Fiber Bragg grating - Revision history

166.186.168.52: hyphen

2015-01-06T13:27:35Z

hyphen

← Older revision		Revision as of 15:27, 6 January 2015
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	~~[[Image:Pb_napoleon~~.~~png\|400px\|right]]~~		Hi there. Let me start by introducing the writer, her title is Sophia. Mississippi is accurate psychic predictions ([http://netwk.hannam.ac.kr/xe/data_2/85669 Discover More Here]) exactly where her home is but her spouse desires them to transfer. Office supervising is what she does for a living. What me and my family members adore is to climb but I'm considering on beginning something certified psychics ([http://ustanford.com/index.php?do=/profile-38218/info/ ustanford.com]) new.<br><br>my weblog; psychic love readings ([http://Fashionlinked.com/index.php?do=/profile-13453/info/ http://Fashionlinked.com/index.php?do=/profile-13453/info/])
	~~'''Napoleon's problem''' is a famous [[compass and straightedge\|compass construction]] problem. In it~~, ~~a [[circle]] and its center are given. The challenge~~ is ~~to divide the circle into four equal arcs using only a [[Compass_(drafting)\|compass]]~~. ~~[[Napoleon]] was known to be an amateur [[mathematician]] but it~~ is ~~not known if he either created or solved the problem.~~ ~~Napoleon's friend the Italian mathematician~~ [~~[Lorenzo Mascheroni]] introduced the limitation of using only a compass (no straight edge) into geometric constructions~~. ~~But actually, the challenge above is easier than the '''real Napoleon's problem''', consisting in finding the center of a given circle with compass alone~~. ~~The following sections will describe solutions to both problems, and the proofs that they work~~.

	~~In 1672, [[Georg Mohr~~]~~] produced a book, "Euclides Danicus", that predated Mascheroni, though the book was only rediscovered in 1928.~~

	~~==Finding the centre of a given circle==~~

	~~=== Construction ===~~
	~~Let (C~~) ~~be the circle, whose centre~~ is to ~~be found~~. ~~Let A be~~ a ~~point on (C)~~.

	~~A circle (C1) centered at A meets (C) at B~~ and B'.

	~~Two circles (C2) centered at B and B', with radius AB, cross again at point C.~~

	~~A circle (C3) centered at C with radius AC meets (C1) at D and D'.~~

	~~Two circles (C4) centered at D and D' with radius AD meet at A, and at O, the sought center of (C).~~

	Note: for this to work the radius of circle (C1) must be neither too small nor too large. More precisely, this radius must be between half and double of the radius of (C): if the radius is greater than the diameter of (~~C), (C1) will not intersect (C); if the radius is shorter than half the radius of (C), point C will be between A and O and (C3) will not intersect (C).~~

	~~=== Proof ===~~
	[~~[Image~~:~~PbNapoleon2.svg\|right\|250px]]~~
	~~The idea behind the proof is to construct, with compass alone, the length b²~~/~~a when lengths ''a'' and ''b'' are known, and a~~/~~2 ≤ b ≤ 2a~~.

	~~In the figure on the right, a circle of radius ''a'' is drawn, centred at O; on it a point A is chosen, from which points B and B' can be determined such that AB and AB' have a length of ''b''~~. Point A' lies opposite A, but does not need to be constructed (it would require a straightedge); similarly point H is the (virtual) intersection of AA' and BB'. Point C can be determined from B and B', using circles of radius ''b''.

	~~Triangle ABA' has a right angle at B and BH is perpendicular to AA', so :~~
	~~: <math>\frac{AH}{AB}~~ = ~~\frac{AB}{AA'}<~~/~~math>~~
	~~Therefore <math>AH = \frac{b^2}{2a}<~~/~~math> and AC = b²~~/a.

	~~In the above construction of the center, such a configuration appears twice :~~
	* points A, B and B' are on the circle (C)~~, radius a{{su\|b=1}} = r ; AB, AB', BC, and B'C are equal to b{{su\|b=1}} = R, so~~ <~~math~~>~~AC = \frac{R^2}{r}~~<~~/math~~>;
	* points A, D and D' are on the circle of centre C, radius <math>a_2 = \frac{R^2}{r}</~~math> ; DA, D'A, DO, and D'O are equal to b{{su\|b=2}} = R, so <math>AO = \frac{R^2}{R^2~~/r}= r</~~math>.~~
	~~Therefore O is the centre of circle (C). <!~~-~~- pending details of how AO is perpendicular to BB' and DD', which places O on the diameter through A. -->~~

	~~==Dividing a given circle into four equal arcs given its centre==~~
	~~[[Image~~:~~Nap4.png\|right\|300px]]~~
	Centred on any point X on circle '''''C''''', draw an arc through O (the centre of '''''C''''') which intersects '''''C''''' at points V and Y. Do the same centred on Y through O, intersecting '''''C''''' at X and Z. Note that the [[Line segment\|segments]] OV, OX, OY, OZ, VX, XY, YZ have the same length, all distances being equal to the radius of the circle '''''C'''''.

	~~Now draw an arc centred on V which goes through Y and an arc centred on Z which goes through X; call where these two arcs intersect T. Note that the distances VY and XZ are <math>\sqrt{3}<~~/~~math> times the radius of the circle '''''C'''''~~.

	~~Put the compass radius equal to the distance OT (<math>\sqrt{2}<~~/~~math> times the radius of the circle '''''C''''') and draw an arc centred on Z which intersects the circle '''''C''''' at U and W~~. ~~UVWZ is a square and the [[arc (geometry)\|arcs]] of '''''C''''' UV, VW, WZ, and ZU are each equal to a quarter of the circumference of '''''C'''''.~~

	=~~=See also==~~
	*[[Napoleon's theorem]]
	*[[Napoleon points]]

	~~[[Category:Euclidean plane geometry]]~~
	~~[[Category:Mathematical problems]]~~
	~~[[Category:Articles containing proofs]]~~
	~~[[Category:Napoleon\|Problem]~~]

en>Monkbot: /* Grating structure */Fix CS1 deprecated date parameter errors

2014-01-18T19:02:41Z

Grating structure: Fix CS1 deprecated date parameter errors

← Older revision		Revision as of 21:02, 18 January 2014
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	~~Hi there~~, ~~I am Sophia~~. ~~She~~ is ~~really fond~~ of ~~caving but she doesn~~'~~t have~~ the ~~time recently~~. ~~Credit authorising~~ is ~~how he tends~~ to ~~make money~~. I'~~ve usually loved living in Alaska~~.~~<br><br>Here~~ is ~~my web page best psychic readings~~ ([~~http~~://~~cspl~~.~~postech~~.ac.kr/~~zboard~~/~~Membersonly~~/~~144571 cspl~~.~~postech~~.ac.kr])		[[Image:Pb_napoleon.png\|400px\|right]]
			'''Napoleon's problem''' is a famous [[compass and straightedge\|compass construction]] problem. In it, a [[circle]] and its center are given. The challenge is to divide the circle into four equal arcs using only a [[Compass_(drafting)\|compass]]. [[Napoleon]] was known to be an amateur [[mathematician]] but it is not known if he either created or solved the problem. Napoleon's friend the Italian mathematician [[Lorenzo Mascheroni]] introduced the limitation of using only a compass (no straight edge) into geometric constructions. But actually, the challenge above is easier than the '''real Napoleon's problem''', consisting in finding the center of a given circle with compass alone. The following sections will describe solutions to both problems, and the proofs that they work.

			In 1672, [[Georg Mohr]] produced a book, "Euclides Danicus", that predated Mascheroni, though the book was only rediscovered in 1928.

			==Finding the centre of a given circle==

			=== Construction ===
			Let (C) be the circle, whose centre is to be found. Let A be a point on (C).

			A circle (C1) centered at A meets (C) at B and B'.

			Two circles (C2) centered at B and B', with radius AB, cross again at point C.

			A circle (C3) centered at C with radius AC meets (C1) at D and D'.

			Two circles (C4) centered at D and D' with radius AD meet at A, and at O, the sought center of (C).

			Note: for this to work the radius of circle (C1) must be neither too small nor too large. More precisely, this radius must be between half and double of the radius of (C): if the radius is greater than the diameter of (C), (C1) will not intersect (C); if the radius is shorter than half the radius of (C), point C will be between A and O and (C3) will not intersect (C).

			=== Proof ===
			[[Image:PbNapoleon2.svg\|right\|250px]]
			The idea behind the proof is to construct, with compass alone, the length b²/a when lengths ''a'' and ''b'' are known, and a/2 ≤ b ≤ 2a.

			In the figure on the right, a circle of radius ''a'' is drawn, centred at O; on it a point A is chosen, from which points B and B' can be determined such that AB and AB' have a length of ''b''. Point A' lies opposite A, but does not need to be constructed (it would require a straightedge); similarly point H is the (virtual) intersection of AA' and BB'. Point C can be determined from B and B', using circles of radius ''b''.

			Triangle ABA' has a right angle at B and BH is perpendicular to AA', so :
			: <math>\frac{AH}{AB} = \frac{AB}{AA'}</math>
			Therefore <math>AH = \frac{b^2}{2a}</math> and AC = b²/a.

			In the above construction of the center, such a configuration appears twice :
			* points A, B and B' are on the circle (C), radius a{{su\|b=1}} = r ; AB, AB', BC, and B'C are equal to b{{su\|b=1}} = R, so <math>AC = \frac{R^2}{r}</math>;
			* points A, D and D' are on the circle of centre C, radius <math>a_2 = \frac{R^2}{r}</math> ; DA, D'A, DO, and D'O are equal to b{{su\|b=2}} = R, so <math>AO = \frac{R^2}{R^2/r}= r</math>.
			Therefore O is the centre of circle (C). <!-- pending details of how AO is perpendicular to BB' and DD', which places O on the diameter through A. -->

			==Dividing a given circle into four equal arcs given its centre==
			[[Image:Nap4.png\|right\|300px]]
			Centred on any point X on circle '''''C''''', draw an arc through O (the centre of '''''C''''') which intersects '''''C''''' at points V and Y. Do the same centred on Y through O, intersecting '''''C''''' at X and Z. Note that the [[Line segment\|segments]] OV, OX, OY, OZ, VX, XY, YZ have the same length, all distances being equal to the radius of the circle '''''C'''''.

			Now draw an arc centred on V which goes through Y and an arc centred on Z which goes through X; call where these two arcs intersect T. Note that the distances VY and XZ are <math>\sqrt{3}</math> times the radius of the circle '''''C'''''.

			Put the compass radius equal to the distance OT (<math>\sqrt{2}</math> times the radius of the circle '''''C''''') and draw an arc centred on Z which intersects the circle '''''C''''' at U and W. UVWZ is a square and the [[arc (geometry)\|arcs]] of '''''C''''' UV, VW, WZ, and ZU are each equal to a quarter of the circumference of '''''C'''''.

			==See also==
			*[[Napoleon's theorem]]
			*[[Napoleon points]]

			[[Category:Euclidean plane geometry]]
			[[Category:Mathematical problems]]
			[[Category:Articles containing proofs]]
			[[Category:Napoleon\|Problem]]

en>ماني: interwiki Persian

2012-07-06T20:21:37Z

interwiki Persian

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