How long is a piece of string?
401023
as long as you want it to be =)
348345
12 inches + 7 over 8 inches long
1166850
a long bow is six feet long.  that's why it's called a long bow, as opposed to just another bow.
23477
Length = 6,125,000 inches = 510,416.66 feet = 96.66 miles\n\nHeight = 2,625,000 inches = 218,750 feet = 41.43 miles\n\nHeh I wonder why you need to know this ;-)
683
Long enough to reach the end of a hole
623638
Remaining length of string\n= 7m - 2.56m\n= 4.44m\n= 444 cm\n\nHow did I do it, below is the working.\n7m is also same as 7.00m.\n\n  7.00\n- 2.56\n------\n  4.44\n------\nCancel the last 0 to 10.\nCancel the middle zero to 9.\nCancel the 7 to 6.\n\n1m = 100cm\n4.44 m = 100 cm * 4.44 = 444cm
1114562
twice the length from one end to the middle
651027
A furlong is a measure of distance within Imperial units and U.S. customary units. Although its definition has varied historically, in modern terms it equals 660 feet or 220 yards, and is therefore equal to 201.168 metres.
829353
the longest chord .... the one that passes through the center ... is a diameter
724952
http://www.onlineconversion.com/length_common.htm
962049
If you are talking about the size usually it is a smaller size such as a 27 or 25 gauge.  And yes as the idiot said before me...it is a sterile one...lol.The length is 1inch to 1 1/2 inches.
959035
It's not "big", it's "long".  About 3 1/3 feet long.  Or 39.4 inches.
104678
How long is a piece of string?\n\nYour question is far too general to be answered.
762503
The suggestion above, to measure from the nut to the bridge, gives you the scale length.  A standard acoustic has a scale length of 24 - 26 inches.  Smaller than that, and you could call it a 3/4. From the above, a 3/4 might measure 18 - 20 in.\n\nAnother way to do it is to measure the body & neck together. A standard steel string measures just over 40 in from head to base - the entire length.  A classical (gut/nylon strings) is generally 38-39 in.  Take 3/4 of that to see where the guitar fits in this scale.\n\nHope this helps.
300240
LongestPenis x 12 characters\n ************ = 12 asterisks (12 across)
823010
As I write this, the question has been posted for 3 hours without an answer, so I'll give it a try.  When you suspend a slack (not taut) string between two points, the string, having weight, will naturally form a parabolic shape between the points.\n\nEarlier I did the math, figuring out how long the string would be if strung across the 8-inch diameter and falling parabolically to a depth of 10 inches in the center.  This is a "length of curve" problem that involves some very messy calculus.\n\nThat's not the only problem.  Having done the math for the longest string -- the one that crosses the diameter -- I realized that the same problem would have to be repeated for each of the chords that will have shorter distances between the endpoints and shorter depths.\n\nAs a result, after reading your question again, I decided to take a new tack, focusing on the "how do you make" part of your question.  There's still some math, but not nearly as much as there would've been otherwise.\n\nTo begin, think of a circular tennis racket with loose strings sinking in the center.  That's the kind of net we're going to design.  For orientation, think of the 8-inch circular rim as the face of a clock. with 12:00, 1:00, ... 11:00 positions.\n\nFor the general shape, long strings will connect the 12:00 and 6:00 positions, falling parabolically to a central depth of 10 inches, and also connect the 3:00 and 9:00 positions, falling to the same depth.\n\nThere can be shorter strings connecting 1:00 and 5:00, 11:00 and 7:00, 10:00 and 2:00, and 8:00 and 4:00, and falling to a shallower depth (less than 10 inches).  These four strings will all be the same length.  Another set of four still-shorter strings will connect 2 and 4, 10 and 8, 11 and 1, and 7 and 5, again all of equal length and falling to a still-shorter depth.\n\nAll strings will be parabolic.  The longer and deeper strings will be in the center, and the shorter and shallower strings (the chords) will be toward the sides.  There will be an odd number of both vertical (e.g., 12-6) and horizontal (e.g., 9-3) strings because there's one central string in each direction and an even number of strings on the sides due to symmetry.\n\nTo form the net, the strings will all be woven as they interlace with each other.  I imagine that your net may be constructed with fishing line.\n\nHaving described the general structure of the net, we now have to do some math to figure out the spacing of the strings.  Going back to the tennis racquet idea, let's suppose we want the strings to be equally spaced <when viewed from above.>\n\nFor an example, suppose the strings are a half-inch apart.  Let the 12-6 line be the y-axis and the 9-3 line be the x-axis.  We want vertical strings at x-values of 0, 0.5, 1.0, 1.5, ... 3.5 (plus and minus) for a total of 15 strings.  There'll also be 15 horizontal strings for a total of 30.\n\nWe've already said that the central strings will attach across diameters (plus and minus 4) and descend to a depth of 10.  In three dimensions, suppose the bottom of the net is at (0,0,0), and the top circle is in the plane z = 10.  The equation of the parabola for the center strings is z = ax^2, passing through (4,10), so 10 = 16a, or a = 5/8, and the final equation in the x-z plane is\n\nz = (5/8) x^2  (Equation 1)\n\nLet's look at the placement of the string at x=3.  In the x-y plane, the circle has radius 4, and by the Pythagorean Theorem, the y-value is sqrt(4^2 - 3^2) = sqrt 7.  This string will be attached at (3, sqrt 7, 10) in 3 dimensions, with the other end at negative sqrt 7 in the y-dimension.  The distance between the attachment points is 2 sqrt 7.\n\nHow far down will this string go?  To get that, we evaluate Equation 1 at x=3:\n\nz = (5/8)(3^2) = 45/8 = 5 5/8 = 5.625\n\nThis means the <height of the netting> at x=3 is z= 5.625 inches.  But the top of the netting is at z=10, so the <drop> is 10 -
930279
Twice as long as it is from one to the middle!  How did I do?  This is the correct answer by the way.