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  • Orienting ship movement in asteroids [closed]

    - by BadSniper
    Possible Duplicate: Move sprite in the direction it is facing? I'm programming asteroids game. I'm trying to give velocity in the direction of ship face. I've tried using velocity.x = velocity.x * cos r, velocity.y = velocity.y * sin r where velocity is a vector and r is the angle rotated. but it's not moving in right direction. Could someone help with this problem? It is supposed to return the vector in which ship is facing. I don't understand the problem.

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  • Placing items randomly in a dynamically generated terrain

    - by Programlocura
    I'm currently working on a 'Tiny wings' like game. I've already asked about the angle of the items in curved lines and i solved (thank you for responses),i'm currently placing the items in random positions, but the terrain it's dynamically and i don't get this working. I've a vector with the points of the terrain, and i'm doing something similar to this (Iterating trough vector): if( _SPManager && i % 15 == 0 && i != 0 ) { if ( _settings.specialPoints && _currentPoints < _settings.specialPoints ) { _SPManager.addPoint( _hillsPosition[i].x , _hillsPosition[i].y ); _currentPoints++; } } But it isn't working as i expected. It isn't displaying the right number of items (Always shows less than i specified). How do i should place the items? Thanks :)

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  • Evolving Architectures &ndash; Part II but Design is emergent

    This is part II of a series on agile architecture. You can read part I here. In the previous installment I provided a definition for software architecture and raised the apparent friction between the up front design implied by software architecture and the YAGNI approach and deferred requirements prompted by agile development in the large. This installment take a look at an additional angle of the problem which is the difference between design and architecture (while...Did you know that DotNetSlackers also publishes .net articles written by top known .net Authors? We already have over 80 articles in several categories including Silverlight. Take a look: here.

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  • How do i rotate a CALayer around a diagonal line?

    - by Mattias Wadman
    Hi. I'm trying to implement a flip animation to be used in board game like iPhone-application. The animation is supposed to look like a game piece that rotates and changes to the color of its back (kind of like an Reversi piece). I've managed to create an animation that flips the piece around its orthogonal axis, but when I try to flip it around a diagonal axis by changing the rotation around the z-axis the actual image also gets rotated (not surprisingly). Instead I would like to rotate the image "as is" around a diagonal axis. I have tried to change layer.sublayerTransform but with no success. Here is my current implementation. It works by doing a trick to resolve the issue of getting a mirrored image at the end of the animation. The solution is to not actually rotate the layer 180 degrees, instead it rotates it 90 degrees, changes image and then rotates it back. + (void)flipLayer:(CALayer *)layer toImage:(CGImageRef)image withAngle:(double)angle { const float duration = 0.5f; CAKeyframeAnimation *diag = [CAKeyframeAnimation animationWithKeyPath:@"transform.rotation.z"]; diag.duration = duration; diag.values = [NSArray arrayWithObjects: [NSNumber numberWithDouble:angle], [NSNumber numberWithDouble:0.0f], nil]; diag.keyTimes = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], [NSNumber numberWithDouble:1.0f], nil]; diag.calculationMode = kCAAnimationDiscrete; CAKeyframeAnimation *flip = [CAKeyframeAnimation animationWithKeyPath:@"transform.rotation.y"]; flip.duration = duration; flip.values = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], [NSNumber numberWithDouble:M_PI / 2], [NSNumber numberWithDouble:0.0f], nil]; flip.keyTimes = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], [NSNumber numberWithDouble:0.5f], [NSNumber numberWithDouble:1.0f], nil]; flip.calculationMode = kCAAnimationLinear; CAKeyframeAnimation *replace = [CAKeyframeAnimation animationWithKeyPath:@"contents"]; replace.duration = duration / 2; replace.beginTime = duration / 2; replace.values = [NSArray arrayWithObjects:(id)image, nil]; replace.keyTimes = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], nil]; replace.calculationMode = kCAAnimationDiscrete; CAAnimationGroup *group = [CAAnimationGroup animation]; group.removedOnCompletion = NO; group.duration = duration; group.timingFunction = [CAMediaTimingFunction functionWithName:kCAMediaTimingFunctionLinear]; group.animations = [NSArray arrayWithObjects:diag, flip, replace, nil]; group.fillMode = kCAFillModeForwards; [layer addAnimation:group forKey:nil]; }

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  • car race game collision condition.

    - by ashok patidar
    car race game in as3 in which car is fix but track is scrolling where ever my car move i want car should always on the track only when it try to move none directional track that time it has to be rotate at some angle so that it can move in track direction only. i am unable to think logic. i need movement like this "http://www.emanueleferonato.com/2007/05/15/create-a-flash-racing-game-tutorial/" track should be scrollable as **http://as3.mindmafya.com/GameAS/ScrollingMaps-1.php pl z provide me suitable solution for that.

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  • Implementing Projectile Motion

    - by DMan
    I've scored the internet for sources and have found a lot of useful information, but they are math sites trying to tell me how to solve what angle an object has to be at to reach y location. However, I'm trying to run a simulation, and haven't found any solid equations that can be implemented to code to simulate a parabolic curve. Can those with some knowledge of physics help me on this?

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  • Rotate point in rectangle

    - by Dested
    I have a point in a rectangle that I need to rotate an arbitrary degree and find the x y of the point. How can I do this using javascript. Below the x,y would be something like 1,3 and after I pass 90 into the method it will return 3,1. |-------------| | * | | | | | |-------------| _____ | *| | | | | | | | | _____ |-------------| | | | | | *| |-------------| _____ | | | | | | | | |* | _____ Basically I am looking for the guts to this method function Rotate(pointX,pointY,rectWidth,rectHeight,angle){ /*magic*/ return {newX:x,newY:y}; }

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  • cgaffinetransformrotate jagged edges

    - by Raj
    I am trying to apply cgaffinetransformrotate transform to uiimageview. However the image edges seems to be jaded. Is there any workaround this problem. If I apply rotate angle to 90% than I don't see these jaded edges. I need to apply smaller angles, this is where I see the problem Thanks,

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  • TFS Template Customization - SharePointPortal site

    - by Adam Jenkin
    I am customizing a Process Template for TFS2008. I am using the "MSF for Agile... v4.2" template as the base template and would like to set the version control settings of the "Project Management" document library todo the following: Major Versions : Enabled Documents must be checked our before they can be edited I'm using the editor GUI provided by the tfs powertools, however it does not appear these settings are available. Is it possible to define these settings in the WssTasks.XML file or do I need to approach this from a different angle.

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  • How do i rotate a CALayer around a diagonal axis?

    - by Mattias Wadman
    Hi, im trying to implement a flip animation to be used in board game like application. The animation is suppose to look like a game piece that rotate and change to the color of its back. I have managed to get a animation that flips around orthogonal axis, but when i try to flip around a diagonal axis by changing the rotation around the z-axis not surprisingly the actual image also gets rotated. Instead i would like to rotate the image as it is around a diagonal axis. I have tried to change layer.sublayerTransform but with no success. Here is the current implementation. It works by doing a trick to resolve the issue of getting a mirrored image at the end of the animation. The solution is to not actually rotate the layer 180 degrees, instead it rotates it 90 degrees, changes image and then rotates it back. + (void)flipLayer:(CALayer *)layer toImage:(CGImageRef)image withAngle:(double)angle { const float duration = 0.5f; CAKeyframeAnimation *diag = [CAKeyframeAnimation animationWithKeyPath:@"transform.rotation.z"]; diag.duration = duration; diag.values = [NSArray arrayWithObjects: [NSNumber numberWithDouble:angle], [NSNumber numberWithDouble:0.0f], nil]; diag.keyTimes = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], [NSNumber numberWithDouble:1.0f], nil]; diag.calculationMode = kCAAnimationDiscrete; CAKeyframeAnimation *flip = [CAKeyframeAnimation animationWithKeyPath:@"transform.rotation.y"]; flip.duration = duration; flip.values = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], [NSNumber numberWithDouble:M_PI / 2], [NSNumber numberWithDouble:0.0f], nil]; flip.keyTimes = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], [NSNumber numberWithDouble:0.5f], [NSNumber numberWithDouble:1.0f], nil]; flip.calculationMode = kCAAnimationLinear; CAKeyframeAnimation *replace = [CAKeyframeAnimation animationWithKeyPath:@"contents"]; replace.duration = duration / 2; replace.beginTime = duration / 2; replace.values = [NSArray arrayWithObjects:(id)image, nil]; replace.keyTimes = [NSArray arrayWithObjects: [NSNumber numberWithDouble:0.0f], nil]; replace.calculationMode = kCAAnimationDiscrete; CAAnimationGroup *group = [CAAnimationGroup animation]; group.removedOnCompletion = NO; group.duration = duration; group.timingFunction = [CAMediaTimingFunction functionWithName:kCAMediaTimingFunctionLinear]; group.animations = [NSArray arrayWithObjects:diag, flip, replace, nil]; group.fillMode = kCAFillModeForwards; [layer addAnimation:group forKey:nil]; }

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  • Archery game programming algorithm

    - by Ricky
    I need the algorithm to animate the arrow based on 2 parameters, angle while shooting and power while drawing the bow. Ive tried to use y=asinx but it works only when shooting in up direction. Doesnt work well while shooting with straight or down direction. Thanks.

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  • boost ublas: rotate 2d vector

    - by AndreasT
    Erm. I hope I am seriously overlooking something. I want to rotate a 2d vector (kartesian) v by a certain angle phi. I can't find a function that generates the appropriate matrix or just performs that function. I know how to do this by hand. I am looking for a ublas utility "something" that does this for me.

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  • Reading Source Code Aloud

    - by Jon Purdy
    After seeing this question, I got to thinking about the various challenges that blind programmers face, and how some of them are applicable even to sighted programmers. Particularly, the problem of reading source code aloud gives me pause. I have been programming for most of my life, and I frequently tutor fellow students in programming, most often in C++ or Java. It is uniquely aggravating to try to verbally convey the essential syntax of a C++ expression. The speaker must give either an idiomatic translation into English, or a full specification of the code in verbal longhand, using explicit yet slow terms such as "opening parenthesis", "bitwise and", et cetera. Neither of these solutions is optimal. On the one hand, an idiomatic translation is only useful to a programmer who can de-translate back into the relevant programming code—which is not usually the case when tutoring a student. In turn, education (or simply getting someone up to speed on a project) is the most common situation in which source is read aloud, and there is a very small margin for error. On the other hand, a literal specification is aggravatingly slow. It takes far far longer to say "pound, include, left angle bracket, iostream, right angle bracket, newline" than it does to simply type #include <iostream>. Indeed, most experienced C++ programmers would read this merely as "include iostream", but again, inexperienced programmers abound and literal specifications are sometimes necessary. So I've had an idea for a potential solution to this problem. In C++, there is a finite set of keywords—63—and operators—54, discounting named operators and treating compound assignment operators and prefix versus postfix auto-increment and decrement as distinct. There are just a few types of literal, a similar number of grouping symbols, and the semicolon. Unless I'm utterly mistaken, that's about it. So would it not then be feasible to simply ascribe a concise, unique pronunciation to each of these distinct concepts (including one for whitespace, where it is required) and go from there? Programming languages are far more regular than natural languages, so the pronunciation could be standardised. Speakers of any language would be able to verbally convey C++ code, and due to the regularity and fixity of the language, speech-to-text software could be optimised to accept C++ speech with a high degree of accuracy. So my question is twofold: first, is my solution feasible; and second, does anyone else have other potential solutions? I intend to take suggestions from here and use them to produce a formal paper with an example implementation of my solution.

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  • Mapscript queryByPoint return no results

    - by lucian.jp
    I have a dynamically generated mapfile made with c# mapscript that is defined like: MAP EXTENT 5.91828 45.63552 5.92346 45.65051 IMAGECOLOR 192 192 192 IMAGETYPE png SIZE 256 256 STATUS ON TRANSPARENT TRUE UNITS METERS NAME "GMAP_TILE" OUTPUTFORMAT NAME "png" MIMETYPE "image/png" DRIVER "GD/PNG" EXTENSION "png" IMAGEMODE "PC256" TRANSPARENT TRUE END SYMBOL NAME "circle" TYPE ELLIPSE FILLED TRUE POINTS 1 1 END END SYMBOL NAME ">" TYPE TRUETYPE ANTIALIAS TRUE CHARACTER ">" GAP -20 FONT "arial" POSITION CC END PROJECTION "proj=merc" "a=6378137" "b=6378137" "lat_ts=0.0" "lon_0=0.0" "x_0=0.0" "y_0=0" "units=m" "k=1.0" "nadgrids=@null" END LEGEND IMAGECOLOR 255 255 255 KEYSIZE 20 10 KEYSPACING 5 5 LABEL SIZE MEDIUM TYPE BITMAP BUFFER 0 COLOR 0 0 0 FORCE FALSE MINDISTANCE -1 MINFEATURESIZE -1 OFFSET 0 0 PARTIALS TRUE END POSITION LL STATUS OFF END QUERYMAP COLOR 255 255 0 SIZE -1 -1 STATUS ON STYLE HILITE END SCALEBAR ALIGN CENTER COLOR 0 0 0 IMAGECOLOR 255 255 255 INTERVALS 4 LABEL SIZE MEDIUM TYPE BITMAP BUFFER 0 COLOR 0 0 0 FORCE FALSE MINDISTANCE -1 MINFEATURESIZE -1 OFFSET 0 0 PARTIALS TRUE END POSITION LL SIZE 200 3 STATUS OFF STYLE 0 UNITS MILES END WEB IMAGEPATH "" IMAGEURL "" QUERYFORMAT text/html LEGENDFORMAT text/html BROWSEFORMAT text/html END LAYER NAME "Troncons" PROJECTION "proj=longlat" "ellps=WGS84" "datum=WGS84" END STATUS DEFAULT TEMPLATE "nofile.html" TOLERANCE 100 TOLERANCEUNITS METERS TYPE LINE UNITS METERS CLASS NAME "Troncons" STYLE ANGLE 360 COLOR 0 0 255 SIZE 5 SYMBOL "circle" WIDTH 5 END STYLE ANGLE 360 COLOR 0 0 0 SIZE 12 SYMBOL ">" WIDTH 1 END END FEATURE POINTS 5.91828 45.63552 5.91876 45.63611 5.91898 45.6364 5.91936 45.63701 5.91952 45.63731 5.91968 45.63762 5.91993 45.63825 5.92003 45.63856 5.92018 45.63919 5.92028 45.63983 5.92031 45.64014 5.92033 45.64046 5.92034 45.64077 5.92034 45.64108 5.92034 45.64171 5.92035 45.64234 5.92035 45.6428 5.92037 45.6433 5.9204 45.64394 5.92046 45.64458 5.92056 45.64522 5.92062 45.64554 5.92069 45.64586 5.92077 45.64617 5.92097 45.64679 5.92122 45.64739 5.92136 45.64769 5.92169 45.64828 5.92207 45.64886 5.92228 45.64914 5.92272 45.64969 5.92321 45.65023 5.92346 45.65051 END END END END I try to queryByPoint to retreive the index of the shape clciked near. In the code below I made a specific test function with fixed point instead of points passed by parameter so I am sure the point I use is actually part of a feature. In my case I use the first point of the only feature contained in mapfile. public string GetTronconId() { //_map is my dynamically created mapObj if (_map != null) for (int i = 0; i < _map.numlayers; i++) { layerObj layer = _map.getLayer(i); // Code never pass this point if (layer.queryByPoint(_map, new pointObj(5.91898, 45.6364, 0, 0), (int) MS_QUERY_MODE.MS_QUERY_MULTIPLE, 100) == (int) MS_RETURN_VALUE.MS_SUCCESS) { int numresults = layer.getNumResults(); if (numresults != 0) { layer.open(); for (int j = 0; j < numresults; j++) { resultCacheMemberObj resultat = layer.getResult(j); shapeObj shape = null; if (layer.getShape(shape, resultat.tileindex, resultat.shapeindex) == (int) MS_RETURN_VALUE.MS_SUCCESS) return shape.getValue(0); } } } } return null; } I have a dummy TEMPLATE set, I do not eveen have to use the tolerance since the point is derectly in a shape, but the queryByPoint keep returning me MS_FAILURE. From my searches on the web everything seem to be OK. Any idea?

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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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  • Mapping a BigInteger to a circle

    - by Martin
    I have a C# system using 160 bit numbers, stored in a BigInteger. I want to display these things on a circle, which means mapping the 0-2^160 range into the 0-2Pi range. How would I do this? The approach that jumps instantly to mind is angle = (number / pow(2, 160)) * TwoPi; However, that has complexities because the division will truncate the result into an integer.

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  • make a 3D spotlight cone matrix

    - by Soubok
    How to create the transformation matrix (4x4) that transforms a cylinder (of height 1 and diameter 1) into a cone that represents my spotlight (position, direction and cutoff angle) ? --edit-- In other words: how to draw the cone that represents my spotlight by drawing a cylinder through a suitable transformation matrix.

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  • fit a ellipse in Python given a set of points xi=(xi,yi)

    - by Gianni
    I am computing a series of index from a 2D points (x,y). One index is the ratio between minor and major axis. To fit the ellipse i am using the following post when i run these function the final results looks strange because the center and the axis length are not in scale with the 2D points center = [ 560415.53298363+0.j 6368878.84576771+0.j] angle of rotation = (-0.0528033467597-5.55111512313e-17j) axes = [0.00000000-557.21553487j 6817.76933256 +0.j] thanks in advance for help import numpy as np from numpy.linalg import eig, inv def fitEllipse(x,y): x = x[:,np.newaxis] y = y[:,np.newaxis] D = np.hstack((x*x, x*y, y*y, x, y, np.ones_like(x))) S = np.dot(D.T,D) C = np.zeros([6,6]) C[0,2] = C[2,0] = 2; C[1,1] = -1 E, V = eig(np.dot(inv(S), C)) n = np.argmax(np.abs(E)) a = V[:,n] return a def ellipse_center(a): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] num = b*b-a*c x0=(c*d-b*f)/num y0=(a*f-b*d)/num return np.array([x0,y0]) def ellipse_angle_of_rotation( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] return 0.5*np.arctan(2*b/(a-c)) def ellipse_axis_length( a ): b,c,d,f,g,a = a[1]/2, a[2], a[3]/2, a[4]/2, a[5], a[0] up = 2*(a*f*f+c*d*d+g*b*b-2*b*d*f-a*c*g) down1=(b*b-a*c)*( (c-a)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) down2=(b*b-a*c)*( (a-c)*np.sqrt(1+4*b*b/((a-c)*(a-c)))-(c+a)) res1=np.sqrt(up/down1) res2=np.sqrt(up/down2) return np.array([res1, res2]) if __name__ == '__main__': points = [(560036.4495758876, 6362071.890493258), (560036.4495758876, 6362070.890493258), (560036.9495758876, 6362070.890493258), (560036.9495758876, 6362070.390493258), (560037.4495758876, 6362070.390493258), (560037.4495758876, 6362064.890493258), (560036.4495758876, 6362064.890493258), (560036.4495758876, 6362063.390493258), (560035.4495758876, 6362063.390493258), (560035.4495758876, 6362062.390493258), (560034.9495758876, 6362062.390493258), (560034.9495758876, 6362061.390493258), (560032.9495758876, 6362061.390493258), (560032.9495758876, 6362061.890493258), (560030.4495758876, 6362061.890493258), (560030.4495758876, 6362061.390493258), (560029.9495758876, 6362061.390493258), (560029.9495758876, 6362060.390493258), (560029.4495758876, 6362060.390493258), (560029.4495758876, 6362059.890493258), (560028.9495758876, 6362059.890493258), (560028.9495758876, 6362059.390493258), (560028.4495758876, 6362059.390493258), (560028.4495758876, 6362058.890493258), (560027.4495758876, 6362058.890493258), (560027.4495758876, 6362058.390493258), (560026.9495758876, 6362058.390493258), (560026.9495758876, 6362057.890493258), (560025.4495758876, 6362057.890493258), (560025.4495758876, 6362057.390493258), (560023.4495758876, 6362057.390493258), (560023.4495758876, 6362060.390493258), (560023.9495758876, 6362060.390493258), (560023.9495758876, 6362061.890493258), (560024.4495758876, 6362061.890493258), (560024.4495758876, 6362063.390493258), (560024.9495758876, 6362063.390493258), (560024.9495758876, 6362064.390493258), (560025.4495758876, 6362064.390493258), (560025.4495758876, 6362065.390493258), (560025.9495758876, 6362065.390493258), (560025.9495758876, 6362065.890493258), (560026.4495758876, 6362065.890493258), (560026.4495758876, 6362066.890493258), (560026.9495758876, 6362066.890493258), (560026.9495758876, 6362068.390493258), (560027.4495758876, 6362068.390493258), (560027.4495758876, 6362068.890493258), (560027.9495758876, 6362068.890493258), (560027.9495758876, 6362069.390493258), (560028.4495758876, 6362069.390493258), (560028.4495758876, 6362069.890493258), (560033.4495758876, 6362069.890493258), (560033.4495758876, 6362070.390493258), (560033.9495758876, 6362070.390493258), (560033.9495758876, 6362070.890493258), (560034.4495758876, 6362070.890493258), (560034.4495758876, 6362071.390493258), (560034.9495758876, 6362071.390493258), (560034.9495758876, 6362071.890493258), (560036.4495758876, 6362071.890493258)] a_points = np.array(points) x = a_points[:, 0] y = a_points[:, 1] from pylab import * plot(x,y) show() a = fitEllipse(x,y) center = ellipse_center(a) phi = ellipse_angle_of_rotation(a) axes = ellipse_axis_length(a) print "center = ", center print "angle of rotation = ", phi print "axes = ", axes from pylab import * plot(x,y) plot(center[0:1],center[1:], color = 'red') show() each vertex is a xi,y,i point plot of 2D point and center of fit ellipse

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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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  • Drawing a Dragons curve in Python

    - by Connor Franzoni
    I am trying to work out how to draw the dragons curve, with pythons turtle using the An L-System or Lindenmayer system. I no the code is something like the Dragon curve; initial state = ‘F’, replacement rule – replace ‘F’ with ‘F+F-F’, number of replacements = 8, length = 5, angle = 60 But have no idea how to put that into code.

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