2020年GeCAA理论第1题-天文摄影
英文题目
1 Astrophotography
An astrophotographer, based on the Equator, uses a good digital camera on a tripod, but with no tracking. The camera has a telescopic lens with focal length of 150mm and aperture (objective diameter) of 84mm. The sensor has effective light collecting diameter of 22.5mm. The photographic target is a star field at the observer’s Zenith.
(a) (2 points) Calculate the field of view (the angular width of the image) which can be captured on the sensor using this equipment.
(b) (5 points) The pixels in the camera’s sensor are separated by a distance of 3.23 μm. What is the maximum possible exposure time for a single frame, so that no star trails appear on the exposed image?
(c) (3 points) For a better-quality image of the star field, the photographer decides to take multiple shots at the exposure time calculated in b) and then to stack them together. The total time for all these shots is 30 s (ignore any time taken to write data to the memory card) What proportion of the total field of view is possible at the higher signal to noise ratio?
中文翻译
1 天文摄影
一位在赤道上的天文摄影师用一部没有跟踪的相机进行摄影。这台相机接在焦距150mm、口径84mm的望远镜后面。相机的感光元件为直径22.5mm。他要拍摄的目标为位于天顶的星场。
（a）（2分）计算这套仪器的视场大小。
（b）（5分）感光元件上像素间距离为3.23μm。计算星点不拖线情况下最大的曝光时间
（c）（3分）为了获得更高质量的星场照片，摄影师决定用（b）中计算出的曝光时间拍摄多张照片并将它们叠加在一起。总曝光时间为30s（忽略所有的读卡写卡时间）。计算拥有最高信噪比的部分占总视场的比例。
官方解答
(a) By simple right angle triangle,
$$FOV = 2 \times tan^{−1} ( \dfrac{sensor width} {2 × Focal Length}) = 2 × tan^{−1} ( \dfrac{22.5} {2 × 150})$$
$$α ≈ 8.58°$$
(b)<nowiki> Number of pixels on the sensor will be given by,
$$N = \dfrac{Sensor width}{pixel width}= \dfrac{22.5}{3.23 × 10^{−3}} = 6965.94$$
As the number of pixels have to be integer, we take $$N = 6966$$
Angular coverage of each pixel will be,
$$Pixel view = \dfrac{8.578°} {6966} = 0.001 23°/pixel ≈ 4.43′′/pixel$$
As the stars complete one full circle in 23.9344 hours,
student loses this one mark if 15° per hour is used
$$t_1 = 1.23 × 10^{−3}\dfrac{23.9344h} {360} = 0.294 s ≈ 0.3 s$$
(in reality this is probably a factor of 10 smaller than the eye would detect)
(c) In 30 seconds, the sky will move by,
$$Δα = \dfrac{360°} {23.9344 × 120} = 0.125 342°$$
$$\therefore \dfrac{ α − Δα}{α}=\dfrac{8.578 − 0.125342}{8.578}= 98.5% $$
(or 8.453° is total field of view in high resolution images from the stack)
Only penalise 15°/hr once in the whole solution.
解答翻译
(a) 根据简单的直角三角形
$$FOV = 2 \times tan^{−1} ( \dfrac{传感器宽度} {2 × 焦距}) = 2 × tan^{−1} ( \dfrac{22.5} {2 × 150})$$
$$FOV ≈ 8.58°$$
(b)<nowiki> 总像素数量由下式给定,
$$N = \dfrac{传感器直径}{像素宽度}= \dfrac{22.5}{3.23 × 10^{−3}} = 6965.94$$
As the number of pixels have to be integer, we take $$N = 6966$$
Angular coverage of each pixel will be,
$$Pixel view = \dfrac{8.578°} {6966} = 0.001 23°/pixel ≈ 4.43′′/pixel$$
As the stars complete one full circle in 23.9344 hours,
student loses this one mark if 15° per hour is used
$$t_1 = 1.23 × 10^{−3}\dfrac{23.9344h} {360} = 0.294 s ≈ 0.3 s$$
(in reality this is probably a factor of 10 smaller than the eye would detect)
(c) In 30 seconds, the sky will move by,
$$Δα = \dfrac{360°} {23.9344 × 120} = 0.125 342°$$
$$\therefore \dfrac{ α − Δα}{α}=\dfrac{8.578 − 0.125342}{8.578}= 98.5% $$
(or 8.453° is total field of view in high resolution images from the stack)
Only penalise 15°/hr once in the whole solution.
