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  • Camera field of view: 3D projections & trigonometry

    - by Thomas O
    Okay, here goes. I have a camera at (Xc, Yc, Zc.) The Xc and Yc coordinates are latitude/longitude, and the Zc coordinate is an altitude in metres. I have a point at (Xp, Yp, Zp) and a field of view on the camera (Th1, Th2) - where Th1 is horizontal FOV and Th2 is vertical FOV. Given this information, I'd like to: test if the point is visible (i.e. in the camera's FOV) project the point as the camera would see it I've figured out already that the camera's horizontal view at any given distance is tan(Th1) * distance, but I don't know how to test if the point is visible. Accuracy is not critical. I would prefer a simple solution over a complicated solution, if it works well enough. The computations will be performed by a small microcontroller, which isn't very fast at things like trig functions. P.S. this is not homework, I'm doing this for some game development. It will be integrated with the real world, hence the latitude/longitude/altitude. It involves flying real RC planes through virtual hoops (or chasing virtual targets), so I have to project the positions of these hoops on a display.

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  • Missing Coordinates. Basic Trigonometry Help.

    - by TheDarkIn1978
    please refer to my quick diagram attached below. what i'm trying to do is get the coordinates of the yellow dots by using the angle from the red dots' known coordinates. assuming each yellow dot is about 20 pixels away from the x:50/y:250 red dot at a right angle (i think that's what it's called) how do i get their coordinates? i believe this is very basic trigonometry and i should use Math.tan(), but they didn't teach us much math in art school.

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  • Good .NET library for fast streaming / batching trigonometry (Atan)?

    - by Sean
    I need to call Atan on millions of values per second. Is there a good library to perform this operation in batch very fast. For example, a library that streams the low level logic using something like SSE? I know that there is support for this in OpenCL, but I would prefer to do this operation on the CPU. The target machine might not support OpenCL. I also looked into using OpenCV, but it's accuracy for Atan angles is only ~0.3 degrees. I need accurate results.

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  • Raycasting "fisheye effect" question

    - by mattboy
    Continuing my exploration of raycasting, I am very confused about how the correction of the fisheye effect works. Looking at the screenshot below from the tutorial at permadi.com, the way I understand the cause of the fisheye effect is that the rays that are cast are distances from the player, rather than the distances perpendicular to the screen (or camera plane) which is what really needs to be displayed. The distance perpendicular to the screen then, in my world, should simply be the distance of Y coordinates (Py - Dy) assuming that the player is facing straight upwards. Continuing the tutorial, this is exactly how it seems to be according to the below screenshot. From my point of view, the "distorted distance" below is the same as the distance PD calculated above, and what's labelled the "correct distance" below should be the same as Py - Dy. Yet, this clearly isn't the case according to the tutorial. My question is, WHY is this not the same? How could it not be? What am I understanding and visualizing wrong here?

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  • How can I draw an arrow at the edge of the screen pointing to an object that is off screen?

    - by Adam Henderson
    I am wishing to do what is described in this topic: http://www.allegro.cc/forums/print-thread/283220 I have attempted a variety of the methods mentioned here. First I tried to use the method described by Carrus85: Just take the ratio of the two triangle hypontenuses (doesn't matter which triagle you use for the other, I suggest point 1 and point 2 as the distance you calculate). This will give you the aspect ratio percentage of the triangle in the corner from the larger triangle. Then you simply multiply deltax by that value to get the x-coordinate offset, and deltay by that value to get the y-coordinate offset. But I could not find a way to calculate how far the object is away from the edge of the screen. I then tried using ray casting (which I have never done before) suggested by 23yrold3yrold: Fire a ray from the center of the screen to the offscreen object. Calculate where on the rectangle the ray intersects. There's your coordinates. I first calculated the hypotenuse of the triangle formed by the difference in x and y positions of the two points. I used this to create a unit vector along that line. I looped through that vector until either the x coordinate or the y coordinate was off the screen. The two current x and y values then form the x and y of the arrow. Here is the code for my ray casting method (written in C++ and Allegro 5) void renderArrows(Object* i) { float x1 = i->getX() + (i->getWidth() / 2); float y1 = i->getY() + (i->getHeight() / 2); float x2 = screenCentreX; float y2 = ScreenCentreY; float dx = x2 - x1; float dy = y2 - y1; float hypotSquared = (dx * dx) + (dy * dy); float hypot = sqrt(hypotSquared); float unitX = dx / hypot; float unitY = dy / hypot; float rayX = x2 - view->getViewportX(); float rayY = y2 - view->getViewportY(); float arrowX = 0; float arrowY = 0; bool posFound = false; while(posFound == false) { rayX += unitX; rayY += unitY; if(rayX <= 0 || rayX >= screenWidth || rayY <= 0 || rayY >= screenHeight) { arrowX = rayX; arrowY = rayY; posFound = true; } } al_draw_bitmap(sprite, arrowX - spriteWidth, arrowY - spriteHeight, 0); } This was relatively successful. Arrows are displayed in the bottom right section of the screen when objects are located above and left of the screen as if the locations of the where the arrows are drawn have been rotated 180 degrees around the center of the screen. I assumed this was due to the fact that when I was calculating the hypotenuse of the triangle, it would always be positive regardless of whether or not the difference in x or difference in y is negative. Thinking about it, ray casting does not seem like a good way of solving the problem (due to the fact that it involves using sqrt() and a large for loop). Any help finding a suitable solution would be greatly appreciated, Thanks Adam

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  • Game Institute Math Courses

    - by W3Geek
    I'm 21 years old and I suck at math, I mean really bad. I don't have the necessary logic to apply it towards programming. I would like to learn the math and logic of applying it. I found Game Institute (http://www.gameinstitute.com) awhile back and heard a lot of praise about them. Are there Math courses any good? Thank you. Edit: My high school was terrible and did not prepare me for any math. I am fairly decent at programming, I just don't have the logic to apply any mathematics to programming, as an example I don't understand the algorithm of finding the size of a user's screen. Yes I have heard of KhanAcademy (http://www.khanacademy.org/) and I have completed a lot of maths on his website but I still don't have the logic to apply any of it to programming.

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  • determine collision angle on a rotating body

    - by jorb
    update: new diagram and updated description I have a contact listener set up to try and determine the side that a collision happened at relative to the a bodies rotation. One way to solve this is to find the value of the yellow angle between the red and blue vectors drawn above. The angle can be found by taking the arc cosine of the dot product of the two vectors (Evan pointed this out). One of my points of confusion is the difference in domain of the atan2 function html canvas coordinates and the Box2d rotation information. I know I have to account for this somehow... SS below questions: Does Box2D provide these angles more directly in the collision information? Am I even on the right track? If so, any hints? I have the following javascript so far: Ship.prototype.onCollide = function (other_ent,cx,cy) { var pos = this.body.GetPosition(); //collision position relative to body var d_cx = pos.x - cx; var d_cy = pos.y - cy; //length of initial vector var len = Math.sqrt(Math.pow(pos.x -cx,2) + Math.pow(pos.y-cy,2)); //body angle - can over rotate hence mod 2*Pi var ang = this.body.GetAngle() % (Math.PI * 2); //vector representing body's angle - same magnitude as the first var b_vx = len * Math.cos(ang); var b_vy = len * Math.sin(ang); //dot product of the two vectors var dot_prod = d_cx * b_vx + d_cy * b_vy; //new calculation of difference in angle - NOT WORKING! var d_ang = Math.acos(dot_prod); var side; if (Math.abs(d_ang) < Math.PI/2 ) side = "front"; else side = "back"; console.log("length",len); console.log("pos:",pos.x,pos.y); console.log("offs:",d_cx,d_cy); console.log("body vec",b_vx,b_vy); console.log("body angle:",ang); console.log("dot product",dot_prod); console.log("result:",d_ang); console.log("side",side); console.log("------------------------"); }

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  • "Walking" along a rotating surface in LimeJS

    - by Dave Lancea
    I'm trying to have a character walk along a plank (a long, thin rectangle) that works like a seesaw, being rotated around a central point by box2d physics (falling objects). I want the left and right arrow keys to move the player up and down the plank, regardless of it's slope, and I don't want to use real physics for the player movement. My idea for achieving this was to compute the coordinate based on the rotation of the plank and the current location "up" or "down" the board. My math is derived from here: http://math.stackexchange.com/questions/143932/calculate-point-given-x-y-angle-and-distance Here's the code I have so far: movement = 0; if(keys[37]){ // Left movement = -3; } if(keys[39]){ // Right movement = 3; } // this.plank is a LimeJS sprite. // getRotation() Should return an angle in degrees var rotation = this.plank.getRotation(); // this.current_plank_location is initialized as 0 this.current_plank_location += movement; var x_difference = this.current_plank_location * Math.cos(rotation); var y_difference = this.current_plank_location * Math.sin(rotation); this.setPosition(seesaw.PLANK_CENTER_X + x_difference, seesaw.PLANK_CENTER_Y + y_difference); This code causes the player to swing around in a circle when they are out of the center of the plank given a slight change in rotation of the plank. Any ideas on how I can get the player position to follow the board position?

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  • Spherical to Cartesian Coordinates

    - by user1258455
    Well I'm reading the Frank's Luna DirectX10 book and, while I'm trying to understand the first demo, I found something that's not very clear at least for me. In the updateScene method, when I press A, S, W or D, the angles mTheta and mPhi change, but after that, there are three lines of code that I don't understand exactly what they do: // Convert Spherical to Cartesian coordinates: mPhi measured from +y // and mTheta measured counterclockwise from -z. float x = 5.0f*sinf(mPhi)*sinf(mTheta); float z = -5.0f*sinf(mPhi)*cosf(mTheta); float y = 5.0f*cosf(mPhi); I mean, this explains that they do, it says that it converts the spherical coordinates to cartesian coordinates, but, mathematically, why? why the x value is calculated by the product of the sins of both angles? And the z by the product of the sine and cosine? and why the y just uses the cosine? After that, those values (x, y and z) are used to build the view matrix. The book doesn't explain (mathematically) why those values are calculated like that (and I didn't find anything to help me to understand it at the first Part of the book: "Mathematical prerequisites"), so it would be good if someone could explain me what exactly happen in those code lines or just give me a link that helps me to understand the math part. Thanks in advance!

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  • Points around a circumference C#

    - by Lautaro
    Im trying to get a list of vectors that go around a circle, but i keep getting the circle to go around several times. I want one circel and the dots to be placed along its circumference. I want the first dot to start at 0 and the last dot to end just before 360. Also i need to be able to calculate the spacing by the ammount of points. List<Vector2> pointsInPath = new List<Vector2>(); private int ammountOfPoints = 5; private int blobbSize = 200; private Vector2 topLeft = new Vector2(100, 100); private Vector2 blobbCenter; private int endAngle = 50; private int angleIncrementation; public Blobb() { blobbCenter = new Vector2(blobbSize / 2, blobbSize / 2) + topLeft; angleIncrementation = endAngle / ammountOfPoints; for (int i = 0; i < ammountOfPoints; i++) { pointsInPath.Add(getPointByAngle(i * angleIncrementation, 100, blobbCenter)); // pointsInPath.Add(getPointByAngle(i * angleIncrementation, blobbSize / 2, blobbCenter)); } } private Vector2 getPointByAngle(float angle, float distance, Vector2 centre) { return new Vector2((float)(distance * Math.Cos(angle) ), (float)(distance * Math.Sin(angle))) + centre ; }

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  • infer half vector length in BRDF

    - by cician
    it's my first question on stack. Is it possible to infer length of the half angle vector for specular lighting from N·L and N·V without the whole view and light vectors? I may be completely off-track, but I have this gut feeling it's possible... Why? I'm working on a skin shader and I'm already doing one texture lookup with N·L+N·E and one texture lookup for specular with N·H+N·V. The latter one can be transformed into N·L+N·E lookup if only I had the half vector length. Doing so could simplify the shader a bit and move some operations into the pre-computed lookup texture. It would make a huge difference since I'm trying to squeeze as much functionality as possible to a single pass mobile version so instruction count matters. Thanks.

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  • How to rotate camera centered around the camera's position?

    - by tnutty
    Currently I am using gluLook at like so: gluLookAt(position.x, position.y, position.z, viewPoint.x, viewPoint.y, viewPoint.z, upVector.x, upVector.y, upVector.z); with the above, don't know if you need more information, how could I change it so that the camera acts like its rotating around itself, instead rotating around its viewpoint. You can see the current code at https://github.com/dchhetri/OpenGL-City/blob/master/opengl_camera.cpp, that class was adapted from codecolony.com.

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  • Compile-time trigonometry in C

    - by lhahne
    I currently have code that looks like while (very_long_loop) { ... y1 = getSomeValue(); ... x1 = y1*cos(PI/2); x2 = y2*cos(SOME_CONSTANT); ... outputValues(x1, x2, ...); } the obvious optimization would be to compute the cosines ahead-of-time. I could do this by filling an array with the values but I was wondering would it be possible to make the compiler compute these at compile-time?

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  • How to use the asin() function in objective c

    - by Daniel Thatcher
    I am trying to use the asin() function in an iOS app to calculate an angle from the y axis. I am using trigonometry, but I must be doing something wrong with the asin() function, as when I try to pass in 0.707.... as asin(rotation) where rotation is a double equivalent to 0.707..., I get around 0.78....., where as my calculator gives me 44.991..., which is about correct from the variables passed in. What am I doing wrong, please can somebody help me?

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  • How to get X,Y,Z rotations of vertices on a sphere at the origin?

    - by Stoff81
    Hey, I have a sphere in my game world and i would like to place a plane at each vertex on this sphere for debugging purposes. The planes should be orientated so that they lie flat against the sphere (perpendicular to the normals). The sphere is located at the origin, so all the vertices are relative to that. If my thinking is correct, i should be able to do this using the positions of the vertices and some simple trigonometry. I have tried a few combinations but have had no joy yet. I would greatly appreciate some help on this. Thanks. Here is my code: float xRot = RADIANS_TO_DEGREES(sinf(vertex.x/PLANET_RADIUS)); float yRot = RADIANS_TO_DEGREES(cosf(vertex.y/PLANET_RADIUS)); glRotatef(xRot, 1.0, 0, 0); glRotatef(yRot, 0, 1.0, 0);

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  • Angle between two 2d vectors, diff between two methods?

    - by Sean Ochoa
    Hey all. I've got this code snippet, and I'm wondering why the results of the first method differ from the results of the second method, given the same input? public double AngleBetween_1(vector a, vector b) { var dotProd = a.Dot(b); var lenProd = a.Len*b.Len; var divOperation = dotProd/lenProd; return Math.Acos(divOperation) * (180.0 / Math.PI); } public double AngleBetween_2(vector a, vector b) { var dotProd = a.Dot(b); var lenProd = a.Len*b.Len; var divOperation = dotProd/lenProd; return (1/Math.Cos(divOperation)) * (180.0 / Math.PI); }

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  • get rotation direction of UIView on touchesMoved

    - by mlecho
    this may sound funny, but i spent hours trying to recreate a knob with a realistic rotation using UIView and some trig. I achieved the goal, but now i can not figure out how to know if the knob is rotating left or right. The most pertinent part of the math is here: - (void)touchesMoved:(NSSet *)touches withEvent:(UIEvent *)event { UITouch *touch = [touches anyObject]; CGPoint pt = [touch locationInView:self]; float dx = pt.x - iv.center.x; float dy = pt.y - iv.center.y; float ang = atan2(dy,dx); //do the rotation if (deltaAngle == 0.0) { deltaAngle = ang; initialTransform = iv.transform; }else { float angleDif = deltaAngle - ang; CGAffineTransform newTrans = CGAffineTransformRotate(initialTransform, -angleDif); iv.transform = newTrans; currentValue = [self goodDegrees:radiansToDegrees(angleDif)]; } } ideally, i could leverage a numeric value to tell me if the rotation is positive or negative.

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  • calculating the potential effect of inaccurate triangle vertex positions on the triangle edge lenght

    - by stingrey
    i'm not sure how to solve the following problem: i have a triangle with each of the three known vertex positions A,B,C being inaccurate, meaning they can each deviate up to certain known radii rA, rB, rC into arbitrary directions. given such a triangle, i want to calculate how much the difference of two specific edge lengths (for instance the difference between lengths of edge a and edge b) of the triangle may change in the worst case. is there any elegant mathematical solution to this problem? the naive way i thought of is calculating all 360^3 angle combinations and measuring the edge differences for each case, which is a rather high overhead.

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  • Atlas style map index for static google map

    - by Ben Holland
    Hello, I'm using a static google map, but really this problem could apply to any maps project. I want to divide a map into multiple quadrants (of say 50x50 pixels) and label the columns as A, B, C.... and the rows as 1, 2, 3... Next I plan to do something like, 1) Find the markers which are the farthest north, east, south, and west 2) Use that info to to define the bounding boxes of each row and column box 3) Classify each marker by its row and column (Example Marker 1 = [A,2]) A few requirements, I don't know the zoom level because I let Google set the zoom level appropriately for me and I would rather not use an algorithm that is dependent on a zoom level. I do however know the locations of all of the markers that are shown on the map. Here is an example of a map that I would like to classify the markers for, static map example link. I found these which look like a good start, Resource 1, Resource 2 But I think I'm still in need of some help getting started. Can anyone help write out some pseudo code or post a few more resources? I'm kind of in a rut at the moment. Thanks! Much appreciated of any help!

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  • Help Understanding Function

    - by Fred F.
    What does the following function perform? public static double CleanAngle(double angle) { while (angle < 0) angle += 2 * System.Math.PI; while (angle > 2 * System.Math.PI) angle -= 2 * System.Math.PI; return angle; } This is how it is used with ATan2. I believe the actually values passed to ATan2 are always positive. static void Main(string[] args) { int q = 1; //'x- and y-coordinates will always be positive values //'therefore, do i need to "clean"? foreach (Point oPoint in new Point[] { new Point(8,20), new Point(-8,20), new Point(8,-20), new Point(-8,-20)}) { Debug.WriteLine(Math.Atan2(oPoint.Y, oPoint.X), "unclean " + q.ToString()); Debug.WriteLine(CleanAngle(Math.Atan2(oPoint.Y, oPoint.X)), "cleaned " + q.ToString()); q++; } //'output //'unclean 1: 1.19028994968253 //'cleaned 1: 1.19028994968253 //'unclean 2: 1.95130270390726 //'cleaned 2: 1.95130270390726 //'unclean 3: -1.19028994968253 //'cleaned 3: 5.09289535749705 //'unclean 4: -1.95130270390726 //'cleaned 4: 4.33188260327232 }

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  • uniform generation of points on 3D box

    - by Myx
    Hello: I would like to generate random points on a 3D box defined by its (minx, miny, minz) and (maxx, maxy, maxz) corners. I was thinking of generating a random point inside of the box and then somehow projecting it onto one of the box sides. However, I don't have explicit plane information for the box sides and this seems like it will not produce a uniform distribution of points since if some sides of the box are bigger than others, those sides should have more points generated on them. Any suggestions are appreciated. Thanks.

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  • C# - Angle between two 2d vectors, diff between two methods?

    - by Sean Ochoa
    Hey all. I've got this code snippet, and I'm wondering why the results of the first method differ from the results of the second method, given the same input? public double AngleBetween_1(vector a, vector b) { var dotProd = a.Dot(b); var lenProd = Len*b.Len; var divOperation = dotProd/lenProd; return Math.Acos(divOperation) * (180.0 / Math.PI); } public double AngleBetween_2(vector a, vector b) { var dotProd = a.Dot(b); var lenProd = Len*b.Len; var divOperation = dotProd/lenProd; return (1/Math.Cos(divOperation)) * (180.0 / Math.PI); }

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  • 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.

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  • How can I work around the fact that in C++, sin(M_PI) 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) edit: In case anyone's curious, this was what I landed on: double sin2(double theta) // in degrees { double s = sin(toRadians(theta)); if (fabs(s - (int)s) < 0.000001) { return floor(s + 0.5); } return s; } where toRadians() is a macro that converts to radians

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