to locate a point on a coordinate grid ,where do you start?
469175
x-coordinate: Count the horizontal distance from the origin (0,0).  If the point is to the right, then the number is positive.  It the point is to the left, then is is negative.\n\ny-coordinate: Count the vertical distance from the origin (0,0).  If the point is to above, then the number is positive.  It the point is to below, then is is negative.\n\nThen list them as an ordered pair (x,y)
853127
Well first of all you will have to know something about how they will be entered and since you are talking points, they will have to be numbers or strings representing numbers. Secondly, you must always know which is the x, y and the z coordinate so there either has to be a test that numbers must pass or you must always specify the specific order.\n\nNow, the way I would tackle this is using an array of points. Since you must have at least 3 to form a quorum, you can automatically have it return that it is not a quorum if you have less than 3 points. \n\nNext, once you have all the points entered or one at a time, you can have it check if it is a quorum. You will need to use a pythagorean theorem that checks the distance between one point and another point on each plane. So I would make sure you create a generic Pythagorean theorem function which returns an integer (distance m between two points).\n\nSo for instance you have the following points (in x, y, z order)...\n\n1,2,3 and point 1,10,3 and point 1,10,8\n\n1) We have three points so it could potentially be a quorum\n2) Point 1 and point 2 are only 8 units away on the y axis (10 minus 2) so far so good.\n3)  Point 2 and point 3 are 5 units away on the z axis (8 minus 3)\n4) This is where the theorem comes in... between points 1 and 3 (8 squared + 5 squared = distance squared)\n\n64 + 25 = 89 squared or about 9.434 units away. \n\nSince all three points are within 30 units it is a quorum. \n\nNow this example was really simple because the points were simple and small, plus point 2 and point 3 were lined up. But lets say you put point 3 at coordinates 1,8,8 you would have to run the theorem for point 1 to point 3 (going down on the z-axis and over on the y-axis) and then point 2 and 3 (going down on the z-axis as well as over on the y-axis).\n\nIt is hard to explain without drawing a picture so I recommend you draw a 3-D cube and plot a few points on paper. Then re-read what I am saying so you can see what I am talking about. \n\nAs you can probably tell from this process it gets exponentially large really quick with the more points you add. So make sure your theorem equation is small and fast and try to limit your inputs to probably something under 30 points or so.\n\nSo you loop through your points in the array, checking each one against point 1. Then go to point 2 and check points 3 onwards against it. Then point 3 checks point 4 onwards against it. You could use a point class to accomplish this but be careful because that could also add some confusion later.\n\nHope this made sense to you. Be sure to draw the picture, it will help you tremendously.\n\nGood luck!
740091
to locate a place, you need to know it's gedesic coordinates:longitude,latitude and height over MSL(mean sea level).MSL is not the same everywhere, each country decides wich level is it: it's defined after tide gauge, the average level is the MSL, tou must also know that different seas don't have the same level. \nHeight over MSL is defined by different surveying techniques: it can be GPS measurements, main levelling or geometric levelling (wich gives elevation accordingly to a height-known point).\nIf cartesians coordinates are used (X,Y,Z),it is more simple because you need to know distance from earth's center, you don't need to know sea level.
1116088
You need to know the length of a line if you want to know the amount that it departs or its longitude. \n\nIf you want to know where on the globe that is... its someplace between seattle, washington and portland, oregon
957617
2d point of no return
513
That's the trouble with answers that give no explanation. Who can you believe?\n\nSince we go in each direction once, the ORDER of the moves doesn't matter. \n\nDraw a grid. Start at 0,0\n10 North means 0,10 You are due north 10 miles. Then we come back south 6 miles, so we're 4 miles north of where we started. \n4 miles east and then 7 west means we're 3 miles west of the starting point. \n\nDraw a right triangle 4 miles up and then 3 miles to the left of the starting point. We want the hypotenuse of the triangle whose sides are 3 and 4. \nTime for Pythagoras: a^2 + b^2 = c^2 \n3^2 + 4^2 = c^2 \n9 + 16 = 25 Take the square root of that for your final distance. Direction: NW
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  }
1022457
Type O or Offset <enter>\nClick on Object to be offset <enter>\nType in distance to offset <enter>\nClick on side to offset \nGame over!!!!
889456
I'm not sure I understand the question, but if you want to translate the vector from its current position to the origin, then just add (4,-5) to its beginning and end points.\n The vector will now be pointing from the origin to (6,1).\nHope this helps.
582203
The use of grid lines on maps (latitude and longitude) was first suggested by the Greek astronomer Hipparcus about 300 years before Christ. He was the first who devised a method for fixing the location of places on earth by observation of the the sun, moon and stars. He worked out the mathematics of spherical trigonometry, which allowed the results of these observations to be plotted on an earth which he knew already to be a sphere.\n\nLatitude can be found easily by measuring the height of the sun at noon. The measurement of longitude is not so simple as you need an accurate clock.
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maps.google.com\n\nCan I search the map on Google Maps using latitude and longitude coordinates?\n\nYou can search for some locations using latitude and longitude coordinates on Google Maps. An example of this can be seen here.\n\nPlease note that the functionality of coordinate searches may be limited at this time.
915176
First of all, latitude first, then longitude.\n\nThere are TWO common ways:\n\nUse the numer of degrees followed by N(orth) or S(outh),\nthen the numer of degrees followed by E(ast) or W(est).\n\nEx) 30*N, 90*W\n\nUse only the number of degrees (positive for North-negative for South), followed by the number of degrees (positive for East-negative for West)\n\nEx) 30*, -90*
1108611
east/west.\n\nto remember, longitutde has lines that are all as "long" as each other (one pole to the other.)
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All GPS can tell you is your location on the earth. GPS cannot tell you where your property corners are, unless you already know the coordinates of those property corners. Your GPS unit does not know where your property corners are until you provide it with the information it needs. \n\nIf you knew the coordinates, which could be either latitude and longitude that mappers use, or state plane coordinates that surveyors use, then you could use GPS to guide you to the corner locations. But since you do not know where the corners are, you have no way of entering the required data into the GPS unit. \n\nYou not only need to get a professional surveyor to survey your property, you also need to tell the surveyor that you need him to provide you with the state plane coordinates of those points, so you can enter them into your GPS unit, then you will be able to find them yourself in the future.
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Please use the search function. This question was asked 5 days ago:\n\nA proper polar coordinate system (one that assigns each point in the plane a unique coordinate) will always leave out a ray of points. For example, the standard polar system that identifies the open set (0, oo) X (0, 2Pi) with R^2 - {0} by the function f(r, t) = (rcos(t), rsin(t)) ( = (x, y) ) does not correspond to any points on the positive x-axis and necessarily leaves out the origin (there is no way to include the origin uniquely). This is usually glossed over in a basic calculus or physics course as one actually uses an atlas of polar coordinate charts (this atlas never includes the origin, but at least covers the plane minus the origin); you use whichever chart is convenient for the problem at hand. Some definitions will disregard the need for a homeomorphism and include the positive x-axis by considering a domain that is not open, (0, oo)X[0, 2Pi)
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I'd think you could find two points with the same single variable like temperature, or something easier to visualize like height, but two variables is questionable. Suppose you had 2 altimeters that you could position remotely (and they automatically stayed at opposite points). Idealize height as being a smooth function of location. \nInitially the altimeters indicate unequal heights. When you move them one way the difference increases. Then move the opposite way to decrease the difference. Move a little bit, then reassess the relative changes, find a new direction if needed, and make another small step toward decreasing difference. Now mentally superpose the two surfaces you are surveying. If there are no height maxima or minima and the surfaces intersect, repeating this procedure should find a pair of zero-difference points. Now you can follow the contour of intersection of the surfaces and the difference remains zero. If the surfaces diverge in every direction there must be an altitude maximum or minimum in one or both survey areas; move until the extreme is off the survey area and there is overlap, and repeat the zero-finding procedure. There is an overwhelming probability that such a pair of areas, and therefore of points, can be found. Earth does not have a situation where one hemisphere's altitude range is completely separate from the other hemisphere's. The same goes for temperature and pressure. \nWhat about a second variable, say pressure? Suppose it too can be mapped to two intersecting surfaces. Project the zero-height-difference and zero-pressure-difference contours onto the same plane. If the two contours intersect anywhere, the two-variables-equal point pair has been found. If they don't, you can't find one here; move to a whole new area.\nBy this logic you're extremely likely to find a pair of one-variable equality points, and pretty likely to find a two-variable pair of points. I know this is not a good mathematical answer, which should be a proof or disproof, but then it may not be provable or disprovable.
88605
I like using Google Earth to get Lat/Long info. Just take it to where you want to go, and the coordinates are in the lower left corner. Free download, and hours of nerdy entertainment.