$r = ycos(theta) - xsin(theta) $ derivation for Hough Transform
$begingroup$
I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.
Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.
Many thanks in advance
polar-coordinates parametric
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
$endgroup$
add a comment |
$begingroup$
I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.
Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.
Many thanks in advance
polar-coordinates parametric
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
$endgroup$
add a comment |
$begingroup$
I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.
Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.
Many thanks in advance
polar-coordinates parametric
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
$endgroup$
I am trying to see how $y = xtan(theta) + dfrac{r}{cos(theta)}$ is made from the graph.
Also how does the derivation work if $(x, y)$ is in the different quadrant? i.e. the $theta$ location stays the same.
Many thanks in advance
polar-coordinates parametric
polar-coordinates parametric
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
edited Jan 6 at 17:20
zcahfg2
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
asked Jan 6 at 17:09
zcahfg2zcahfg2
387213
387213
add a comment |
add a comment |
1 Answer
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$begingroup$
Starting with $y=mx+c$,
m is the slope of the line =$tan(theta)$
We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
So that
$$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
$$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
$$ c=r/cos(theta))$$
Replacing $m$ and $c$ in $y=mx+c$, you get
$$y=xtan(theta)+r/cos(theta)$$
The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).
answered Jan 6 at 18:46
Al-CAl-C
386
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Starting with $y=mx+c$,
m is the slope of the line =$tan(theta)$
We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
So that
$$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
$$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
$$ c=r/cos(theta))$$
Replacing $m$ and $c$ in $y=mx+c$, you get
$$y=xtan(theta)+r/cos(theta)$$
The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).
answered Jan 6 at 18:46
Al-CAl-C
386
$endgroup$
add a comment |
$begingroup$
Starting with $y=mx+c$,
m is the slope of the line =$tan(theta)$
We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
So that
$$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
$$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
$$ c=r/cos(theta))$$
Replacing $m$ and $c$ in $y=mx+c$, you get
$$y=xtan(theta)+r/cos(theta)$$
The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).
answered Jan 6 at 18:46
Al-CAl-C
386
$endgroup$
add a comment |
$begingroup$
Starting with $y=mx+c$,
m is the slope of the line =$tan(theta)$
We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
So that
$$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
$$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
$$ c=r/cos(theta))$$
Replacing $m$ and $c$ in $y=mx+c$, you get
$$y=xtan(theta)+r/cos(theta)$$
The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).
answered Jan 6 at 18:46
Al-CAl-C
386
$endgroup$
Starting with $y=mx+c$,
m is the slope of the line =$tan(theta)$
We also have, with $rge0$: $y=rcos(theta)$ and $x=-rsin(theta)$
So that
$$ rcos(theta)= tan(theta)(-rsin(theta))+c$$
$$ c=r(cos(theta)+sin^2(theta)/cos(theta))$$
$$ c=r/cos(theta))$$
Replacing $m$ and $c$ in $y=mx+c$, you get
$$y=xtan(theta)+r/cos(theta)$$
The parameter $theta$ in the range $(0,pi)$, measured from the plus side of the $x$ axis. In this derivation, $r$ will be positive (when located on the minus side of the $x$ axis) or negative (when located on the plus side of the $x$ axis).
answered Jan 6 at 18:46
Al-CAl-C
386
answered Jan 6 at 18:46
Al-CAl-C
386
answered Jan 6 at 18:46
Al-CAl-C
386
answered Jan 6 at 18:46
Al-CAl-C
386
386
add a comment |
add a comment |
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