How to convert a straight line into polar coordinates?
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I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format
edit: maybe I am not wording it clearly, this is the question word for word:
geometry polar-coordinates
edited Dec 1 '18 at 20:56
user376343
3,7883827
asked Nov 30 '18 at 17:25
dshawndshawn
324
$endgroup$
add a comment |
$begingroup$
I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format
edit: maybe I am not wording it clearly, this is the question word for word:
geometry polar-coordinates
edited Dec 1 '18 at 20:56
user376343
3,7883827
asked Nov 30 '18 at 17:25
dshawndshawn
324
$endgroup$
$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15
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thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02
2
$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16
$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57
add a comment |
$begingroup$
I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format
edit: maybe I am not wording it clearly, this is the question word for word:
geometry polar-coordinates
edited Dec 1 '18 at 20:56
user376343
3,7883827
asked Nov 30 '18 at 17:25
dshawndshawn
324
$endgroup$
I'm trying to go through a simple exercise for the Hough transform where I have a simple straight line in the form of $;y=-x+5;$ and I want to obtain polar coordinates $;(rho,theta);$, I know polar coordinates can be represented by $;rho = x⋅cos(theta) + y⋅sin(theta),;$ what are the steps I'm supposed to take to solve this problem? I have searched around and couldn't really find any examples I can follow in this exact format
edit: maybe I am not wording it clearly, this is the question word for word:
geometry polar-coordinates
geometry polar-coordinates
edited Dec 1 '18 at 20:56
user376343
3,7883827
asked Nov 30 '18 at 17:25
dshawndshawn
324
edited Dec 1 '18 at 20:56
user376343
3,7883827
asked Nov 30 '18 at 17:25
dshawndshawn
324
edited Dec 1 '18 at 20:56
user376343
3,7883827
edited Dec 1 '18 at 20:56
user376343
3,7883827
edited Dec 1 '18 at 20:56
user376343
3,7883827
3,7883827
asked Nov 30 '18 at 17:25
dshawndshawn
324
asked Nov 30 '18 at 17:25
dshawndshawn
324
asked Nov 30 '18 at 17:25
dshawndshawn
324
324
$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15
$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02
2
$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16
$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57
add a comment |
$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15
$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02
2
$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16
$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57
$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15
$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15
$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02
$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02
2
2
$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16
$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16
$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57
$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
After the definition of the Hough transform is
$rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$
$theta=frac pi4;$ is the angle between $x-$ axis and $OA.$
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
$endgroup$
add a comment |
$begingroup$
Using polar coordinates, a line is represented as
$$ax+by+c=arhocostheta+brhosintheta+c=0$$
or
$$rho=-frac c{acostheta+bsintheta}.$$
With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten
$$rho=frac p{cos(theta-theta_0)},$$
where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.
For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
$endgroup$
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
After the definition of the Hough transform is
$rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$
$theta=frac pi4;$ is the angle between $x-$ axis and $OA.$
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
$endgroup$
add a comment |
$begingroup$
After the definition of the Hough transform is
$rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$
$theta=frac pi4;$ is the angle between $x-$ axis and $OA.$
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
$endgroup$
add a comment |
$begingroup$
After the definition of the Hough transform is
$rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$
$theta=frac pi4;$ is the angle between $x-$ axis and $OA.$
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
$endgroup$
After the definition of the Hough transform is
$rho$ equal to the distance of the origin to the line. The nearest point on the line (to the origin) is $;P=left(frac{5}{2},frac{5}{2}right);$ so $rho=frac{5sqrt 2}{2}.$
$theta=frac pi4;$ is the angle between $x-$ axis and $OA.$
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
answered Dec 1 '18 at 21:17
user376343user376343
3,7883827
3,7883827
add a comment |
add a comment |
$begingroup$
Using polar coordinates, a line is represented as
$$ax+by+c=arhocostheta+brhosintheta+c=0$$
or
$$rho=-frac c{acostheta+bsintheta}.$$
With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten
$$rho=frac p{cos(theta-theta_0)},$$
where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
Using polar coordinates, a line is represented as
$$ax+by+c=arhocostheta+brhosintheta+c=0$$
or
$$rho=-frac c{acostheta+bsintheta}.$$
With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten
$$rho=frac p{cos(theta-theta_0)},$$
where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
Using polar coordinates, a line is represented as
$$ax+by+c=arhocostheta+brhosintheta+c=0$$
or
$$rho=-frac c{acostheta+bsintheta}.$$
With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten
$$rho=frac p{cos(theta-theta_0)},$$
where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
$endgroup$
Using polar coordinates, a line is represented as
$$ax+by+c=arhocostheta+brhosintheta+c=0$$
or
$$rho=-frac c{acostheta+bsintheta}.$$
With $theta_0:=tandfrac ba$ and $p:=-dfrac c{sqrt{a^2+b^2}}$, it can be rewritten
$$rho=frac p{cos(theta-theta_0)},$$
where $theta_0$ is the direction of the normal to the line, and $p$ the distance from the line to the origin.
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
edited Nov 30 '18 at 23:20
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
answered Nov 30 '18 at 23:13
Yves DaoustYves Daoust
127k673226
127k673226
add a comment |
add a comment |
$begingroup$
You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.
For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
$endgroup$
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
add a comment |
$begingroup$
You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.
For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
$endgroup$
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
add a comment |
$begingroup$
You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.
For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
$endgroup$
You can write any point $(x,y)$ on the line as $(rcos theta, rsin theta)$, where $r = sqrt{x^2+y^2}$ and $theta = tan^{-1}(y/x)$.
For example, consider the point $(4,3)$, which is on the line. You have $r = sqrt{4^2+3^2} = 5$ and $theta = tan^{-1}(3/4) = 0.6435$. This gives $cos theta = 0.8$ and $sin theta = 0.6$. You can see that $x = r cos theta$ and $y = r sin theta$.
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
edited Dec 2 '18 at 3:41
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
answered Nov 30 '18 at 17:39
Aditya DuaAditya Dua
1,16918
1,16918
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
add a comment |
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
is it $tan ^{-1}?$
$endgroup$
– user376343
Dec 1 '18 at 20:58
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
$begingroup$
@user376343 Yes, fixed it. Thanks,
$endgroup$
– Aditya Dua
Dec 2 '18 at 3:41
add a comment |
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$begingroup$
Are you sure that’s what $p$ is supposed to be?
$endgroup$
– amd
Nov 30 '18 at 20:15
$begingroup$
thats just the format, it says a equation that is y=mx+b can be expressed in polar coordinates like that equation. it says the hough transform of a line turns into a point and I need to find the point in (p,θ) space
$endgroup$
– dshawn
Nov 30 '18 at 23:02
2
$begingroup$
"I know polar coordinates can be represented by p=x⋅cos(θ)+y⋅sin(θ)": er, no.
$endgroup$
– Yves Daoust
Nov 30 '18 at 23:16
$begingroup$
Well the issue is this equation is given to me by the question, so I don't know am I misreading the question? I know the p is a greek letter and not a normal p. I will post the full question just in case
$endgroup$
– dshawn
Dec 1 '18 at 1:57