I have two coordinates.   Lat:53.307718   Long:-6.230189  Lat:53.307708   Long:-6.229988   .  I then use the distance formula   distance = SQRT((Y2-Y1)*(Y2-Y1) +(X2-X1)*(X2-X1))   .  I put it points in.   distance = SQRT((53.307708-53.307718)*(53.307708-53.307718) +(-6.229988--6.230189)*(-6.229988--6.230189))   and it get a value of 0.000201  What is the factor to convert that point to convert to meters?  I but the two points into this website  and found the distance to be 12.92 meters. But when i do it then for the next point the factor isnt the same.
123829
I am trying to measure distances between points and writing the calculated measure between these points in the attribute table.   The points represents a vehicle's location based on GPS data according to existence location in time aspect. I could add the longitude and latitude data from Excel to a shape layer.       After creating shapefile, I could be able to measure the euclidean distances between the nodes, but if I will do the calculations manually, I might need to have immortal life.   How can I calculate automatically the distances between the nodes and write the calculations in my table or Excel or CSV data.  the distance will calculate according to the id's.  I mean the first distance will be for FID 1 and 2; the second will be for 2-3, third 3-4 and etc.  Please help me :)
71810
Problem:  $F = \frac{(G m_1 m_2)}{r^2}$  An object of unknown mass is placed directly between the Earth and the Moon such that the forces of gravity acting on the object are equal from both the Earth and the Moon. The distance from the Moon to the object is exactly one-tenth of the distance from the Moon to Earth. Using the above formula, determine the distance from the Moon to the Earth.  Given is:  The factor $G≈6.67∙10−11\mathbf{N(\frac{m^2}{kg^2})}$ is a universal constant.  Mass of Earth $≈5.972∙10^{24}\mathbf{kg}$   Mass of the Moon as $≈7.348∙10^{22}\mathbf{kg}$  Mass of unknown object = $m$  My solution:  $F = \frac{(6.67 *10^-11) m (7.348*10^{24})}{R^2}$  $F = \frac{(6.67 *10^-11) (5.92* 10^22) m}{10R^2}$  As gravitational force acting on the object is equal   $\frac{(5.92\cdot 10^22)}{100r^2} = \frac{(7.348*10^{24})}{r^2}$  $R = 1.24 \cdot 10^4 KM$  Is this a viable solution to the problem? Please provide feedback. Thank you!
113727
I'm looking for help calculating the distance between a sequence of points that are in a single Shapefile in QGIS.  Below is what my data look like and a blank distance column I've added to show how I'd like the distance to look.  I want to know the distance between point 1 and 2, 2 and 3, etc. I'd like distance to be in meters or Km, but currently my Shapefile is in a projection whose units are decimal degrees.  ID  LAT         LON         TIME        DISTANCE 1   10.08527    124.59833   21:24:37    0 2   10.08523    124.59830   21:25:07    ? 3   10.08526    124.59832   21:25:37    ? 4   10.08526    124.59831   21:26:07    ?   A number of people have asked similar questions, but none quite get at what I want to do.  This post is close, but this is in PostGIS, not QGIS    This post got me part of the way there, but since I'm new to QGIS the answer doesn't provide enough detail for me.  For example, once I've installed the GRASS plugin, I think I need to save/import the Shapefile with my series of GPS points into GRASS so I can use the v.distance module, but I don't know how to do that.       Is the GRASS v.distance module the only way to go?  Or is there a more straight forward way?  If v.distance is the only way would someone be able to point me to or explain more step-by-step how to do this?
18814
I have latitude and longitude as 19.0649070739746 and 73.1308670043945 respectively.   In this case both coordinates are 13 decimal places long, but sometimes I also get coordinates which are 6 decimal places long.   Do fewer decimal points affect accuracy, and what does every digit after the decimal place signify?
37637
I see that the Point type is made to store geometric coordinates, however I don't see what problem it is trying to solve, or what advantages it offers over simply using two numerical columns. It's 16 bytes, so is it just an abstraction for two 8 byte floats? Or is it a string containing some kind of geohash that optimizes the performance of certain queries?  Let's say I want to get all the points for a specific viewport on a Leaflet map... with numerical columns representing latitude and longitude, I'd just filter for all the rows that have a latitude number between the N/S bounds and a longitude number between the E/W bounds, easy. I looked into using Point here, but the syntax is cumbersome and it seems to me like unnecessary overhead for development and for database performance... is it?
80381
Could anyone provide with a way to calculate latitude and longitude from DEM, for each raster cell?   In previous posts about this, I have seen that some suggest calculating lat and long in certain sample point locations. This doesn't work for me, as I need complete rasters.   Also, I use ArcGIS 10.2 version so $$XMAP, $$YMAP does not work for me.   Is there any other way inside ArcGIS to get lat and long?
48219
The function  should calculate distances between two points.  When I am trying to calculate distance between points having 100 meters between them  p1=new esri.geometry.Point(3997030.6690969253, 7444299.320646087, new esri.SpatialReference({ wkid: 102113 })); Object p2=new esri.geometry.Point(3996951.455397143, 7444142.154020177, new esri.SpatialReference({ wkid: 102113 })); Object esri.geometry.getLength(p1, p2) 176.00045037719127   I am getting 176 which is wrong. Projection is Web Mercator (WKID 102113).    @Krystian found that getLength uses just Pythagoras theorem to calculate length. So it gives length in projection meters. How to obtain distance in real meters?
90017
I'm trying to figure out how to mathematically derive the common points of two intersecting circles on earth's surface given a center Lat/Lon and a radius for each point.  For example, given:   Lat/Lon (37.673442, -90.234036) Radius 107.5 NM Lat/Lon (36.109997, -90.953669) Radius 145 NM   I should find two intersection points with one of them being (36.948, -088.158).  It would be trivially easy to solve this on a flat plane but I don't have any experience solving equations on an imperfect sphere such as the earth' surface.
23039
I tried to solve the following exercises, so I want to ask you if my answers are correct.     1) Given the coordinates system $(O'; X'' Y'')$ asociated to the    basis $B=[{b_1=\frac{1}{\sqrt{2}}; \frac{1}{\sqrt{2}};  &gt; b_2=\frac{-1}{\sqrt{2}}; \frac{1}{\sqrt{2}}}]$ with origin in    $O'=(2;5)$ and the system $(O;XY)$ asociated to the standar basis, find the intersection between $Q: y''^2=-4 \sqrt{2} x''$    and the straight line in $(O,XY)$ $V=(0;5)+t(-1;1) t\epsilon R$    Well, first I tried to write $Q$ in terms of $X$ and $Y$, and not in terms of $X''$ and $Y''$. I used the formula:  $(\begin{matrix}  X''\\  Y'' \\  \end{matrix})$ =$(C)_B$ × $(\begin{matrix}  X-2\\  Y-5 \\  \end{matrix})$   Then I replaced those values in the equation of $Q$ and I got $Q: \frac{(-x+y-3)^2}{2}=-4 \sqrt{2} \frac{(x+y-7)}{\sqrt{2}}$   After that, I found the intersection between $Q$ and $V$ using equations and I got that the intersection points in $(O;XY)$ are $P_1(3;2)$ and $P_2(-1;6)$      2)Given the coordinates system $(O,X'Y')$ asociated to the basis    $B=[{b_1=\frac{1}{\sqrt{2}}; \frac{1}{\sqrt{2}};  &gt; b_2=\frac{-1}{\sqrt{2}}; \frac{1}{\sqrt{2}}}]$ with origin in    $O=(0;0)$ and the ellipse $9X'^2+5Y'^2-3X'+2Y'-10=0$ find the    coordinates of the centre of the ellipse in $(O,XY)$    So, here I found the centre of the ellipse in $(O,X'Y')$ and it was $(\frac{1}{6}; - \frac{1}{5})$, then, doing something similar to what I've done before I found out that the centre in $(O,XY)$ was $( \frac{11}{30 \sqrt{2}};$ $\frac{-1}{30 \sqrt{2}}$)
80614
I'm trying to find the points within a certain distance, in meters, from another point.  I'm using the function ST_Buffer, but I don't get how to express the distance in meters.  I'm trying with something like this:  select 1 as cartodb_id, ST_Buffer(   ST_Transform(     ST_GeomFromText(       POINT(-58.38145 -34.60368)'       , 4326     )     , 3857   )   , 500 ) as the_geom_webmercator   I have the coordinates from google maps (in wgs84, 4326) then I transform them to webmercator (3857) and then I pass 500 as parameter, and I get something that looks like the desired area, but after comparing it with what google maps says it falls short for a non neglectable distance.  Is this the right way to achieve it?   note: you can try the sql query in cartodb  --  From this question:  I came with the following code (I use SRID 31997 to calculate the buffer, and then go back to webmercator)  select 1 as cartodb_id, ST_Transform(   ST_Buffer(     ST_Transform(       ST_GeomFromText(         'POINT(-58.427185 -34.624592)'         , 4326       )       , 31997     )     , 2000   ), 3857 ) as the_geom_webmercator   Now if falls short for no more than 20 meters. I thought mercator (EPSG 3857) was in metric units.
114818
I was wondering if anyone had any insight or information on how one might go about determining point coordinates in an n-dimensional space from point pair distances.  For example, say I start with a table of pairwise Euclidean distances between points for 10 points $\{a, \ldots, j\}$.  This would have a distance for each distinct pair of points, a through j, so 45 separate distances.  I want to be able to produce an n-dimensional vector for each point so that the Euclidean distance between each of the points matches the point pair distances from my original table.  Any insight on the subject would be much appreciated.
118322
I have a parametric equation   x= 24 sin (2πt) y= -24 cos(2πt)+35  This parametric equation is related to a diagram of a Ferris wheel, and is the equation of motion of a point A on the diagram which is attached.    I need to find the cartesian equation of the path of A, since the original equation given on the diagram is a parametric equation and find the distance between O and A when t=0.25.  I have found the cartesian equation and it is , but I don't know how to find the distance between O and A on the diagram (A is one of the smaller double Ferris wheel). Do I assume that since O is stationary it is on the origin of the plane, and I just work out how far away from the origin is A at t=0.25, or am I just approaching this wrong? It has been a long time since I have done vectors, so I am sorry if I sound completely wrong. Has the distance formula got anything to do with it?  Then next I have to state the position vector of point B at time t minutes. I have no idea how to approach this, so this will be of additional help.  Correct me if I am wrong in my thinking somewhere.  Thanks
103090
I am a beginner at ArcGIS and have a technical question.  I need to assign a projected coordinate system to my boundary/polygon data in order to calculate square miles.  There are many to choose from.  Which should I be using?  Here are my geographic coordinate system details:    Angular Unit: Degree (0.017453292519943295) Prime Meridian: Greenwich (0.000000000000000000) Datum: D_North_American_1983                         Spheroid: GRS_1980                          Semimajor Axis: 6378137.000000000000000000                         Semiminor Axis: 6356752.314140356100000000                          Inverse Flattening: 298.257222101000020000   Thanks in advance for any consideration.
47762
I have a set of x,y coordinates and wish to find a new coordinate pair which is as far away from any point in the initial set, while also being within the convex hull of the original points.   The current process I have to determine the new point is as follows:   Generate every possible midpoint between the original coordinate pairs. Compute the distance between the midpoints and every original point. Find the smallest distance calculated in Step 2 for each midpoint. Find the midpoint with the largest distance from Step 3.   If multiple points have the same maximum distance to the nearest original set of points, only one of those points is needed (for my purpose) and can be selected arbitrarily.  Is there a better method for finding these coordinates?
52632
There are several algorithms to calculate distance between two latitude/longitude points.     Hubeny     But there are not so many algorithms (actually I can't finally find it) which can calculate latitude/longitude of point B from another latitude/longitude point A with bearing and distance.      I'm glad to know if there are any algorithms which is lesser accurate, but faster than Vincenty's formulae.
56624
I have a $3$-D sphere of radius $R$, centered at the origin.   $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ are two points on the sphere.   The Euclidean distance is easy to calculate, but what if I were to restrict myself to traveling on the surface of the sphere?    Two approaches come to mind: use  in some way, or simply use trigonometry: calculate the angle between the two points and get a distance from that.    Will both/either of these methods work? Which would be easier?  Somewhat related to . Maybe it will inspire someone to go answer it!
104030
Usually when on Earth we can use as an approximation:  g = a = 9.18 $\frac{m}{s^2}$ s = vi + 1/2 a$t^{2}$  However, how would I determine the distance traveled in situations where the the acceleration due to gravity is changing non-negligibly because the distances are so large? I know that the following equation has to be used somehow:  a = $Gmr^{-2}$  I am not sure exactly how though. For example, if given the information that an object is at a certain distance and is being attracted by a larger object of a given mass, the method should calculate where the object will be at a given time.