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title: 'CE203 Lab4: Basic error propagation calculation'

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CE203 Lab4: Basic error propagation calculation
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1. Given the discussion in the lecture, can you argue that the propagation of error/uncertainty formula for a non-linear model is more general and can be used for a linear model?
- Tip: Think about the definition of linear independence.
- Answer:
    - In page 10, it assumes linear independence and gets the equation: ![](https://i.imgur.com/vAjwgjy.png) 
    where $E$ is $\sigma$
    If we consider three variables only, you will get equation(1):![](https://i.imgur.com/7YVKjQ5.png)
    - In page 19, you can find this equation(2):![](https://i.imgur.com/HE8AuJU.png)
    - Obvious, when the equation is linear independence, $\sigma_{xy},\sigma_{xz},\sigma_{yz}$ are all zero, and equation(2) will be the same as equation(1). In conclusion, equation(1) is a simplified version of equation(2) for linear independent functions.
2. The distance from the nadir point of an image to the bottom of a building is $x_1$ with a standard deviation $\sigma_{x_1}=\pm2\ \mu m$. For the same building, the relief displacement is $x_2$ with a standard deviation $\sigma_{x_2}=\pm3\ \mu m$. What is the standard deviation of the <font color=red>radial distance</font> from the nadir point to the top of the building?
- Figure: ![](https://i.imgur.com/h99oWAW.png)
- Tips: Always find the relationship between variables
- Answer:
    - Step 1: find the relasionship between variables: $R=x_1+x_2$.
    - Step 2: judge that whether the variables are linearly independent or not: Yes as covariance is not given.
    - Step 3: use error propagation formula:
    $\sigma_{R}^2=(\frac{ \partial f}{\partial x_1}\sigma_{x_1})^2+(\frac{ \partial f}{\partial x_2}\sigma_{x_2})^2=\sigma_{x_1}^2+\sigma_{x_2}^2=4\mu m^2+9\mu m^2=13\mu m^2$
    $\sigma_R=\pm\sqrt{13}\mu m$
    ($f=x_1+x_2$)
3. 	The distance from the nadir point of an image to the bottom of a building is $x_1$ with a standard deviation $\sigma_{x_1}=\pm2\ \mu m$. For the same building, the distance from the nadir point to its top is $x_2$ with a standard deviation $\sigma_{x_2}=\pm3\ \mu m$. What is the standard deviation of the relief displacement of that building? Comment on the derived standard deviations in problems 2 and 3.
- Figure:![](https://i.imgur.com/D2YIMzi.png)
- Answer
    - Step 1: find the relasionship between variables: $d=-x_1+x_2$.
    - Step 2: judge that whether the variables are linearly independent or not: Yes as covariance is not given.
    - Step 3: use error propagation formula:
    $\sigma_{d}^2=(\frac{ \partial f}{\partial x_1}\sigma_{x_1})^2+(\frac{ \partial f}{\partial x_2}\sigma_{x_2})^2=(-1)^2\sigma_{x_1}^2+\sigma_{x_2}^2=4\mu m^2+9\mu m^2=13\mu m^2$
    $\sigma_d=\pm\sqrt{13}\mu m$
    ($f=-x_1+x_2$)
    - Comment:Given two equations: a=x+y, b=x-y.
    $\sigma_a=\sigma_b$ The error of x+y is the same as the error of x-y.
4. Given a focal length of 6.0", a ground distance between two points of 5,536', and a corresponding photo distance between the same two points of 4.92"; calculate the standard deviation of the estimated flying height caused by an error in the ground distance of ±1.0'.
- Figure:![](https://i.imgur.com/vsU8osp.png)
- Answer:
    - step 1: find the relasionship between variables:$f:H=r:R\rightarrow H=\frac{fR}{r}$
    - step 2: judge that whether the variables are linearly independent or not: Yes as covariance is not given.
    - step 3: use error propagation formula
    Only R has error. The other two variables are constants. $\sigma_H^2=(\frac{\partial H}{\partial R}\sigma_R)^2= (\frac{f}{r}\sigma_R)^2\approx 1.22'$
    $\sigma_H\approx \pm1.11'$
5. 	A vertical image, with a pixel size of 1.266 μm, was taken from a flying height of 1,000 m above the ground. The radial distance from the principal point of that image to the top of a building was measured as 1500 pixels. The radial distance from the principal point of that image to the bottom of the same building was measured as 1450 pixels. The standard deviation for these measurements is $\pm4$ pixels. What is the height of that building as well as its standard deviation?	
- ![](https://i.imgur.com/VWzPCep.png)
- answer of height
    - $h=1000(m)\times(\frac{50 (pixels)}{1500(pixels)})=\frac{100}{3}(m)$
- Error Propagation:
    - The error are from $r_a$ and $d$ according to the quesiton. Assume $r_a$ and $d$ are linearly independent as the covariance is not given.
    - $\sigma_d^2=16+16=32 (pixels^2)$ (see Q2 if you cannot understand)
    - $\sigma_h^2=(\frac{\partial h}{\partial d}\sigma_d)^2+(\frac{\partial h}{\partial r_a}\sigma_{r_a})^2=(\frac{H}{r_a}\sigma_d)^2+(-\frac{Hd}{r_a^2}\sigma_{r_a})^2=\frac{128}{9}+\frac{16}{2025}\approx 14.23 (m^2)$ (14.22 if students ignore the second term)
    - $\sigma_h\approx \pm3.77 (m)$ 
6. A vertical photograph was taken with a scale of 1:24,000. The distance between two points on the image is 2.41 inches. The standard deviation of the distance measurement is ±0.008 inches. What is the corresponding object distance between these points as well as its standard deviation?
- answer:
    - $d= 2.41\times 24000=57840 (inches)$
     ($d$ is the disance)
    - $\sigma_d^2=(24000\sigma^2)$
     ($\sigma=\pm0.008 (inches)$)
     $\rightarrow \sigma_d\approx \pm1.24 (inches)$
7. 	What is the area – together with its standard deviation – covered by an aerial photograph taken at a height of $12000\pm{0.5}^\prime$above the average ground with a $4\pm0.01''\times4\pm0.01''$ square on an image acquired by a camera with a focal length of $6\pm{0.003}''$?
- area:
    - $l_1=4''\times\frac{12000'}{6''}=8000'=l_2$
    - $area=A=l_1\times l_2=64000000\ (ft^2)$
- error
    - $l_1=a\frac{b}{c}=l_2$
    - $\sigma_{l_1}^2=\sigma_{l_2}^2=(\frac{\partial l_1}{\partial a}\sigma_a)^2+(\frac{\partial l_1}{\partial b}\sigma_b)^2+(\frac{\partial l_1}{\partial c}\sigma_c)^2=(\frac{b}{c}\sigma_a)^2+(\frac{a}{c}\sigma_b)^2+(-\frac{ab}{c^2}\sigma_c)^2$
    $=(\frac{12000'}{6''}0.01'')^2+(\frac{4''}{6''}0.5')^2+(\frac{4''\times12000'}{6''\times6''}0.003'')^2=416\frac{1}{9} (ft^2)$
    - $area=A=l_1\times l_2$
    - $\sigma_{A}^2=(\frac{\partial A}{\partial l_1}\sigma_{l_1})^2+(\frac{\partial A}{\partial l_2}\sigma_{l_2})^2=2(l_2\sigma_{l_1})^2$
    - $\sigma_A\approx\pm4707759.81\ (ft^2)$
8. You are working on a construction company that is interested in deriving quantitative estimates (e.g., positions, distances, and object heights) over a building site with dimensions of $200m\times500m$. The company is using an Unmanned Aerial vehicle (UAV) equipped with a digital camera. What would be the pros/cons of using images from flying heights of 50 m versus 100 m to derive these measurements in terms of uncertainty?  
- Answer:
    - let $l_1$ = length of the building in images when flightheight = 50m
    $l_2$ = length of the building in images when flightheight = 100m
    Assuming $l_1$ and $l_2$ have the same $\sigma_l$ and fightheight is errorless
    $f$=focal length with $\sigma_f$
    $L$= lengh of the building
    $L=l_1\times 50/f$ or $l_2\times100/f$
    - When flight height = 50:
    $\sigma_L^2=(50\sigma_l/f)^2+(50l_1\sigma_l/f^2)^2$
    - When flight height = 100:
    $\sigma_L^2=(100\sigma_l/f)^2+(100l_2\sigma_l/f^2)^2$
    - let $L$ =500, $l_1$ will become $10f$ and $l_2$ will become $5f$
    $50m: \sigma_L^2=(550\sigma_l/f)^2$
    $100m: \sigma_L^2=(600\sigma_l/f)^2$
    - let $L$ = 200, $l_1$ will become $4f$ and $l_2$ will become $2f$
    $50m:\sigma_L^2=(250\sigma_L/f)^2$
    $100m:\sigma_L^2=(300)\sigma_L/f)^2$
    - Although lower flight height results in lower uncertainty, it causes larger occlusions.
     ![](https://i.imgur.com/hm6y15V.png)
9. 	For the height-based estimation using the x-parallax equation (refer to the Figure below), what is the standard deviation of $H-h_A$ assuming that $B$ and $c$ are errorless (i.e., constants) and the standard deviations for $x_{a_l}$ and $x_{a_r}$ are $\sigma_l$ and $\sigma_r$, respectively?
$B=100m, focal\  length = 0.12m, \sigma_l=\sigma_r=\pm2\ \mu m, x_{a_r}=.02m, x_{a_l}=-0.015m$
![](https://i.imgur.com/Bxh4wLB.png)
- let $z=H-h_A$
- Value of $z$:
    - $p_x=|x_{a_l}|+|x_{a_r}|=0.035\ (m)$
    - $z=Bc/p_x=100\times 0.12/0.035=2400/7\ (m)\approx342.86\ (m)$
- $\sigma_z$
    - $\sigma_{p_x}^2=\sigma_l^2+\sigma_r^2=8\ (\mu m^2)$
    - $\sigma_z^2=(\frac{\partial z}{\partial {p_x}}\sigma_{p_x})^2=(-\frac{Bc}{p_x^2}\sigma_{p_x})^2=78367.35\ (\mu m^2)$
    - $\sigma_z=\pm 279.94\ (\mu m^)$