How do you approximate an arc with straight lines?
823299
How about drawing a bunch of straight lines of differing length, from a fixed point, so that the tips create the image of an arc?
902740
In addition to what “Spoonish1” told you I suggest you to write a program calculating minimal distance between 2 straight lines in 3D. Be this distance =0, then and only then you may state that 2 lines intersect. A classic line equation in 3D looks like:\n(y-y1)/(y2-y1)=(x-x1)/(x2-x1)=(z-z1)/(z2-z1) – just as a matter of fact. Has your desire faded? Or more help?
266904
i don't know i don't think it's possible\n\ni tried 1010-10=1000\n\nso u c cant just do it with one straight line
557771
Draw a straight line.\nDraw a good cuve (mirror), using a compass or a protractor(for later mark a point as centre and place the protractor on this point coinciding with the axis.)\nThe curve should be of equal length on each side of principle axis.\nNow mark the pole, radius and focal point (twice the radius).\nIn case u use protractor, take focal lenght as twice the distance between the marked pont and pole.\n\nRemember how the rays reflect/refract, and accordingly draw two incident and two reflected/refracted rays.\nGive the standard sign convention...I hope plane mirror and prism are easier.
407911
So you are doing some numerical analysis... if you have a spreadsheet program you can use that. Use the trapezoid rule or Simpson's rule for some small increment in x.  Your spreadsheet program is able to evaluate arctan.  The second question can be evaluated with Newton's method.
458048
Or you can do this:\n\nhttp://www.xahlee.org/SpecialPlaneCurves_dir/Parabola_dir/parabolaTangents.png\n\nto form a parabola out of many straight lines.
441297
Actually there are 2 answers to this. One method uses a technique called Rhumbline(sp?) (or straight line) another uses something called the great circle.\n\nI once wrote a program to do this for hurricane tracking.\nThe method I used was the straight line. For what i was doing this was fine however for navigation the great circle\nmethod is usually prefered. This code is old C (not even C++). Hope this helps you. Note: this forum destroys my coding style (all my indentations are striped).\n\n/*\n * Constants and macros used to \n * calculate the projected position of\n * the storm.\n */\n\n#define PI            3.14159265358979323846\n\n#define DEG_PER_RAD  (180.0 / (float) PI)  /* Ratio of Degrees to radians */\n\n#define DEG_TO_RAD(theta) ((theta) / DEG_PER_RAD)\n\n/*\n * The following does a straight line projection\n * of the position of a storm given current \n * position, heading, speed, and a time interval.\n * For further information \n * see "Computers for Sea and Sky".\n */\n\nvoid project_pos (start_lat,\n                  start_lon,\n                  heading,\n                  speed,\n                  interval,\n                  proj_lat,\n                  proj_lon)\n\n  float   start_lat;\n  float   start_lon;\n  int     heading;\n  int     speed;\n  float   interval;\n  float * proj_lat;\n  float * proj_lon;\n  {\n\n    float rad_lat;\n    float rad_lon;\n    float rad_heading;\n    float rad_of45;\n\n    rad_of45    = DEG_TO_RAD (45);\n    rad_lat     = DEG_TO_RAD (start_lat);\n    rad_lon     = DEG_TO_RAD (start_lon);\n    rad_heading = DEG_TO_RAD (heading);\n\n    *proj_lat = (interval * speed * cos (rad_heading) / 60.0) + start_lat;\n\n    if (*proj_lat > 90.0)\n      {\n        *proj_lat = 180.0 - *proj_lat;\n      }\n    else if (*proj_lat < -90.0)\n      {\n        *proj_lat = -180.0 - *proj_lat;\n      }\n\n    switch (heading)\n      {\n        case  90:\n        case 270:\n\n          *proj_lon = start_lon -\n                      ((interval * speed * sin (rad_heading)) /\n                       (60.0 * cos (rad_lat)));\n          break;\n\n        default:\n\n          *proj_lon = start_lon -\n                      (DEG_PER_RAD * tan (rad_heading) *\n                        ((log (fabs (tan (rad_of45 +\n                          (DEG_TO_RAD (*proj_lat) / 2.0))))) -\n                        (log (fabs (tan (rad_of45 + (rad_lat / 2.0)))))));\n\n          if (*proj_lon > 180.0)\n            {\n              *proj_lon -= 360.0;\n            }\n          else if (*proj_lon < -180.0)\n            {\n              *proj_lon += 360.0;\n            }\n\n          break;\n      }\n  }
247689
Use analytical trig (hint: sin function).  That should give you the answer you are looking for.
639826
1. Draw a square.\n2. Bisect the square top to bottom by drawing a line through its midpoint.\n3. Draw a diagonal line from the base of the bisecting line (at the bottom center of the square) to the outer, upper corner of the square. \n4. Using the diagonal line as the radius, drop an arc from the outer, upper corner of the square down to the baseline.\n5. Draw a line straight up from the point of intersection of the arc and the baseline. Extend the top edge of the square to meet this line and form a rectangle - it's golden.
760652
Set the compass arc length to over half of the line length.  Scribe and arc from each end of the line.  Draw a line between where the arc's cross that will not only be the middle of the line but the bisecting line will be perpendicular to the line.
855656
I'm assuming you're talking about cutting a circle into quadrants? This might be the long way to do it, but I would first find the center of the circle. Here's a web page for how to do that:\n\nhttp://www.makeitsolar.com/science-fair-ideas/90-find-circle-center.htm\n\nThen you draw your diameter. (there's half)\n\nFinally, find and draw a line perpendicular to your diameter that crosses through the center of the circle.\n\nLike I said - probably the long way to do it, but I'm a very visual person when it comes to geometry. I like to draw everything out.
1113346
Get a compass (not the navigation kind...the adjustable point-and-pencil kind), and set it at some length greater than 1/2 the segment's length.  Draw an arc through the segment, above and below the segment, from both endpoints of the segment.  There are two points of intersection for these arcs that you just drew: one above and one below the original segment.  When you connect these intersection points, you have bisected the segment with your new line, which is perpendicular to the original.
18831
A straight line.\n(Now I'm expecting you to come with a funny answer to this common sense question)
1078788
For any line length X, it can be divided into 7 equal parts like so:\nStart with your segment --we'll call it AB-- and at point A, draw a ray going towards the top of your paper.  Then, with your compass, mark off seven equidistant points, starting at A (put your compass on A, make a mark on the ray; put your compass on that mark, make another on the ray, etc.) Then, draw a segment from your 7th point to point B.  Then just draw 6 more parallel lines (constructing congruent arcs and such) and your segment AB is divided.
934709
Plot 3 points, connect them with lines. you get a triangle\n\nList most acute (sharp) angle 1st, most obtuse (blunt) angle last, and remaining one in the middle.\n\nIt looks like C will be the most obtuse
151606
if all the lines are equal in every way (duplicate) then you can put a measurment (dimension) from the existing line to a datum or border whichever you choose and then put a callout like 3@12.00' (3 lines at 12.00 feet)in the dimensionwith a arrow pointing at the line.\n\njeff
161309
I don't work with arccotangents a lot, but I'll give it a shot.\n\n1.  Write down y=arccot(x).\n2.  Draw a picture of a right triangle with base x, height 1, and angle y (between the base and hypotenuse).  The hypotenuse will therefore be sqrt(1+x^2).  This triangle represents the relationship between x and y and will be important later on.\n3.  Take the cotangent of both sides to give x=cot(y).\n4.  Rewrite cot(y) as cos(y)/sin(y).\n5.  Implicitly differentiate the function.  I hope you know how to do this.  If you don't, the basic rule is to differentiate both x and y terms as you normally would, except all terms with a y in them are multiplied by y' (where y' is dy/dx).  Here's how it's done with implicit differentiation and the quotient rule:\nx = cos(y)/sin(y)\n1 = (sin(y)*(-sin(y)*y') - cos(y)*cos(y)*y')/sin^2(y)\n1 = (-sin^2(y)*y' - cos^2(y)*y')/sin^2(y)\n1 = y'*(-sin^2(y) - cos^2(y))/sin^2(y)\n1 = -y'/sin^2(y)    (This is thanks to the identity sin^2(y)+cos^2(y)=1)\n6.  Now get the derivative term-- the y' term-- on one side of the equation:\ny' = -sin^2(y)\n7.  Look at your drawing of the triangle.  To get a proper value for the sin, you must divide the length of each side by sqrt(1-x^2) so that the hypotenuse is length 1.  Now the base of the triangle is x/sqrt(1+x^2) and the height is 1/sqrt(1+x^2).\n8.  Remember that the sin of a function is the length of a side opposite the angle on a right triangle with hypotenuse 1.  In other words, the sin of y is 1/sqrt(1+x^2)\n9.  Substitute that into your equation\ny' = -sin^2(y)\nto give\ny' = -1/(1+x^2)\n\nAnd you're done!