Campanus Cusps is a topic in astrology and in astronomy. I know that there exists iteration methods to do this. There may be explicit equations too. I want to know any mathod
125115
Why don't you look it up on the Internet???
166208
Ah a question right up my alley...Moon - no you cant see the mirror, it would take 6 seconds for a light beam to get from here to there and back. Look for the crater Tycho (most notible because of the "ray" pattern) it is in the NE corner of the moon (you will see it in the SW corner because telescopes invert the image. Several other craters and mountains, make note of shadow positions. Using a lunar chart, find the Sea of Tranquility (Mare Tranquillium) where we first set down in 1969.\n\nSun - Using a white card held up to the eyepiece (some scopes have a bracket that can do this) Line the scope up so that it points to the sun DO NOT LOOK THRU THE SCOPE USE THE SHADOW ON THE GROUND. Focus the suns image on the white card and follow sunspots.\n\nOther planets Mars - track mars throughout the year and note the stars behind it.  As you see in just a few months mars will appear to go backwards in the sky then go forward again,  This is called retrograde.\n\nYou know of jupiter and saturn ring tilt, great spot moons etc.\n\nStar clouds and star clusters - you will need a star chart and not all are visable at the same time - M42 Orion Nebulae in his belt, M31 Andromada galaxy, the only object outside our galaxy visable to the naked eye located just above the second of third pair of stars coming off the great square pegasus - in the big dipper, second star in the handle should see a double star, this was used to test ship navigators in olden times, they had to be able to see the two stars with the naked eye.\n\nThat should keep you busy for one night LOL just kidding...email me anytime you need help
202153
ABOUT THE BINOMIAL THEOREM I AM TEEMING WITH A LOT OF NEWS\n\nWITH MANY CHEERFUL FACTS ABOUT THE SQUARE ON THE HYPOTENEUSE\n\n(W. S. Gilbert, I am The Very Model of a Modern Major General)\n\nI am assuming we are dealing with what we know is a PRIMITIVE Pythagorean triple ie one like (3,4,5) or (5,12,13) or (8,15,17) or (20,21,29) in which there is no common factor between the three sides, NOT one like (6,8,10) which is divisible by 2.\n\nThere are tables published of the smallest triples, in ascending order of hypooteneuse, so you may be able to look it up and not need to calculate it. \n\nhttp://www.math.rutgers.edu/~erowland/tripleslist-long.html\n\nHowever, two points.  \n\nNot every integer can be a hypoteneuse. There are gaps in the table eg after (3,4,5) the next smallesr hypoteneuse is 13 (in (5, 12, 13). There are no primitive triples with a hypoteneuse of 6, 7, 8, 9, 10, 11 or 12. And therefore if you were given a hypoteneuse of 12, there are no integer solutions to it.\n\nSecondly, there are cases where a number is hypoteneuse to more than one triple. 65 is the lowest such number and 85 is the next.\n\n(16, 63, 65) and (33, 56, 65)\n(13, 84, 85) and (36, 77, 85)\n\nThese are known as siblings. Both give correct solutions and you can't say one is right and the other is wrong. Here are a couple of special cases, with more than two solutions:\n\n(47, 1104, 1105)\n(744, 817, 1105)\n(576, 943, 1105) \n(264, 1073, 1105)\n\nThere are EIGHT siblings with a hypoteneuse of 32045, SIXTEEN siblings with a hypoteneuse of 1185665, THIRTY-TWO siblings with a hypoteneuse of 48612265, AND SIXTY FOUR siblings with a hypoteneuse of 2576450045.\n\nHere are rthe eight siblings with a hypoteneuse of 32045:\n\n(2277, 31964, 32045) \n(8283, 30956, 32045)\n(17253, 27004, 32045) \n(21093, 24124, 32045)\n(22244, 23067, 32045)\n(15916, 27813, 32045)\n(6764, 31323 , 32045)\n(716, 32037, 32045)\n\nLacking access to tables, for small numbers, you may be able to work it out, by trial and error, using the following properties\n\n(a) the hypoteneuse must be the biggest side and must be an odd number (to be a primitive triple)\n\n(b) one of the legs will be an odd number and the other an even number\n\n(c) As a result, the difference between the two small sides will always be an odd number. In fact it is either a prime number (eg 1, 7 and 17 feature, but 11, 13 and 19 do not) or a power of a prime number (49 is the lowest such).\n\n(d) the sum of the two legs must be bigger than the hypoteneuse\n\n(e) one and only one of the legs must be divisible by 3\n\n(f)  one and only one of the legs must be divisible by 4\n\n(g)  one and only one side (which might be the hypoteneuse) must be divisible by 5\n\n(h) follows from (e) (f) and (g)\n\n(h) the product of the three sides is always divisible by 60\n\nThus if you knew the hypoteneuse and could see it was not divisible by 5, then it follows the legs must, between them, be divisible BY 3 4 AND 5\n\nSo if you knew the hypoteneuse is 61, it seems obvious to try one of the legs as 60 and see if that works (it does, (11, 60, 61) is a triple. \n\nIn fact you can work out the third member is 11 by adding 60 and 61 to see if it is a perfect square (it is 121) and taking the square root of it, if so. (another property of triples whose two largest sides differ by 1).\n\nGiven that 17 is a hypoteneuse and knowing one of the other sides must be 3 x 4 or 3 x 5 (4 x 5 would be larger than 17) you have a choice of (12, x, 17) where x = 5 or 15 (has to be odd as 12 is even) or (y, 15, 17) where y is 4, 8, 12 or 16)\n\nIt turns out to be (8, 15, 17)\n\nAgain, knowing a property of all triples where the hypoteneuse and the larger leg differ by 2, there is a shortcut, that leads to 8 ... take the average of the two longer sides, multiply by 4 and take the square root. (thus 16 => 64 => 8)\n\nThis holds true for \n(12, 35, 37)\n(16, 63, 65)\n(20, 99, 101) etc\n\nSo it is possible to make intelligent guesses at the
22552
Photometric (O-C) diagrams are the tool of choice for representing period variability. For eclipsing stars the ordinate in such diagram is generally formed from observed (O) timings of minimum light from which is subtracted the time calculated ( C ) from some convenient pdarkredictive ephemeris. The abcissa is conventionally the cycle count of orbital revolution or some other measure of time
273729
False. One star in a constellation could be a only a few dozen light years away from us, while on right next to it in the constellation might be a galaxy millions of light-years away.  \n\nSecond answer: True, the zodiac is the set of constellations along the elliptic.\n\nThird answer: False, the seasons are caused by the tilt of the earth's axis, not the precession (the movement of the earth's polar axis around the ecliptic pole, a cycle that takes almost 26,000 years).\n\nFouth answer: True.  At winter soltice, the Sun is as far south as it's going to get.
180718
So you are doing some numerical analysis... if you have a spreadsheet program you can use that. Use the trapezoid rule or Simpson's rule for some small increment in x.  Your spreadsheet program is able to evaluate arctan.  The second question can be evaluated with Newton's method.
368520
Scientists using NASA's Spitzer Space Telescope have discovered and directly imaged a small brown dwarf star, 50 times the mass of Jupiter, orbiting with a planet around a Sun-like star. Such an arrangement has never before been seen but might be common, the scientists say, leading to solar systems with distorted planetary orbits.\n\nKevin Luhman of Penn State University is the lead author on a report describing this discovery, which will be published in The Astrophysical Journal. The discovery concerns a class of the coldest brown dwarfs, called T dwarfs.\n\n\nThe detailed research paper on T dwarf - Ref\nhttp://www.science.psu.edu/alert/Luhman9-2006-2.htm
151190
This sounds like a homework question?  Maybe I shouldn't give you the whole answer, so you can learn something on your own.\n\nYou are describing a Cassegrain telescope.  You can use the mirror formula to calculate the image plane for the first optic (the concave mirror) and then use that as the object plane for the second optic (the convex mirror) to get your final focal image plane.\n\nMirror Equation:  (1/So) + (1/Si) = -2/R\n\nSo = object plane (distance from object to mirror)\nSi = image plane (distance from image to mirror)\nR  = radius of curvature of mirror\n\nThe hardest part of these calculations is to use the right sign convention and to remember that the stars (initial object plane) can be treated as at infinity.
193662
You have to find the Least Common Multiple.\nFirst we know that it's going to be 84 or more years because Uranus takes 84 years to complete one orbit.\n So we know that the other planets' time length to make one orbit has to go into 84. \nFirst I did Earth: 84/1 = 84. \nNow I need to do Jupiter: 84/12 = 7. \nNext I do Saturn: 84/30 = 2 remainder 24. \nSo now I need to find a common multiple of 30 and 84. \n84 x 2 = 168 (no). 84 x 3 = 252 (no). 84 x 4 = 336 (no). 84 x 5 = 420 (yes).\n\n  14 \n  30)420 \n  -30 \n  --- \n  120 \n  -120 \n  --- \n  0 \n\nSo 420 is the number that goes into all the time lengths for the planets. Therefore we find out out that they will align once in 420 years.\nIf you knew the last time they were all in a line. You add that year + 420 and you get the next year when they will alll be in a line.
517016
Kepler had something to say about this type of problem.\nGMT^2=4*pi^2*a^3
120488
every one that passes by (-1,6). there are infinite equations. You must give some more clues.\n\n Let me explan myself: \nif your equation were x=-1 (completely horizontal)\nif your equation were y=6 (completely vertical)\nand so on f you have a crcle, a hperbolic, a sinoidal, a straight line with some inclination.... there are just a lot!!!!\n\nhope i help\n\nthanks for the details\n\nJust substitute x=-1 on each equaton, and the one that results on y=6 is the one you are looking for.
170379
All these problems are available in coordinate geometry text books.  For the problem 1)\n\nThe equation of the circle with A(x1,y1)and B(x2,y2) as end points of diameter is (x-x1)(x-x2) + (y-y1)(y-y2) = 0\n\nSimply substitute A(2, -3) and B(8,5) wou get \n(x-2)(x-8) + (y+3)(y-5) = 0\nthat is, x^2 - 10x+16 + y^2 -2y -15 =0\nOr, x^2 + y^2 - 10x -2y + 1 =0\n\nThis is the required equation.\n\n2) In standard form of parabola y^2 = 4ax, the origin (0,0) is vertex, S(a,0) is focus and x + a = 0 is the equation of directrix.  In the present case, the focus is at (2,3).  this means the coordinates of vertex are (0,3)and the equation of diretrix is x + 2 = 0.  Therefore the equation of the parabola is (y-3)^2 = 4. 2. (x-2)\nOr y^2 -6y -8x + 25 =  0\n\nSimilarly attempt the problems 3 & 4
59168
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
467188
can you give examples (such as the witches equation)
588522
Actually, astrology is no longer considered a science, so scientists as such would not have a lot to say about the mystical significance of planetary conjunctions.\nAstronomy (the scientific study of the universe and space) is quite separate and distinct from astrology.
376799
i don't know anything about this book or author.  it is bogus so forget it.\n\nthere is one star (sun), eight planets (mercury, venus, earth, mars, jupiter, uranus, and neptune), hundreds of dwarf, ball-shaped bodies ("1 ceres", "134340 pluto", and "136199 eris" for example), and tens of thousands small irregularly-shaped bodies in the solar system, and there is no hint of another star or any more planets.\n\nhttp://en.wikipedia.org/wiki/Extrasolar_planet\nhttp://exoplanet.eu/catalog.php\n\nin 2006,  astronomers found evidence for a planet orbiting a star 21,500 ± 3,300 away.  the star is cataloged as OGLE-2005-BGL-390L.  the planet, OGLE-2005-BGL-390Lb, is 5.5 earth masses, but astronomers have also found more than 200 others.
663802
Ugh I cannot believe people actually ask this retarded questions.\nTell you what sparky, go outside at night, when there are plenty of stars. Now, take a picture of a friend with a flash or using an external light (as they did on the moon) Know what you'll find when the photos come back? NO STARS!\nWhen you understand what FOCAL LENGTH means, you'll know why the photos on the moon have no stars in them. I hope the website helps you understand...