Universal formula to calculate rotating by angle












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I am generating roads and buildings that belong to them and since I want the streets to be rotated and then connected with each other, I need to rotate both them and their respective buildings. Is there a formula that would be able to translate (for example) my (x,y) point into point A given the (x,y) as the topleft corner of a building, (x1,y1) as the start of the street and alpha being the angle I am rotating by?



enter image description here






trigonometry transformation rotations







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    0












    $begingroup$


    I am generating roads and buildings that belong to them and since I want the streets to be rotated and then connected with each other, I need to rotate both them and their respective buildings. Is there a formula that would be able to translate (for example) my (x,y) point into point A given the (x,y) as the topleft corner of a building, (x1,y1) as the start of the street and alpha being the angle I am rotating by?



    enter image description here






    trigonometry transformation rotations







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I am generating roads and buildings that belong to them and since I want the streets to be rotated and then connected with each other, I need to rotate both them and their respective buildings. Is there a formula that would be able to translate (for example) my (x,y) point into point A given the (x,y) as the topleft corner of a building, (x1,y1) as the start of the street and alpha being the angle I am rotating by?



      enter image description here






      trigonometry transformation rotations







      share|cite|improve this question









      $endgroup$




      I am generating roads and buildings that belong to them and since I want the streets to be rotated and then connected with each other, I need to rotate both them and their respective buildings. Is there a formula that would be able to translate (for example) my (x,y) point into point A given the (x,y) as the topleft corner of a building, (x1,y1) as the start of the street and alpha being the angle I am rotating by?



      enter image description here






      trigonometry transformation rotations




      trigonometry transformation rotations






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 17 '18 at 20:04









      Andrius NaruševičiusAndrius Naruševičius

      15315




      15315






















          1 Answer
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          $begingroup$

          Yes, there is:



          First, typically in trig we assume $y$ increases when you go up on paper, but you have that flipped. But that's ok, since the angle is typically measured counter-clockwise, while your $alpha$ is measured clock-wise, so the math will actually work out just the same.



          That is, in general, if you have a point that is at $(x,y)$, when assuming the rotation point is the origin $(0,0)$, then if you rotate by $alpha$ degrees, then the new point will be at $(x',y')$, where:



          $$x'=xcdot cos alpha - y cdot sin alpha$$



          $$y'=ycdot cos alpha + x cdot sin alpha$$



          Now, applied to your case, since your point of rotation is not $(0,0)$, but is $(x1,y1)$, we first need to find the points relative to the rotation point $(x1,y1)$, so we need to shift the coordinate system by subtracting$1$ and $y1$. For example your original point $(x,y)$ would be considered at $(x,y)_S=(x-x1,y-y1)$, assuming $(x1,y1)$ is at $(0,0)$.



          $$x_A=(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          OK, but again this is relative to taking $(x1,y1)$ as the origin, so to get the $x$ and $y$ back relative to your coordinate system, we need to add $x1$ and $y1$ back, i.e. we get:



          $$x_A=x1+(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=y1+(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          As another example, point $D$ ends up at:



          $$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$



          $$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
            $endgroup$
            – Andrius Naruševičius
            Dec 17 '18 at 20:45










          • $begingroup$
            @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
            $endgroup$
            – Bram28
            Dec 17 '18 at 20:50











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          1 Answer
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          1 Answer
          1






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          active

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          active

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          1












          $begingroup$

          Yes, there is:



          First, typically in trig we assume $y$ increases when you go up on paper, but you have that flipped. But that's ok, since the angle is typically measured counter-clockwise, while your $alpha$ is measured clock-wise, so the math will actually work out just the same.



          That is, in general, if you have a point that is at $(x,y)$, when assuming the rotation point is the origin $(0,0)$, then if you rotate by $alpha$ degrees, then the new point will be at $(x',y')$, where:



          $$x'=xcdot cos alpha - y cdot sin alpha$$



          $$y'=ycdot cos alpha + x cdot sin alpha$$



          Now, applied to your case, since your point of rotation is not $(0,0)$, but is $(x1,y1)$, we first need to find the points relative to the rotation point $(x1,y1)$, so we need to shift the coordinate system by subtracting$1$ and $y1$. For example your original point $(x,y)$ would be considered at $(x,y)_S=(x-x1,y-y1)$, assuming $(x1,y1)$ is at $(0,0)$.



          $$x_A=(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          OK, but again this is relative to taking $(x1,y1)$ as the origin, so to get the $x$ and $y$ back relative to your coordinate system, we need to add $x1$ and $y1$ back, i.e. we get:



          $$x_A=x1+(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=y1+(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          As another example, point $D$ ends up at:



          $$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$



          $$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
            $endgroup$
            – Andrius Naruševičius
            Dec 17 '18 at 20:45










          • $begingroup$
            @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
            $endgroup$
            – Bram28
            Dec 17 '18 at 20:50
















          1












          $begingroup$

          Yes, there is:



          First, typically in trig we assume $y$ increases when you go up on paper, but you have that flipped. But that's ok, since the angle is typically measured counter-clockwise, while your $alpha$ is measured clock-wise, so the math will actually work out just the same.



          That is, in general, if you have a point that is at $(x,y)$, when assuming the rotation point is the origin $(0,0)$, then if you rotate by $alpha$ degrees, then the new point will be at $(x',y')$, where:



          $$x'=xcdot cos alpha - y cdot sin alpha$$



          $$y'=ycdot cos alpha + x cdot sin alpha$$



          Now, applied to your case, since your point of rotation is not $(0,0)$, but is $(x1,y1)$, we first need to find the points relative to the rotation point $(x1,y1)$, so we need to shift the coordinate system by subtracting$1$ and $y1$. For example your original point $(x,y)$ would be considered at $(x,y)_S=(x-x1,y-y1)$, assuming $(x1,y1)$ is at $(0,0)$.



          $$x_A=(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          OK, but again this is relative to taking $(x1,y1)$ as the origin, so to get the $x$ and $y$ back relative to your coordinate system, we need to add $x1$ and $y1$ back, i.e. we get:



          $$x_A=x1+(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=y1+(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          As another example, point $D$ ends up at:



          $$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$



          $$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
            $endgroup$
            – Andrius Naruševičius
            Dec 17 '18 at 20:45










          • $begingroup$
            @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
            $endgroup$
            – Bram28
            Dec 17 '18 at 20:50














          1












          1








          1





          $begingroup$

          Yes, there is:



          First, typically in trig we assume $y$ increases when you go up on paper, but you have that flipped. But that's ok, since the angle is typically measured counter-clockwise, while your $alpha$ is measured clock-wise, so the math will actually work out just the same.



          That is, in general, if you have a point that is at $(x,y)$, when assuming the rotation point is the origin $(0,0)$, then if you rotate by $alpha$ degrees, then the new point will be at $(x',y')$, where:



          $$x'=xcdot cos alpha - y cdot sin alpha$$



          $$y'=ycdot cos alpha + x cdot sin alpha$$



          Now, applied to your case, since your point of rotation is not $(0,0)$, but is $(x1,y1)$, we first need to find the points relative to the rotation point $(x1,y1)$, so we need to shift the coordinate system by subtracting$1$ and $y1$. For example your original point $(x,y)$ would be considered at $(x,y)_S=(x-x1,y-y1)$, assuming $(x1,y1)$ is at $(0,0)$.



          $$x_A=(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          OK, but again this is relative to taking $(x1,y1)$ as the origin, so to get the $x$ and $y$ back relative to your coordinate system, we need to add $x1$ and $y1$ back, i.e. we get:



          $$x_A=x1+(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=y1+(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          As another example, point $D$ ends up at:



          $$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$



          $$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$






          share|cite|improve this answer











          $endgroup$



          Yes, there is:



          First, typically in trig we assume $y$ increases when you go up on paper, but you have that flipped. But that's ok, since the angle is typically measured counter-clockwise, while your $alpha$ is measured clock-wise, so the math will actually work out just the same.



          That is, in general, if you have a point that is at $(x,y)$, when assuming the rotation point is the origin $(0,0)$, then if you rotate by $alpha$ degrees, then the new point will be at $(x',y')$, where:



          $$x'=xcdot cos alpha - y cdot sin alpha$$



          $$y'=ycdot cos alpha + x cdot sin alpha$$



          Now, applied to your case, since your point of rotation is not $(0,0)$, but is $(x1,y1)$, we first need to find the points relative to the rotation point $(x1,y1)$, so we need to shift the coordinate system by subtracting$1$ and $y1$. For example your original point $(x,y)$ would be considered at $(x,y)_S=(x-x1,y-y1)$, assuming $(x1,y1)$ is at $(0,0)$.



          $$x_A=(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          OK, but again this is relative to taking $(x1,y1)$ as the origin, so to get the $x$ and $y$ back relative to your coordinate system, we need to add $x1$ and $y1$ back, i.e. we get:



          $$x_A=x1+(x-x1)cdot cos alpha - (y-y1) cdot sin alpha$$



          $$y_A=y1+(y-y1)cdot cos alpha + (x-x1) cdot sin alpha$$



          As another example, point $D$ ends up at:



          $$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$



          $$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 17 '18 at 20:32

























          answered Dec 17 '18 at 20:20









          Bram28Bram28

          63.1k44793




          63.1k44793












          • $begingroup$
            This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
            $endgroup$
            – Andrius Naruševičius
            Dec 17 '18 at 20:45










          • $begingroup$
            @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
            $endgroup$
            – Bram28
            Dec 17 '18 at 20:50


















          • $begingroup$
            This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
            $endgroup$
            – Andrius Naruševičius
            Dec 17 '18 at 20:45










          • $begingroup$
            @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
            $endgroup$
            – Bram28
            Dec 17 '18 at 20:50
















          $begingroup$
          This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
          $endgroup$
          – Andrius Naruševičius
          Dec 17 '18 at 20:45




          $begingroup$
          This is amazing, I actually managed to make it work from first try: imgur.com/a/HQZonnK Thank you so much!!!
          $endgroup$
          – Andrius Naruševičius
          Dec 17 '18 at 20:45












          $begingroup$
          @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
          $endgroup$
          – Bram28
          Dec 17 '18 at 20:50




          $begingroup$
          @AndriusNaruševičius Ha! I was afraid (and in fact fully expected) that I forgot a negation somewhere, for that's so easy to do ... so it's surprising to me too it works on the first try! :) Thanks for your feedback and glad I could help!
          $endgroup$
          – Bram28
          Dec 17 '18 at 20:50


















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