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IOAA 2019, Теория — Задача 6. The height of the chimney of Tiszaújváros power plant

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Всички материали от IOAA 2019

Theory problem sheet · 20 т.

Условие

The European Copernicus Earth-observation programme operates two Sentinel-2 remote sensing satellites. These satellites orbit the Earth on Sun-synchronous polar orbits at about 800 km altitude. They pass over a given area once every few days, always taking images at the same local time (accurate to within a few minutes). The cameras are sensitive to 13 different optical and near-infrared spectral bands. The resolution of the images is 10 meters.

The 3rd tallest building in Hungary is the chimney of a power plant near the town Tiszaújváros. You can see two Sentinel-2 satellite images. One is from June 29, another one is from December 16, close to the summer and winter solstices, respectively. The orientation of the images is as normal, i.e. north is up and east is to the right.

False-colour Sentinel-2 satellite image of the Tiszaújváros power plant area with the Tisza river, taken on 29 June 2018; short shadows, scale bar 100 m lower left.

False-colour Sentinel-2 satellite image of the same power plant area taken on 16 December 2018; a long chimney shadow is visible, scale bar 100 m lower left.

The estimated shadow lengths, based on the images above and the scales given in the lower-left corners, of them are $x_1 = 125\ \mathrm{m}$ and $x_2 = 780\ \mathrm{m}$ . Answer the following questions:

a) On which date do we expect the shadow to be longer? Mark the letter of your answer with × on the answer sheet.

(A) on June 29
(B) on December 16 [1 т.]

b) At which time of the day did the Sentinel-2 satellites fly over this area? Mark the letter of your answer with × on the answer sheet.

(A) early morning (B) late morning (C) early afternoon (D) late afternoon [1 т.]

c) Based on the given shadow lengths, estimate the height of this chimney. For this calculation only, assume that the satellite images were taken at local noon. [16 т.]

d) What could affect the accuracy of the chimney's derived height (more than one choice is possible)? Mark the letter of your answer with × on the answer sheet.

(A) The oblate spheroid shape of the Earth.
(B) Limited resolution of the satellite images and the ill-defined edge of the shadow.
(C) The elevation of the base of the tower above sea level.
(D) Seasonal variation in the tilt of the Earth's rotational axis.
(E) Taking into account the effect of atmospheric refraction. [2 т.]

Решение

a)

(B) the one on December 16 (1 p)
At a given moment of the day, the Sun elevation is lower in winter on the northern hemisphere.

b)

(B) late in the morning (1 p)
The shadows point slightly west of north. In fact these satellite images are taken at around 9:45 UTC (10:45 Central European Time).

c)

The measured shadow lengths are $x_1 = 125\ \mathrm{m}$ (epoch 1, summer solstice) and $x_2 = 780\ \mathrm{m}$ (epoch 2, winter solstice).
Let the unknown height of the chimney be $h$ . The chimney and its shadow form the legs of a right-angled triangle and thus

\text{(6.1)}\quad \tan\alpha_1 = h/x_1

\text{(6.2)}\quad \tan\alpha_2 = h/x_2,\qquad \text{(1 p)}

where $\alpha_1$ and $\alpha_2$ are the Sun elevation angles at epoch 1 and epoch 2, respectively. (1 p)
At the time of the summer and winter solstices, the Earth is located at opposite sides of the Sun during its orbital revolution. Since the Equator is tilted by $\varepsilon \approx 23.5^\circ$ to the orbital plane, at a given point of the surface, and at a given time of the day, the Sun elevation angles will differ by $\alpha_1 - \alpha_2 = 2\varepsilon$ (here we ignore refraction and that the satellite images were not taken on the exact solstice days). (4 p)
Notice that, to a good approximation, $\alpha_1 - \alpha_2 \approx 45^\circ$ . Assuming $\alpha_1 = \alpha_2 + 45^\circ$ makes the solution a lot easier. Using the trigonometric identity

\text{(6.3)}\quad \tan\alpha_1 = \tan(\alpha_2 + 45^\circ) = \frac{\tan\alpha_2 + \tan 45^\circ}{1 - \tan\alpha_2\tan 45^\circ},

and substituting $\tan 45^\circ = 1$ , Eq. (6.3) becomes

\text{(6.4)}\quad \tan\alpha_1 = \frac{1 + \tan\alpha_2}{1 - \tan\alpha_2}.\qquad \text{(4 p)}

Substituting $\tan\alpha_1$ and $\tan\alpha_2$ from Eqs. (6.1) and (6.2) into Eq. (6.4), one gets a quadratic equation for the unknown $h$ :

\text{(6.5)}\quad \frac{h}{x_1} = \frac{1 + \dfrac{h}{x_2}}{1 - \dfrac{h}{x_2}}.\qquad \text{(2 p)}

Reducing this equation to the standard format

\text{(6.6)}\quad h^2 + h(x_1 - x_2) + x_1 x_2 = 0,

and according to the quadratic formula, the two roots are obtained as

\text{(6.7)}\quad h_{1,2} = \frac{x_2 - x_1 \pm \sqrt{(x_1 - x_2)^2 - 4x_1x_2}}{2}.\qquad \text{(1 p)}

Now with the measured values of $x_1 = 125\ \mathrm{m}$ and $x_2 = 780\ \mathrm{m}$ , the two roots for $h$ are $h_1 = 426\ \mathrm{m}$ and $h_2 = 229\ \mathrm{m}$ . (1 p)

Obviously, only one of the solutions is valid for our case.
One may simply notice that $h_1 \approx 426\ \mathrm{m}$ seems just too high for a chimney (would in fact be comparable to the Empire State Building in New York, with 443 m; the Eiffel tower in Paris is only 324 m high).
Alternatively, substituting $h_1 = 426\ \mathrm{m}$ into Eq. (6.2), we obtain the Sun elevation angle at around the winter solstice, $\alpha_2 = \arctan(h_1/x_2) \approx 28.7^\circ$ . However, at the geographic latitudes of Hungary, the Sun does not rise above $\approx 19^\circ$ at this time of the year, so the larger value obtained for $h$ is clearly invalid. (2 p)
Note that according to the official data, the Tiszaújváros power plant has a chimney as high as $h = 250\ \mathrm{m}$ .

d)

(B) and (E) (2 p)

Оригинал в Архива: TH-problem-final.pdf · официални решения: TH-solution-final.pdf · допълнителни данни: TH-anssheet.pdf