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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
asked Jan 1 at 19:51
David W.David W.
6034820
add a comment |
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
asked Jan 1 at 19:51
David W.David W.
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
add a comment |
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
asked Jan 1 at 19:51
David W.David W.
6034820
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
asked Jan 1 at 19:51
David W.David W.
6034820
asked Jan 1 at 19:51
David W.David W.
6034820
asked Jan 1 at 19:51
David W.David W.
6034820
asked Jan 1 at 19:51
David W.David W.
6034820
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
add a comment |
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
add a comment |
1 Answer
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oldest
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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>answered Jan 1 at 20:14
blexblex
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
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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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>answered Jan 1 at 20:14
blexblex
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
add a comment |
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>answered Jan 1 at 20:14
blexblex
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
add a comment |
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>answered Jan 1 at 20:14
blexblex
12.4k12245
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>answered Jan 1 at 20:14
blexblex
12.4k12245
edited Jan 1 at 20:28
answered Jan 1 at 20:14
blexblex
12.4k12245
answered Jan 1 at 20:14
blexblex
12.4k12245
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
add a comment |
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
add a comment |
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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