This DirectX 7 function: http://msdn.microsoft.com/en-us/library/windows/desktop/gg426183(v=vs.85).aspx
The parameters include an LPRECT as well as X and Y co-ordinates. My confusion lies in which take priority; the top left co-ordinate of the rectangle, or the X & Y co-ordinates.
Maybe I'm being a total idiot and they serve different purposes, but I don't understand how. Please be patient with me; I'm certainly not as experienced as 99% of you guys, but I'm getting there slowly.
Thanks.
This was a moronic question.
The rect is the part of the image that you want to render, and the x and y co-ordinates are where you want to render to. The source rectangle, and the destination co-ordinates.
Sorry for being such a fool.
Related
i've been looking around to find the best way to have wrap arround edges of a world for the bodies.
I managed to use this topic to do it :
How do I make a Box2D wrap around world?
Using SetTransform() i could make it appear on the other X/Z side.
Now Let's say i have an object for example a simple box 10x10.
If half the box goes beyond upper Y edge i want the portion that goes beyond to appear below with the other part of the box that is still visibile on the upper part to stay there.
To summarize i want a "real wrap arround edge like this used to be done in old games.
I hope i was clear enough...
Edit :
I've added a picture to explain what i mean :
Thanks
I have not used Box2d but I have resolved this problem before in Processing, hopefully the logic translates easily
In Processing a rectangle is drawn as rect(x, y, rectangleWidth, rectangleHeight) at position x, y which represents the top left corner of the rectangle. The rectangle's width and height point right and down from x and y respectively
The idea is to draw the rectangle normally unless the bottom would be off the bottom edge of the viewport (because the rectangle's height points down based on y). If the rectangle is off the bottom edge of the viewport then you instead draw two partial rectangles
In the code below height is the height of your viewport
if (y < height-10)
rect(x, y, 10, 10); // Normal condition, entire rectangle
else {
rect(x, 0, 10, 10-(height-y)); // Top partial rectangle
rect(x, y, 10, height-y); // Bottom partial rectangle
}
Finally i googled a bit found this article on Unity :
http://gamedevelopment.tutsplus.com/articles/create-an-asteroids-like-screen-wrapping-effect-with-unity--gamedev-15055
On Box2D I created 8 ghosts bodies which were positionned as defined in the article.
During Box2D Steps i added some logics to check where my original body is positionned.
When it goes to the edges the Ghost Body appears on the other edge. It works also when the original Body goes on a corder. 4 ghosts bodies will appear on each edges.
I'm using the Kinect for Windows SDK (v1.8)
I'm successfully reading motion from the Kinect but I'm now wondering how to get the absolute position of the joint I'm tracking (right hand)
I'm using the following function, NuiTransformSkeletonToDepthImage, but the values range from 100-200 in both x and y coordinates.
Any suggestions for how to transform the returned coordinates to screen coordinates?
Got a feeling I'm missing something really obvious though...
Thank you in advance
Ok, after quite a bit of looking around and trial and error; I found the solution.
When you call NuiTransformSkeletonToDepthImage(); you can specify a Kinect resolution using the NUI_IMAGE_RESOLUTION enum. I believe the values are; 80x60, 320x240, 640x480, 1280x960
Depending on what resolution you decide to use; you'll need to divide the returned x and y coords by the corresponding resolution. Then multiply those by your window size and you should have the coordinates in your window.
I believe you can also specify the resolution when creating an instance of the Kinect Sensor interface.
Example as my explanation isn't too concise:
float x, y
const Vector4 fromPos = skeleton.SkeletonPositions[NUI_SKELETON_POSITION_HAND_RIGHT];
NuiTransformSkeletonToDepthImage(fromPos, &x, &y, NUI_IMAGE_RESOLUTION_320x240);
x = x * ((float)WINDOW_WIDTH / 320.0f);
y = y * ((float)WINDOW_HEIGHT / 240.0f);
Any questions; feel free to ask.
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I want to imitate the well-known HTML's great great RECTANGLE. I mean all of the characteristic of the rectangles like borders, border-radius, triangulated quad on corners, etc. I don't like to use any other libraries except mine. I would like to create this one for the sake of learning and experience, and also to use it for the future as a GUI system. I am working on this concept of shuffled rectangles.
It is composed of:
4 GL_TRIANGLES as quadrilateral on corners
4 arcs on corners
4 borders on all sides
And one front big rectangle on the front
And these are the outputs I made so far :)
w/o border-radius
w/ border-radius
So the things I am really really confused with are
Border-sizes
Border-locations
Is it the X, Y's or the W, H's
When to draw or not front-most rectangle
Anything I don't know yet.
...Please comment other things I should include here for clarification. Thanks!
Edit:
Hmm..., Okay as for a minimal question only. I just wanted to implement this stuffs and programmatically compute their values as I change a single attributes of the rectangle.
border-radii-sizes
border-sides
I'm putting too much images here, please please understand me :(
*left-border
*corner
I was thinking of that kind of rectangles positioning and it's really difficult in my head to compute for their coordinates or the sizes base on the set of attributes I'm gonna define on the designing part. For example, If I define the border-radius-top-left to have a value of 50% with its counter part of border-size-left with a certain value, what would be the formula I must consider. Or, must I need to add any component/ private attributes in order to make this thing happen?
Yey!! I have figured it out!!
Please let me SO to discuss my [problem solved] here.
Sorry for my unclassified art :) I made it colorful for property separation.
Explanation:
Arcs w/c serves as corner-radii.
The formula for points (x, y) will be automatically generated here
corner-radii-points (x, y) are the target.
points (x, y) Should be the only one adjusting based on the given radii values.
Curved part of this should be static in position.
Fake borders these are the inner-side-borders.
Properties of this such as [x, y, width, height] will depend on corner-radii-points(x, y) and points(x, y).
Inner quad w/c is the inner-rectangle
This will just serves as cover
Properties of this such as [x1, y1, x2, y2](this is a polygon so I labeled it like that) will depend on points (x, y)
Now I can simply do this:
Designing Part:
int w3 = rect.width >> 3;
int h3 = rect.height >> 3;
rect.setBorderRadius(C_NW, w3, h3);
rect.setBorderRadius(C_NE, w3, h3);
rect.setBorderRadius(C_SE, w3, h3);
rect.setBorderRadius(C_SW, w3, h3);
rect.setColors(C_NW, cc::getColor(COLORS::RED));
rect.setColors(C_NE, cc::getColor(COLORS::GREEN));
rect.setColors(C_SE, cc::getColor(COLORS::BLUE));
rect.setColors(C_SW, cc::getColor(COLORS::YELLOW));
rect.setBorderColor(B_TOP, cc::getColor(COLORS::WHITE));
rect.setBorderColor(B_RIGHT, cc::getColor(COLORS::WHITE));
rect.setBorderColor(B_BOTTOM, cc::getColor(COLORS::GRAY3));
rect.setBorderColor(B_LEFT, cc::getColor(COLORS::GRAY3));
rect.setBorderSize(B_TOP, 20);
rect.setBorderSize(B_RIGHT, 20);
rect.setBorderSize(B_BOTTOM, 20);
rect.setBorderSize(B_LEFT, 20);
Results:
rect is the one with Lightning McQueen image.
Those are the formulas I evaluate base on trial and error.
Also thanks to Sir Mark Garcia for helping me by demonstrating some diagrams, and suggested to use MS Paint for visualization :)
Next problem is masking as you can see that there are non-curved borders and corner radius at the same time, but I won't focus on that at this moment.
If ever someone is interested in this kind of rectangle implementation, you can PM me here and I'll give you the source code.
I'm Processing batches of Images and scanning them for a pattern of pixels. I can get the program to work with SetCursorPos() and it works perfect, but I'm adapting the tech to live motion feed. SerCursorPos() does not work with that window, but mouse_event does work.
My Issue: Where it was easy to pass the X,Y of the Bitmap to SetCursorPos, I've tried a few formulas to convert and provide the x, y to dX and dY for mouse_event, but they seem to be off by 70-100 coordinates or pixels. For Example I've tried using the HWND's RECT as a base for width and hieght applied against 65535; and that didn't work. Should I be using a GetSystemMetric of the window? Any help would be appreciated. Should I use a DirectX alternative?
At least if I understand what you're asking, it sounds like you're looking for ClientToScreen (and possibly doing some compensation for the position of the bitmap inside the client area).
This is quite complicated to explain, so I will do my best, sorry if there is anything I missed out, let me know and I will rectify it.
My question is, I have been tasked to draw this shape,
(source: learnersdictionary.com)
This is to be done using C++ to write code that will calculate the points on this shape.
Important details.
User Input - Centre Point (X, Y), number of points to be shown, Font Size (influences radius)
Output - List of co-ordinates on the shape.
The overall aim once I have the points is to put them into a graph on Excel and it will hopefully draw it for me, at the user inputted size!
I know that the maximum Radius is 165mm and the minimum is 35mm. I have decided that my base [Font Size][1] shall be 20. I then did some thinking and came up with the equation.
Radius = (Chosen Font Size/20)*130. This is just an estimation, I realise it probably not right, but I thought it could work at least as a template.
I then decided that I should create two different circles, with two different centre points, then link them together to create the shape. I thought that the INSIDE line will have to have a larger Radius and a centre point further along the X-Axis (Y staying constant), as then it could cut into the outside line.*
*(I know this is not what it looks like on the picture, just my chain of thought as it will still give the same shape)
So I defined 2nd Centre point as (X+4, Y). (Again, just estimation, thought it doesn't really matter how far apart they are).
I then decided Radius 2 = (Chosen Font Size/20)*165 (max radius)
So, I have my 2 Radii, and two centre points.
This is my code so far (it works, and everything is declared/inputted above)
for(int i=0; i<=n; i++) //output displayed to user
{
Xnew = -i*(Y+R1)/n; //calculate x coordinate
Ynew = pow((((Y+R1)*(Y+R1)) - (Xnew*Xnew)), 0.5); //calculate y coordinate
AND
for(int j=0; j<=n; j++)//calculation for angles and output displayed to user
{
Xnew2 = -j*(Y+R2)/((n)+((0.00001)*(n==0))); //calculate x coordinate
Ynew2 = Y*(pow(abs(1-(pow((Xnew2/X),2))),0.5));
if(abs(Ynew2) <= R1)
cout<<"\n("<<Xnew2<<", "<<Ynew2<<")"<<endl;
I am having the problem drawing the crescent moon that I cannot get the two circles to have the same starting point?
I have managed to get the results to Excel. Everything in that regard works. But when i plot the points on a graph on Excel, they do not have the same starting points. Its essentially just two half circles, one smaller than the other (Stops at the Y axis, giving the half doughnut shape).
If this makes sense, I am trying to get two parts of circles to draw the shape as such that they have the same start and end points.
If anyone has any suggestions on how to do this, it would be great, currently all I am getting more a 'half doughnut' shape, due to the circles not being connected.
So. Does anyone have any hints/tips/links they can share with me on how to fix this exactly?
Thanks again, any problems with the question, sorry will do my best to rectify if you let me know.
Cheers
Formular for points on a circle:
(x-h)^2+(y-k)^2=r^2
The center of the circle is at (h/k)
Solving for y
2y1 = k +/- sqrt( -x^2 + 2hx +r^2 - h^2)
So now if the inner circle has its center # h/k , the half-moon will begin # h and will stretch to h - r2
Now you need to solve the endpoint formular for the inner circle and the outter circle and plot it. Per x you should receive 4 points (solve the equation two times, each with two solutions)
I did not implement it, but this would be my train of thought...