Evaluate the eccentricity of the elliptical section of a right circular cone
$begingroup$
A right circular cone, with the apex angle $alpha=60^{o}$, is thoroughly cut with a smooth plane inclined at an acute angle $theta=70^{o}$ with its geometrical axis to generate an elliptical section (As shown in the diagram) .
How to calculate the eccentricity of the elliptical section generated?
Is there any set formula to calculate eccentricity for the similar case i.e. eccentricity, $e$ as the function $alpha$ & $theta$ i.e. $e=f(alpha, theta)$?
geometry conic-sections solid-geometry
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
$endgroup$
add a comment |
$begingroup$
A right circular cone, with the apex angle $alpha=60^{o}$, is thoroughly cut with a smooth plane inclined at an acute angle $theta=70^{o}$ with its geometrical axis to generate an elliptical section (As shown in the diagram) .
How to calculate the eccentricity of the elliptical section generated?
Is there any set formula to calculate eccentricity for the similar case i.e. eccentricity, $e$ as the function $alpha$ & $theta$ i.e. $e=f(alpha, theta)$?
geometry conic-sections solid-geometry
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
$endgroup$
$begingroup$
Where's the diagram? It might not be that important but, since you said it...
$endgroup$
– user228113
May 12 '15 at 16:07
$begingroup$
Hint The eccentricity of an ellipse is $sqrt{1-left({bover a}right)^2}$, where $age b$ are the lengths of its axes... I think $bover a$ can most likely be found knowing only $alpha$ and $theta$.
$endgroup$
– user228113
May 12 '15 at 16:14
$begingroup$
As far as i remember there really was a formula for e in terms of $theta$ and $alpha $ only. I recall reading like this somewhere.
$endgroup$
– Mann
May 12 '15 at 16:38
add a comment |
$begingroup$
A right circular cone, with the apex angle $alpha=60^{o}$, is thoroughly cut with a smooth plane inclined at an acute angle $theta=70^{o}$ with its geometrical axis to generate an elliptical section (As shown in the diagram) .
How to calculate the eccentricity of the elliptical section generated?
Is there any set formula to calculate eccentricity for the similar case i.e. eccentricity, $e$ as the function $alpha$ & $theta$ i.e. $e=f(alpha, theta)$?
geometry conic-sections solid-geometry
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
$endgroup$
A right circular cone, with the apex angle $alpha=60^{o}$, is thoroughly cut with a smooth plane inclined at an acute angle $theta=70^{o}$ with its geometrical axis to generate an elliptical section (As shown in the diagram) .
How to calculate the eccentricity of the elliptical section generated?
Is there any set formula to calculate eccentricity for the similar case i.e. eccentricity, $e$ as the function $alpha$ & $theta$ i.e. $e=f(alpha, theta)$?
geometry conic-sections solid-geometry
geometry conic-sections solid-geometry
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
edited Aug 23 '15 at 7:07
Paramanand Singh
50.1k556163
50.1k556163
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
asked May 12 '15 at 16:06
Harish Chandra RajpootHarish Chandra Rajpoot
29.6k103771
29.6k103771
$begingroup$
Where's the diagram? It might not be that important but, since you said it...
$endgroup$
– user228113
May 12 '15 at 16:07
$begingroup$
Hint The eccentricity of an ellipse is $sqrt{1-left({bover a}right)^2}$, where $age b$ are the lengths of its axes... I think $bover a$ can most likely be found knowing only $alpha$ and $theta$.
$endgroup$
– user228113
May 12 '15 at 16:14
$begingroup$
As far as i remember there really was a formula for e in terms of $theta$ and $alpha $ only. I recall reading like this somewhere.
$endgroup$
– Mann
May 12 '15 at 16:38
add a comment |
$begingroup$
Where's the diagram? It might not be that important but, since you said it...
$endgroup$
– user228113
May 12 '15 at 16:07
$begingroup$
Hint The eccentricity of an ellipse is $sqrt{1-left({bover a}right)^2}$, where $age b$ are the lengths of its axes... I think $bover a$ can most likely be found knowing only $alpha$ and $theta$.
$endgroup$
– user228113
May 12 '15 at 16:14
$begingroup$
As far as i remember there really was a formula for e in terms of $theta$ and $alpha $ only. I recall reading like this somewhere.
$endgroup$
– Mann
May 12 '15 at 16:38
$begingroup$
Where's the diagram? It might not be that important but, since you said it...
$endgroup$
– user228113
May 12 '15 at 16:07
$begingroup$
Where's the diagram? It might not be that important but, since you said it...
$endgroup$
– user228113
May 12 '15 at 16:07
$begingroup$
Hint The eccentricity of an ellipse is $sqrt{1-left({bover a}right)^2}$, where $age b$ are the lengths of its axes... I think $bover a$ can most likely be found knowing only $alpha$ and $theta$.
$endgroup$
– user228113
May 12 '15 at 16:14
$begingroup$
Hint The eccentricity of an ellipse is $sqrt{1-left({bover a}right)^2}$, where $age b$ are the lengths of its axes... I think $bover a$ can most likely be found knowing only $alpha$ and $theta$.
$endgroup$
– user228113
May 12 '15 at 16:14
$begingroup$
As far as i remember there really was a formula for e in terms of $theta$ and $alpha $ only. I recall reading like this somewhere.
$endgroup$
– Mann
May 12 '15 at 16:38
$begingroup$
As far as i remember there really was a formula for e in terms of $theta$ and $alpha $ only. I recall reading like this somewhere.
$endgroup$
– Mann
May 12 '15 at 16:38
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $beta$ so that $beta = alpha/2$ (this will help simplify the calculations later).
Clearly in $Delta AOC$ we have $$frac{OA}{sin angle OCA} = frac{AC}{sin angle AOC}$$ so that $$b = AC = dfrac{sin theta}{sin(theta + beta)}tag{1}$$ Similarly $$c = AB = frac{sin theta}{sin(theta - beta)}tag{2}$$ and $$a = BC = frac{bsin 2beta}{sin(theta - beta)} = frac{sinthetasin 2beta}{sin(theta + beta)sin(theta - beta)}tag{3}$$ Let the incircle of $Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $Delta ABC$ then we know that $$tan frac{C}{2} = frac{r}{CD}$$ so that $$CD = frac{r}{tan(C/2)} = dfrac{r}{tanleft(dfrac{pi - theta - beta}{2}right)} = rtanleft(frac{theta + beta}{2}right)tag{4}$$ Note further that the inradius $r$ is given by the formula
begin{align}
r &= frac{Delta}{s}notag\
&= frac{bcsin 2beta}{a + b + c}notag\
&= dfrac{dfrac{sin^{2}thetasin 2beta}{sin(theta + beta)sin(theta - beta)}}{dfrac{sintheta{sin(theta + beta) + sin(theta - beta) + sin 2beta}}{sin(theta + beta)sin(theta - beta)}}notag\
&= frac{sinthetasin 2beta}{2sinthetacosbeta + 2sinbetacosbeta}notag\
&= frac{2sinthetasinbetacosbeta}{2cosbeta{sintheta + sinbeta}}notag\
&= frac{sinthetasinbeta}{sintheta + sinbeta}tag{5}
end{align} If $e$ is the eccentricity then we know that $$e = frac{text{distance of focus from center}}{text{length of semi major axis}} = frac{(a/2) - CD}{a/2} = frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as
begin{align}
e &= dfrac{a - 2rtanleft(dfrac{theta + beta}{2}right)}{a}notag\
&= 1 - frac{2r}{a}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{2sinthetasinbeta}{sintheta + sinbeta}cdotfrac{sin^{2}theta - sin^{2}beta}{2sinthetasinbetacosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{sintheta - sinbeta}{cosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2cosdfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2sindfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}notag\
&= 1 - frac{cos beta - cos theta}{cos beta}notag\
&= frac{cos theta}{cos beta} = frac{costheta}{cos(alpha/2)}tag{6}
end{align}
Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < beta = alpha / 2 < theta leq pi/2$. If $beta = alpha/2 = theta$ then the section becomes a parabola and if $beta = alpha/2 > theta$ then the section becomes a hyperbola. For the angles given in question we have $e = cos 70^{circ}/cos 30^{circ} = 0.39493084ldots$.
Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = costheta/cosbeta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
$begingroup$
@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 6:49
$begingroup$
But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
$begingroup$
@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $beta$ so that $beta = alpha/2$ (this will help simplify the calculations later).
Clearly in $Delta AOC$ we have $$frac{OA}{sin angle OCA} = frac{AC}{sin angle AOC}$$ so that $$b = AC = dfrac{sin theta}{sin(theta + beta)}tag{1}$$ Similarly $$c = AB = frac{sin theta}{sin(theta - beta)}tag{2}$$ and $$a = BC = frac{bsin 2beta}{sin(theta - beta)} = frac{sinthetasin 2beta}{sin(theta + beta)sin(theta - beta)}tag{3}$$ Let the incircle of $Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $Delta ABC$ then we know that $$tan frac{C}{2} = frac{r}{CD}$$ so that $$CD = frac{r}{tan(C/2)} = dfrac{r}{tanleft(dfrac{pi - theta - beta}{2}right)} = rtanleft(frac{theta + beta}{2}right)tag{4}$$ Note further that the inradius $r$ is given by the formula
begin{align}
r &= frac{Delta}{s}notag\
&= frac{bcsin 2beta}{a + b + c}notag\
&= dfrac{dfrac{sin^{2}thetasin 2beta}{sin(theta + beta)sin(theta - beta)}}{dfrac{sintheta{sin(theta + beta) + sin(theta - beta) + sin 2beta}}{sin(theta + beta)sin(theta - beta)}}notag\
&= frac{sinthetasin 2beta}{2sinthetacosbeta + 2sinbetacosbeta}notag\
&= frac{2sinthetasinbetacosbeta}{2cosbeta{sintheta + sinbeta}}notag\
&= frac{sinthetasinbeta}{sintheta + sinbeta}tag{5}
end{align} If $e$ is the eccentricity then we know that $$e = frac{text{distance of focus from center}}{text{length of semi major axis}} = frac{(a/2) - CD}{a/2} = frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as
begin{align}
e &= dfrac{a - 2rtanleft(dfrac{theta + beta}{2}right)}{a}notag\
&= 1 - frac{2r}{a}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{2sinthetasinbeta}{sintheta + sinbeta}cdotfrac{sin^{2}theta - sin^{2}beta}{2sinthetasinbetacosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{sintheta - sinbeta}{cosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2cosdfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2sindfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}notag\
&= 1 - frac{cos beta - cos theta}{cos beta}notag\
&= frac{cos theta}{cos beta} = frac{costheta}{cos(alpha/2)}tag{6}
end{align}
Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < beta = alpha / 2 < theta leq pi/2$. If $beta = alpha/2 = theta$ then the section becomes a parabola and if $beta = alpha/2 > theta$ then the section becomes a hyperbola. For the angles given in question we have $e = cos 70^{circ}/cos 30^{circ} = 0.39493084ldots$.
Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = costheta/cosbeta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
$begingroup$
@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 6:49
$begingroup$
But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
$begingroup$
@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
|
show 2 more comments
$begingroup$
Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $beta$ so that $beta = alpha/2$ (this will help simplify the calculations later).
Clearly in $Delta AOC$ we have $$frac{OA}{sin angle OCA} = frac{AC}{sin angle AOC}$$ so that $$b = AC = dfrac{sin theta}{sin(theta + beta)}tag{1}$$ Similarly $$c = AB = frac{sin theta}{sin(theta - beta)}tag{2}$$ and $$a = BC = frac{bsin 2beta}{sin(theta - beta)} = frac{sinthetasin 2beta}{sin(theta + beta)sin(theta - beta)}tag{3}$$ Let the incircle of $Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $Delta ABC$ then we know that $$tan frac{C}{2} = frac{r}{CD}$$ so that $$CD = frac{r}{tan(C/2)} = dfrac{r}{tanleft(dfrac{pi - theta - beta}{2}right)} = rtanleft(frac{theta + beta}{2}right)tag{4}$$ Note further that the inradius $r$ is given by the formula
begin{align}
r &= frac{Delta}{s}notag\
&= frac{bcsin 2beta}{a + b + c}notag\
&= dfrac{dfrac{sin^{2}thetasin 2beta}{sin(theta + beta)sin(theta - beta)}}{dfrac{sintheta{sin(theta + beta) + sin(theta - beta) + sin 2beta}}{sin(theta + beta)sin(theta - beta)}}notag\
&= frac{sinthetasin 2beta}{2sinthetacosbeta + 2sinbetacosbeta}notag\
&= frac{2sinthetasinbetacosbeta}{2cosbeta{sintheta + sinbeta}}notag\
&= frac{sinthetasinbeta}{sintheta + sinbeta}tag{5}
end{align} If $e$ is the eccentricity then we know that $$e = frac{text{distance of focus from center}}{text{length of semi major axis}} = frac{(a/2) - CD}{a/2} = frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as
begin{align}
e &= dfrac{a - 2rtanleft(dfrac{theta + beta}{2}right)}{a}notag\
&= 1 - frac{2r}{a}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{2sinthetasinbeta}{sintheta + sinbeta}cdotfrac{sin^{2}theta - sin^{2}beta}{2sinthetasinbetacosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{sintheta - sinbeta}{cosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2cosdfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2sindfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}notag\
&= 1 - frac{cos beta - cos theta}{cos beta}notag\
&= frac{cos theta}{cos beta} = frac{costheta}{cos(alpha/2)}tag{6}
end{align}
Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < beta = alpha / 2 < theta leq pi/2$. If $beta = alpha/2 = theta$ then the section becomes a parabola and if $beta = alpha/2 > theta$ then the section becomes a hyperbola. For the angles given in question we have $e = cos 70^{circ}/cos 30^{circ} = 0.39493084ldots$.
Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = costheta/cosbeta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
$begingroup$
@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 6:49
$begingroup$
But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
$begingroup$
@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
|
show 2 more comments
$begingroup$
Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $beta$ so that $beta = alpha/2$ (this will help simplify the calculations later).
Clearly in $Delta AOC$ we have $$frac{OA}{sin angle OCA} = frac{AC}{sin angle AOC}$$ so that $$b = AC = dfrac{sin theta}{sin(theta + beta)}tag{1}$$ Similarly $$c = AB = frac{sin theta}{sin(theta - beta)}tag{2}$$ and $$a = BC = frac{bsin 2beta}{sin(theta - beta)} = frac{sinthetasin 2beta}{sin(theta + beta)sin(theta - beta)}tag{3}$$ Let the incircle of $Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $Delta ABC$ then we know that $$tan frac{C}{2} = frac{r}{CD}$$ so that $$CD = frac{r}{tan(C/2)} = dfrac{r}{tanleft(dfrac{pi - theta - beta}{2}right)} = rtanleft(frac{theta + beta}{2}right)tag{4}$$ Note further that the inradius $r$ is given by the formula
begin{align}
r &= frac{Delta}{s}notag\
&= frac{bcsin 2beta}{a + b + c}notag\
&= dfrac{dfrac{sin^{2}thetasin 2beta}{sin(theta + beta)sin(theta - beta)}}{dfrac{sintheta{sin(theta + beta) + sin(theta - beta) + sin 2beta}}{sin(theta + beta)sin(theta - beta)}}notag\
&= frac{sinthetasin 2beta}{2sinthetacosbeta + 2sinbetacosbeta}notag\
&= frac{2sinthetasinbetacosbeta}{2cosbeta{sintheta + sinbeta}}notag\
&= frac{sinthetasinbeta}{sintheta + sinbeta}tag{5}
end{align} If $e$ is the eccentricity then we know that $$e = frac{text{distance of focus from center}}{text{length of semi major axis}} = frac{(a/2) - CD}{a/2} = frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as
begin{align}
e &= dfrac{a - 2rtanleft(dfrac{theta + beta}{2}right)}{a}notag\
&= 1 - frac{2r}{a}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{2sinthetasinbeta}{sintheta + sinbeta}cdotfrac{sin^{2}theta - sin^{2}beta}{2sinthetasinbetacosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{sintheta - sinbeta}{cosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2cosdfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2sindfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}notag\
&= 1 - frac{cos beta - cos theta}{cos beta}notag\
&= frac{cos theta}{cos beta} = frac{costheta}{cos(alpha/2)}tag{6}
end{align}
Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < beta = alpha / 2 < theta leq pi/2$. If $beta = alpha/2 = theta$ then the section becomes a parabola and if $beta = alpha/2 > theta$ then the section becomes a hyperbola. For the angles given in question we have $e = cos 70^{circ}/cos 30^{circ} = 0.39493084ldots$.
Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = costheta/cosbeta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
$endgroup$
Let the point where the plane of ellipse intersects the axis of the cone be called $O$ (notice the angle $theta$ in figure in question) and let the vertex of the cone be denoted by $A$. Let $OA$ be the unit of our measurement so that $OA = 1$. Let other vertices of the triangular section be called $B, C$ (with $B$ lying on left of $O$). Let $AB = c, BC = a, CA = b$. Also let half the apex angle at $A$ be denoted by $beta$ so that $beta = alpha/2$ (this will help simplify the calculations later).
Clearly in $Delta AOC$ we have $$frac{OA}{sin angle OCA} = frac{AC}{sin angle AOC}$$ so that $$b = AC = dfrac{sin theta}{sin(theta + beta)}tag{1}$$ Similarly $$c = AB = frac{sin theta}{sin(theta - beta)}tag{2}$$ and $$a = BC = frac{bsin 2beta}{sin(theta - beta)} = frac{sinthetasin 2beta}{sin(theta + beta)sin(theta - beta)}tag{3}$$ Let the incircle of $Delta ABC$ touch $BC$ on point $D$ so that $D$ is a focus of the ellipse. $BC$ is the major axis of ellipse and and if know the distance $CD$ we effectively know the the eccentricity of the ellipse. If $r$ is the inradius of $Delta ABC$ then we know that $$tan frac{C}{2} = frac{r}{CD}$$ so that $$CD = frac{r}{tan(C/2)} = dfrac{r}{tanleft(dfrac{pi - theta - beta}{2}right)} = rtanleft(frac{theta + beta}{2}right)tag{4}$$ Note further that the inradius $r$ is given by the formula
begin{align}
r &= frac{Delta}{s}notag\
&= frac{bcsin 2beta}{a + b + c}notag\
&= dfrac{dfrac{sin^{2}thetasin 2beta}{sin(theta + beta)sin(theta - beta)}}{dfrac{sintheta{sin(theta + beta) + sin(theta - beta) + sin 2beta}}{sin(theta + beta)sin(theta - beta)}}notag\
&= frac{sinthetasin 2beta}{2sinthetacosbeta + 2sinbetacosbeta}notag\
&= frac{2sinthetasinbetacosbeta}{2cosbeta{sintheta + sinbeta}}notag\
&= frac{sinthetasinbeta}{sintheta + sinbeta}tag{5}
end{align} If $e$ is the eccentricity then we know that $$e = frac{text{distance of focus from center}}{text{length of semi major axis}} = frac{(a/2) - CD}{a/2} = frac{a - 2CD}{a}$$ We can now evaluate the expression for $e$ as
begin{align}
e &= dfrac{a - 2rtanleft(dfrac{theta + beta}{2}right)}{a}notag\
&= 1 - frac{2r}{a}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{2sinthetasinbeta}{sintheta + sinbeta}cdotfrac{sin^{2}theta - sin^{2}beta}{2sinthetasinbetacosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - frac{sintheta - sinbeta}{cosbeta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2cosdfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}tanleft(dfrac{theta + beta}{2}right)notag\
&= 1 - dfrac{2sindfrac{theta + beta}{2}sindfrac{theta - beta}{2}}{cos beta}notag\
&= 1 - frac{cos beta - cos theta}{cos beta}notag\
&= frac{cos theta}{cos beta} = frac{costheta}{cos(alpha/2)}tag{6}
end{align}
Looking at the above simple result it appears that there is perhaps a much simpler way to find $e$. Note that for the conic section to be an ellipse it is essential that $0 < beta = alpha / 2 < theta leq pi/2$. If $beta = alpha/2 = theta$ then the section becomes a parabola and if $beta = alpha/2 > theta$ then the section becomes a hyperbola. For the angles given in question we have $e = cos 70^{circ}/cos 30^{circ} = 0.39493084ldots$.
Note: If we are willing to consider the case of parabola/hyperbola mentioned above then also the formula $e = costheta/cosbeta$ remains true (eccentricity for parabola is $1$ and that of a hyperbola is $>1$). Note that in this case we don't have a triangle $ABC$ and hence there is no incircle whose intersection with the plane of conic section gives us the focus. However we do have a circle (actually in 3D its a sphere) which touches the generators of the cone emanating from apex $A$ as well the plane of the conic section and the point where it touches the plane of conic section is the focus of the conic section.
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
edited Aug 23 '15 at 11:42
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
answered Aug 23 '15 at 6:30
Paramanand SinghParamanand Singh
50.1k556163
50.1k556163
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
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– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
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@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
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– Paramanand Singh
Aug 23 '15 at 6:49
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But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
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– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
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– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
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@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
|
show 2 more comments
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
$begingroup$
@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 6:49
$begingroup$
But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
$begingroup$
@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
$begingroup$
Thanks for your effort. Could you please reduce your result to $$e=frac{costheta}{cosalpha}$$ ? Or I will check it soon.
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:46
$begingroup$
@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 6:49
$begingroup$
@HarishChandraRajpoot; Your result looks problematic because if $alpha > pi/2$ then eccentricity comes out to be negative.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 6:49
$begingroup$
But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
But let me tell you the restriction of $alpha$ & $theta$ which must be held true to make $e$ always positive as follows $$0<alpha<thetaleqfrac{pi}{2}$$
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 6:52
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
$begingroup$
Your individual restrictions are true but in addition there must always be $alpha<theta$. OK, for instance, kindly check it for any values such that $alpha>thetaleq frac{pi}{2}$ will such a case exist $alpha>theta$.?
$endgroup$
– Harish Chandra Rajpoot
Aug 23 '15 at 7:02
$begingroup$
@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
$begingroup$
@HarishChandraRajpoot: check the figure you have drawn. Clearly $alpha > theta$. Please read my previous comment. Your comments assume $alpha$ to be half of angle at the vertex of the cone. My answer and the figure in your question say that $alpha$ is the full angle at the vertex of the cone. there is no further constraint on $alpha, theta$ apart from the one I mention.
$endgroup$
– Paramanand Singh
Aug 23 '15 at 7:05
|
show 2 more comments
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$begingroup$
Where's the diagram? It might not be that important but, since you said it...
$endgroup$
– user228113
May 12 '15 at 16:07
$begingroup$
Hint The eccentricity of an ellipse is $sqrt{1-left({bover a}right)^2}$, where $age b$ are the lengths of its axes... I think $bover a$ can most likely be found knowing only $alpha$ and $theta$.
$endgroup$
– user228113
May 12 '15 at 16:14
$begingroup$
As far as i remember there really was a formula for e in terms of $theta$ and $alpha $ only. I recall reading like this somewhere.
$endgroup$
– Mann
May 12 '15 at 16:38