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Can a spiral have its centroid at the origin?
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A spiral is a curve $gamma$ with the polar equation $r=f(theta)$ where $f$ is a continuous positive strictly monotone function on some interval $[a, b]$, $-infty<a<b<infty$. Best known examples are the logarithmic spiral and the Archimedean spiral.
Problem: Find a spiral whose centroid is the origin of the coordinate system.
Progress so far: We want $$int_gamma x,ds = int_gamma y ,ds = 0 tag1$$
Note that $x = f(theta)costheta$, $y = f(theta)sintheta$, and $ds = sqrt(f'(theta))^2 + f(theta)^2,dtheta$. Thus, we need the function
$$g(theta) = f(theta) sqrt(f'(theta))^2 + f(theta)^2 $$
to be orthogonal to both $cos theta$ and $sintheta$ on the interval $[a, b]$, meaning $$int_a^b g(theta)costheta,dtheta = int_a^b g(theta)sintheta,dtheta = 0tag2$$ A natural way to satisfy (2) is to take $[a, b] = [0, 2pi]$ and $g$ to be constant (say $gequiv 1$ as scaling does not matter). However this fails, because solving the equation $gequiv 1$ for $f$ (as an autonomous ODE) yields $f(theta) = sqrtsin 2theta$ (up to a shift), which is not even defined, let alone monotone, on any interval of length $2pi$.
Note: It is not required for $[a, b]$ to have length $2pi$ or a multiple of $2pi$; it can be any nontrivial finite interval.
multivariable-calculus polar-coordinates plane-curves centroid
asked Jul 25 at 19:23
user357151
13.8k31140
 |Â
show 2 more comments
up vote
7
down vote
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A spiral is a curve $gamma$ with the polar equation $r=f(theta)$ where $f$ is a continuous positive strictly monotone function on some interval $[a, b]$, $-infty<a<b<infty$. Best known examples are the logarithmic spiral and the Archimedean spiral.
Problem: Find a spiral whose centroid is the origin of the coordinate system.
Progress so far: We want $$int_gamma x,ds = int_gamma y ,ds = 0 tag1$$
Note that $x = f(theta)costheta$, $y = f(theta)sintheta$, and $ds = sqrt(f'(theta))^2 + f(theta)^2,dtheta$. Thus, we need the function
$$g(theta) = f(theta) sqrt(f'(theta))^2 + f(theta)^2 $$
to be orthogonal to both $cos theta$ and $sintheta$ on the interval $[a, b]$, meaning $$int_a^b g(theta)costheta,dtheta = int_a^b g(theta)sintheta,dtheta = 0tag2$$ A natural way to satisfy (2) is to take $[a, b] = [0, 2pi]$ and $g$ to be constant (say $gequiv 1$ as scaling does not matter). However this fails, because solving the equation $gequiv 1$ for $f$ (as an autonomous ODE) yields $f(theta) = sqrtsin 2theta$ (up to a shift), which is not even defined, let alone monotone, on any interval of length $2pi$.
Note: It is not required for $[a, b]$ to have length $2pi$ or a multiple of $2pi$; it can be any nontrivial finite interval.
multivariable-calculus polar-coordinates plane-curves centroid
asked Jul 25 at 19:23
user357151
13.8k31140
I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case.
– Mark Bennet
Jul 25 at 19:39
1
@MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.)
– Brian Tung
Jul 25 at 19:41
3
For $[a,b] = [0,2pi]$ (and likewise for multiples of $2pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $theta$ between $0$ and $pi$, the spiral point corresponding to $theta+pi$ is farther below the $x$-axis than the point corresponding to $theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $theta$ in $(pi,2pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider.
– Blue
Jul 25 at 19:42
(continued) Note that your spirals are inherently "left heavy" over the interval $(0,pi)$. Therefore, if the vertical-balancing $theta$ (from my previous comment) is between $pi$ and $3pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant.
– Blue
Jul 25 at 19:53
1
@Blue That is the intuition I was trying to capture.
– Mark Bennet
Jul 25 at 20:11
 |Â
show 2 more comments
up vote
7
down vote
favorite
up vote
7
down vote
favorite
A spiral is a curve $gamma$ with the polar equation $r=f(theta)$ where $f$ is a continuous positive strictly monotone function on some interval $[a, b]$, $-infty<a<b<infty$. Best known examples are the logarithmic spiral and the Archimedean spiral.
Problem: Find a spiral whose centroid is the origin of the coordinate system.
Progress so far: We want $$int_gamma x,ds = int_gamma y ,ds = 0 tag1$$
Note that $x = f(theta)costheta$, $y = f(theta)sintheta$, and $ds = sqrt(f'(theta))^2 + f(theta)^2,dtheta$. Thus, we need the function
$$g(theta) = f(theta) sqrt(f'(theta))^2 + f(theta)^2 $$
to be orthogonal to both $cos theta$ and $sintheta$ on the interval $[a, b]$, meaning $$int_a^b g(theta)costheta,dtheta = int_a^b g(theta)sintheta,dtheta = 0tag2$$ A natural way to satisfy (2) is to take $[a, b] = [0, 2pi]$ and $g$ to be constant (say $gequiv 1$ as scaling does not matter). However this fails, because solving the equation $gequiv 1$ for $f$ (as an autonomous ODE) yields $f(theta) = sqrtsin 2theta$ (up to a shift), which is not even defined, let alone monotone, on any interval of length $2pi$.
Note: It is not required for $[a, b]$ to have length $2pi$ or a multiple of $2pi$; it can be any nontrivial finite interval.
multivariable-calculus polar-coordinates plane-curves centroid
asked Jul 25 at 19:23
user357151
13.8k31140
A spiral is a curve $gamma$ with the polar equation $r=f(theta)$ where $f$ is a continuous positive strictly monotone function on some interval $[a, b]$, $-infty<a<b<infty$. Best known examples are the logarithmic spiral and the Archimedean spiral.
Problem: Find a spiral whose centroid is the origin of the coordinate system.
Progress so far: We want $$int_gamma x,ds = int_gamma y ,ds = 0 tag1$$
Note that $x = f(theta)costheta$, $y = f(theta)sintheta$, and $ds = sqrt(f'(theta))^2 + f(theta)^2,dtheta$. Thus, we need the function
$$g(theta) = f(theta) sqrt(f'(theta))^2 + f(theta)^2 $$
to be orthogonal to both $cos theta$ and $sintheta$ on the interval $[a, b]$, meaning $$int_a^b g(theta)costheta,dtheta = int_a^b g(theta)sintheta,dtheta = 0tag2$$ A natural way to satisfy (2) is to take $[a, b] = [0, 2pi]$ and $g$ to be constant (say $gequiv 1$ as scaling does not matter). However this fails, because solving the equation $gequiv 1$ for $f$ (as an autonomous ODE) yields $f(theta) = sqrtsin 2theta$ (up to a shift), which is not even defined, let alone monotone, on any interval of length $2pi$.
Note: It is not required for $[a, b]$ to have length $2pi$ or a multiple of $2pi$; it can be any nontrivial finite interval.
multivariable-calculus polar-coordinates plane-curves centroid
asked Jul 25 at 19:23
user357151
13.8k31140
edited Jul 25 at 21:21
asked Jul 25 at 19:23
user357151
13.8k31140
asked Jul 25 at 19:23
user357151
13.8k31140
asked Jul 25 at 19:23
user357151
13.8k31140
13.8k31140
I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case.
– Mark Bennet
Jul 25 at 19:39
1
@MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.)
– Brian Tung
Jul 25 at 19:41
3
For $[a,b] = [0,2pi]$ (and likewise for multiples of $2pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $theta$ between $0$ and $pi$, the spiral point corresponding to $theta+pi$ is farther below the $x$-axis than the point corresponding to $theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $theta$ in $(pi,2pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider.
– Blue
Jul 25 at 19:42
(continued) Note that your spirals are inherently "left heavy" over the interval $(0,pi)$. Therefore, if the vertical-balancing $theta$ (from my previous comment) is between $pi$ and $3pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant.
– Blue
Jul 25 at 19:53
1
@Blue That is the intuition I was trying to capture.
– Mark Bennet
Jul 25 at 20:11
 |Â
show 2 more comments
I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case.
– Mark Bennet
Jul 25 at 19:39
1
@MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.)
– Brian Tung
Jul 25 at 19:41
3
For $[a,b] = [0,2pi]$ (and likewise for multiples of $2pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $theta$ between $0$ and $pi$, the spiral point corresponding to $theta+pi$ is farther below the $x$-axis than the point corresponding to $theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $theta$ in $(pi,2pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider.
– Blue
Jul 25 at 19:42
(continued) Note that your spirals are inherently "left heavy" over the interval $(0,pi)$. Therefore, if the vertical-balancing $theta$ (from my previous comment) is between $pi$ and $3pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant.
– Blue
Jul 25 at 19:53
1
@Blue That is the intuition I was trying to capture.
– Mark Bennet
Jul 25 at 20:11
I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case.
– Mark Bennet
Jul 25 at 19:39
I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case.
– Mark Bennet
Jul 25 at 19:39
1
1
@MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.)
– Brian Tung
Jul 25 at 19:41
@MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.)
– Brian Tung
Jul 25 at 19:41
3
3
For $[a,b] = [0,2pi]$ (and likewise for multiples of $2pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $theta$ between $0$ and $pi$, the spiral point corresponding to $theta+pi$ is farther below the $x$-axis than the point corresponding to $theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $theta$ in $(pi,2pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider.
– Blue
Jul 25 at 19:42
For $[a,b] = [0,2pi]$ (and likewise for multiples of $2pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $theta$ between $0$ and $pi$, the spiral point corresponding to $theta+pi$ is farther below the $x$-axis than the point corresponding to $theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $theta$ in $(pi,2pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider.
– Blue
Jul 25 at 19:42
(continued) Note that your spirals are inherently "left heavy" over the interval $(0,pi)$. Therefore, if the vertical-balancing $theta$ (from my previous comment) is between $pi$ and $3pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant.
– Blue
Jul 25 at 19:53
(continued) Note that your spirals are inherently "left heavy" over the interval $(0,pi)$. Therefore, if the vertical-balancing $theta$ (from my previous comment) is between $pi$ and $3pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant.
– Blue
Jul 25 at 19:53
1
1
@Blue That is the intuition I was trying to capture.
– Mark Bennet
Jul 25 at 20:11
@Blue That is the intuition I was trying to capture.
– Mark Bennet
Jul 25 at 20:11
 |Â
show 2 more comments
1 Answer
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I can find an approximation which convinces me that one could find (numerically) one that satisfies your conditions. My first cut at a "spiral" is drawn below. It consists of the line segment from the origin to $(2,0)$, an arc of a circle from $(2,0)$ to $(2 cos theta, 2sin theta)$ and a line segment tangent to the circle of length $L$. I find that for $theta = 4.95, Lapprox 0.786$ the $y$ centroid is correct and the $x$ centroid is very slightly right of center. Shortening the initial segment can move the $x$ centroid back without impacting $y$.
I realize my straight line and circle do not meet the monotonic increase of $r$ with $theta$ but they were easy to calculate and we can approach them arbitrarily closely.
answered Jul 26 at 4:29
Ross Millikan
275k21186351
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1 Answer
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I can find an approximation which convinces me that one could find (numerically) one that satisfies your conditions. My first cut at a "spiral" is drawn below. It consists of the line segment from the origin to $(2,0)$, an arc of a circle from $(2,0)$ to $(2 cos theta, 2sin theta)$ and a line segment tangent to the circle of length $L$. I find that for $theta = 4.95, Lapprox 0.786$ the $y$ centroid is correct and the $x$ centroid is very slightly right of center. Shortening the initial segment can move the $x$ centroid back without impacting $y$.
I realize my straight line and circle do not meet the monotonic increase of $r$ with $theta$ but they were easy to calculate and we can approach them arbitrarily closely.
answered Jul 26 at 4:29
Ross Millikan
275k21186351
add a comment |Â
up vote
0
down vote
I can find an approximation which convinces me that one could find (numerically) one that satisfies your conditions. My first cut at a "spiral" is drawn below. It consists of the line segment from the origin to $(2,0)$, an arc of a circle from $(2,0)$ to $(2 cos theta, 2sin theta)$ and a line segment tangent to the circle of length $L$. I find that for $theta = 4.95, Lapprox 0.786$ the $y$ centroid is correct and the $x$ centroid is very slightly right of center. Shortening the initial segment can move the $x$ centroid back without impacting $y$.
I realize my straight line and circle do not meet the monotonic increase of $r$ with $theta$ but they were easy to calculate and we can approach them arbitrarily closely.
answered Jul 26 at 4:29
Ross Millikan
275k21186351
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I can find an approximation which convinces me that one could find (numerically) one that satisfies your conditions. My first cut at a "spiral" is drawn below. It consists of the line segment from the origin to $(2,0)$, an arc of a circle from $(2,0)$ to $(2 cos theta, 2sin theta)$ and a line segment tangent to the circle of length $L$. I find that for $theta = 4.95, Lapprox 0.786$ the $y$ centroid is correct and the $x$ centroid is very slightly right of center. Shortening the initial segment can move the $x$ centroid back without impacting $y$.
I realize my straight line and circle do not meet the monotonic increase of $r$ with $theta$ but they were easy to calculate and we can approach them arbitrarily closely.
answered Jul 26 at 4:29
Ross Millikan
275k21186351
I can find an approximation which convinces me that one could find (numerically) one that satisfies your conditions. My first cut at a "spiral" is drawn below. It consists of the line segment from the origin to $(2,0)$, an arc of a circle from $(2,0)$ to $(2 cos theta, 2sin theta)$ and a line segment tangent to the circle of length $L$. I find that for $theta = 4.95, Lapprox 0.786$ the $y$ centroid is correct and the $x$ centroid is very slightly right of center. Shortening the initial segment can move the $x$ centroid back without impacting $y$.
I realize my straight line and circle do not meet the monotonic increase of $r$ with $theta$ but they were easy to calculate and we can approach them arbitrarily closely.
answered Jul 26 at 4:29
Ross Millikan
275k21186351
answered Jul 26 at 4:29
Ross Millikan
275k21186351
answered Jul 26 at 4:29
Ross Millikan
275k21186351
answered Jul 26 at 4:29
Ross Millikan
275k21186351
275k21186351
add a comment |Â
add a comment |Â
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I would observe that a trivial spiral with a single point at the origin satisfies the conditions - you might want to exclude this case.
– Mark Bennet
Jul 25 at 19:39
1
@MarkBennet: Why would the centroid of a single point not be at that point? (ETA: I think it's a reasonable program for addressing the question at any rate.)
– Brian Tung
Jul 25 at 19:41
3
For $[a,b] = [0,2pi]$ (and likewise for multiples of $2pi$), no spiral fitting your definition can have its centroid on the $x$-axis: For each $theta$ between $0$ and $pi$, the spiral point corresponding to $theta+pi$ is farther below the $x$-axis than the point corresponding to $theta$ is above the axis. Thus, over such intervals, the spiral is inherently "bottom heavy". Since you allow arbitrary intervals, however, you can take any spiral you like and find the (unique) $theta$ in $(pi,2pi)$ at which vertical balance is achieved. Of course, there's also the $y$-axis to consider.
– Blue
Jul 25 at 19:42
(continued) Note that your spirals are inherently "left heavy" over the interval $(0,pi)$. Therefore, if the vertical-balancing $theta$ (from my previous comment) is between $pi$ and $3pi/2$, then the lower-left portion only contributes more "left-ness", so that the centroid cannot be on the $y$-axis. Thus, you seek, more-specifically, a spiral that ends somewhere in the Fourth Quadrant.
– Blue
Jul 25 at 19:53
1
@Blue That is the intuition I was trying to capture.
– Mark Bennet
Jul 25 at 20:11