how do you make a net with a radius of 4 in and height of 10 in?
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 -
481334
A net is a drawing where you "flatten" the cuboid from a three dimensional representation into a two dimensional representation of each face of the cuboid. I believe that the question wants you to label the top, bottom, front, back, and each side of the original cuboid as you draw the net.
120304
Just get a piece of plastic sheeting..Suspend it off the ground 6 inches..Put a weight of some type in the middle, to form a downward cone..thats about it..You might also look it up in your Boy Scout Hand Book..its a good read.
557989
You can't. It depends on which way the diameter is turned in the circle. There are an infinite amount of solutions. Tell your teacher to be more specific.\n\ntosmith
364081
10 feet from the floor to the hoop....are u asking how high it is? the net itself is only about a foot in length.
909019
You'll get the same strength(within reason) and the same dimensions using this method.I can't believe noone has answered this question yet.Good luck on the other site!
752012
try   funbasedlearning.com
984598
Search Google and SourceForge :)\n\nRawlyn.
796040
material needed=2pir(h+r)\nvolume=pir^2h\npir^2h=1000cc (given)\nh=1000/(pir^2)\nsubstituing h in the equation 2pir(h+r)\nmaterial=2pir(1000/pir^2+r)\n=2pir*1000/pir^2+2pir^2\ndA/dt=-2000/r^2+4pir\nsetting this to zero\n-2000+4pir^3=0\n4pir^3=2000\nr^3=2000/4pi\nr=cube root of 500/3.14\n=5.4 approx\nsubstituting\nh=1000/(3.14*5.4*5.4)\n=10.9 aapprox\nso the radius is 5.4 cm approx\nand the height 10.9 approx
24463
Its easy to make one yourself.\n\nhttp://www.graphic.org/venbas.html
890237
4x10 , 8x5 it amounts to 4ft x10ft
868064
you can create numbers and shapes out of cardboard\ncreate cone/hat out of nice colored paper.\nand of course put a string trough with needle
813209
perimeter= 4(L+W+H)\narea= length times the width times the height
282242
that's a real vague question, try to be more specific or at least say what king of structure do u want? (steel, conc, ..?)\nanyways this is an exremely simple example.\n\nlets' say u want to design a tension member -carrying tension force = 100 tons, -between 2 supports and this member is a steel member of yield stress= 1.4\nso to get the area u need :\n100/1.4= 71.43 cm square let's say again u want it to be I-shaped steel then u'll need  I number 400 i.e IPE 400
238652
Assuming you have a rectangular object, then L x H x W will work. For other shapes you need something else. See sources.
323670
measure each diagonal, come up with the area of one triangle X it by 4, then work out the area of the base simple!
727989
You could try lots of things, like:\nbeads\nfoam balls with pieces of wood to hold the sides together\npaper mache \nclay \n\nReally anything that you can mold or put together would work.