Quick ways of finding principle curvature in the Cartesian plane
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Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?
What about the mean curvature?
polar-coordinates curvature
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
$endgroup$
add a comment |
$begingroup$
Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?
What about the mean curvature?
polar-coordinates curvature
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
$endgroup$
1
$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04
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I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46
$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47
add a comment |
$begingroup$
Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?
What about the mean curvature?
polar-coordinates curvature
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
$endgroup$
Say I have some polar curve $r=f(theta)$ in the Cartesian plane (smooth and twice-differential) and I've found a formula for the curvature, $k$. Does a clever trick exist (that isn't merely finding the maxima and minima of $k$) that allows us to find the principle curvature of our curve?
What about the mean curvature?
polar-coordinates curvature
polar-coordinates curvature
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
asked Dec 17 '18 at 21:01
Jonathan LowJonathan Low
635
635
1
$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04
$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46
$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47
add a comment |
1
$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04
$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46
$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47
1
1
$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04
$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04
$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46
$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46
$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47
$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.
Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
So mean curvature is $k_a=2(pi/2)/lapprox0.41$
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
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add a comment |
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1 Answer
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$begingroup$
Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.
Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
So mean curvature is $k_a=2(pi/2)/lapprox0.41$
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
$endgroup$
add a comment |
$begingroup$
Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.
Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
So mean curvature is $k_a=2(pi/2)/lapprox0.41$
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
$endgroup$
add a comment |
$begingroup$
Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.
Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
So mean curvature is $k_a=2(pi/2)/lapprox0.41$
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
$endgroup$
Since the curvature $k=dtheta/dl$, then average curvature is $k_a=Deltatheta/Delta l$. So you break you curve into segments at the points where $k=0$ or $k=infty$. For each segment you calculate $|Deltatheta|$ — angle the tangent of the curve rotated. You sum all $|Delta theta|$ and divide by total length of the curve.
Example. Let's calculate the average curvature for $y=sin x$ from $0$ to $2pi$. Curvature is zero for $x=pi$. So we consider two segments. At first segment the tangent rotated from $theta_0=pi/4$ to $theta_pi=-pi/4$, so $|Deltatheta_1|=pi/2$. It's the same for the second segment $|Deltatheta_2|=pi/2$. The length of the curve is $$l=int_0^{2pi}sqrt{1+cos^2x}dx=4sqrt{2}E(1/2)approx7.64$$
So mean curvature is $k_a=2(pi/2)/lapprox0.41$
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
answered Dec 17 '18 at 22:19
Vasily MitchVasily Mitch
2,6691312
2,6691312
add a comment |
add a comment |
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$begingroup$
What do you mean by principle curvature? Principle curvature exists for n-dimensional surfaces where n>2. One dimensional curves have only one curvature at given point.
$endgroup$
– Vasily Mitch
Dec 17 '18 at 21:04
$begingroup$
I'm sorry, I think I've misunderstood the concept - I thought it meant where curvature is maximal - I was wrong. Sorry about that.
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:46
$begingroup$
However, does there happen to be a way to find the average curvature for a 2-dimensional graph? - or a quick way of finding maxima and minima?
$endgroup$
– Jonathan Low
Dec 17 '18 at 21:47