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# Trigonometry, Rotation

Hi experts,

I'm looking for some help with trigonometry, more specifically rotating 4 points around a centre point.

I have a rectangle,  with centre point 0,0 and want to rotate this counter-clockwise a variable number of degrees.  Say 45 degrees for example.  The 4 points on the rectangle are 6,3 : -6,3 : -6,-3 : 6,3.

I know this involves sin and cos but I cannot figure out how to use these functions.

So, simply put - how do i work out the new coordinates for a rotation of 45 degrees for the above coordinates.

Any help would be appreciated.

Eoin.
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inema
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1 Solution

Commented:
New_Y = Old_Y * Sin ( angle )
New_X = Old_X * Cos ( angle )
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Commented:
So, rotating the point (6,3) CCW by 45°, would be result in the point:

( 6*Cos(45), 3*Sin(45) )  =~  ( 4.2, 2.12 )
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Commented:
Close but not quite, InteractiveMind.

New_Y = Old_Y * cos( angle ) + old_x * sin(angle)
New_X = Old_X * Cos ( angle ) - old_y*sin(angle)

Notice that when angle=0, the cos is 1 and the sin is 0, so new_x = old_x and new_y = old_y
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Author Commented:

Radius, from 0,0 to 6,3, using pythagoras' = 6.7
sqrt(6^2 + 3^2)
sqrt(45) = ~6.7

new_X = radius * cos (angle)
new_X = 6.7 * cos(45)
new_X = 2.1

new_Y = radius * sin (angle)
new_Y = 6.7 * sin(45)
new_Y = 6.4

so 6, 3 rotates to 2.1, 6.4 ?
Is this correct?

The other numbers are confusing me because of all this quadrant and negative number stuff, anyone have any ideas?

Thanks again,
Eoin
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Author Commented:

First i got the previous angle using Tan (angle) = Opposite / Adjacent

Tan (angle) = 3/6
Tan (angle) = 0.5
Angle = ~27 degrees

So new angle to work with is 27+45 = 72 degrees

I then used

Does this make sense? It looks correct on a poorly drawn excel spreadsheet i made :)
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Commented:
inema, my expressions work.

New_Y = Old_Y * cos + old_x * sin
New_X = Old_X * Cos - old_y*sin

new_x^2 + new_y^2 = (old_y*cos + old_x*sin)^2 + (old_x*cos - old_y*sin)^2
= old_y^2 * cos^2 + 2*old_x*old_y*cos*sin + old_x^2*sin^2 + old_x^2*cos^2 - 2*old_x*old_y*cos*sin + old_y^2*sin^2
= old_y^2*(cos^2 + sin^2) + (2 - 2)*(old_x*old_y*cos_sin) + old_x*(cos^2 + sin^2)
new_x^2 + new_y^2 = old_y^2 + old_x^2
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Freelance programmer / ConsultantCommented:
You ought to have a look at polar and cartesian co-ordinates.  At present you are in cartesian co-ordinates (point specified by distance on x and y axes).  Polar co-ordinates are the distance and angle from a defined point/axis.  To rotate in polar co-ordinates about the center point all you do is add the angles together.

eg.
if the point is 2.5, 30 degrees after rotating by 45 degrees the point is now 2.5, 75 degrees.
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Author Commented:
I used your expression for the top right point (6,3) and it worked.  However when I try the same expression for negative numbers the results are off.

I read a bit about quadrants and that they change positives and negatives depending on which one you are working in.  Can anyone elaborate?

AndyAinscow,
Thanks for your input, however I don't think polar co-ordinates will be beneficial as I have a cartesian map to work with - it might be one unneccessary complication too many ;)
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Freelance programmer / ConsultantCommented:
Two functions - one converts cartesian to polar and one converts polar to cartesian
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Author Commented:
okay, after doing some research on the polar/cartesian conversions i've found that this is what i was doing with my original calculations.

Thanks for the help guys.  Points go to Andy.
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