|
Hello,
i have set of 2D points and i want to offset with a given distance (like offset command in AutoCAD)
i do not know how to deal with corners. i have searched on Internet, there are advanced methods like straight skeletons etc. but my polyline is not self crossing and no holes in it.
Is there any simple method? any references?
Best Regards.
Bekir
|
|
|
|
|
Why can't you just iterate through the points and add your offset to each one? It seems like this should translate the corners along with everything else.
|
|
|
|
|
|
beko wrote: Just translating the edges with the offset distance will not work.
Why not?
beko wrote: I have found an algorithm not tested throughly, but it is working at least for my very simple 3 points polyline test
What kind of test you did?
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
|
|
|
|
|
Hello,
At the end, it is a translation however it must be implemented with respect to the normals, as far as i read through if it is a parametric curve then it is rather complicated. If you like to see how much it can further get complicated, you can have a look at the link below.
http://timeguy.com/cradek-files/emc/liu2007_line_arc_offset.pdf[^]
for the test, i have just created a polyline with three points and used the algorithm from Eduardo in my earlier reply, and compared it with AutoCAD output 
Best Regards.
|
|
|
|
|
That paper seems to give a good description of how to do it. It's complicated, you will have lots of cases - you'll just have to get used to it.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
|
|
|
|
|
There is no simple solution to this problem, the best known general solution, is to create capsules between consecutive point pairs of your polyline where the capsule width is the offset, then union the capsule outlines. The capsule may be centered around the edge or it may be biased towards one side of the edge it is really up to the end requirement. For the union operation use something like Alan Murta's Generic Polygon Clipper library.
If you know your polyline is enclosed and represents a convex polygon, then simply calculate the centroid, then offset each edge by the product of the desired offset amount and the vector composed of the mid-point of the edge at hand minus the centroid.
A simple example can be found here:
http://www.codeproject.com/KB/recipes/Wykobi.aspx
|
|
|
|
|
Thank you for the all answers,
I will have a look at it.
Best Regards.
|
|
|
|
|
For 0 < a, b < 1, if it makes a difference.
Thanks!
|
|
|
|
|
It is only periodic is a/b is rational and then it is fairly easy to work out.
Peter
"Until the invention of the computer, the machine gun was the device that enabled humans to make the most mistakes in the smallest amount of time."
|
|
|
|
|
If you consider the wave form:
y(t) = A*sin(w*t + q)
A is the amplitude of the wave (height up the y axis)
q is the phase shift of the wave (left/right shift along the x axis)
t is a variable
w is the angular frequency
Angular frequency equals to 2*PI*f -> where f is the frequency of the wave.
The period of the wave (t) is 1/f. (t = 1/f)
A Cosine wave is just a Sine wave with a 90 degree phase shift.
Cheers,
|
|
|
|
|
Since cos(ax) sin(bx) = 1/2 sin((a+b)x) + 1/2 cos((a-b)x) should be (a+b)/(a-b) rationale.
If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.
This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke
|
|
|
|
|
im trying to convert Quaternion to Euler but i dont have appropriate code, i searched on lots of sites but their converting code do not match with converting back, mean, if i convert from Quaternion to Euler then i convert back the result value to Quaternion, i got different output, here is my code
public static float[] ToEuler(double tmp_x, double tmp_y, double tmp_z,double tmp_w)
{
float[] Euler = new float[3];
double sqr_X = tmp_x * tmp_x;
double sqr_Y = tmp_y * tmp_y;
double sqr_Z = tmp_z * tmp_z;
double sqr_W = tmp_w * tmp_w;
const double radian = 180 / Math.PI;
double check = tmp_x * tmp_y + tmp_z * tmp_w;
double unit = sqr_X + sqr_Y + sqr_Z + sqr_W;
if (check > 0.499 * unit)
{
Euler[0] = 0;
Euler[1] = (float)(radian * Math.Atan2(tmp_x, tmp_w));
Euler[2] = (float)90;
return Euler;
}
else if (check < -0.499 * unit)
{
Euler[0] = 0;
Euler[1] = (float)(-radian * Math.Atan2(tmp_x, tmp_w));
Euler[2] = (float)-90;
return Euler;
}
Euler[0] = (float)(radian * (Math.Atan2(2 * (tmp_x * tmp_w + tmp_y * tmp_z), (-sqr_X - sqr_Y + sqr_Z + sqr_W))));
Euler[1] = (float)(radian * (Math.Asin(-2 * (tmp_x * tmp_z - tmp_y * tmp_w))));
Euler[2] = (float)(radian * (Math.Atan2(2.0 * (tmp_x * tmp_y + tmp_z * tmp_w), (sqr_X - sqr_Y - sqr_Z + sqr_W))));
return Euler;
}
public static float[] ToEulerA(double x, double y, double z, double w)
{
double sqw = w * w;
double sqx = x * x;
double sqy = y * y;
double sqz = z * z;
double unit = sqx + sqy + sqz + sqw;
double test = x * y + z * w;
double RX = 0;
double RY = 0;
double RZ = 0;
if (test > 0.499 * unit)
{
RX = 2 * Math.Atan2(x, w);
RY = Math.PI / 2;
RZ = 0;
}
else if (test < -0.499 * unit)
{
RX = -2 * Math.Atan2(x, w);
RY = -Math.PI / 2;
RZ = 0;
}
else
{
RX = Math.Atan2(2 * y * w - 2 * x * z, sqx - sqy - sqz + sqw);
RY = Math.Asin(2 * test / unit);
RZ = Math.Atan2(2 * x * w - 2 * y * z, -sqx + sqy - sqz + sqw);
}
RX = 180 / Math.PI * RX;
RY = 180 / Math.PI * RY;
RZ = 180 / Math.PI * RZ;
return new float[] { (float)RX, (float)RY, (float)RZ };
}
im not sure which one is correct or both are incorrect...
any idea to solve this ?
thanks
TVMU^P[[IGIOQHG^JSH`A#@`RFJ\c^JPL>;"[,*/|+&WLEZGc`AFXc!L
%^]*IRXD#@GKCQ`R\^SF_WcHbORY87??6?N8?BcRAV\Z^&SU~%CSWQ@#2
W_AD`EPABIKRDFVS)EVLQK)JKSQXUFYK[M`UKs*$GwU#(QDXBER@CBN%
Rs0~53%eYrd8mt^7Z6]iTF+(EWfJ9zaK-i?TV.C\y<p?jxsg-b$f4ia>
--------------------------------------------------------
128 bit encrypted signature, crack if you can
|
|
|
|
|
|
unfortunately i got code from that site, that site converter have some incorrect conversion, eg. at the end of page you will find a java converter, write real = 1, i = 1, and output will be 116.56505117707799, now put this 116.56505117707799 in Euler to Quaternion converter on another page and now result is completely different, why ?
TVMU^P[[IGIOQHG^JSH`A#@`RFJ\c^JPL>;"[,*/|+&WLEZGc`AFXc!L
%^]*IRXD#@GKCQ`R\^SF_WcHbORY87??6?N8?BcRAV\Z^&SU~%CSWQ@#2
W_AD`EPABIKRDFVS)EVLQK)JKSQXUFYK[M`UKs*$GwU#(QDXBER@CBN%
Rs0~53%eYrd8mt^7Z6]iTF+(EWfJ9zaK-i?TV.C\y<p?jxsg-b$f4ia>
--------------------------------------------------------
128 bit encrypted signature, crack if you can
|
|
|
|
|
You could just reference Microsoft.DirectX 
|
|
|
|
|
Hi,
Can anyone point me to algorithms for drawing a line/spiral/ellipse, if its a c++ code then its my lucky day otherwise anything will do.
Also i need to draw a gradient as well and i cant use GradientFill api (lucky me )
Any help at all will be much appreciated
Thanks
C++ where friends have access to your private members !
|
|
|
|
|
Monty2 wrote: Can anyone point me to algorithms for drawing a line/spiral/ellipse, if its a c++ code then its my lucky day otherwise anything will do.
Is something like this[^] sufficient? It's pretty nice - I've used it in one of my projects.
Less in your favour is this[^] implementation in C#.
|
|
|
|
|
Well i am sorry if i was not clear on the question i actually need to implement functions like
DrawLine(POINT ptFrom,POINT ptTo);
DrawGradient(CRect rc,COLORREF FromColor,COLORREF ToColor);
For some reasons i cant use the stock functions provided (i need to get all the points in the line)
C++ where friends have access to your private members !
|
|
|
|
|
|
Monty2 wrote: For some reasons i cant use the stock functions provided
Care to give us a clue to why so we know better what to suggest?
If you don't have the data, you're just another a**hole with an opinion.
|
|
|
|
|
Tim Craig wrote: Care to give us a clue to why so we know better what to suggest?
Fair enough, let me explain it further
I need to draw N rectangles side by side (their centers lying on a line or trajectory).
Works like this, user gives two points (left and right) then i need to draw a series of rectangles from the first point to the second with the rectangles centers lying on the line.
i figured for that i need to calculate the points that will lie on that line, hence the requirement for an algorithm that will give me the points that will lie on a line(x1,y1,x2,y2)
Secondly user will provide the color of the first rectangle and the last and i need to fill the rest of the rects with the gradient color, hence the requirement for a gradient algorithm
Here is a sample image (it uses sine wave i think instead of a line)
Sample[^]
And lastly thanks for your valuable time
C++ where friends have access to your private members !
|
|
|
|
|
Well, this might be a place to get started. Bresenham's algorithm[^] certainly picks pixels to draw straight lines and some other curves. You might also check some of the 3D drawing class libraries, like VRML or the library that originated at SGI, as they often implement a process they call "extrude" where they sweep a shape through space that follows a curve, a cylinder is an example of extruding a circle along a straight line. I don't think they use pixels generally but you might get some ideas.
If you don't have the data, you're just another a**hole with an opinion.
|
|
|
|
|
Thanks a bunch for your ideas the Bresenham should work perfectly i think
C++ where friends have access to your private members !
|
|
|
|
|
For example, used for video conference or video capture.
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
