Calc help - shortest length along spool that is covered?
960889
Circumference of spool = 2PiR\n\nSince the length of the cable is l, the number of times the cable can be wrapped around the spool = l/(2PiR)\n\nEach "wrap" will cover the spool by 2r.\n\nTherefore the total length that can be covered = 2r*l/(2PiR)\n=rl/(PiR)
104678
How long is a piece of string?\n\nYour question is far too general to be answered.
253154
l = length\nw = width\n\nl = 8w\n\nlw = 128\n(8w)(w) = 128\n8w² = 128\nw² = 16\nw = 4\nl = 8(4)\nl = 32\n\nWidth is 4 feet.
968006
The D=8" pulley is rotating at 120 rpm, so you can use its circumference (C) and its rotation speed (W) to find the length of rope (or whatever) that passes over it every minute:\n\nL = W*C = W*πD = 120*π*8 = 3015.9"\n \nNow, this must equal the length of rope passing over the smaller pulley (because its all 1 rope). So use the same equation as before, but with the smaller pulley's circumference (c) and rotation speed (w):  \n\nL = w*c = w*π*d\n\nSolve this for the diameter of the smaller pulley (d):\n\nd = L / (w*π) = (3015.9) / (150*π) = 6.4"\n\nThat's the long way.  It boils down to the (much simpler) expression:\n\nd*w = D*W\n\nFrom which you can get the smaller diameter:\n\nd = (D*W) / w = (8*120) / 150 = 6.4"\n\nBut you have to understand the first way in order to understand the easier way...
724952
http://www.onlineconversion.com/length_common.htm
683260
A length of a single wire:\nL^2 = 72*72 + 54*54 (from a right triangle)\nso L = 90 ft\nFour sets of wires -> he needs 360 ft
928881
how longs a bit of string
337805
lowest to highest... thickest to thinnest: e a d g... as far as gauge goes, they're usually sized about 20 hundredths of an inch apart (ie. lowest to highest .105, .085, .065, .045).
1066144
Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
236391
1S = 1T + 1B\n2S +1T = 8B\n\nrewrite 1st equation as 1S-1T=1B and add 2nd equation. get\n3S=9B, so S=3B\n\n3 buttons balance 1 spool
1057756
lets say length= L\nwidth=W\nperimeter=P=22=2L+2W\nL=W+7\n\n22=2(W+7)+2W\n22=4W+14\n8=4W\n2=W\n9=L
109636
How long is a piece of string! comes to mind
316497
If a curve in n dimensions is given parametrically as:\n\n( f1(s), f2(s), ..., fn(s) ) for a <= s <= b\n\nthen you can approximate the arc length by dividing the domain [a,b] into many small pieces a = s0, s1, s2, ..., sk = b and summing the lengths of the chords (secants? straight line segments. whatever) between s_{i} and s_{i+1}.\n\nThe length of the curve is defined as the least upper bound of all such sums (if that set has an upper bound).\n\nUnder suitable hypotheses you can compute this as the definite integral of ``ds'' as s varies from a to b, using the Euclidean rule\n\n(ds)^2 = (df1(s)/ds)^2 + ... + (dfn(s)/ds)^2\n\ni.e., your length is the definite integral from a to b of the square root of the sum of the squares of the derivative of each coordinate function.\n\nFor example for a circle of radius one at the origin parameterized as ( cos(t), sin(t) ) as t varies from a = 0 to b = 2 pi you get ... 2 pi ... which is what you would expect.  :-)  :-)\n\nOr for the arc length of ( (s^3)/3 - s, s^2 ) from s = 5 to s = 8 the length is the integral from 5 to 8 of\n\nsqrt( (deriv (s^3)/3 - s w.r.t. s)^2 + (deriv s^2 w.r.t. s)^2 ) ds\n\n= sqrt( (s^2 - 1)^2 + (2s)^2 ) ds\n= sqrt( s^4 - 2s^2 + 1 + 4s^2 ) ds\n= sqrt( s^4 + 2s^2 + 1 ) ds\n= (s^2 + 1) ds\n\ni.e., integral from s=5 to s=8 of (s^2 + 1) ds.
918906
google it http://www.webopedia.com/TERM/s/spooling.html
93089
This seems like a homework problem, so I'll give a few hints.\n\nThe arclength is radius times the angle subtended.\n\nTo find the angle subtended, consider the isosceles triangle created inside the circle where the chord is the hypotenuse and the radius of the circle is the two other legs. \n\nSince the isosceles triangle created is not a right triangle you need to use a specific formula to find the angle "C" subtended by the chord.
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
385800
1 foot (12 inches) in a straight line