Mixing
Determining the angle degree of an arc in ellipse?
$begingroup$
Is it possible to determine the angle in degree of an arc in ellipse by knowing the arc length, ellipse semi-major and semi-minor axis ?
If I have an arc length at the first quarter of an ellipse and I want to know the angle of it, what is the data that I will need it to use it and what is the exact method to use it?
Please take a look at this picture:
Actually I know how to determine the arc length of a ellipse here. but I want to do the obverse.
geometry differential-geometry circles conic-sections
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
$endgroup$
add a comment |
$begingroup$
Is it possible to determine the angle in degree of an arc in ellipse by knowing the arc length, ellipse semi-major and semi-minor axis ?
If I have an arc length at the first quarter of an ellipse and I want to know the angle of it, what is the data that I will need it to use it and what is the exact method to use it?
Please take a look at this picture:
Actually I know how to determine the arc length of a ellipse here. but I want to do the obverse.
geometry differential-geometry circles conic-sections
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
$endgroup$
1
$begingroup$
Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral.
$endgroup$
– robjohn♦
Jul 4 '13 at 13:14
add a comment |
$begingroup$
Is it possible to determine the angle in degree of an arc in ellipse by knowing the arc length, ellipse semi-major and semi-minor axis ?
If I have an arc length at the first quarter of an ellipse and I want to know the angle of it, what is the data that I will need it to use it and what is the exact method to use it?
Please take a look at this picture:
Actually I know how to determine the arc length of a ellipse here. but I want to do the obverse.
geometry differential-geometry circles conic-sections
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
$endgroup$
Is it possible to determine the angle in degree of an arc in ellipse by knowing the arc length, ellipse semi-major and semi-minor axis ?
If I have an arc length at the first quarter of an ellipse and I want to know the angle of it, what is the data that I will need it to use it and what is the exact method to use it?
Please take a look at this picture:
Actually I know how to determine the arc length of a ellipse here. but I want to do the obverse.
geometry differential-geometry circles conic-sections
geometry differential-geometry circles conic-sections
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
edited Apr 13 '17 at 12:20
Community♦
1
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
asked Jul 4 '13 at 13:10
Mohammad FakhreyMohammad Fakhrey
60231120
60231120
1
$begingroup$
Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral.
$endgroup$
– robjohn♦
Jul 4 '13 at 13:14
add a comment |
1
$begingroup$
Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral.
$endgroup$
– robjohn♦
Jul 4 '13 at 13:14
1
1
$begingroup$
Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral.
$endgroup$
– robjohn♦
Jul 4 '13 at 13:14
$begingroup$
Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral.
$endgroup$
– robjohn♦
Jul 4 '13 at 13:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Parameterization of an ellipse by angle from the center is
$$
gamma(phi)=(cos(phi),sin(phi))frac{ab}{sqrt{a^2sin^2(phi)+b^2cos^2(phi)}}
$$
and
$$
left|gamma'(phi)right|=absqrt{frac{a^4sin^2(phi)+b^4cos^2(phi)}{left(a^2sin^2(phi)+b^2cos^2(phi)right)^3}}
$$
and integrating $left|gamma'(phi)right|$ gets extremely messy.
So instead, we use $theta$, where
$$
btan(theta)=atan(phi)
$$
Then
$$
gamma(theta)=(acos(theta),bsin(theta))
$$
and
$$
left|gamma'(theta)right|=sqrt{a^2sin^2(theta)+b^2cos^2(theta)}
$$
Now, integrating $left|gamma'(theta)right|$ is a lot simpler.
$$
intleft|gamma'(theta)right|,mathrm{d}theta
=b,mathrm{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)
$$
However, to go from from arc length to angle, we still need to invert the Elliptic integral.
Solution of the problem given using Mathematica
Here we get the solution to 30 places.
In= Phi[a_,b_,s_,opts:OptionsPattern] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]
In= Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]
Out= 0.87052028743193111752524449959
Thus, $phidoteq0.87052028743193111752524449959text{ radians}$.
Let me describe the algorithm a bit.
FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]
inverts the elliptic integral to get $theta$ from $a$, $b$, and $s$.
Now, we want to find $phi$ so that $btan(theta)=atan(phi)$. However, simply using $tan^{-1}left(frac batan(theta)right)$ will only return a result in $left(-fracpi2,fracpi2right)$. To get the correct value, we use the relation
$$
begin{align}
tan(phi-theta)
&=frac{tan(phi)-tan(theta)}{1+tan(phi)tan(theta)}\
&=frac{frac batan(theta)-tan(theta)}{1+frac batan(theta)tan(theta)}\
&=frac{btan(theta)-atan(theta)}{a+btan(theta)tan(theta)}\
phi
&=theta+tan^{-1}left(frac{(b-a)tan(theta)}{a+btan^2(theta)}right)
end{align}
$$
This is why we have
t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
$endgroup$
1
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
1
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
add a comment |
$begingroup$
The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b,E(theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumference. Also, for those more interested readers, it is easy to show that $b,E(theta,1-a^2/b^2) = a,E(theta,1-b^2/a^2)$, as found at Wolfram World webpage http://mathworld.wolfram.com/Ellipse.html. This change rule is useful for avoiding negative values of parameters. F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
$endgroup$
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Parameterization of an ellipse by angle from the center is
$$
gamma(phi)=(cos(phi),sin(phi))frac{ab}{sqrt{a^2sin^2(phi)+b^2cos^2(phi)}}
$$
and
$$
left|gamma'(phi)right|=absqrt{frac{a^4sin^2(phi)+b^4cos^2(phi)}{left(a^2sin^2(phi)+b^2cos^2(phi)right)^3}}
$$
and integrating $left|gamma'(phi)right|$ gets extremely messy.
So instead, we use $theta$, where
$$
btan(theta)=atan(phi)
$$
Then
$$
gamma(theta)=(acos(theta),bsin(theta))
$$
and
$$
left|gamma'(theta)right|=sqrt{a^2sin^2(theta)+b^2cos^2(theta)}
$$
Now, integrating $left|gamma'(theta)right|$ is a lot simpler.
$$
intleft|gamma'(theta)right|,mathrm{d}theta
=b,mathrm{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)
$$
However, to go from from arc length to angle, we still need to invert the Elliptic integral.
Solution of the problem given using Mathematica
Here we get the solution to 30 places.
In= Phi[a_,b_,s_,opts:OptionsPattern] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]
In= Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]
Out= 0.87052028743193111752524449959
Thus, $phidoteq0.87052028743193111752524449959text{ radians}$.
Let me describe the algorithm a bit.
FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]
inverts the elliptic integral to get $theta$ from $a$, $b$, and $s$.
Now, we want to find $phi$ so that $btan(theta)=atan(phi)$. However, simply using $tan^{-1}left(frac batan(theta)right)$ will only return a result in $left(-fracpi2,fracpi2right)$. To get the correct value, we use the relation
$$
begin{align}
tan(phi-theta)
&=frac{tan(phi)-tan(theta)}{1+tan(phi)tan(theta)}\
&=frac{frac batan(theta)-tan(theta)}{1+frac batan(theta)tan(theta)}\
&=frac{btan(theta)-atan(theta)}{a+btan(theta)tan(theta)}\
phi
&=theta+tan^{-1}left(frac{(b-a)tan(theta)}{a+btan^2(theta)}right)
end{align}
$$
This is why we have
t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
$endgroup$
1
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
1
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
add a comment |
$begingroup$
Parameterization of an ellipse by angle from the center is
$$
gamma(phi)=(cos(phi),sin(phi))frac{ab}{sqrt{a^2sin^2(phi)+b^2cos^2(phi)}}
$$
and
$$
left|gamma'(phi)right|=absqrt{frac{a^4sin^2(phi)+b^4cos^2(phi)}{left(a^2sin^2(phi)+b^2cos^2(phi)right)^3}}
$$
and integrating $left|gamma'(phi)right|$ gets extremely messy.
So instead, we use $theta$, where
$$
btan(theta)=atan(phi)
$$
Then
$$
gamma(theta)=(acos(theta),bsin(theta))
$$
and
$$
left|gamma'(theta)right|=sqrt{a^2sin^2(theta)+b^2cos^2(theta)}
$$
Now, integrating $left|gamma'(theta)right|$ is a lot simpler.
$$
intleft|gamma'(theta)right|,mathrm{d}theta
=b,mathrm{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)
$$
However, to go from from arc length to angle, we still need to invert the Elliptic integral.
Solution of the problem given using Mathematica
Here we get the solution to 30 places.
In= Phi[a_,b_,s_,opts:OptionsPattern] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]
In= Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]
Out= 0.87052028743193111752524449959
Thus, $phidoteq0.87052028743193111752524449959text{ radians}$.
Let me describe the algorithm a bit.
FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]
inverts the elliptic integral to get $theta$ from $a$, $b$, and $s$.
Now, we want to find $phi$ so that $btan(theta)=atan(phi)$. However, simply using $tan^{-1}left(frac batan(theta)right)$ will only return a result in $left(-fracpi2,fracpi2right)$. To get the correct value, we use the relation
$$
begin{align}
tan(phi-theta)
&=frac{tan(phi)-tan(theta)}{1+tan(phi)tan(theta)}\
&=frac{frac batan(theta)-tan(theta)}{1+frac batan(theta)tan(theta)}\
&=frac{btan(theta)-atan(theta)}{a+btan(theta)tan(theta)}\
phi
&=theta+tan^{-1}left(frac{(b-a)tan(theta)}{a+btan^2(theta)}right)
end{align}
$$
This is why we have
t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
$endgroup$
1
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
1
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
add a comment |
$begingroup$
Parameterization of an ellipse by angle from the center is
$$
gamma(phi)=(cos(phi),sin(phi))frac{ab}{sqrt{a^2sin^2(phi)+b^2cos^2(phi)}}
$$
and
$$
left|gamma'(phi)right|=absqrt{frac{a^4sin^2(phi)+b^4cos^2(phi)}{left(a^2sin^2(phi)+b^2cos^2(phi)right)^3}}
$$
and integrating $left|gamma'(phi)right|$ gets extremely messy.
So instead, we use $theta$, where
$$
btan(theta)=atan(phi)
$$
Then
$$
gamma(theta)=(acos(theta),bsin(theta))
$$
and
$$
left|gamma'(theta)right|=sqrt{a^2sin^2(theta)+b^2cos^2(theta)}
$$
Now, integrating $left|gamma'(theta)right|$ is a lot simpler.
$$
intleft|gamma'(theta)right|,mathrm{d}theta
=b,mathrm{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)
$$
However, to go from from arc length to angle, we still need to invert the Elliptic integral.
Solution of the problem given using Mathematica
Here we get the solution to 30 places.
In= Phi[a_,b_,s_,opts:OptionsPattern] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]
In= Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]
Out= 0.87052028743193111752524449959
Thus, $phidoteq0.87052028743193111752524449959text{ radians}$.
Let me describe the algorithm a bit.
FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]
inverts the elliptic integral to get $theta$ from $a$, $b$, and $s$.
Now, we want to find $phi$ so that $btan(theta)=atan(phi)$. However, simply using $tan^{-1}left(frac batan(theta)right)$ will only return a result in $left(-fracpi2,fracpi2right)$. To get the correct value, we use the relation
$$
begin{align}
tan(phi-theta)
&=frac{tan(phi)-tan(theta)}{1+tan(phi)tan(theta)}\
&=frac{frac batan(theta)-tan(theta)}{1+frac batan(theta)tan(theta)}\
&=frac{btan(theta)-atan(theta)}{a+btan(theta)tan(theta)}\
phi
&=theta+tan^{-1}left(frac{(b-a)tan(theta)}{a+btan^2(theta)}right)
end{align}
$$
This is why we have
t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
$endgroup$
Parameterization of an ellipse by angle from the center is
$$
gamma(phi)=(cos(phi),sin(phi))frac{ab}{sqrt{a^2sin^2(phi)+b^2cos^2(phi)}}
$$
and
$$
left|gamma'(phi)right|=absqrt{frac{a^4sin^2(phi)+b^4cos^2(phi)}{left(a^2sin^2(phi)+b^2cos^2(phi)right)^3}}
$$
and integrating $left|gamma'(phi)right|$ gets extremely messy.
So instead, we use $theta$, where
$$
btan(theta)=atan(phi)
$$
Then
$$
gamma(theta)=(acos(theta),bsin(theta))
$$
and
$$
left|gamma'(theta)right|=sqrt{a^2sin^2(theta)+b^2cos^2(theta)}
$$
Now, integrating $left|gamma'(theta)right|$ is a lot simpler.
$$
intleft|gamma'(theta)right|,mathrm{d}theta
=b,mathrm{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)
$$
However, to go from from arc length to angle, we still need to invert the Elliptic integral.
Solution of the problem given using Mathematica
Here we get the solution to 30 places.
In= Phi[a_,b_,s_,opts:OptionsPattern] := Block[ {t}, t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)] /. FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]]
In= Phi[4, 2.9`32, 3.31`32, WorkingPrecision->30]
Out= 0.87052028743193111752524449959
Thus, $phidoteq0.87052028743193111752524449959text{ radians}$.
Let me describe the algorithm a bit.
FindRoot[b EllipticE[t,(b^2-a^2)/b^2] == s, {t, 0}, opts]
inverts the elliptic integral to get $theta$ from $a$, $b$, and $s$.
Now, we want to find $phi$ so that $btan(theta)=atan(phi)$. However, simply using $tan^{-1}left(frac batan(theta)right)$ will only return a result in $left(-fracpi2,fracpi2right)$. To get the correct value, we use the relation
$$
begin{align}
tan(phi-theta)
&=frac{tan(phi)-tan(theta)}{1+tan(phi)tan(theta)}\
&=frac{frac batan(theta)-tan(theta)}{1+frac batan(theta)tan(theta)}\
&=frac{btan(theta)-atan(theta)}{a+btan(theta)tan(theta)}\
phi
&=theta+tan^{-1}left(frac{(b-a)tan(theta)}{a+btan^2(theta)}right)
end{align}
$$
This is why we have
t+ArcTan[(b-a)Tan[t]/(a+b Tan[t]^2)]
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
edited Jul 5 '13 at 14:50
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
answered Jul 4 '13 at 14:40
robjohn♦robjohn
270k27313641
270k27313641
1
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
1
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
add a comment |
1
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
1
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
1
1
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
$begingroup$
Can you please show me how to solve the example in the picture ? (I want to understand not just solve the example) note: this isn't a homework
$endgroup$
– Mohammad Fakhrey
Jul 5 '13 at 7:32
1
1
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
$begingroup$
@MohammadFakhrey: I've added the example, but that will only show you how to solve.
$endgroup$
– robjohn♦
Jul 5 '13 at 14:53
add a comment |
$begingroup$
The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b,E(theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumference. Also, for those more interested readers, it is easy to show that $b,E(theta,1-a^2/b^2) = a,E(theta,1-b^2/a^2)$, as found at Wolfram World webpage http://mathworld.wolfram.com/Ellipse.html. This change rule is useful for avoiding negative values of parameters. F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
$endgroup$
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
add a comment |
$begingroup$
The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b,E(theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumference. Also, for those more interested readers, it is easy to show that $b,E(theta,1-a^2/b^2) = a,E(theta,1-b^2/a^2)$, as found at Wolfram World webpage http://mathworld.wolfram.com/Ellipse.html. This change rule is useful for avoiding negative values of parameters. F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
$endgroup$
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
add a comment |
$begingroup$
The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b,E(theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumference. Also, for those more interested readers, it is easy to show that $b,E(theta,1-a^2/b^2) = a,E(theta,1-b^2/a^2)$, as found at Wolfram World webpage http://mathworld.wolfram.com/Ellipse.html. This change rule is useful for avoiding negative values of parameters. F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
$endgroup$
The nice answer by robjohn demands a comment. His result for the integral of the arc-length of the elipse (with major axis $2a$ and minor axis $2b$) is $b,E(theta,1-a^2/b^2)$, but this is only a quarter of the complete elliptic circumference. Also, for those more interested readers, it is easy to show that $b,E(theta,1-a^2/b^2) = a,E(theta,1-b^2/a^2)$, as found at Wolfram World webpage http://mathworld.wolfram.com/Ellipse.html. This change rule is useful for avoiding negative values of parameters. F. M. S. Lima, University of Brasilia (fabio-at-fis.unb.br)
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
edited Feb 10 '16 at 15:19
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
answered Feb 5 '16 at 22:07
Dr. Fabio M. S. LimaDr. Fabio M. S. Lima
765
765
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
add a comment |
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
$begingroup$
Between $theta=0$ and $theta=2pi$, $b,text{EllipticE}left(theta,frac{b^2-a^2}{b^2}right)$ gives the entire circumference. Perhaps I am missing something.
$endgroup$
– robjohn♦
Feb 6 '16 at 0:45
add a comment |
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Just as to get the arc length from the angle you need to use an elliptic integral, to get the angle from the arc length you need to invert the elliptic integral.
$endgroup$
– robjohn♦
Jul 4 '13 at 13:14