Rotate rerctangle in Bitmap

raju_shiva

Hi sir, I have origin axis of bitmap and x1,x2,y1,y2. I want to draw a rectangle and rotate it with the given degree.I am trying with this code,but i am not getting. Can some one help me

double x1,x2,y1,y2;
sf=(1024/31); //Its a bitmap image
A1 = 10.00; //these are values i am reading from hardware
A2 = 10.00;
B1 = 10.00;
B2 = 10.00;

x1 = (A1 * sf);
x2 = (A2 * sf);
y1 = (B1 * sf);
y2 = (B2 * sf);
//To find the center origin of Bitmap
center.x=(((pCellInfo->rcBitmapRect.right-pCellInfo->rcBitmapRect.left)/2)+pCellInfo->rcBitmapRect.left);
center.y=(((pCellInfo->rcBitmapRect.bottom-pCellInfo->rcBitmapRect.top)/2)+pCellInfo->rcBitmapRect.top);

int a = 45;
float Angle = ( 3.142 * a ) / 180;

x1 = x1 * cos(Angle) + y1 * sin(Angle);
y1 = -x1 * sin(Angle) + y1 * cos(Angle);
x2 = x2 * cos(Angle) + y2 * sin(Angle);
y2 = -x2 * sin(Angle) + y2 * cos(Angle);

//I am using the right formula to claculate the axis for rotation
Origin(525,454); // center origin(center.x,center.y);
MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL);

LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

If i test with this hard code values,i am getting the rectangle rotated
MoveToEx(pCellInfo->hDC,525,289,NULL);// Angle 45 for reference
LineTo(pCellInfo->hDC,360,454);
LineTo(pCellInfo->hDC,525,619);
LineTo(pCellInfo->hDC,690,454);
LineTo(pCellInfo->hDC,525,289);

how can i get the same values Thanks Raj

CPallini

raju_shiva wrote:

i am not getting

What does it mean, exactly (i.e. please elaborate)?

raju_shiva wrote:

A1 = 10.00; //these are values i am reading from hardware A2 = 10.00; B1 = 10.00; B2 = 10.00;

How do you hope to get a rectangle from these values? :)

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
[My articles]

Peter_in_2780

When you asked this same question[^] yesterday, part of my reply said:

Look at the actual values you are passing to MoveToEx and LineTo.

Until you do that, I'm not interested in helping any more, and I don't think too many others will be either.

Software rusts. Simon Stephenson, ca 1994.

raju_shiva

CPallini wrote:

. please elaborate)?

These are left,top,right,bottam values for the rectangle i.e Rectangle(__in HDC hdc, __in int left, __in int top, __in int right, __in int bottom); Thanks Raj

raju_shiva

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); // Here i will find the new axis from the center axis
// i.e (center.x point - x1,center.y point-y2)
//so that i get the start point from where i can start to draw rectangle
LineTo(pCellInfo->hDC,center.x-x1,center.y+y1); //From here i will draw the rectangle

I hope you got it,what i am trying to do. If i am doing wrong,please let me know, Thanks for your reply Raj

CPallini

{10,10,10,10} as {left,top,right,bottom} values give an empty rectangle. :)

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
[My articles]

raju_shiva

CPallini wrote:

{10,10,10,10} as {left,top,right,bottom} values give an empty rectangle

I am not passing the values directly. As it is a bitmap image,on that i am drawing the rectangle I am multiplying it :

x1 = (A1 * sf); //i.e (10 * 33.03)
x2 = (A2 * sf); //i.e (10 * 33.03)
y1 = (B1 * sf); //i.e (10 * 33.03)
y2 = (B2 * sf); //i.e (10 * 33.03)

where sf is sf=(1024/31); //Its a bitmap image Then while drawing the rectangle,i start as (center.x-left)

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); //(center.x-left)
//(center.y-bottom)

I hope you got it. Thanks Raj

CPallini

raju_shiva wrote:

x1 = (A1 * sf); //i.e (10 * 33.03) x2 = (A2 * sf); //i.e (10 * 33.03) y1 = (B1 * sf); //i.e (10 * 33.03) y2 = (B2 * sf); //i.e (10 * 33.03)

I see no rectangle here. :)

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
[My articles]

raju_shiva

CPallini wrote:

I see no rectangle here

After getting all the values Here i am drawing the rectangle

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL);
LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

Thanks Raj

CPallini

OK (sorry if I didn't get you). Now, what is the problem with your code (expected behaviour vs observed one)?. :)

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
[My articles]

raju_shiva

Now i want to rotate it for the given degree.How can i do it. I am bit confused please help me Thanks Raj

CPallini

You are defining the rectangle via offsets from center, that is

P0={-x1,-y2}, P1{-x1, y1}, P2={x2,y1}, P3={x2,-y2}

hence, if you wan't rotate with angle phi around the center, than you should compute:

Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

i.e.:

P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}
P1R =...
P2R =...
P3R =...

and then connect the center+PiR points the way you did before. :)

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
[My articles]

raju_shiva

CPallini wrote:

P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}

I am confused with it??? Now suppose my values are:

x1 = 330,x2 = 330,y1=330,y2=330
Origin(center.x,cebter.y) = Origin(525,454);

I have to do the get the new x1,x2,y1,y2 before calling MoveToEx and LineTo Am i right?? i.e

Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

for each x1,x2,y1,y2. Then call the function

MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); // where x1 = Pr1,x2=Pr2....y2 = Pr4
LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

Thanks Raj

CPallini

Something like this

// rotate around center
double phi = atan(1.0) * 2/3; // 30 degrees
int x[4];
int y[4];
x[0] = -x1 * cos(phi) - y2 * sin(phi);
y[0] = x1 * sin(phi) - y2 * cos(phi);

x[1] = -x1 * cos(phi) + y1 * sin(phi);
y[1] = x1 * sin(phi) + y1 * cos(phi);

x[2] = x2 * cos(phi) + y1 * sin(phi);
y[2] = -x2 * sin(phi) + y1 * cos(phi);

x[3] = x2 * cos(phi) - y2 * sin(phi);
y[3] = -x2 * sin(phi) - y2 * cos(phi);

for (int i=0; i<4; i++)
{
x[i] += center.x;
y[i] += center.y;
}

MoveToEx(pCellInfo->hDC,x[3],y[3],NULL);
for (int i=0; i<4; i++)
{
LineTo(pCellInfo->hDC, x[i],y[i]);
}

I suppose. :)

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
[My articles]

Peter_in_2780

When I said "look at the actual values", I didn't mean the expressions, I meant the actual values. In other words, either add some code to print out the values when you call the function or put a breakpoint on the call and inspect the values in your debugger. Then you should be able to figure out what is wrong and work back to where your problem is.

Software rusts. Simon Stephenson, ca 1994.

raju_shiva

Thanks a lot ,its working fine. Thanks Raj

raju_shiva

Thank u peter for your reply.Its working fine now Raj

CPallini

You are welcome. :)

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
[My articles]

Rick York

The code looks correct but I find it to be slightly offensive to me sensibilities. :) I'm just kidding. I usually make a function or method when I see a sequence of code repeated more than twice. Something like this is generic enough that I definitely would. Of course, it's not your job to put this is into a function. That is an exercise for the reader. ;)

CPallini

Rick York wrote:

he code looks correct but I find it to be slightly offensive to me sensibilities. Smile I'm just kidding. I usually make a function or method when I see a sequence of code repeated more than twice. Something like this is generic enough that I definitely would.

Nah, that's childish. You've to set up a linear algebra library, with vectors and rotation matrices, properly overloading the multiplication operator. Of course that's an exercise for you. ;P :laugh: On my defence, I wouldn't write such code repetitions in my own software (well, maybe I do in quick and dirty junk programs), the goal there was being as explicit as possible (the OP had doubts on the steps to take) :)

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
[My articles]