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  • Java: Calculate distance between a large number of locations and performance

    - by Ally
    I'm creating an application that will tell a user how far away a large number of points are from their current position. Each point has a longitude and latitude. I've read over this article http://www.movable-type.co.uk/scripts/latlong.html and seen this post http://stackoverflow.com/questions/837872/calculate-distance-in-meters-when-you-know-longitude-and-latitude-in-java There are a number of calculations (50-200) that need carried about. If speed is more important than the accuracy of these calculations, which one is best?

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  • how to fast compute distance between high dimension vectors

    - by chyojn
    assume there are three group of high dimension vectors: {a_1, a_2, ..., a_N}, {b_1, b_2, ... , b_N}, {c_1, c_2, ..., c_N}. each of my vector can be represented as: x = a_i + b_j + c_k, where 1 <=i, j, k <= N. then the vector is encoded as (i, j, k) wich is then can be decoded as x = a_i + b_j + c_k. my question is, if there are two vector: x = (i_1, j_1, k_1), y = (i_2, j_2, k_2), is there a method to compute the euclidian distance of these two vector without decode x and y.

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  • "Distance" between colours in PHP

    - by Phil
    I'm looking for a function that can accurately represent the distance between two colours as a number or something. For example I am looking to have an array of HEX values or RGB arrays and I want to find the most similar colour in the array for a given colour eg. I pass a function a RGB value and the 'closest' colour in the array is returned

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  • VMware annonce VMware View 4.6, la solution de virtualisation bureautique améliore l'accès à distance grâce à la prise en charge du PCoIP

    VMware annonce VMware View 4.6 La solution de virtualisation bureautique améliore l'accès à distance grâce à la prise en charge du PCoIP VMware, le spécialiste des logiciels de virtualisation et d'infrastructures de Cloud Computing, annonce la disponibilité générale de la version 4.6 de VMware View, une solution qui offre un accès à distance qui prend désormais en charge le protocole PCoIP. Pour mémoire, la prise en charge du PCoIP permet de simplifier et de sécuriser les processus de connexion à distance et d'authentification sur le poste de travail, et, ajoute Vmware, « en libérant l'entreprise de l'obligation de posséder un VPN SSL ». « La prise en char...

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  • Optimising SQL distance query

    - by Alex
    I'm running an MySQL query that returns results based on location. However I have noticed recently that its really slowing down my PHP app. I used CodeIgniter and the profiler shows the query taking 4.2seconds. The geoname table has 500,000 rows. I have some indexes on the key columns, how else can speed up this query? Here is my SQL: SELECT `products`.`product_name`, `geoname`.`geonameid`, `geoname`.`latitude`, `geoname`.`longitude`, `products`.`product_id`, AVG(ratings.vote) as rating, count(comments.comment_id) as total_comments, (6371 * acos(cos(radians(38.7666667)) * cos(radians(geoname.latitude)) * cos(radians(geoname.longitude) - radians(-3.3833333)) + sin(radians(38.7666667)) * sin(radians(geoname.latitude)))) AS distance FROM (`foods`) JOIN `geoname` ON `geoname`.`geonameid` = `products`.`geoname_id` LEFT JOIN `ratings` ON `ratings`.`var_id` = `products`.`product_id` LEFT JOIN `comments` ON `comments`.`var_id` = `products `.`product_id` WHERE `products`.`product_id` != 82 GROUP BY `products`.`product_id` HAVING `distance` < 99 ORDER BY `distance` LIMIT 10

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  • REST API for driving distance?

    - by Mark
    Is there a service that will give me the driving distance between two addresses? Apparently Google Maps API requires you to display a map, which I don't want to do (on that particular page), and I'd like to just snag the data and save it to my DB after a user submits a form, rather than waiting for JS to do it's thing. If it's relevant, this is going into a Django app. I discovered that CloudMade offers a Python API, which is nice, except their latest dev release has a bug in it (can't use the API object), but more importantly, it's support for Canada is awful/worthless (couldn't find directions from any major city around here!).

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  • How to find distance between two geopoints in c using calculateDistance(const CWaypoint& wp)

    - by Harsha
    void getAllDataByPointer(string *pname,double *platitude, double *plongitude); void getAllDataByReference(string &pname,double &platitude, double &plongitude); double calculateDistance(const CWaypoint& wp); void print(int format); bool less(const CWaypoint& wp_right); CWaypoint add(const CWaypoint& wp_right); These are the functions I am using. I have the values as output but how to call the latitude values of two different cities so that I can use the following formula distance = ERADIUS * (acos(sin(latitude_1)*sin(latitude_2) + cos(latitude_1) * cos(latitude_2)*cos(longitude_2 - longitude_1)));

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  • Combinatorial optimisation of a distance metric

    - by Jose
    I have a set of trajectories, made up of points along the trajectory, and with the coordinates associated with each point. I store these in a 3d array ( trajectory, point, param). I want to find the set of r trajectories that have the maximum accumulated distance between the possible pairwise combinations of these trajectories. My first attempt, which I think is working looks like this: max_dist = 0 for h in itertools.combinations ( xrange(num_traj), r): for (m,l) in itertools.combinations (h, 2): accum = 0. for ( i, j ) in itertools.izip ( range(k), range(k) ): A = [ (my_mat[m, i, z] - my_mat[l, j, z])**2 \ for z in xrange(k) ] A = numpy.array( numpy.sqrt (A) ).sum() accum += A if max_dist < accum: selected_trajectories = h This takes forever, as num_traj can be around 500-1000, and r can be around 5-20. k is arbitrary, but can typically be up to 50. Trying to be super-clever, I have put everything into two nested list comprehensions, making heavy use of itertools: chunk = [[ numpy.sqrt((my_mat[m, i, :] - my_mat[l, j, :])**2).sum() \ for ((m,l),i,j) in \ itertools.product ( itertools.combinations(h,2), range(k), range(k)) ]\ for h in itertools.combinations(range(num_traj), r) ] Apart from being quite unreadable (!!!), it is also taking a long time. Can anyone suggest any ways to improve on this?

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  • Are there software options (preferabbly .NET) for doing distance and speed analysis of footballers moving on video?

    - by Anonymous Type
    Editing Question for Clarity Thanks for feedback so far, very insightful. I'm not sure how far along this part of the software community is, and what if any libraries exist for me to leverage from. Heres what I'm trying to do. Problem: Take an existing video of a game of rugby league. The Rugby League field is 100 metres long, 70 metres wide, and has white line markings every 10 metres running along the width of the field, as well as along the sidelines. Each side has 13 players on the field. Players on each team have identical jerseys that normally constrast strongly against background colours (green/brown field colour) and the referee's colour (usually yellow) and the designated water runner (orange). All players have a unique number in thick white lettering on their backs for identification. Video is taken with a high definition camera. Currently only one camera is used (2D) and existing video does not contain a foreground object of fixed spatial dimensions (as suggested in one answer for comparision measurements, however I could add this to future filming sessions if it is worthwhile). The player's do not run in a straight line 50% of the time but will go sideways on on a diagonal to the play the ball. The distance measured always starts from the spot of the previous "tackle", which ends where the player stops forward movement. It is not always possible to determine the players number from the video (facing other direction, sunlight, others standing in the way of the camera). But this isn't important as the software could allow for manual inputting of unknown "runs" at a later point after analysis. Determine the distance between two points (i.e. where the player started his "run" and where he finished it). I'm guessing that this would be quite doable if I manually marked the start and end point in the video. But how would I use landmarks in the background to determine the distance (assuming the person taking the video has kept it from jerking around). Question: Do software packages or libraries exist that are specialised enough to assist with writing analysis software to determine a sports persons distance travelled based on video taken of the performance?

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  • Find location using only distance and range?

    - by pinnacler
    Triangulation works by checking your angle to three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and it's to my right at 90 degrees." Repeat 2 more times for different targets and angles. Trilateration works by checking your distance from three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and I'm 100 meters away from that." Repeat 2 more times for different targets and ranges. But both of those methods rely on knowing WHAT you're looking at. Say you're in a forest and you can't differentiate between trees, but you know where key trees are. These trees have been hand picked as "landmarks." You have a robot moving through that forest slowly. Do you know of any ways to determine location based solely off of angle and range, exploiting geometry between landmarks? Note, you will see other trees as well, so you won't know which trees are key trees. Ignore the fact that a target may be occluded. Our pre-algorithm takes care of that. 1) If this exists, what's it called? I can't find anything. 2) What do you think the odds are of having two identical location 'hits?' I imagine it's fairly rare. 3) If there are two identical location 'hits,' how can I determine my exact location after I move the robot next. (I assume the chances of having 2 occurrences of EXACT angles in a row, after I reposition the robot, would be statistically impossible, barring a forest growing in rows like corn). Would I just calculate the position again and hope for the best? Or would I somehow incorporate my previous position estimate into my next guess? If this exists, I'd like to read about it, and if not, develop it as a side project. I just don't have time to reinvent the wheel right now, nor have the time to implement this from scratch. So if it doesn't exist, I'll have to figure out another way to localize the robot since that's not the aim of this research, if it does, lets hope it's semi-easy.

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  • Find location using only distance and bearing?

    - by pinnacler
    Triangulation works by checking your angle to three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and it's to my right at 90 degrees." Repeat 2 more times for different targets and angles. Trilateration works by checking your distance from three KNOWN targets. "I know the that's the Lighthouse of Alexandria, it's located here (X,Y) on a map, and I'm 100 meters away from that." Repeat 2 more times for different targets and ranges. But both of those methods rely on knowing WHAT you're looking at. Say you're in a forest and you can't differentiate between trees, but you know where key trees are. These trees have been hand picked as "landmarks." You have a robot moving through that forest slowly. Do you know of any ways to determine location based solely off of angle and range, exploiting geometry between landmarks? Note, you will see other trees as well, so you won't know which trees are key trees. Ignore the fact that a target may be occluded. Our pre-algorithm takes care of that. 1) If this exists, what's it called? I can't find anything. 2) What do you think the odds are of having two identical location 'hits?' I imagine it's fairly rare. 3) If there are two identical location 'hits,' how can I determine my exact location after I move the robot next. (I assume the chances of having 2 occurrences of EXACT angles in a row, after I reposition the robot, would be statistically impossible, barring a forest growing in rows like corn). Would I just calculate the position again and hope for the best? Or would I somehow incorporate my previous position estimate into my next guess? If this exists, I'd like to read about it, and if not, develop it as a side project. I just don't have time to reinvent the wheel right now, nor have the time to implement this from scratch. So if it doesn't exist, I'll have to figure out another way to localize the robot since that's not the aim of this research, if it does, lets hope it's semi-easy.

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  • WPF drag distance threshold

    - by justin984
    I have a program with two WPF treeviews that allow dragging and dropping between the two. The problem is, it can be annoying to open / close items on the treeviews because moving the mouse just one pixel while holding the left mouse button triggers the drag / drop functionality. Is there some way to specify how far the mouse should move before it's considered a drag / drop? Thanks!

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  • Java2D Distance Collision Detection

    - by Trizicus
    My current setup is only useful once collision has been made; obviously there has to be something better than this? public boolean CollisionCheck(Rectangle rect1, Rectangle rect2) { if(rect1.intersects(rect2)) { return true; } return false; } How can I do preemptive collision detection?

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  • Get the distance between two geo points

    - by Chmouel Boudjnah
    Hi, I want to make a apps that check what's the nearest place from where the user is. I can easily get the location of the user and I have a list of places with latitude and longitude. What would be the best way to know the nearest place of the list against the current position. I could not find anything in the google APIs. I am worried I need to resort to my calculate and have to do math to calculate it. What do you guys think ? Cheers and thanks for reading or replying.

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  • calculate distance with linq or subsonic C# MVC

    - by minus4
    i have this MySQL statement from a search page, the user enters there postcode and it finds the nearest stiocklist within 15 MIles of the entered postcode. SELECT * , ( ( ACOS( SIN( "+SENTLNG +" * PI( ) /180 ) * SIN( s_lat * PI( ) /180 ) + COS( " + SENTLNG +" * PI( ) /180 ) * COS( s_lat * PI( ) /180 ) * COS( ( " + SENTLANG + " - s_lng ) * PI( ) /180 ) ) *180 / PI( ) ) *60 * 1.1515 ) AS distance_miles FROM new_stockists WHERE s_lat IS NOT NULL HAVING distance_miles <15 ORDER BY distance_miles ASC LIMIT 0 , 15 but now i am using linq and subsonic and not got a clue how do do this in linq or subsonic your help would be much appreciated, please also not that i have to sent in a dynamic from address, thats the postcode mentioned at the top of the page, i do a call to google to get then lng and lat from them for the postcode given.

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  • calculate distance with linq or subsonic

    - by minus4
    i have this MySQL statement from a search page, the user enters there postcode and it finds the nearest stiocklist within 15 MIles of the entered postcode. SELECT * , ( ( ACOS( SIN( "+SENTLNG +" * PI( ) /180 ) * SIN( s_lat * PI( ) /180 ) + COS( " + SENTLNG +" * PI( ) /180 ) * COS( s_lat * PI( ) /180 ) * COS( ( " + SENTLANG + " - s_lng ) * PI( ) /180 ) ) *180 / PI( ) ) *60 * 1.1515 ) AS distance_miles FROM new_stockists WHERE s_lat IS NOT NULL HAVING distance_miles <15 ORDER BY distance_miles ASC LIMIT 0 , 15 but now i am using linq and subsonic and not got a clue how do do this in linq or subsonic your help would be much appreciated, please also not that i have to sent in a dynamic from address, thats the postcode mentioned at the top of the page, i do a call to google to get then lng and lat from them for the postcode given.

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  • Find a node in a Graph that minimizes the distance between two other nodes

    - by Andrés
    Here is the thing. I have a directed weighted graph G, with V vertices and E edges. Given two nodes in the graph, let's say A, and B, and given the weight of an edge A-B denoted as w(A, B), I need to find a node C so that max(w(A, C), w(B, C)) is minimal among all possibilities. By possibilities I mean all the values C can take. I don't know if it is completely clear, if it's not, I'll try to be more precise. Thanks in advance.

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  • Connecting grouped dots/points on a scatter plot based on distance

    - by ToNoY
    I have 2 sets of depth point measurements, for example: > a depth value 1 2 2 2 4 3 3 6 4 4 8 5 5 16 40 6 18 45 7 20 58 > b depth value 1 10 10 2 12 20 3 14 35 I want to show both groups in one figure plotted with depth and with different symbols as you can see here plot(a$value, a$depth, type='b', col='green', pch=15) points(b$value, b$depth, type='b', col='red', pch=14) The plot seems okay, but the annoying part is that the green symbols are all connected (though I want connected lines also). I want connection only when one group has a continued data points at 2 m interval i.e. the symbols should be connected with a line from 2 to 8 m (green) and then group B symbols should be connected from 10-14 m (red) and again group A symbols should be connected (green), which means I do NOT want to see the connection between 8 m sample with the 16 m for group A. An easy solution may be dividing the group A into two parts (say, A-shallow and A-deep) and then plotting A-shallow, B, and A-deep separately. But this is completely impractical because I have thousands of data points with hundreds of groups i.e. I have to produce many depth profiles. Therefore, there has to be a way to program so that dots are NOT connected beyond a prescribed frequency/depth interval (e.g. 2 m in this case) for a particular group of samples. Any idea?

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