|
You can calculate the semi-major axis length. Have a look at:http://en.wikipedia.org/wiki/Semimajor_axis[^].
If you start from the centre of the ellipse (which in your example is the centre of your rectagle) and add this length, you get the point on the circumference of the ellipse where you can start (and end) drawing your vertical line.
Starting from there, you can adjust the angle to find the perfect spot for your verical lines. 
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Can you give me the equation to find the semi-major axis length, knowing the four points I got? My head just in the verge of exploding reading the article. I'm really not that good at math, honest.
I also read and tried the code from http://en.wikipedia.org/wiki/Ellipse[^], following the links in wikipedia. This function to generate an ellipse is almost the same as DrawClosedCurve() or DrawEllipse(), using that I still can't get the outer most point to put the line.
|
|
|
|
|
Valeriant wrote: I also read and tried the code from http://en.wikipedia.org/wiki/Ellipse[^], following the links in wikipedia. This function to generate an ellipse is almost the same as DrawClosedCurve() or DrawEllipse(), using that I still can't get the outer most point to put the line.
Yep that's a good starting point too. See where it makes you go through the whole circumference:
for (var i = 0; i < 360; i += 360 / steps)
You can use that to get the coordinates to the interesting points, using the angle.
I'll try to put together some code when I have some spare time, but it won't be soon I'm afraid.
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Appreciate the time, will wait your help.
I did try each points generated from the steps, but none are exactly the outer most two. And formulating the angle, I cannot. 
|
|
|
|
|
Please consider that the code I'm presenting here is quite raw, and you'll probably want to clean it up and make a few changes.
I decided to give you a start-from-scratch idea, which is easier to implement. Please remember that drawing a good looking circle or ellipse is not necessarily an easy task. Have a look on the web, you'll see there are a bunch of good algorithms to do that.
The simplest way of drawing an ellipse is something like what you found on wikipedia. In my example you'll see a very similar approach. I just added rotation, which you need in order to draw your cylinder, and optimized the code (you don't have to go through the whole 360 degrees, but only half of it, cause the rest is specular).
private class EllipseOuterXPoints
{
internal double left_x = double.MaxValue;
internal double left_y = double.MaxValue;
internal double right_x = double.MinValue;
internal double right_y = double.MinValue;
internal void CheckPoint(double x, double y)
{
if (x < left_x)
{
left_x = x;
left_y = y;
}
if (x > right_x)
{
right_x = x;
right_y = y;
}
}
}
private EllipseOuterXPoints DrawEllipse(Graphics g, double center_x, double center_y, double semi_minor, double semi_major, double angle)
{
EllipseOuterXPoints OuterPoints = new EllipseOuterXPoints();
double beta = -angle * (Math.PI / 180);
double sin_beta = Math.Sin(beta);
double cos_beta = Math.Cos(beta);
double increment = (Math.PI * 2) / (semi_major * 2);
double sin_alpha;
double cos_alpha;
double x1 = 0;
double y1 = 0;
double x2 = 0;
double y2 = 0;
double prev_x1 = 0;
double prev_y1 = 0;
double prev_x2 = 0;
double prev_y2 = 0;
double first_x1 = 0;
double first_y1 = 0;
Pen mypen = new Pen(Color.Black);
for (double alpha = 0; alpha < Math.PI; alpha += increment)
{
sin_alpha = Math.Sin(alpha);
cos_alpha = Math.Cos(alpha);
x1 = center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta));
y1 = center_y + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta));
sin_alpha = Math.Sin(alpha * 2);
cos_alpha = Math.Cos(alpha * 2);
x2 = center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta));
y2 = center_y + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta));
if (alpha == 0)
{
first_x1 = x1;
first_y1 = y1;
}
else
{
g.DrawLine(mypen, (float)prev_x1, (float)prev_y1, (float)x1, (float)y1);
g.DrawLine(mypen, (float)prev_x2, (float)prev_y2, (float)x2, (float)y2);
}
OuterPoints.CheckPoint(x1, y1);
OuterPoints.CheckPoint(x2, y2);
prev_x1 = x1;
prev_y1 = y1;
prev_x2 = x2;
prev_y2 = y2;
}
g.DrawLine(mypen, (float)x2, (float)y2, (float)first_x1, (float)first_y1);
return OuterPoints;
}
First point you should note is the "magic" taking place here:
double increment = (Math.PI * 2) / (semi_major * 2);
This is what allows you to draw a decent-looking ellipse. The number of points you calculate on the circumference is derived by a brute empirical calculation based on the major axis length. Not that fine, not that beautiful, but it works.
Secondly, note that while I draw the ellipse I keep a record of the leftmost and rightmost points (this is based only on the X axis, since I didn't manage rotation on the Y axis!). These are the outer points you will connect to create your cylinder.
So, just call this code inside a Paint event with something like:
private void picCanvas_Paint(object sender, PaintEventArgs e)
{
EllipseOuterXPoints OuterPoints1 = DrawEllipse(e.Graphics, picCanvas.Size.Width / 2, picCanvas.Size.Height / 4, picCanvas.Size.Height / 8, picCanvas.Size.Width / 4, 10);
EllipseOuterXPoints OuterPoints2 = DrawEllipse(e.Graphics, picCanvas.Size.Width / 2, (picCanvas.Size.Height / 4) * 3, picCanvas.Size.Height / 8, picCanvas.Size.Width / 4, 10);
Pen mypen = new Pen(Color.Black);
e.Graphics.DrawLine(mypen, (float)OuterPoints1.left_x, (float)OuterPoints1.left_y, (float)OuterPoints2.left_x, (float)OuterPoints2.left_y);
e.Graphics.DrawLine(mypen, (float)OuterPoints1.right_x, (float)OuterPoints1.right_y, (float)OuterPoints2.right_x, (float)OuterPoints2.right_y);
}
And there is your (hopefully) good looking cylinder!
Have a look at the result here[^].
I think this is all you need to get you started. Good luck!
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Thank you for your time! I've try to implement it, and it does get the left and right points! Very cool, and lotsa math there for me.
I wonder if you could help me more, on filling the cylinder. Here is my code: http://rapidshare.com/files/368251744/IsoCylinder.rar[^]. Just as you please and no rush, maybe if you have time again  . Thank you.
|
|
|
|
|
Glad it helped you. I'll try and have a look at your code as soon as I can.
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
I tried to do some cleanup, but be aware that my code is still not that polished.
I did some experiments on the track I first suggested you: using the angle to get the outer points for both ellipses. This simple version works just fine, except for rotation angles not in the range from -10 to +10 degrees you'll need to compensate (i.e. add an offset to the angles 0 and PI I'm using here to get the outer points):
private void FillEllipse(Graphics g, Color line_color, Color fill_color, double center_x, double center_y, double semi_minor, double semi_major, double angle)
{
Pen mypen = new Pen(line_color);
SolidBrush mybrush = new SolidBrush(fill_color);
double PI2 = Math.PI * 2;
double beta = -angle * (Math.PI / 180);
double sin_beta = Math.Sin(beta);
double cos_beta = Math.Cos(beta);
double increment = (Math.PI * 2) / (semi_major * 2);
double sin_alpha;
double cos_alpha;
int x;
int y;
List<Point> points = new List<Point>();
for (double alpha = 0; alpha < PI2; alpha += increment)
{
sin_alpha = Math.Sin(alpha);
cos_alpha = Math.Cos(alpha);
x = (int)(center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta)));
y = (int)(center_y + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta)));
points.Add(new Point(x, y));
}
Point[] point_array = points.ToArray();
g.FillPolygon(mybrush, point_array);
g.DrawPolygon(mypen, point_array);
}
private void FillCylinder(Graphics g, double center_x, double center_y, double height, double semi_minor, double semi_major, double angle)
{
Pen mypen = new Pen(Color.Black);
SolidBrush mybrush = new SolidBrush(Color.LightGray);
double semi_height = height / 2;
double beta = -angle * (Math.PI / 180);
double sin_beta = Math.Sin(beta);
double cos_beta = Math.Cos(beta);
double sin_alpha = Math.Sin(Math.PI);
double cos_alpha = Math.Cos(Math.PI);
int left_x1 = (int)(center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta)));
int left_y1 = (int)(center_y - semi_height + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta)));
int left_x2 = (int)(center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta)));
int left_y2 = (int)(center_y + semi_height + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta)));
sin_alpha = Math.Sin(0);
cos_alpha = Math.Cos(0);
int right_x1 = (int)(center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta)));
int right_y1 = (int)(center_y - semi_height + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta)));
int right_x2 = (int)(center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta)));
int right_y2 = (int)(center_y + semi_height + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta)));
FillEllipse(g, Color.Black, Color.LightGray, center_x, center_y + semi_height, semi_minor, semi_major, angle);
Point p_left_1 = new Point(left_x1, left_y1);
Point p_right_1 = new Point(right_x1, right_y1);
Point p_left_2 = new Point(left_x2, left_y2);
Point p_right_2 = new Point(right_x2, right_y2);
Point[] mypoints = new Point[] { p_left_1, p_right_1, p_right_2, p_left_2 };
g.FillPolygon(mybrush, mypoints);
FillEllipse(g, Color.Black, Color.LightGray, center_x, center_y - semi_height, semi_minor, semi_major, angle);
g.DrawLine(mypen, left_x1, left_y1, left_x2, left_y2);
g.DrawLine(mypen, right_x1, right_y1, right_x2, right_y2);
}
private void picCanvas_Paint(object sender, PaintEventArgs e)
{
FillCylinder(e.Graphics, picCanvas.Size.Width / 2, picCanvas.Size.Height / 2, picCanvas.Size.Height / 2, picCanvas.Size.Height / 8, picCanvas.Size.Width / 4, 10);
}
See the result here[^].
Of course you can also choose to continute with the "brute force" approach I showed earlier for getting the outer points, which is ok for any rotation angle. In that case I suggest you calculate the points for both ellipses and store them in point arrays, so that you can also get the outer points. Then, you can proceed to drawing:
private class EllipseOuterXPoints
{
internal int left_x = int.MaxValue;
internal int left_y = int.MaxValue;
internal int right_x = int.MinValue;
internal int right_y = int.MinValue;
internal void CheckPoint(int x, int y)
{
if (x < left_x)
{
left_x = x;
left_y = y;
}
if (x > right_x)
{
right_x = x;
right_y = y;
}
}
}
private Point[] GetEllipsePoints(double center_x, double center_y, double semi_minor, double semi_major, double angle, EllipseOuterXPoints OuterPoints)
{
double PI2 = Math.PI * 2;
double major = semi_major * 2;
double increment = (Math.PI * 2) / major;
Point[] points = new Point[(int)major];
double beta = -angle * (Math.PI / 180);
double sin_beta = Math.Sin(beta);
double cos_beta = Math.Cos(beta);
double sin_alpha;
double cos_alpha;
int x;
int y;
int i = 0;
for (double alpha = 0; alpha < PI2; alpha += increment)
{
sin_alpha = Math.Sin(alpha);
cos_alpha = Math.Cos(alpha);
x = (int)(center_x + ((semi_major * cos_alpha * cos_beta) - (semi_minor * sin_alpha * sin_beta)));
y = (int)(center_y + ((semi_major * cos_alpha * sin_beta) + (semi_minor * sin_alpha * cos_beta)));
points[i++] = new Point(x, y);
OuterPoints.CheckPoint(x, y);
}
return points;
}
private void FillCylinder_BruteForce(Graphics g, double center_x, double center_y, double height, double semi_minor, double semi_major, double angle)
{
Pen mypen = new Pen(Color.Black);
SolidBrush mybrush = new SolidBrush(Color.LightGray);
double semi_height = height / 2;
EllipseOuterXPoints OuterPoints1 = new EllipseOuterXPoints();
EllipseOuterXPoints OuterPoints2 = new EllipseOuterXPoints();
Point[] points1 = GetEllipsePoints(center_x, center_y - semi_height, semi_minor, semi_major, angle, OuterPoints1);
Point[] points2 = GetEllipsePoints(center_x, center_y + semi_height, semi_minor, semi_major, angle, OuterPoints2);
g.FillPolygon(mybrush, points2);
g.DrawPolygon(mypen, points2);
Point p_left_1 = new Point(OuterPoints1.left_x, OuterPoints1.left_y);
Point p_right_1 = new Point(OuterPoints1.right_x, OuterPoints1.right_y);
Point p_left_2 = new Point(OuterPoints2.left_x, OuterPoints2.left_y);
Point p_right_2 = new Point(OuterPoints2.right_x, OuterPoints2.right_y);
Point[] mypoints = new Point[] { p_left_1, p_right_1, p_right_2, p_left_2 };
g.FillPolygon(mybrush, mypoints);
g.FillPolygon(mybrush, points1);
g.DrawPolygon(mypen, points1);
g.DrawLine(mypen, OuterPoints1.left_x, OuterPoints1.left_y, OuterPoints2.left_x, OuterPoints2.left_y);
g.DrawLine(mypen, OuterPoints1.right_x, OuterPoints1.right_y, OuterPoints2.right_x, OuterPoints2.right_y);
}
private void picCanvas_Paint(object sender, PaintEventArgs e)
{
FillCylinder_BruteForce(e.Graphics, picCanvas.Size.Width / 2, picCanvas.Size.Height / 2, picCanvas.Size.Height / 2, picCanvas.Size.Height / 8, picCanvas.Size.Width / 4, 45);
}
See the result here[^].
Well, this should give you enough meat to chew I hope. Enjoy!
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Thanks! I'm gonna try your code, tell you the truth though, I have no idea of the math whatsoever. 
But I might be able to chew the meat slowly, maybe add some salt or recook it.
Thanks again for taking the time to help. Appreciate it.
|
|
|
|
|
My pleasure. Good luck!
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
IMHO you should consider this for an article - A1 stuff 
I don't speak Idiot - please talk slowly and clearly
'This space for rent'
Driven to the arms of Heineken by the wife
|
|
|
|
|
Thanks Andy, really appreciated.
I will consider writing an article on this. Maybe when I'm on my holidays next month, then hopefully I'll get enough free time. If I don't overplay my PS3, that is hehe. 
Now I "only" need to get good at writing articles... ARGH! 
2+2=5 for very large amounts of 2
(always loved that one hehe!)
|
|
|
|
|
Hi all,
Is there anyone could point out any way in bypassing a firewall via compression techniques? would appreciate if any available algorithm (java) is provided as well.
What i want is for the software to basically compress the files or sites in order to bypass the firewall instead of hiding the IP address. Is it possible?
Tried google but couldnt get the answer i m looking for.
Thanks
|
|
|
|
|
|
What is a good algorithm to calculate if a square is intersecting a rectangle?
I noticed in Windows forms the picture box Rectangle had an "intersect" function but I would like to program something on those lines with an algorithm instead.modified on Monday, March 8, 2010 7:52 PM
|
|
|
|
|
There is no PictureBox.Intersect; maybe you are confusing with Rectangle.Intersect?
are the rectangles always orthogonal (parallel to the axes)? (if so, use Rectangle.Interset)
or under an arbitrary but fixed angle? or a variable angle?
|
|
|
|
|
Luc Pattyn wrote: There is no PictureBox.Intersect; maybe you are confusing with Rectangle.Intersect?
Ah yes I confused my self that is what I meant, a rectangle intersection
|
|
|
|
|
I figured out a solution my self, wasn't as tricky as I had thought, being 2am didn't help though:
public bool HitTest(Sprite paddle)
{
bool left = _ball.Position.X + _ball.Width > paddle.Position.X;
bool right = _ball.Position.X < paddle.Position.X + paddle.Width;
bool up = _ball.Position.Y + _ball.Height > paddle.Position.Y;
bool down = _ball.Position.Y < paddle.Position.Y + paddle.Height;
return (left && right && up && down);
}
|
|
|
|
|
Does anyone have any experience with writing parallel algorithms. I am teaching my self CUDA; Nvidia's C language with parallel extensions for G80 (Tesla) and GT200 (Tesla) GUP's. I am having a hard time with the concepts of thinking of threads in so many different dimensions. Is there a good source of parallel design architecture anyone knows of that might help me better understand this concept? With CUDA you literally have to think in 5 dimensions at once just to get decent data parallelism at CGMA (compute to global memory access) above 1.0.
The book I am reading is: Programming Massively Parallel Processors. It is based on: SIMT (single-instruction, multiple-thread) where as CPU's are based on: SIDM (single-instruction, multiple-data). and the book tries to cover both topics at once, which is a little confusing since CUDA supports running on CPU or GPU in parallel data access.
The book is intended for undergraduates and I am about at that level, but have no formal training in a university, only professional mentoring. Can some one point me in the right direction? I am a visual learner, the book is full of diagrams, but all the diagrams seem to show data and events flowing in every direction at once.
Currently I have covered: History of GPU computing, Introduction to CUDA (Matrix Multiplication Example), CUDA Threads (Same Examples), CUDA Memories (same examples), and now working on: Performance Considerations; this chapter is kind of mind blowing!
~TheArch
|
|
|
|
|
have you done any non-CUDA multi-threading ?
|
|
|
|
|
yes, I have worked extensively with threading and even use to debug parallel threading architecture on Java UI's what took advantage of multiple processors on super spark servers. threading it no problem. I fully understand the concept, however on a GPU it is a bit different, the developer is forced to use the parallel design-constructs of the GPU, and to get the correct memory bandwidth ratios of processing data-parallelism you are forced to design the code so it takes advantage of the hardware, simple loops with nested loops and control flow logic seriously complicates the process.
this constraint is a difficult concept to understand, grasp, and implement. Just for matrix multiplication you have to think in three dimensions just to get a global memory ratio of 1.0, which barely makes use of the full processing power of the GPu, stream processors are also used hand have limitations on the number of current threads that can execute a 'block' a quadrant of GPU cores sharing global memory.
so to make full use of a G80 GPU the code has to be written in such a fashion (a dependency caused by manufacturing costs and lithography techniques) to process the three dimensions of the multiplication in what I consider a strange and cumbersome way, where each matrix (G1, G1, R1) are further divided to make use if faster but smaller shared memory which cuts the bandwidth in half. so now you are thinking in six dimensions and also have to consider how many threads a stream processor can handle for the process just to achieve 748Mb band-with and full GPU utilization.
it is algorithms like I have no experience designing.
but I have a strong desire to learn, even if the approach seems unintuitive and abstract!
------Modification-------
I wrote the response on the bus, couldn't hardly see the screen I was typing into, I think I am going to put a sugestion in the site sugestions box to make it wasier to respond to threads using a mobile device, I use to be a mobile device programmer. I could do it in a peice of cake.
-------------------------
~TheArch
|
|
|
|
|
ah...
unfortunately, this sounds like it's well beyond anything i've done.
good luck !
|
|
|
|
|
Hi,
So I am coding CUDA stuff since a year now.
I started to program the examples that they showed in their documentation.
I found that really helpful. I also liked this article :
[^]
Actually, I would not recommend any of the books, because the documentation from Nvidia is enough, but maybe it helps you.
Have fun, cheers
You have the thought that modern physics just relay on assumptions, that somehow depends on a smile of a cat, which isn’t there.( Albert Einstein)
|
|
|
|
|
I have a mapping program. Upon the map are laid polygons that represent rooms in a building.
The program must tell which room a tracked asset is in, AND if the asset is not within any of the rooms, it must tell which is the closest room to its location.
I have already implemented the code that can tell which polygon contains a given point, BUT I can't see how to figure out the closest polygon to a given point if the point is not within any polygon.
In other words, if the point is not within any of the polygons, how do I figure out which polygon it is closest to?
|
|
|
|
|
Do you mean closest to the walls or closest to the doors?
Because if it is about tracking an asset, it should be closest to the door.
This could be one approach:
You could create a virtual circle around the point and find all polygon falling on it. if there are none, increase the size of circle. eg a circle of 10, 20,40,80 and so on. when you do encounter one polygon, thats your winner. In case you get more than one, reduce the circle slowly till you get a single.
|
|
|
|
|
General 
News 
Suggestion 
Question 
Bug 
Answer 
Joke 
Praise 
Rant 
Admin 
Submit an article or tip
