Artificial Light
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In our eReader you can find the full English version of the book. Read Artificial Light Online - link to read the book on full screen. Our eReader also allows you to upload and read Pdf, Txt, ePub and fb2 books. In the Mini eReder on the page below you can quickly view all pages of the book - Read Book Artificial Light
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A real practical limitation which still exists is thecurvature of the earth, and the distance an object may be seen by theeye at sea-level depends upon the height of the object. The relation isapproximately expressed thus, -- Range in nautical miles = 8/7 square root of Height of object infeet. For example, the top of a tower 100 feet high is visible to an eyeat sea-level a distance of 8/7 square root of 100 = 80/7 = 11. 43miles. Now if the eye is 49 feet above sea-level, a similar computationw...ill show how far away it may be seen by the original eye at sea-level. This is 8/7 square root of 49 = 8 miles. Hence an eye 49 feet abovesea-level will be able to see the top of the 100-foot tower at adistance of 11. 43 + 8 or 19. 43 nautical miles. Under these conditions animaginary line drawn from the top of the tower to the eye will be justtangent to the spherical surface of the sea at a distance of 8 milesfrom the eye and 11. 43 miles from the tower.
The luminous intensity of a light-source or of the beam of light isdirectly responsible for the range.
What to read after Artificial Light?
You can find similar books in the "Read Also" column, or choose other free books by Luckiesh, Matthew, B. 1883 to read onlineMoreLess
To quickly assess the difficulty of the text, read a short excerpt:
A real practical limitation which still exists is thecurvature of the earth, and the distance an object may be seen by theeye at sea-level depends upon the height of the object. The relation isapproximately expressed thus, -- Range in nautical miles = 8/7 square root of Height of object infeet. For example, the top of a tower 100 feet high is visible to an eyeat sea-level a distance of 8/7 square root of 100 = 80/7 = 11. 43miles. Now if the eye is 49 feet above sea-level, a similar computationw...ill show how far away it may be seen by the original eye at sea-level. This is 8/7 square root of 49 = 8 miles. Hence an eye 49 feet abovesea-level will be able to see the top of the 100-foot tower at adistance of 11. 43 + 8 or 19. 43 nautical miles. Under these conditions animaginary line drawn from the top of the tower to the eye will be justtangent to the spherical surface of the sea at a distance of 8 milesfrom the eye and 11. 43 miles from the tower.
The luminous intensity of a light-source or of the beam of light isdirectly responsible for the range.
What to read after Artificial Light?
You can find similar books in the "Read Also" column, or choose other free books by Luckiesh, Matthew, B. 1883 to read onlineMoreLess
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