I'm trying to calculate bounds for square with particular radius and when center of square is known(longitude, latitude). But I'm getting into troubles with it.
I've tried to use haversine formula from here :
But I'm getting into troubles when radius is pretty big.
Currently to find
1). latitude delta in radians I use:
radiusInMeters / EARTH_RADIUS_METERS
2). longitude delta in radians I use:
2.0 * | arcsin( |sin(radiusInMeters/(EARTH_RADIUS_METERS*2.0))
/ |cos(latitudeStart)| | ) |
These formulas I got from haversine formula.
Could please someone point me to the exact generic formula for calculations which will be good for big and small distances for my case.
Also how should be handled situations when radius exceeds -180˚/180˚ on longitude or -90˚/90˚ on latitude?
UPDATE
Some clarifications. Let's say that I'm staying in some particular point with coordinates (lon, lat), where lon is -113˚ and lat 50˚.
I would like to query points in some radius of interest from database. For that I need to calculate bounds of a "square". (Then filter-out stuff that's not in interesting radius). Formula above works fine on small distances(let's say 'till 100 kilometers(63 miles). But the more I go from equatorial point the poles, the more rounding errors I get.)
Thanks in advance
Related
First of all, thanks for reading. I have Vincenty working in Excel(VBA) already, and want to do this in Excel, but this is a math question, not a coding question. By the way, I'll readily cop up front to the fact that ellipsoids are way over my head.
I'm looking to calculate accurate direct distance between two objects, given their latitude, longitude and altitude. Vincenty was an interesting start, but two issues:
(a) Vincenty is the distance along the ellipsoid, and I would need the chord length.
(b) Vincenty doesn't account for elevation, and the distance between points increases as elevation increases.
It would be easy to take Vincenty as my horizontal distance, and use the elevation difference to solve for the slope, but that doesn't seem accurate.
Maybe this should just be solving for the line between points on concentric circles (i.e. the lower elevation versus the higher elevation) except what Earth radius to use? I mean, it's an ellipsoid, so...?
My distances will typically be 10 - 40 miles, but millimeter precision is required.
Point me in the right direction? Thanks! ~Mike
I have 5 {x,y} points randomly placed on a grid
Each of the points do not know the {x,y} coordinates of the other points
Each of the points do know the distance of each of the other points from their {x,y} position
Each of the points exchanges this distance information with every other point
So every point knows every distance of every other point
Using this distance information every point can calculate (by finding the angles) triangles for every other point using itself as a reference point
Example, point 1 can calculate the following triangles:
1-2-3,
1-2-4,
1-2-5,
1-3-4,
1-3-5,
1-4-5,
and using the distance data recieved from the other points it can also calculate
2-3-4,
2-3-5,
2-4-5,
3-4-5
I would like to build a map of the location of every other point relative to a single point
How should I go about doing this? I am asuming it would be some kind of triangulation algorithm but these mainly seem to compute the location of a point from three other points, not the other way around where the other points {x,y} coordinates are discovered based on only the distance information.
I have tried plotting the two possible triangles for every 3 triangle points and then rotating them on a fixed known point to try and align them, but I think this avenue will end up with too many possibilities and errors
Ultimately I would like every point to end up with {x,y} coordinates of every other point relative to itself
You know the distance from one point to every other, dij. Thus, point 2 lies in a circumference of center point 1 and radius = d12. Point 3 lies in a circumference of center point 1 and R=d13 and it also lies in another circumference of center point 2 and R=d23.
See this picture:
I've set point 2 in X-axis for simplicity.
As you see, point 3 is on the intersection of two cicrcumferences centered at P1 and P2. There is a second intersection, P3a. Let's choose the one that is upwards and continue.
For P4 we can use three circumferences, centered at P1, P2 and P3. Again we get two solutions.
The same process can be done with the rest of points. For Pn you have n-1 circumferences.
I'm sure you can find the maths for circle-circle intersection.
Some remarks must be observed:
1) The construction is simpler if you first sort the points by distance to P1.
2) Not all distances generate a solution. For example, increase d13 an there's no intersection between the two circumferences for P3. Or increase d14 and now the three circumferences don't intersect in just the two expected points 4 and 4a.
3) This fact can be overworked by considering the average of intersections and the distance from each solution to this average. You can set a tolerance in these distances and tell if the average is a solution or else some dij is wrong. Since two solutions are possible, you must consider two averages.
4) The two possible triangulations are symmetric, over X-axis in the case I've drawn.
The real solution is obtained by a rotation around P1. To calculate the angle of rotation you need the {x,y} coordinates of another point.
I'm currently drawing up a mock database schema with two tables: Booking and Waypoint.
Booking stores the taxi booking information.
Waypoint stores the pickup and drop off points during the journey, along with the lat lon position. Each sequence is a stop in the journey.
How would I calculate the distance between the different stops in each journey (using the lat/lon data) in Excel?
Is there a way to programmatically define this in Excel, i.e. so that a formula can be placed into the mileage column (Booking table), lookup the matching sequence (via bookingId) for that journey in the Waypoint table and return a result?
Example 1:
A journey with 2 stops:
1 1 1 MK4 4FL, 2, Levens Hall Drive, Westcroft, Milton Keynes 52.002529 -0.797623
2 1 2 MK2 2RD, 55, Westfield Road, Bletchley, Milton Keynes 51.992571 -0.72753
4.1 miles according to Google, entry made in mileage column in Booking table where id = 1
Example 2:
A journey with 3 stops:
6 3 1 MK7 7DT, 2, Spearmint Close, Walnut Tree, Milton Keynes 52.017486 -0.690113
7 3 2 MK18 1JL, H S B C, Market Hill, Buckingham 52.000674 -0.987062
8 3 1 MK17 0FE, 1, Maids Close, Mursley, Milton Keynes 52.040622 -0.759417
27.7 miles according to Google, entry made in mileage column in Booking table where id = 3
If you want to find the distance between two points just use this formula and you will get the result in Km, just convert to miles if needed.
Point A: LAT1, LONG1
Point B: LAT2, LONG2
ACOS(COS(RADIANS(90-Lat1)) *COS(RADIANS(90-Lat2)) +SIN(RADIANS(90-Lat1)) *SIN(RADIANS(90-lat2)) *COS(RADIANS(long1-long2)))*6371
Regards
Until quite recently, accurate maps were constructed by triangulation, which in essence is the application of Pythagoras’s Theorem. For the distance between any pair of co-ordinates take the square root of the sum of the square of the difference in x co-ordinates and the square of the difference in y co-ordinates. The x and y co-ordinates must however be in the same units (eg miles) which involves factoring the latitude and longitude values. This can be complicated because the factor for longitude depends upon latitude (walking all round the North Pole is less far than walking around the Equator) but in your case a factor for 52o North should serve. From this the results (which might be checked here) are around 20% different from the examples you give (in the second case, with pairing IDs 6 and 7 and adding that result to the result from pairing IDs 7 and 8).
Since you say accuracy is not important, and assuming distances are small (say less than 1000 miles) you can use the loxodromic distance.
For this, compute the difference of latitutes (dlat) and difference of longitudes (dlon). If there were any chance (unlikely) that you're crossing meridian 180º, take modulo 360º to ensure the difference of longitudes is between -180º and 180º. Also compute average latitude (alat).
Then compute:
distance= 60*sqrt(dlat^2 + (dlon*cos(alat))^2)
This distance is in nautical miles. Apply conversions as needed.
EXPLANATION: This takes advantage of the fact that one nautical mile is, by definition, always equal to one minute-arc of latitude. The cosine corresponds to the fact that meridians get closer to each other as they approach the poles. The rest is just application of Pythagoras theorem -- which requires that the relevant portion of the globe be flat, which is of course only a good approximation for small distances.
It all depends on what the distance is and what accuracy you require. Calculations based on "Earth locally flat" model will not provide great results for long distances but for short distance they may be ok. Models assuming Earth is a perfect sphere (e.g. Haversine formula) give better accuracy but they still do not produce geodesic grade results.
See Geodesics on an ellipsoid for more details.
One of the high accuracy (fraction of a mm) solutions is known as Vincenty's formulae. For my Excel VBA implementation look here https://github.com/tdjastrzebski/Vincenty-Excel
Understanding that it would only be an estimate...
How what decimal constant would be able to be used to find a point X Miles away from a point in latitude and longitude to facilitate creating a lat long bounding box.
Unfortunately there is no such simple constant. As you go farther north, the "walking distance" between lines of longitude becomes smaller and smaller. If you were right next to the north pole, you could walk in a circle around it, covering almost no distance at all, and yet you'd still have touched every line of longtiude.
What you need is the great-circle distance between two points on a sphere.
Not sure what you are trying to do. You want a create a bounding box based on distance from a point? Are you looking for a way to calculate new lat/long from given lat/long using a distance in miles?
You can use the manhattan function to get an approximation of distance (realizing that lat/long are based on a spheroid approximating the earth, and truly calculating this requires more math), calculating x and y values with a forumla to follow
Manhattan function:
sqrt(x*x + y*y)
X and Y from Lat/Long:
x = 69.1 * (lat2 - lat1)
y = 53 * (lon2 - lon1)
Still, that method is pretty error-prone.
There's the great circle distance formula too, gotta use some trig for it, but it's probably worth it since you can get pretty good error in approximation depending upon what part of the spheroid (e.g. the lat/long) you start at.
Check out this page:
http://www.meridianworlddata.com/Distance-Calculation.asp
If i have a position specified in latitude and longitude, then it can cover a box, depending on how many digits of accuracy are in the position.
For example, given a position of 55.0° N, 3.0° W (i.e. to 1 decimal place), and assuming a truncation (as opposed to rounding), this could cover anything that's 55.01° to 55.09°. This would cover the area in this box: http://www.openstreetmap.org/?minlat=55.0&maxlat=55.1&maxlon=-3.0&minlon=-3.1&box=yes
Is there anyway to calculate the area of that box? I only want to do this once, so a simple website that provides this calculation would suffice.
The main reason I want to do this is because I have a position to a very high number of decimal places, and I want to see how precise it is.
While the Earth isn't exactly spherical you can treat it as such for these calculations.
The North/South calculation is relatively simple as there are 180° from pole to pole and the distance is 20,014 km (Source) so one degree == 20014/180 = 111.19 km.
The East/West calculation is more difficult as it depends on the latitude. The Equatorial distance is 40,076 km (Source) so one degree = 40076/360 = 111.32 km. The circumference at the poles is (by definition) 0 km. So you can calculate the circumference at any latitude by trigonometry (circumference * sin(latitude)).
I have written an online geodesic area calculator which is available at
http://geographiclib.sf.net/cgi-bin/Planimeter. Enter in the four
corners of your box (counter clockwise) and it will give its area.