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?
trigonometry transformation rotations
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
$endgroup$
add a comment |
$begingroup$
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?
trigonometry transformation rotations
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
$endgroup$
add a comment |
$begingroup$
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?
trigonometry transformation rotations
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
$endgroup$
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?
trigonometry transformation rotations
trigonometry transformation rotations
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
asked Dec 17 '18 at 20:04
Andrius NaruševičiusAndrius Naruševičius
15315
15315
add a comment |
add a comment |
1 Answer
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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:
$$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$
$$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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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:
$$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$
$$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$
$$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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:
$$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$
$$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
$endgroup$
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:
$$x_D=x1+(x-x1)cdot cos alpha - (y+h-y1) cdot sin alpha$$
$$y_D=y1+(y+h-y1)cdot cos alpha + (x-x1) cdot sin alpha$$
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
edited Dec 17 '18 at 20:32
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
answered Dec 17 '18 at 20:20
Bram28Bram28
63.1k44793
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
add a comment |
$begingroup$
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
add a comment |
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