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| 04.24.2008 at 09:13AM PDT, ID: 23350809 |
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Zooming / Sizing in drawing App
Asked by Dodsworth in Microsoft Visual Basic.Net, Game Programming, Theory
Tags: vb.net
I have a little (but growing) drawing app that allows a collection of shape objects to be moved around the screen.
I need some pointers as to how to zoom in and out.
My shape class stores the location and size of each shape, so I guess I need to add some sort of multiplier to the drawing functions. I also need o consider the size of the user screen
Any ideas ?Start Free Trial
I need some pointers as to how to zoom in and out.
My shape class stores the location and size of each shape, so I guess I need to add some sort of multiplier to the drawing functions. I also need o consider the size of the user screen
Any ideas ?Start Free Trial
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| 04.24.2008 at 09:39AM PDT, ID: 21432748 |
Often times you can just use one of the "Transform" methods:
http://msdn2.microsoft.com
After performing the scaling operation...drawing with the same coordinates will look smaller or bigger...
http://msdn2.microsoft.com
After performing the scaling operation...drawing with the same coordinates will look smaller or bigger...
| 04.24.2008 at 11:53PM PDT, ID: 21437464 |
That kind of works in so much as my shapes are bigger but I have clip and invalidate and hittest code that nolonger works.
I really need some math on the coordinates.
I really need some math on the coordinates.
| 05.03.2008 at 07:27AM PDT, ID: 21492458 |
Hi,
If yours is a bitmap drawing, then each zoom-in is a rescaling of image, which has bad results when magnification is large.
Assuming vector drawing for the shapes, you can define a zoomfactor and render the entire set of shapes each time you zoom or displace the viewport. A sample pseudocode follows.
In the example the orign is set in the middle of the viewport. This simplify the computation of scaling an object, in general doing by moving it to the orign, multiply its vertices by a factor and moving it back to its previous place. Assuming the orign in the center of the viewport we can simply multiply its coordinates by the desired factor.
Of course, when drawing each object you can check first if it is inside the viewport, then clipping it or even skiping it, but I think you already is taking care about that.
Jose
Open in New Window
If yours is a bitmap drawing, then each zoom-in is a rescaling of image, which has bad results when magnification is large.
Assuming vector drawing for the shapes, you can define a zoomfactor and render the entire set of shapes each time you zoom or displace the viewport. A sample pseudocode follows.
In the example the orign is set in the middle of the viewport. This simplify the computation of scaling an object, in general doing by moving it to the orign, multiply its vertices by a factor and moving it back to its previous place. Assuming the orign in the center of the viewport we can simply multiply its coordinates by the desired factor.
Of course, when drawing each object you can check first if it is inside the viewport, then clipping it or even skiping it, but I think you already is taking care about that.
Jose
1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: |
shape reactangle(P1,P2)
viewport V(V1,V2)
zoomFactor Z // default Z=1
displacement D(dx,dy) // defauly dx=0, dy=0
function render(shapes, Z, D)
{
foreach shape
s1.x = P1.x * Z - dx
s1.y = P1.y * Z - dy
s2.x = P2.x * Z - dx
s2.y = P2.y * Z - dy
draw shape(s1,s2)
}
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| 05.03.2008 at 07:35AM PDT, ID: 21492490 |
An additional comment.
If you want the center of the viewport to be static in the zooming, then the lines
s1.x = P1.x * Z - dx
etc.
should be replaced by
(s1.x = P1.x - dx) * Z
etc.
Jose
If you want the center of the viewport to be static in the zooming, then the lines
s1.x = P1.x * Z - dx
etc.
should be replaced by
(s1.x = P1.x - dx) * Z
etc.
Jose
| 05.03.2008 at 09:26AM PDT, ID: 21492822 |
If your program adopts orign O(0,0) at left lower or left upper then it would be a huge task to change everithing. This is bad idea.
If it is the case, then you can use the procedure described before. A sample Windows C++ code to draw a rectangle is provided below.
(point - displacement) moves the object to the orign,
*Z scale the object
+displacement moves the object back
Jose
Open in New Window
If it is the case, then you can use the procedure described before. A sample Windows C++ code to draw a rectangle is provided below.
(point - displacement) moves the object to the orign,
*Z scale the object
+displacement moves the object back
Jose
1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: |
void DrawRectangle(HDC hdc,POINT point,HPEN pencolor,
int x1, int y1, int x2, int y2, // shape end points
float Z
{
int dx,dy; // displacement
dx=Width/2; dy=Height/2; // middle size of the viewport
MoveToEx(hdc,(x1-dx)*Z+dx, (y1-dy)*Z+dy, &point);
LineTo(hdc, (x2-dx)*Z+dx, (y1-dy)*Z+dy);
LineTo(hdc, (x2-dx)*Z+dx, (y2-dy)*Z+dy);
LineTo(hdc, (x1-dx)*Z+dx, (y2-dy)*Z+dy);
LineTo(hdc, (x1-dx)*Z+dx, (y1-dy)*Z+dy);
}
|
| 05.10.2008 at 03:53AM PDT, ID: 21538460 |
Sorry Jose, I'm not understanding the viewport and centering concept...
"In the example the orign is set in the middle of the viewport. This simplify the computation of scaling an object, in general doing by moving it to the orign, multiply its vertices by a factor and moving it back to its previous place. Assuming the orign in the center of the viewport we can simply multiply its coordinates by the desired factor."
Can you explain a bit more
"In the example the orign is set in the middle of the viewport. This simplify the computation of scaling an object, in general doing by moving it to the orign, multiply its vertices by a factor and moving it back to its previous place. Assuming the orign in the center of the viewport we can simply multiply its coordinates by the desired factor."
Can you explain a bit more
| 05.10.2008 at 03:48PM PDT, ID: 21540816 |
No problem, this can be little bit confuse...
When the viewport center is coincident to the orign (0,0), we can arrange the objects such that we can scale objects simply by multiplying the vertices of the objects by scale factors, say FX and FY.
When the origin is located at a viewport corner, to scale objects, we first move them to the orign, then scale them, and move them back.
The first approach avoids the two moves.
In the sample code you can notice in line 10, the expression:
(x2-dx)*Z+dx
what means this expression?
First, we move x2 to the orign:
tmp = x2-dx
then we scale, by multiplying by Z
tmp = tmp * Z
and finally we move it back
tmp = tmp + dx
If the viewport center is the orign, then we can let dx=0 and dy=0.
That way, the function
LineTo(hdc, (x2-dx)*Z+dx, (y1-dy)*Z+dy);
can be reduced to
LineTo(hdc, (x2 - 0)*Z+0, (y1 - 0)*Z+0);
that is the same as
LineTo(hdc, x2*Z, y1*Z);
If you are looking for a general code, which uses the window viewport (orign at upper left corner), then it is better to adopt approach 2.
At figures 1 and 2 we can watch the result after zooming in.
Jose
Open in New Window
When the viewport center is coincident to the orign (0,0), we can arrange the objects such that we can scale objects simply by multiplying the vertices of the objects by scale factors, say FX and FY.
When the origin is located at a viewport corner, to scale objects, we first move them to the orign, then scale them, and move them back.
The first approach avoids the two moves.
In the sample code you can notice in line 10, the expression:
(x2-dx)*Z+dx
what means this expression?
First, we move x2 to the orign:
tmp = x2-dx
then we scale, by multiplying by Z
tmp = tmp * Z
and finally we move it back
tmp = tmp + dx
If the viewport center is the orign, then we can let dx=0 and dy=0.
That way, the function
LineTo(hdc, (x2-dx)*Z+dx, (y1-dy)*Z+dy);
can be reduced to
LineTo(hdc, (x2 - 0)*Z+0, (y1 - 0)*Z+0);
that is the same as
LineTo(hdc, x2*Z, y1*Z);
If you are looking for a general code, which uses the window viewport (orign at upper left corner), then it is better to adopt approach 2.
At figures 1 and 2 we can watch the result after zooming in.
Jose
1: 2: 3: 4: 5: 6: 7: 8: 9: 10: 11: 12: 13: 14: 15: 16: 17: 18: 19: 20: |
Code to call the DrawRectangles function:
DrawRectangles(hdc,point,penBlue,100,50, 250,135,zoom)
DrawRectangles(hdc,point,penRed,180,97,350,258,zoom)
Code of DrawRectangles function
void DrawRectangle(HDC hdc,POINT point,HPEN pencolor, x1, y1, x2, y2, double Z)
{
dx=Width/2; dy=Height/2; <-- equivalent to put orign at center of the viewport
x1 = (int)((x1-dx+offsetx)*Z)+dx; y1 = (int)((y1-dy)*Z)+dy;
x2 = (int)((x2-dx+offsetx)*Z)+dx; y2 = (int)((y2-dy)*Z)+dy;
SelectObject(hdc, pencolor);
MoveToEx(hdc,x1-1, y1, &point);
LineTo( hdc, x2, y1);
LineTo( hdc, x2, y2);
LineTo( hdc, x1, y2);
LineTo( hdc, x1, y1);
}
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