements} = 1 \end{equation*}

    \begin{equation*} r = \text{ number of equations} = n = 1 \end{equation*}

3. Required

Next, re-state what has been required of you. In this case we are asked to:

  • Derive the functional model
  • Put it in one of the three main forms
  • Linearize it

And then, as shown below, solve the problem.

4. The functional model

Normally I’d want to see your derivation (or logic behind) the functional model, but we did this in Example 2 so I won’t repeat it here.

The result was as follows:

    \begin{equation*} l_{true}^2 - (E_B-E_A)^2 + (N_B-N_A)^2 = 0 \end{equation*}

which we can rewrite in the general form \mathbf{l}_{true} - \mathbf{F(\mathbf{x})} =\mathbf{0} as follows:

    \begin{equation*} \boxed{l_{true} - \sqrt{(E_B-E_A)^2 + (N_B-N_A)^2 } = 0} \end{equation*}

And to keep things simple in the following, we’ll define the following shorthand:

    \begin{equation*} d_{AB} = [ (E_B-E_A)^2 + (N_B-N_A)^2 ]^{\frac{1}{2}} \end{equation*}

E.g. so that our functional model can be written as follows:

    \begin{equation*} l_i - d_{AB} = 0 \end{equation*}

5. The linearized model

Next I would write that we know that the general linearized form of the functional model for observation equations is as follows:

    \begin{equation*} \mathbf{A}\boldsymbol{\delta} - \mathbf{e} + \mathbf{w} = 0 \end{equation*}

Where \mathbf{w} and \mathbf{A} are given as follows:

    \begin{align*} \underset{r\times 1}{\mathbf{w} } & = \mathbf{F}(\mathbf{x}^0,\mathbf{l}_{measured}) \\ &= \begin{bmatrix} f_1(\mathbf{x}^0,\mathbf{l}_{measured}) \\[2ex] f_2(\mathbf{x}^0,\mathbf{l}_{measured}) \\[2ex] \vdots \\[2ex] f_r(\mathbf{x}^0,\mathbf{l}_{measured}) \\ \end{bmatrix} \end{align*}


    \begin{align*} \underset{r\times u}{\mathbf{A} } & = \left.\frac{d\mathbf{F}}{d\mathbf{x}}\right|_0 \\ & = \begin{bmatrix} \left.\dfrac{df_1}{dx_1}\right|_0 & \left.\dfrac{df_1}{dx_2}\right|_0 & \cdots &\left.\dfrac{df_1}{dx_u}\right|_0\\[2ex] \left.\dfrac{df_2}{dx_1}\right|_0 & \left.\dfrac{df_2}{dx_2}\right|_0 & \cdots &\left.\dfrac{df_2}{dx_u}\right|_0\\[2ex] \vdots & \vdots & & \vdots \\[2ex] \left.\dfrac{df_r}{dx_1}\right|_0 & \left.\dfrac{df_r}{dx_2}\right|_0 & \cdots &\left.\dfrac{df_r}{dx_u}\right|_0\\ \end{bmatrix} \end{align*}

In our case r = n = 1, and we get the following for the misclosure vector, \mathbf{w}:

    \begin{align*} \mathbf{w} & = f(\mathbf{x}^0,\mathbf{l}_{measured}) \\ & = l_{measured} - d_{AB}^0 \\ \end{align*}

which is based on values we were given or can easily computed from what we were given.

And we get the following for the design matrix, \mathbf{A}, by taking the partial derivatives of the left side of the equation in the box above:

    \begin{equation*} \mathbf{A} = \begin{bmatrix} \left.\dfrac{df}{dE_A}\right|_0 & \left.\dfrac{df}{dN_A}\right|_0 &\left.\dfrac{df}{dE_B}\right|_0 & \left.\dfrac{df}{dN_B}\right|_0 \end{bmatrix} \end{equation*}

Each of the partial derivatives is computed as in the following example:

    \begin{align*} \left.\dfrac{df}{dN_A}\right|_0 & = \dfrac{d}{dN_A^0}(l - \sqrt{(E_B^0-E_A^0)^2 + (N_B^0-N_A^0)^2 }) \\ & = -\dfrac{1}{2}\dfrac{2(N_B^0-N_A^0)(-1)}{\sqrt{(E_B^0-E_A^0)^2 + (N_B^0-N_A^0)^2 }} \\ & = \dfrac{N_B^0-N_A^0}{d_{AB}^0} \end{align*}

Similarly, we get the following for the other partial derivatives:

    \begin{align*} \left.\dfrac{df}{dE_A}\right|_0 & = \dfrac{E_B^0-E_A^0}{d_{AB}^0} \end{align*}

    \begin{align*} \left.\dfrac{df}{dN_B}\right|_0 & = -\dfrac{N_B^0-N_A^0}{d_{AB}^0} \end{align*}

    \begin{align*} \left.\dfrac{df}{dE_B}\right|_0 & = -\dfrac{E_B^0-E_A^0}{d_{AB}^0} \end{align*}

Which are all based on values we were given or can easily compute from what we were given.

So, finally, we can substitute what we’ve found into the general linearized model \mathbf{A}\boldsymbol{\delta} - \mathbf{e} + \mathbf{w} = 0 to obtain:

    \begin{equation*} \boxed{ \begin{bmatrix} \dfrac{E_B^0-E_A^0}{d_{AB}^0} & \dfrac{N_B^0-N_A^0}{d_{AB}^0} &  -\dfrac{E_B^0-E_A^0}{d_{AB}^0} & -\dfrac{N_B^0-N_A^0}{d_{AB}^0} \end{bmatrix} \begin{bmatrix} \delta E_A \\ \delta N_A \\ \delta E_B \\ \delta N_B \end{bmatrix} + e_{AB} + l_{measured} - d_{AB}^0 = 0} \end{equation*}

Which provides the linearized form of the equation that we were asked to find.


Here we know all but the corrections \boldsymbol{\delta} and e_{AB} which we seek to estimate. But this form of the equation is linear. And it is based on our available approximate values, unlike the original:

    \begin{equation*} l_{true} - \sqrt{(E_B-E_A)^2 + (N_B-N_A)^2 } = 0 \end{equation*}

which was non-linear and was in terms of the unknown values l_{true} and \mathbf{x}.