I need to sew a slip cover for a sphere, how do i work out the size and shape of the segments needed?
492510
Depending on how you want the seams to run, you could always model it after a soccer ball.\n\nIts covering is composed of hexagons and pentagons.\nMeasure the length of the sides of the shapes, measure the diameter of the soccer ball, then scale it up to the size of your sphere.\n\nEven simpler, you could model it after the two piece cover of a baseball.\n\nIt all depends on how tight a fit you want.\n\n\nIf you want to cover it with identical strips that are wide in the middle and taper to a point at each end you could make a pattern as follows:\n\nDecide how many segments you want.\nMeasure the circumference of your sphere.\nOn your pattern. draw a line that has length half the circumference of the sphere. Label the ends N and S (for North and South poles.)\n\nAt the mid point of that line, draw a second line that crosses it at right angles. Label its ends X and Y.\nTake the length of the circumference and divide its length by the number of segments you want.  Measure that distance on the X-Y line from the N-S line. Label those end points E and W.\nDraw the lines to connect N-E, N-W, S-E, S-W. You now have two skinny triangles base to base.\nThis is a crude pattern for a segment.\n\nIf you want a segment with curved sides and a better fit, there is a point on X-Y that is the center of a circle that goes through N, E, and S. It's probably easiest to determine it by trial and error rather than computing the number.\n\nOnce you find it you can draw the arc through N, E, and S to get a segment with curved sides.\n\nNote: Don't forget to add a seam allowance to the basic pattern.
978173
You need to start off with a six-pointed star (constructed with two equilateral triangles).\n\ntriangle AEC and triangle FBD\n\nNow connect A to D, then connect B to E, and finally C to F. If you \ndrew everything correctly, those three segments should meet at a point \nin the center of the star.  \n\nThe 3 line segments you just drew plus the six sides of the star make \n9 lines. There are 5 points of intersection on each line, and the \ntotal number of intersections is 19.
956315
The simplest way is to cut it into five slices, each with an arc of 72 degrees (or 2*pi/5 radians). You could probably also figure out a way to cut it into strips so that each strip has equal area, but that would be significantly more complicated.
946003
this looks like the closest thing that I could find that looks like how my old technical manual shows on drawing threads:\n\nhttp://honolulu.hawaii.edu/distance/aec80/threads/20c-ctl.htm\n\nI'll look for more tommorrow\n\naddon\nif you want a 3d thread look around here:\nhttp://discussion.autodesk.com/search.jspa?numResults=25&source=forumlist%7C39&q=3d+thread+spring&inputEntered=true&objID=c39\n\n3d springs are easy to do in 2007 there is a helix command that lets you make a 3d helix that you can use for the path of an extrusion
865940
I will try, but it would have been easier if you had copied the problem from your homework exactly - this version is confusing.\n\nThe side of the shed to be used is 20m long, correct?  If this is not the case, substitute the correct value into the work below marked by an asterisk (*).\n\na) The width of the sheep pen is 20m*.  This means two sides of the rectangle total 40m, and the other two sides must be a total of 160-40+20* = 140 m.  So the pen is 20m(wide) by 70m (high).  \nGoing back to check, you get 70+70+20 = 160 (the fourth side is the 20m shed wall) and this is correct.\n\nb) If I remember my geometry correctly, the maximum area would come from a square.  You have a total of 180m for all four sides (160m of fence plus 20m* of shed).  To make a square, each side must be 180/4 or 45m.  This area would be 2025 sq m, vs. 1400 in a).
223084
I've had past expierences with this. cut the size you need. cut lines about 3-4 inches long and 1 apart from eachother.\ntie them together and you've got your self a blanket!
706050
Ok... the first thing to realize is that you are finding the maximum area with the least amount of materials.  To do that you should put the shed at a corner of the rectange, that way two sides of the shed act as part of the fence.  Assume that the shed is not "usable area" for the shep pen.  Accordingly, you need to solve the problem twice, once with using the shed as part of one side, and once using the shed as a corner.\n\nOk, the first thing you need to do is show that the maximum area of a rectangle occurs when the lenght and width are equal.  That is the only part that needs any type of equation analysis, and it really doesn't require calculus.\n\nA=L*W\nAssuming that the width is at most as long as the L, then the equation goes to:\nA=L*(x*L), where x a percentage, ranging from 0 to 1.\nClearly, the area (A) will be maximum when x=1.  That shows that in your problem, the pen will have the maximum area when the length and width of the rectangles are equal.\n\nSo, you know that the fence is 160m., and you have 4 sides, s1, s2, s3 and s4.\nSo:\n160m=s1+s2+s3+s4\n\nNow, if you use the shed as part of one side, namely s4, then the perimeter equation for the shed is:\n180m=s1+s2+s3+s4+20m\n\nSince the maximum area occurs when the lenght and width are equal:\ns1=s2=s3=s4-20m, so substituting s2, s3 and s4 for s1 in the permiter equation, you get:\n4s1+20m=180m\nSolve for s1,\ns1=45m (call this the length)\ns4=45-20=25m\nin this case, the area will be S1*S1, since the lenght and width of the pen will be equal.  \nSo the area is 45*45=2025 sq. m.\n\nNow, put the shed in a corner of the sheep pen.\nSo in our example, we'll make the shed part of s1 and s4.\n\nso the primeter equation will be:\ns1+20m+s2+s3+s4+20m=200m\nNow, s1=s2-20, s3=s2 and s4=s2-20\nSo substituting for s1, s2 and s4, you get:\ns2-20m+s2+s3+s4-20m+40m=200m\nsolve for s2,\ns2=50m=s3 and s1=30m (s2-20m)\nSo max area is 50m*50m=2500 sq. m\nBut, now you need to reduce the area by the area of the shed (400 sq. m), so final usable area is:\n2100 sq. m which is more than using the shed as part of only 1 side.
550617
Take a rope ( that doesn't elongate/stretch easily), wind round both pulleys (just the way belt goes)and mark one end where it touches the other. Measure this and the size in inches is the size of the belt. As to whether 'A' or 'B' or 'C', it is based on cross-section. The smallest is 'A'. I hope I am understood.
1041033
first fold the sheet in half, get a pair of scissors and cut about an inch in from the side starting from the folded edge cutting towards the open edge, but stop about 1 1/2 inces from the open edge. then make a similar cut next to the first one but go from the open edge to the folded edge (remember not to cut all the way through), then you just reapeat alternating as you go until you run out of space (make sure you end with a cut from the folded edge to open edge (trim the end off if it if theres too much extra). now all you need to do is cut along the first fold (the one from when you folded the paper in half), but don't cut through the first or last bit, you need those to hold the ring together\n\nsounds more complicated than it is, you'll see what I mean when you try for yourself  (try it on a scrap)
650879
Use string, paper, and a ruler.  The string can represent the perimeter, and you can draw a grid on the circle to represent the units of area.  Just don't get too in depth on the origins of pi.
850259
here:http://www.cut-the-knot.org/Curriculum/Geometry/AreaOfParallelogram.shtml
503
That depends on what L and P are.  If you plug the permitter and length in, you can get width W.\n\nIf you want a formula that will yield W, it's: (P-2L)/2 = W
253184
Wow, thats a tuff one. Have you considered cardboard or wire and fabric? How much time do you have because this seems like a time consuming task. You could look on the internet fo ideas that might help you out. It depends on what you want it to look like.
830321
I think you should get some plastic PVC pipe and related parts as needed and you can cut them to the size you actually want. So as to not make it so flimsy you should probably use about 1 1/2 to 2 inch pipe. Some kind of fish net or other netting will do .\n\nIf you check out some internet sites for golf nets you can view a pic to use as a patter I suppose.\n\nHere are a couple of good web sites.\n\nhttp://www.sportnetting.com/?src=looksmart\n\nhttp://innovagolf.com/gonetgoprnet2.html\n\nHave fun with your project.\n\nLeon
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
578272
Try downloading a I-mesh.\n\nThey may have it.
337947
You need a stringing machine, then it depends on what kind of machine you have. Would really need more details.\nStringing pattern can normally be figured out or taken from the previous set of strings.