Thank.
Better. Seems that angles are close to but not equal to pi/2.
It may be because the plot box is not a square: the length of
x-axis is not the same as the length of y-axis.
Even curves y = x and y = 1-x look like not orthogonal but
they should since multiplication of their slopes is -1.
-james
plot(1:30,1:30,xlim=c(1,30),ylim=c(1,30),type="n", main = "Rotated
rectangle looks like a parallelogram", asp=1)
## a rect at (10,20) with w = 3 and h = 2
x = 10
y = 20
w = 3
h = 2
x1=x-w
y1=y-h
x2=x+w
y2=y-h
x3=x+w
y3=y+h
x4=x-w
y4=y+h
polygon(c(x1,x2,x3,x4),c(y1,y2,y3,y4),border="blue")
##Rotate it at (10,20) by 45 degree
theta = 45/180*pi
x.rotated = c(10 + (x1-10)*cos(theta)-(y1-20)*sin(theta),
10 + (x2-10)*cos(theta)-(y2-20)*sin(theta),
10 + (x3-10)*cos(theta)-(y3-20)*sin(theta),
10 + (x4-10)*cos(theta)-(y4-20)*sin(theta))
y.rotated = c(20 + (x1-10)*sin(theta)+(y1-20)*cos(theta),
20 + (x2-10)*sin(theta)+(y2-20)*cos(theta),
20 + (x3-10)*sin(theta)+(y3-20)*cos(theta),
20 + (x4-10)*sin(theta)+(y4-20)*cos(theta))
polygon(x.rotated,y.rotated,border="red")
> Your transformation assumes that the x- and y-axes are on the
> same scale. Add 'asp = 1' to your plot() call to set the
> appropriate aspect ratio.
>
> -Peter Ehlers
>
>
> On 2010-06-09 10:13, g...@ucalgary.ca wrote:
>> Rectangle R centered at (x,y) with width 2w and height 2h is given by
>>
>> x1=x-w
>> y1=y-h
>> x2=x+w
>> y2=y-h
>> x3=x+w
>> y3=y+h
>> x4=x-w
>> y4=y+h
>> polygon(c(x1,x2,x3,x4),c(y1,y2,y3,y4))
>>
>> Rotating a point (u,v) at (0,0) by theta degree is given by matrix
>> [cos(theta),-sin(theta)
>> sin(theta),cos(theta)]
>> so we have a new point
>> (u*cos(theta)-v*sin(theta),u*sin(theta)+v*cos(theta)).
>>
>> Hence rotated R by theta at (x,y) is given by
>>
>> x.rotated = c(x + (x1-x)*cos(theta)-(y1-y)*sin(theta),
>> x + (x2-x)*cos(theta)-(y2-y)*sin(theta),
>> x + (x3-x)*cos(theta)-(y3-y)*sin(theta),
>> x + (x4-x)*cos(theta)-(y4-y)*sin(theta))
>> y.rotated = c(y + (x1-x)*sin(theta)+(y1-y)*cos(theta),
>> y + (x2-x)*sin(theta)+(y2-y)*cos(theta),
>> y + (x3-x)*sin(theta)+(y3-y)*cos(theta),
>> y + (x4-x)*sin(theta)+(y4-y)*cos(theta))
>>
>> polygon(x.rotated,y.rotated)
>>
>> But it turns out to be a parallelogram with angles not equal to 90,
>> not a rectangle. See R code below.
>>
>> Any way to improve this so that the rotated rectangle looks like a
>> rectangle? Thanks,
>>
>> -james
>>
>>
>> plot(1:10,1:10,xlim=c(1,20),ylim=c(1,40),type="n", main = "Rotated
>> rectangle looks like a ")
>> ## a rect at (10,20) with w = 3 and h = 2
>> x = 10
>> y = 20
>> w = 3
>> h = 2
>> x1=x-w
>> y1=y-h
>> x2=x+w
>> y2=y-h
>> x3=x+w
>> y3=y+h
>> x4=x-w
>> y4=y+h
>> polygon(c(x1,x2,x3,x4),c(y1,y2,y3,y4),border="blue")
>>
>> ##Rotate it at (10,10) by 45 degree
>> theta = 45/180*pi
>> x.rotated = c(10 + (x1-10)*cos(theta)-(y1-20)*sin(theta),
>> 10 + (x2-10)*cos(theta)-(y2-20)*sin(theta),
>> 10 + (x3-10)*cos(theta)-(y3-20)*sin(theta),
>> 10 + (x4-10)*cos(theta)-(y4-20)*sin(theta))
>> y.rotated = c(20 + (x1-10)*sin(theta)+(y1-20)*cos(theta),
>> 20 + (x2-10)*sin(theta)+(y2-20)*cos(theta),
>> 20 + (x3-10)*sin(theta)+(y3-20)*cos(theta),
>> 20 + (x4-10)*sin(theta)+(y4-20)*cos(theta))
>>
>> polygon(x.rotated,y.rotated,border="red")
>>
>>
>>> On 06/04/2010 01:21 AM, g...@ucalgary.ca wrote:
>>>> boxed.labels draw text with box well.
>>>> But, the box cannot be shadowed and srt = 45 seems not to work:
>>>> text is rotated but the box does not.
>>>> polygon.shadow can rotate and shadow but have to calculate its
>>>> dimensions,
>>>> based on the text length and size.
>>>> Do you have any other way to draw text with rotated and shadowed box?
>>>
>>> The srt argument was intended to allow the user to rotate the text in
>>> 90
>>> degree increments, and the box just changes shape to fit whatever is
>>> in
>>> it. The underlying function that draws the box (rect) doesn't have a
>>> rotation argument. It would be possible to write a special function
>>> using polygon, just do the calculations for box size and then rotate
>>> the
>>> text with srt= and the polygon by transforming the coordinates of the
>>> vertices, as long as the default justification (center) is used. I
>>> can't
>>> do this right at the moment, but if you are really stuck I might be
>>> able
>>> to do it in the near future.
>>>
>>> Jim
>>>
>>>
>
>
>
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