Calculation lat/long- points for a pentagon around center












1















I'm looking for a way to retrieve five points that would make a pentagon around a given center and a given distance.



Something like this:



getPentagonPoints = (latlng, distance) => {

  var pentagonLatLng = ;

  //magic calculations goes here

  return pentagonLatLng;

}





javascript math latitude-longitude







share|improve this question























  • math.stackexchange.com/questions/1990504/…

    – nem035
    Jan 1 at 20:03











  • @nem035 I don't think that's directly applicable, since it's putting the figure on a plane, not a sphere.

    – Robert Dodier
    Jan 2 at 23:42
















1















I'm looking for a way to retrieve five points that would make a pentagon around a given center and a given distance.



Something like this:



getPentagonPoints = (latlng, distance) => {

  var pentagonLatLng = ;

  //magic calculations goes here

  return pentagonLatLng;

}





javascript math latitude-longitude







share|improve this question























  • math.stackexchange.com/questions/1990504/…

    – nem035
    Jan 1 at 20:03











  • @nem035 I don't think that's directly applicable, since it's putting the figure on a plane, not a sphere.

    – Robert Dodier
    Jan 2 at 23:42














1












1








1








I'm looking for a way to retrieve five points that would make a pentagon around a given center and a given distance.



Something like this:



getPentagonPoints = (latlng, distance) => {

  var pentagonLatLng = ;

  //magic calculations goes here

  return pentagonLatLng;

}





javascript math latitude-longitude







share|improve this question














I'm looking for a way to retrieve five points that would make a pentagon around a given center and a given distance.



Something like this:



getPentagonPoints = (latlng, distance) => {

  var pentagonLatLng = ;

  //magic calculations goes here

  return pentagonLatLng;

}





javascript math latitude-longitude




javascript math latitude-longitude






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 1 at 19:51









David W.David W.

6034820




6034820













  • math.stackexchange.com/questions/1990504/…

    – nem035
    Jan 1 at 20:03











  • @nem035 I don't think that's directly applicable, since it's putting the figure on a plane, not a sphere.

    – Robert Dodier
    Jan 2 at 23:42



















  • math.stackexchange.com/questions/1990504/…

    – nem035
    Jan 1 at 20:03











  • @nem035 I don't think that's directly applicable, since it's putting the figure on a plane, not a sphere.

    – Robert Dodier
    Jan 2 at 23:42

















math.stackexchange.com/questions/1990504/…

– nem035
Jan 1 at 20:03





math.stackexchange.com/questions/1990504/…

– nem035
Jan 1 at 20:03













@nem035 I don't think that's directly applicable, since it's putting the figure on a plane, not a sphere.

– Robert Dodier
Jan 2 at 23:42





@nem035 I don't think that's directly applicable, since it's putting the figure on a plane, not a sphere.

– Robert Dodier
Jan 2 at 23:42












1 Answer
1






active

oldest

votes


















0














You can do it like this:






function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>








share|improve this answer





















  • 1





    I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

    – Robert Dodier
    Jan 2 at 23:49











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1 Answer
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oldest

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes









0














You can do it like this:






function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>








share|improve this answer





















  • 1





    I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

    – Robert Dodier
    Jan 2 at 23:49
















0














You can do it like this:






function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>








share|improve this answer





















  • 1





    I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

    – Robert Dodier
    Jan 2 at 23:49














0












0








0







You can do it like this:






function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>








share|improve this answer















You can do it like this:






function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>








function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>





function getPentagonPoints(latlng, distance) {

  var pentagonLatLng = ;

  for (let i = 0; i < 5; i++) {

    // Pi * 2 is 360 degrees, in radians

    // We add 270 degrees (Pi * 1.5 radians) to start at the top

    const angle = Math.PI * 2 / 5 * i + Math.PI * 1.5; // in radians

    const lng = distance * Math.cos(angle) + latlng.lng;

    const lat = distance * Math.sin(angle) + latlng.lat;

    pentagonLatLng.push({lat, lng});

  }

  return pentagonLatLng;

}



// Just for the demo

const canvas = document.querySelector('canvas');

const ctx = canvas.getContext('2d');



const latLng = { lat: 90, lng: 200 };

const points = getPentagonPoints(latLng, 50);



points.forEach((point, i) => {

  if (i === 0) { ctx.moveTo(point.lng, point.lat); }

  ctx.lineTo(point.lng, point.lat);

  if (i === points.length - 1) { ctx.lineTo(points[0].lng, points[0].lat); }

});



ctx.stroke();

<canvas width="400" height="160"></canvas>






share|improve this answer














share|improve this answer



share|improve this answer








edited Jan 1 at 20:28

























answered Jan 1 at 20:14









blexblex

12.4k12245




12.4k12245








  • 1





    I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

    – Robert Dodier
    Jan 2 at 23:49














  • 1





    I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

    – Robert Dodier
    Jan 2 at 23:49








1




1





I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

– Robert Dodier
Jan 2 at 23:49





I think that's OK for a pentagon in the plane, but not on a sphere. (Since OP mentioned latitude and longitude it seems a sphere is implied.) It's probably OK for small pentagons not near the poles, but large pentagons and ones near the poles will be distorted. Maybe a way to go about it is to draw the figure on the equator and then rotate it to the desired latitude. Just guessing here.

– Robert Dodier
Jan 2 at 23:49




















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