Search Results

Search found 74 results on 3 pages for 'trigonometry'.

Page 2/3 | < Previous Page | 1 2 3  | Next Page >

• Calculating bounding box a certain distance away from a lat/long coordinate in Java

  - by Bryce Thomas

  Given a coordinate (lat, long), I am trying to calculate a square bounding box that is a given distance (e.g. 50km) away from the coordinate. So as input I have lat, long and distance and as output I would like two coordinates; one being the south-west (bottom-left) corner and one being the north-east (top-right) corner. I have seen a couple of answers on here that try to address this question in Python, but I am looking for a Java implementation in particular. Just to be clear, I intend on using the algorithm on Earth only and so I don't need to accommodate a variable radius. It doesn't have to be hugely accurate (+/-20% is fine) and it'll only be used to calculate bounding boxes over small distances (no more than 150km). So I'm happy to sacrifice some accuracy for an efficient algorithm. Any help is much appreciated. Edit: I should have been clearer, I really am after a square, not a circle. I understand that the distance between the center of a square and various points along the square's perimeter is not a constant value like it is with a circle. I guess what I mean is a square where if you draw a line from the center to any one of the four points on the perimeter that results in a line perpendicular to a side of the perimeter, then those 4 lines have the same length.

  Read the article

• Choosing circle radius to fully fill a rectangle

  - by Andy

  Hi, the pixman image library can draw radial color gradients between two circles. I'd like the radial gradient to fill a rectangular area defined by "width" and "height" completely. Now my question, how should I choose the radius of the outer circle? My current parameters are the following: A) inner circle (start of gradient) center pointer of inner circle: (width*0.5|height*0.5) radius of inner circle: 1 color: black B) outer circle (end of gradient) center pointer of outer circle: (width*0.5|height*0.5) radius of outer circle: ??? color: white How should I choose the radius of the outer circle to make sure that the outer circle will entirely fill my bounding rectangle defined by width*height. There shall be no empty areas in the corners, the area shall be completely covered by the circle. In other words, the bounding rectangle width,height must fit entirely into the outer circle. Choosing outer_radius = max(width, height) * 0.5 as the radius for the outer circle is obviously not enough. It must be bigger, but how much bigger? Thanks!

  Read the article

• Rotation towards an object in 3d space

  - by retoucher

  hello, i have two coordinates on a 2d plane in 3d space, and am trying to rotate one coordinate (a vector) to face the other coordinate. my vertical axis is the y-axis, so if both of the coordinates are located flat on the 2d plane, they would both have a y-axis of 0, and their x and z coordinates determine their position length/width-wise on the plane. right now, i'm calculating the angle like so (language agnostic): angle = atan2(z2-z1,x2-x1); and am rotating/translating in space like so: pushMatrix(); rotateY(angle); popMatrix(); this doesn't seem to be working though. are my calculations/process correct?

  Read the article

• How to find relation between change in latitudes at centre of map and top/bottom

  - by Imran

  Hi, Im having little trouble finding a relation between the movement at centre and edge of a circle, Im doing for panning world map,my map extent is 180,89:-180,-89, my map pans by adding change(dx,dY) to its extents and not its centre. Now a situation has arrrised where I have to move the map to a specific centre, to calculate the change in longitudes is very easy and simple, but its the change in lattitudes that has caused problem. It seems the change in centreY of map is more than the change at edge of the mapY, or simply if I have to move the map centre from 0long,0lat to 73long,33lat, for dX I simply get 73, but for dY apparently it looks 33 but if i add 33 to top of map that is 89 , it will be 122 which is incorrect since Latitudes are between 90 and -90 . It seems a case a projection of a circle on 2D plane where the edge of circle since is moving backward due to angle expereinces less change and the centre expereinces more change, now is there a relation between these two factors? I tried converting the difference between OriginY and destinationY into radians and then add to Top and Bottom of Map, but it did'nt really work for me. Please note that the map is project on a virtual canvas whose width starts from 256 and increases by 256*2^z , z=0 is default and whole world is visible at that extent of canvas code: public void moveMapTo(double destinationLongitude,double destinationLattitude) // moves map to the new centre { double dXLong=destinationLongitude-centreLongitude; double atanhsinO = atanh(Math.sin(destinationLattitude * Math.PI / 180.00)); double atanhsinD = atanh(Math.sin(centreLatitude * Math.PI / 180.00)); double atanhCentre = (atanhsinD + atanhsinO) / 2; double latitudeSpan =destinationLattitude - centreLatitude; double radianOfCentreLatitude = Math.atan(Math.sinh(atanhCentre)); double dXLat=latitudeSpan / Math.cos(radianOfCentreLatitude); dXLat*=getLattitudeSpan()*(Math.PI/180); <--- HERE IS THE PORBLEM System.out.println("dxLong:"+dXLong+"_dxLat:"+dXLat); mapLeft+=dXLong; mapRight+=dXLong; mapTop+=dXLat; mapBottom+=dXLat; } ////latitude span function private double getLattitudeSpan() { double latitudeSpan = mapTop - mapBottom; latitudeSpan = latitudeSpan / Math.cos(radianOfCentreLatitude); return Math.abs(latitudeSpan); } //ht

  Read the article

• Raphael SVG VML Implement Multi Pivot Points for Rotation

  - by Cody N

  Over the last two days I've effectively figured out how NOT to rotate Raphael Elements. Basically I am trying to implement a multiple pivot points on element to rotate it by mouse. When a user enters rotation mode 5 pivots are created. One for each corner of the bounding box and one in the center of the box. When the mouse is down and moving it is simple enough to rotate around the pivot using Raphael elements.rotate(degrees, x, y) and calculating the degrees based on the mouse positions and atan2 to the pivot point. The problem arises after I've rotated the element, bbox, and the other pivots. There x,y position in the same only there viewport is different. In an SVG enabled browser I can create new pivot points based on matrixTransformation and getCTM. However after creating the first set of new pivots, every rotation after the pivots get further away from the transformed bbox due to rounding errors. The above is not even an option in IE since in is VML based and cannot account for transformation. Is the only effective way to implement element rotation is by using rotate absolute or rotating around the center of the bounding box? Is it possible at all the create multi pivot points for an object and update them after mouseup to remain in the corners and center of the transformed bbox?

  Read the article

• Difficulty with projectile's tracking code

  - by RCIX

  I wrote some code for a projectile class in my game that makes it track targets if it can: if (_target != null && !_target.IsDead) { Vector2 currentDirectionVector = this.Body.LinearVelocity; currentDirectionVector.Normalize(); float currentDirection = (float)Math.Atan2(currentDirectionVector.Y, currentDirectionVector.X); Vector2 targetDirectionVector = this._target.Position - this.Position; targetDirectionVector.Normalize(); float targetDirection = (float)Math.Atan2(targetDirectionVector.Y, targetDirectionVector.X); float targetDirectionDelta = targetDirection - currentDirection; if (MathFunctions.IsInRange(targetDirectionDelta, -(Info.TrackingRate * deltaTime), Info.TrackingRate * deltaTime)) { Body.LinearVelocity = targetDirectionVector * Info.FiringVelocity; } else if (targetDirectionDelta > 0) { float newDirection = currentDirection + Info.TrackingRate * deltaTime; Body.LinearVelocity = new Vector2( (float)Math.Cos(newDirection), (float)Math.Sin(newDirection)) * Info.FiringVelocity; } else if (targetDirectionDelta < 0) { float newDirection = currentDirection - Info.TrackingRate * deltaTime; Body.LinearVelocity = new Vector2( (float)Math.Cos(newDirection), (float)Math.Sin(newDirection)) * Info.FiringVelocity; } } This works sometimes, but depending on the relative angle to the target projectiles turn away from the target instead. I'm stumped; can someone point out the flaw in my code?

  Read the article

• Java Math.cos() Method Does Not Return 0 When Expected

  - by dimo414

  Using Java on a Windows 7 PC (not sure if that matters) and calling Math.cos() on values that should return 0 (like pi/2) instead returns small values, but small values that, unless I'm misunderstanding, are much greater than 1 ulp off from zero. Math.cos(Math.PI/2) = 6.123233995736766E-17 Math.ulp(Math.cos(Math.PI/2)) = 1.232595164407831E-32 Is this in fact within 1 ulp and I'm simply confused? And would this be an acceptable wrapper method to resolve this minor inaccuracy? public static double cos(double a){ double temp = Math.abs(a % Math.PI); if(temp == Math.PI/2) return 0; return Math.cos(a); }

  Read the article

• How can I work around the fact that in C++, sin(3.14159265) is not 0?

  - by Adam Doyle

  In C++, const double Pi = 3.14159265; cout << sin(Pi); // displays: 3.58979e-009 it SHOULD display the number zero I understand this is because Pi is being approximated, but is there any way I can have a value of Pi hardcoded into my program that will return 0 for sin(Pi)? (a different constant maybe?) In case you're wondering what I'm trying to do: I'm converting polar to rectangular, and while there are some printf() tricks I can do to print it as "0.00", it still doesn't consistently return decent values (in some cases I get "-0.00") The lines that require sin and cosine are: x = r*sin(theta); y = r*cos(theta); BTW: My Rectangular - Polar is working fine... it's just the Polar - Rectangular Thanks! edit: I'm looking for a workaround so that I can print sin(some multiple of Pi) as a nice round number to the console (ideally without a thousand if-statements)

  Read the article

• Inner angle between two lines

  - by ocell

  Hi folks, I have two lines: Line1 and Line2. Each line is defined by two points (P1L1(x1, y1), P2L1(x2, y2) and P1L1(x1, y1), P2L3(x2, y3)). I want to know the inner angle defined by these two lines. For do it I calculate the angle of each line with the abscissa: double theta1 = atan(m1) * (180.0 / PI); double theta2 = atan(m2) * (180.0 / PI); After to know the angle I calculate the following: double angle = abs(theta2 - theta1); The problem or doubt that I have is: sometimes I get the correct angle but sometimes I get the complementary angle (for me outer). How can I know when subtract 180º to know the inner angle? There is any algorithm better to do that? Because I tried some methods: dot product, following formula: result = (m1 - m2) / (1.0 + (m1 * m2)); But always I have the same problem; I never known when I have the outer angle or the inner angle! Thanks in advance for reading my trouble and for your time! Oscar.

  Read the article

• Perturb vector by some angle

  - by Myx

  Hello: I have a unit vector in 3D space whose direction I wish to perturb by some angle within the range 0 to theta, with the position of the vector remaining the same. What is a way I can accomplish this? Thanks.

  Read the article

• Calculating rotation in > 360 deg. situations

  - by danglebrush

  I'm trying to work out a problem I'm having with degrees. I have data that is a list of of angles, in standard degree notation -- e.g. 26 deg. Usually when dealing with angles, if an angle exceeds 360 deg then the angle continues around and effectively "resets" -- i.e. the angle "starts again", e.g. 357 deg, 358 deg, 359 deg, 0 deg, 1 deg, etc. What I want to happen is the degree to continue increasing -- i.e. 357 deg, 358 deg, 359 deg, 360 deg, 361 deg, etc. I want to modify my data so that I have this converted data in it. When numbers approach the 0 deg limit, I want them to become negative -- i.e. 3 deg, 2 deg, 1 deg, 0 deg, -1 deg, -2 deg, etc. With multiples of 360 deg (both positive and negative), I want the degrees to continue, e.g. 720 deg, etc. Any suggestions on what approach to take? There is, no doubt, a frustratingly simple way of doing this, but my current solution is kludgey to say the least .... ! My best attempt to date is to look at the percentage difference between angle n and angle n - 1. If this is a large difference -- e.g. 60% -- then this needs to be modified, by adding or subtracting 360 deg to the current value, depending on the previous angle value. That is, if the previous angle is negative, substract 360, and add 360 if the previous angle is positive. Any suggestions on improving this? Any improvements?

  Read the article

• Actionscript 3 3D sizing problem

  - by Simon

  I have an application where I can have an image moving towards the eye. When the image enlarged, I would like to have it resized as 635 px where initial size was 220 px. I have the image with starting position as 0 in z axis. I am wanting to calculate the distance from starting position to the resized image. I have already calculate the distance by hand but when I tried to put it on flash the result is not what i wanted. I am sure that the value I calculated was correct. I know it may be hard to understand my problem. Please help.

  Read the article

• Function to get X, Y position of an object orbiting a point, given a distance and angle in radians?

  - by Jake Petroules

  I am trying to code a function for a camera that orbits a point. Assume a 3d coordinate plane where Z is up. Ignore Z. Let's say the camera's position starts at (0, 0, z). The object to orbit is at, say (50, 50, z). So we have a distance of ~70 units. Calling the function with {(50, 50, z), 70, x} where x is the position in orbit, in radians, should return where the position of the camera should be. I believe this involves cos and tan but my trig isn't that great... point3d getCameraPosition(point3d objectPosition, float distance, float rotationRadians) { // ??? }

  Read the article

• How can I correctly calculate the direction for a moving object?

  - by Jakub Hampl

  I'm solving the following problem: I have an object and I know its position now and its position 300ms ago. I assume the object is moving. I have a point to which I want the object to get. What I need is to get the angle from my current object to the destination point in such a format that I know whether to turn left or right. The idea is to assume the current angle from the last known position and the current position. I'm trying to solve this in MATLAB. I've tried using several variations with atan2 but either I get the wrong angle in some situations (like when my object is going in circles) or I get the wrong angle in all situations. Examples of code that screws up: a = new - old; b = dest - new; alpha = atan2(a(2) - b(2), a(1) - b(1); where new is the current position (eg. x = 40; y = 60; new = [x y];), old is the 300ms old position and dest is the destination point. Edit Here's a picture to demonstrate the problem with a few examples: In the above image there are a few points plotted and annotated. The black line indicates our estimated current facing of the object. If the destination point is dest1 I would expect an angle of about 88°. If the destination point is dest2 I would expect an angle of about 110°. If the destination point is dest3 I would expect an angle of about -80°.

  Read the article

• Built in raytracing?

  - by acidzombie24

  Relating to this question i was wondering if .NET has any libs (or a function) i can use to detect if one point collides with another. I am not sure what angles i should use but is there some function like this func(point src, rect target, angle, distanceOfVision, listPointOrRectOfWalls) Pretty unlikely but i dont know a formula or how to start. Its a quick and dirty prototype. I am thinking of writing the func about but dropping angle and just make line of sight a rectangle and check if any points is between src and target.

  Read the article

• How to interpret situations where Math.Acos() reports invalid input?

  - by Sean Ochoa

  Hey all. I'm computing the angle between two vectors, and sometimes Math.Acos() returns NaN when it's input is out of bounds (-1 input && input 1) for a cosine. What does that mean, exactly? Would someone be able to explain what's happening? Any help is appreciated! Here's me method: public double AngleBetween(vector b) { var dotProd = this.Dot(b); var lenProd = this.Len*b.Len; var divOperation = dotProd/lenProd; // http://msdn.microsoft.com/en-us/library/system.math.acos.aspx return Math.Acos(divOperation) * (180.0 / Math.PI); }

  Read the article

• How do i render and detect a line of sight?

  - by acidzombie24

  If you look at the top right you'll see on a radar an enemy unit line of sight. I was wondering what is the most efficient or easiest way (little code, fairly accurate. doesnt need to be perfect) to detect if something is in your line of sight? I may or may not need to render it (i likely wont). I dont know the formula nor used any math libs/namespaces in C#

  Read the article

• 3D coordinate of 2D point given camera and view plane

  - by Myx

  I wish to generate rays from the camera through the viewing plane. In order to do this, I need my camera position ("eye"), the up, right, and towards vectors (where towards is the vector from the camera in the direction of the object that the camera is looking at) and P, the point on the viewing plane. Once I have these, the ray that's generated is: ray = camera_eye + t*(P-camera_eye); where t is the distance along the ray (assume t = 1 for now). My question is, how do I obtain the 3D coordinates of point P given that it is located at position (i,j) on the viewing plane? Assume that the upper left and lower right corners of the viewing plane are given. NOTE: The viewing plane is not actually a plane in the sense that it doesn't extend infinitely in all directions. Rather, one may think of this plane as a widthxheight image. In the x direction, the range is 0--width and in the y direction the range is 0--height. I wish to find the 3D coordinate of the (i,j)th element, 0

  Read the article

• Looking for calculator source code, BSD-licensed

  - by Horace Ho

  I have an urgent project which need many functions of a calculator (plus a few in-house business rule formulas). As I won't have time to re-invent the wheel so I am looking for source code directly. Requirements: BSD licensed (GPL won't help) in c/c++ programming language 32-bit CPU minimum dependency on platform API/data structure best with both RPN and prefix notation supported emulator/simulator code also acceptable (if not impossible to add custom formula) with following functions (from wikipedia) Scientific notation for calculating large numbers floating point arithmetic logarithmic functions, using both base 10 and base e trigonometry functions (some including hyperbolic trigonometry) exponents and roots beyond the square root quick access to constants such as pi and e plus hexadecimal, binary, and octal calculations, including basic Boolean math fractions optional statistics and probability calculations complex numbers programmability equation solving

  Read the article

• how to transform child elements position into a world position

  - by MrGreg

  So Im making a 2d space game and I have a bunch of spaceships that have turrets. Objects have a position and orientation, the ships being in world coordinates while the turrets are children and coordinates are relative to their parents. How do I efficiently calculate the position of a turret in world coordinates (i.e. when it fires and I need to know where to place a bullet in the world)? Calculating the turrets orientation is trivial - I just add the turrets relative angle to its parents. For position though, I guess I could do a bunch of trigonometry but this MUST be a common problem with a good/fast general solution? Should I be relearning how to do matrix math again? :) btw - Im creating the game in javascript+canvas but its the math/algorithm im interested in here Cheers, Greg

  Read the article

• OpenGL Get Rotated X and Y of quad

  - by matejkramny

  I am developing a game in 2D using LWJGL library. So far I have a rotating box. I have done basic Rectangle collision, but it doesn't work for rotated rectangles. Does OpenGL have a function that returns the vertices of rotated rectangle? Or is there another way of doing this using trigonometry? I had researched how to do this and everything I found was using some matrix that I don't understand so I am asking if there is another way of doing this. For clarification, I am trying to find out the true (rotated) X,Y of each point of the rectangle. Let's say, the first point of a rectangle (top,left) has x=10 y=10.. Width and height is 100 pixels. When I rotate the rectangle using glRotatef() the x and y stay the same. The rotation is happening inside OpenGL. I need to extract the x,y of the rectangle so I can detect collisions properly.

  Read the article

• Elastic Collision Formula in Java

  - by Shijima

  I'm trying to write a Java formula based on this tutorial: 2-D elastic collisions without Trigonometry. I am in the section "Elastic Collisions in 2 Dimensions". In step 1, it mentions "Next, find the unit vector of n, which we will call un. This is done by dividing by the magnitude of n". My below code represents the normal vector of 2 objects (I'm using a simple array to represent the normal vector), but I am not really sure what the tutorial means by dividing the magnitude of n to get the un. int[] normal = new int[2]; normal[0] = ball2.x - ball1.x; normal[1] = ball2.y - ball1.y; Can anyone please explain what un is, and how I can calculate it with my array in Java?

  Read the article

• How do I find a unit vector of another in Java?

  - by Shijima

  I'm writing a Java formula based on this tutorial: 2-D elastic collisions without Trigonometry. I am in the section "Elastic Collisions in 2 Dimensions". Part of step 1 says: Next, find the unit vector of n, which we will call un. This is done by dividing by the magnitude of n. My below code represents the normal vector of 2 objects (I'm using a simple array to represent the normal vector). int[] normal = new int[2]; normal[0] = ball2.x - ball1.x; normal[1] = ball2.y - ball1.y; I am unsure what the tutorial means by dividing the magnitude of n to get the un. What is un? How can I calculate it with my Java array?

  Read the article

• How to change the coordinate origin in Flash's stage with Actionscript?

  - by Petruza

  I think I did this before but can't find the code. Flash as many other graphical frameworks use the top-left corner as the coordinate origin (0,0) because it's how the underlying memory model is by convention. But it would be really simpler for my calculations if the origin was in the center of the stage, because all the game revolves around the center and uses a lot of trigonometry, angles, etc. Is there some built-in method like Stage::setOrigin( uint, uint ); or something like that?

  Read the article

• Find distance between two points using MKMapKit

  - by mag725

  Hi, I'm attempting to find the euclidean distance in meters between two points on an MKMapView using iPhone OS 3.2. The problem is that I have these coordinates in terms of latitude and longitude, which, mathematically provides me enough data to find the distance, but it's going to take some tricky trigonometry. Is there any simpler solution? Thanks!

  Read the article

< Previous Page | 1 2 3  | Next Page >