Hello, , I am rotating a Rectangle (90 degree corners) to an angle from zero to 360 degrees, for a computer programming graphics operation, I'll call this the Rotated Rectangle.
I need to find the width and height of a "Bounds" rectangle around the rotated rectangle (containing all), , this Bounds Rectangle will have sides that are parralell to the x and y axis and will have it's width and height equal to the Rotated Rectangle at 0 degrees and 180 degrees.
The data that I start with is the width and height of the Rotated Rectangle and the Angle of Rotation, what I need is a formular or way to get the Width and Height of the Bounds Rect, after the rotation.
Lets say "Ang" is the angle of rotation in degrees, and the rrWidth and rrHeight is the Width and Height of the Rotated Rectangle, so bWidth and bHeight are the results I need, as the width and height of the Bounds Rectangle. . .
so I need some math like
bWidth = Cos Ang x rrWidth
bHeight = Sin Ang x rrHeight
but incorperating the the sides of the Rotated Rectangle to get a container size for the rotated rectangle
a picture may help to see the spatial relations, look at -
http://www.angelfire.com/h
I can get some "Partial" solutions, for a single quadrent of the rotation, but since the corners of the rotated rect change their left and right and upper and lower reference, I can not seem to get a good way to do this
any help will be apreciated !
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Hi Slick812,
bWidth = fabs( Cos( Ang ) * rrWidth ) + fabs( Sin( Ang ) * rrHeight )
bHeight = fabs( Cos( Ang ) * rrHeight ) + fabs( Sin( Ang ) * rrWidth )
Test this out and I think you will find that this is the solution that you are looking for.
P.S. fabs stands for floating point absolute value if you are using C++. * is multiplication.
I tried that many times and it is good for the width in the upper right quadrent, however it is not good for the bounds height, at least in the ways that I have tried it, I also got the angle of the Right side of the rotated rect and tried to get that into the formula, but I lost it as it went into a diferent quadrent
Gregory_Scot_Greer, thanks, your formulas seem to work, I will have time to fully test them later for all quadrents
also, I meant to ask, but forgot
would you know a way (formula) to get an "Offset" to paint this rotated rect onto the bounding rect (bitmap),
at 0 degrees it would be x = 0 and y = 0
at 90 degrees it would be x = 0 and y = rrWidth
at 180 degrees it would be x = rrWidth and y = -rrHeight
??
Sorry Slick812,
You have not supplied enough information for your second question, so I can only give you a general answer. If you want absolute coordinate mapping on the bounding rectangle, you will have to specify which point the rotating rectangle is rotating about. If you want relative coordinates, it would require that the formula change as you rotate to other quadrants i.e.
the lower left corner of the rotating rectangle would be found at:
When you are rotating for 0 to 90:
bX = Sin( Ang ) * rrHeight
bY = 0;
From 90 to 180, the point would be:
bX = ( Sin( Ang ) * rrHeight ) - ( Cos( Ang ) * rrWidth )
bY = -Cos( Ang ) * rrHeight
From 180 to 270, the point would be:
bX = -( Cos( Ang ) * rrWidth )
bY = -( Sin( Ang ) * rrWidth ) - ( Cos( Ang ) * rrHeight )
From 270 to 360, the point would be:
bX = 0
bY = -( Sin( Ang ) * rrWidth )
If you could give a better description of the mapping that you are attempting to make, I could give you a better formula.
Gregory_Scot_Greer , , I have given that info in the GIF image, the x = 0 and y = 0 , is the rotational center for the rotated rectangle, the x = 0 and the y = 0 is the Top Left cormer of the rotated rect, , and the TOP line of the rotated rect is "Angle" measurement, which is the x axis at degree = 0
it seems that your attempt may do the job?
to late tonight for me to run it, I'll set it up tomorrow, thanks
_TAD_ , I have used that type of calculation before, but in this method, I Do NOT have the rotated rectangle's "corner" co-ordinates after the rotation,
The data that I start with is the width and height of the Rotated Rectangle and the Angle of Rotation, so I would have to calculate the four corner x and y co-ordinates and then do a min and max from that, I do not need the co-ordinates, so I would rather not figure them if possible
OK, Gregory_Scot_Greer, I tried both your formulas (From 90 to 180, the point would be, , , , and the last post) for the OffSet and they Did NOT give me the correct palcement of the rotated Rect on the bitmap, not even close. . .
So I have had it, all I can take with trigonometry, (banging head on monitor) and I am out of time, , so I decided to just let the user drag and drop the rotated rect into the container. . .
I'd like to THANK every one for their time and comments. .
I will give the points to Gregory_Scot_Greer, since he was the only one to post a formula, and I am using the one he gave for the bounds
GwynforWeb, I can offer points for additional Info, but the accepted ansewer, seems to give me the width and height close enough to give me a workable container, but I did not have the time to do the trig and acually check for for acuarcy, if you want to know more about what I need, I use the windows
SetWorldTransform(hDC, XForm1);
to set the transformation for the device context
and the XForm1 is set like
XForm1.eM11 := Cos1;
XForm1.eM12 := -Sin1;
XForm1.eM21 := Sin1;
XForm1.eM22 := Cos1;
if Angle < 91 then
begin
XForm1.eDx := 0.0;
XForm1.eDy := Sin1 * Size1.cx;
end else
if Angle < 181 then
begin
XForm1.eDx := -(Cos1 * Size1.cx);
XForm1.eDy := -(Cos1 * Size1.cy);
end else
if Angle < 271 then
begin
XForm1.eDx := -(Cos1 * Size1.cx);
XForm1.eDy := 0.0;
end else
begin
XForm1.eDy := -(Cos1 * Size1.cx);
XForm1.eDx := (Cos1 * Size1.cy);
end;
but this only works in the first rotation quadrent
if Angle < 91 then
so I do not really use the placement that this provides
The variables Cos1 and Sin1 are the Sin and Cos values from the angle used
I do not have time or inclination now to spend very much more time on this, but if you have info that will work the first time I try it, I can give 200 points for a width and height formula, maybe more if you have anything for the SetWorldTransform( )
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by: grg99Posted on 2004-06-17 at 12:32:07ID: 11338097
Think of it this way: the width is going to be the cosine of the diagonal's angle, the height is the sine.
The diagonal's starting angle is arctan( height / width ).
That's about it!