How to find position in lat,long given direction,speed ,time and initial position?
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  }
9563
Latitude: 47.64709  North\nLongitude:   122.33396 West
1049196
Global positioning system
797532
Check Google Earth.  You can download it off of the web, and it will give you the exact location you are looking for.
194218
Very simple, on the website below.\n\nhttp://www.indo.com/distance/\n\nIt works with city names or lat/long whichever you prefer.
943735
a) T = dL / dt = 16\nL = 16 t \n\nI don't think you can do any better than that since they don't ask for the angular momentum at a particular moment in time, and it certainly is not constant.\n\nb) L = Iw\nw = L / I = 16 t / I = 16(33 * 10^-3) / (1.2 * 10^-3) = 440 rad/s
750320
If it takes 6 secs to reach the max, it will take an additional 6 secs to reach your starting point.  In this case it is 2560.  Then it is a matter of finding the time from that point to going to the ground.\n1st lets correct the orginal equations (acceleration due to gravity is down/negative)\nh = -16t^2 + 192t + 2560\n2nd - set the equation to zero since we want to find the time to ground.\n0 = -16t^2 + 192t + 2560\n16t^2 - 192t = 2560\nt^2 - 12t = 160\n(t^2 - 12t + 36) - 36 = 160\n(t - 6)^2 = 196\nt - 6 = 14\nt = 20\nConclusion: 6 seconds to reach max ht of 3136, another 6 seconds to reach starting point of 2560 and another 8 seconds to hit the ground.
778162
Step 1: find y '\nStep 2: fill in x=1\nThat gives the right answer 2\n\nTh
1154822
either google Earth\n           or\nmapquest\nthese are my favorite places to get directions\ngood luck
91781
Do you mean lattitude and longitude?\n\ntry http://earth.google.com
700119
Velocity:\nv=d/t\n\nd=distance\nt=time\n\nAcceleration:\na=(vf^2-vi^2)/2d\n\nvf=final velocity\nvi=initial velocity
140408
You need to define your abbreviations.\n\nIf you are referring to Automatic Direction Finding try:\n\nwww.simrad.com    They make a Doppler ADF
694191
S = southbound speed\nN = northbound speed\nd = distance\nx = time from 8am\nS(x) + N(x-1) = d\n\nS*2 + N*1 = 500  => N = 500 - 2S\nS*3 + N*2 = 850 = 3S + (1000 -4S) = 1000 - S\nS = 150, N = 200
468833
Latitude: 42.12 N, Longitude: 71.18 W
859591
Longitude and Latitude!
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*
791525
questions are very easy. You must try them by yourself. I will only give hints-\n1> Instantaneous accl  = dv/dt\nAvg accl = (v2-v1)/(t2-t1)\n\n2> draw the vectors, take the components separately along x and y axes.\nremember accl component along x axis increases velocity along x axis, and accl component along y axis increases velocity along y axis.\n3> use the basic kinematics equations like v^2=u^2+2as etc