|
Top Document: [sci.astro] Time (Astronomy Frequently Asked Questions) (3/9) Previous Document: C.10 Why isn't the earliest Sunrise (and latest Sunset) on the longest day of the year? Next Document: C.12 What is the time delivered by a GPS receiver? See reader questions & answers on this topic! - Help others by sharing your knowledge John Horton Conway (the Princeton mathematician who is responsible for "the Game of Life") wrote a book with Guy and Berlekamp, _Winning Ways_, that describes in Volume 2 a number of useful calendrical rules. One of these is an easy "in your head" algorithm for calculating the phase of the Moon, good to a day or better depending on whether you use his refinements or not. In the 20th century, calculate the remainder upon dividing the last two digits of the year by 19; if greater than 9, subtract 19 from this to get a number between -9 and 9. Multiply the result by 11 and reduce modulo 30 to obtain a number between -29 and +29. Add the day of the month and the number of the month (except for Jan and Feb use 3 and 4 for the month number instead of 1 and 2). Subtract 4. Reduce modulo 30 to get a number between 0 and 29. This is the age of the Moon. Example: What was the phase of the Moon on D-Day (June 6, 1944)? Answer: 44/19=2 remainder 6. 6*11=66, reduce modulo 30 to get 6. Add 6+6 to this and subtract 4: 6+6+6-4=14; the Moon was (nearly) full. I understand that the planners of D-day did care about the phase of the Moon, either because of illumination or because of tides. I think that Don Olsen recently discussed this in _Sky and Telescope_ (within the past several years). In the 21st century use -8.3 days instead of -4 for the last number. Conway also gives refinements for the leap year cycle and also for the slight variations in the lengths of months; what I have given should be good to +/- a day or so. User Contributions:Comment about this article, ask questions, or add new information about this topic:Top Document: [sci.astro] Time (Astronomy Frequently Asked Questions) (3/9) Previous Document: C.10 Why isn't the earliest Sunrise (and latest Sunset) on the longest day of the year? Next Document: C.12 What is the time delivered by a GPS receiver? Part0 - Part1 - Part2 - Part3 - Part4 - Part5 - Part6 - Part7 - Part8 - Single Page [ Usenet FAQs | Web FAQs | Documents | RFC Index ] Send corrections/additions to the FAQ Maintainer: jlazio@patriot.net
Last Update March 27 2014 @ 02:11 PM
|
with stars, then every direction you looked would eventually end on
the surface of a star, and the whole sky would be as bright as the
surface of the Sun.
Why would anyone assume this? Certainly, we have directions where we look that are dark because something that does not emit light (is not a star) is between us and the light. A close example is in our own solar system. When we look at the Sun (a star) during a solar eclipse the Moon blocks the light. When we look at the inner planets of our solar system (Mercury and Venus) as they pass between us and the Sun, do we not get the same effect, i.e. in the direction of the planet we see no light from the Sun? Those planets simply look like dark spots on the Sun.
Olbers' paradox seems to assume that only stars exist in the universe, but what about the planets? Aren't there more planets than stars, thus more obstructions to light than sources of light?
What may be more interesting is why can we see certain stars seemingly continuously. Are there no planets or other obstructions between them and us? Or is the twinkle in stars just caused by the movement of obstructions across the path of light between the stars and us? I was always told the twinkle defines a star while the steady light reflected by our planets defines a planet. Is that because the planets of our solar system don't have the obstructions between Earth and them to cause a twinkle effect?
9-14-2024 KP