Mixing
Diameter of an Ellipse at an Angle
$begingroup$
A standard ellipse with semi-major axis $a$, semi-minor $b$ has a "diameter" of $2a$ in one dimension ($phi=0$) and $2b$ in the other ($phi=pi/2$). Is there a function to find the diameter for an arbitrary angle $phi$? By "diameter", I mean the distance between two parallel tangents perpendicular to $phi$. In everyday terms, how wide is the shadow cast by an ellipse when viewed from any angle.
Secondary related question: what's the proper mathematical term for what I mean by "diameter" above; projecting the shadow of the ellipse into 1D and finding the length?
conic-sections
edited Feb 2 at 3:29
Andrei
13.7k21230
asked Feb 2 at 3:07
James AshtonJames Ashton
61
$endgroup$
add a comment |
$begingroup$
A standard ellipse with semi-major axis $a$, semi-minor $b$ has a "diameter" of $2a$ in one dimension ($phi=0$) and $2b$ in the other ($phi=pi/2$). Is there a function to find the diameter for an arbitrary angle $phi$? By "diameter", I mean the distance between two parallel tangents perpendicular to $phi$. In everyday terms, how wide is the shadow cast by an ellipse when viewed from any angle.
Secondary related question: what's the proper mathematical term for what I mean by "diameter" above; projecting the shadow of the ellipse into 1D and finding the length?
conic-sections
edited Feb 2 at 3:29
Andrei
13.7k21230
asked Feb 2 at 3:07
James AshtonJames Ashton
61
$endgroup$
$begingroup$
Regarding the word “diameter”, there are objects known as shapes of constant diameter, where “diameter” means exactly the same thing you asked about. So I think “diameter” is the word you want, but you do need to define it carefully.
$endgroup$
– David K
Feb 3 at 4:34
add a comment |
$begingroup$
A standard ellipse with semi-major axis $a$, semi-minor $b$ has a "diameter" of $2a$ in one dimension ($phi=0$) and $2b$ in the other ($phi=pi/2$). Is there a function to find the diameter for an arbitrary angle $phi$? By "diameter", I mean the distance between two parallel tangents perpendicular to $phi$. In everyday terms, how wide is the shadow cast by an ellipse when viewed from any angle.
Secondary related question: what's the proper mathematical term for what I mean by "diameter" above; projecting the shadow of the ellipse into 1D and finding the length?
conic-sections
edited Feb 2 at 3:29
Andrei
13.7k21230
asked Feb 2 at 3:07
James AshtonJames Ashton
61
$endgroup$
A standard ellipse with semi-major axis $a$, semi-minor $b$ has a "diameter" of $2a$ in one dimension ($phi=0$) and $2b$ in the other ($phi=pi/2$). Is there a function to find the diameter for an arbitrary angle $phi$? By "diameter", I mean the distance between two parallel tangents perpendicular to $phi$. In everyday terms, how wide is the shadow cast by an ellipse when viewed from any angle.
Secondary related question: what's the proper mathematical term for what I mean by "diameter" above; projecting the shadow of the ellipse into 1D and finding the length?
conic-sections
conic-sections
edited Feb 2 at 3:29
Andrei
13.7k21230
asked Feb 2 at 3:07
James AshtonJames Ashton
61
edited Feb 2 at 3:29
Andrei
13.7k21230
asked Feb 2 at 3:07
James AshtonJames Ashton
61
edited Feb 2 at 3:29
Andrei
13.7k21230
edited Feb 2 at 3:29
Andrei
13.7k21230
edited Feb 2 at 3:29
Andrei
13.7k21230
13.7k21230
asked Feb 2 at 3:07
James AshtonJames Ashton
61
asked Feb 2 at 3:07
James AshtonJames Ashton
61
asked Feb 2 at 3:07
James AshtonJames Ashton
61
61
$begingroup$
Regarding the word “diameter”, there are objects known as shapes of constant diameter, where “diameter” means exactly the same thing you asked about. So I think “diameter” is the word you want, but you do need to define it carefully.
$endgroup$
– David K
Feb 3 at 4:34
add a comment |
$begingroup$
Regarding the word “diameter”, there are objects known as shapes of constant diameter, where “diameter” means exactly the same thing you asked about. So I think “diameter” is the word you want, but you do need to define it carefully.
$endgroup$
– David K
Feb 3 at 4:34
$begingroup$
Regarding the word “diameter”, there are objects known as shapes of constant diameter, where “diameter” means exactly the same thing you asked about. So I think “diameter” is the word you want, but you do need to define it carefully.
$endgroup$
– David K
Feb 3 at 4:34
$begingroup$
Regarding the word “diameter”, there are objects known as shapes of constant diameter, where “diameter” means exactly the same thing you asked about. So I think “diameter” is the word you want, but you do need to define it carefully.
$endgroup$
– David K
Feb 3 at 4:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
This is quite easy to work out if you use homogeneous coordinates of points and lines and a matrix to represent the ellipse, and then work with the dual conic. Setting $mathbf x=(x,y,1)^T$, we can write the equation of your standard-position ellipse as $mathbf x^TCmathbf x=0$, where $C=operatorname{diag}(1/a^2,1/b^2,-1)$. (Multiply it out for yourself to verify.) At the same time, by setting $mathbf l=(a,b,c)^T$, the general equation $ax+by+c=0$ of a line can be written $mathbf l^Tmathbf x=0$, i.e., lines can also be represented as vectors of real numbers. A major advantage of this representation in my mind is that there are no special cases to worry about. Vertical lines, which are usually problematic if you’re basing things on slope, are accommodated in this framework with no fuss.
Now, if $C$ is the matrix of a nondegenerate conic, then tangent lines to the conic satisfy the dual equation $mathbf l^TC^{-1}mathbf l=0$. Also, all lines that have a normal that makes an angle of $phi$ with the positive $x$-axis are of the form $(cosphi,sinphi,tau)^T$, where the signed distance of the line from the origin is given by $tau$. By symmetry, then, the distance between two parallel tangents is $2|tau|$. Expanding the dual equation, we have $$begin{bmatrix}cosphi&sinphi&tauend{bmatrix} begin{bmatrix}a^2&0&0\0&b^2&0\0&0&-1end{bmatrix} begin{bmatrix}cosphi\sinphi\tauend{bmatrix} = a^2cos^2phi+b^2sin^2phi-tau^2=0,$$ from which the distance between the two tangents is simply $$2sqrt{a^2cos^2phi+b^2sin^2phi}.$$
answered Feb 3 at 1:11
amdamd
31.7k21052
$endgroup$
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
add a comment |
$begingroup$
The equation for an eclipse in polar coordinates
$r(theta) = frac {ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
When $theta = 0, r=a$, and when $theta = fracpi 2, r = b$
So the formula you’re looking for (where $d = 2r$) is
$r(theta) = frac {2ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
EDIT: misunderstood the question
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
$endgroup$
2
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
add a comment |
$begingroup$
My previous answer was wrong, because I used the distance from the tangent point to the origin. What I should have used was the closest approach of the tangent line to the origin. Per distance from a point to a line, the line
$$Ax+By+C=0$$
approaches within
$$frac{C}{sqrt{A^2+B^2}}$$
of the origin. Per ellipse tangent slope as parameter, the ellipse
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
has a tangent line at slope $m$ of
$$y=mxpmsqrt{m^2a^2+b^2}quadimpliesquad mx-ypmsqrt{m^2a^2+b^2}=0,.$$
Substituting gives the origin distance of the ellipse tangent of
$$sqrt{frac{m^2a^2+b^2}{m^2+1}},.$$
We simply double the above to find the width of the ellipse projected at angle $phi$ such that the slope $m=tanphi$:
$$2sqrt{frac{a^2tan^2phi+b^2}{tan^2phi+1}},.$$
Noting that $tanphicosphi=sinphi$ we can multiply denominator and numerator by $cos^2phi$ to give:
$$2sqrt{frac{a^2sin^2phi+b^2cos^2phi}{sin^2phi+cos^2phi}} \= 2sqrt{a^2sin^2phi+b^2cos^2phi}$$
and so remove the singularities at angles like $phi=fracpi2$ which were an artefact of the parameterisation.
answered Feb 3 at 1:09
James AshtonJames Ashton
61
$endgroup$
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096989%2fdiameter-of-an-ellipse-at-an-angle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is quite easy to work out if you use homogeneous coordinates of points and lines and a matrix to represent the ellipse, and then work with the dual conic. Setting $mathbf x=(x,y,1)^T$, we can write the equation of your standard-position ellipse as $mathbf x^TCmathbf x=0$, where $C=operatorname{diag}(1/a^2,1/b^2,-1)$. (Multiply it out for yourself to verify.) At the same time, by setting $mathbf l=(a,b,c)^T$, the general equation $ax+by+c=0$ of a line can be written $mathbf l^Tmathbf x=0$, i.e., lines can also be represented as vectors of real numbers. A major advantage of this representation in my mind is that there are no special cases to worry about. Vertical lines, which are usually problematic if you’re basing things on slope, are accommodated in this framework with no fuss.
Now, if $C$ is the matrix of a nondegenerate conic, then tangent lines to the conic satisfy the dual equation $mathbf l^TC^{-1}mathbf l=0$. Also, all lines that have a normal that makes an angle of $phi$ with the positive $x$-axis are of the form $(cosphi,sinphi,tau)^T$, where the signed distance of the line from the origin is given by $tau$. By symmetry, then, the distance between two parallel tangents is $2|tau|$. Expanding the dual equation, we have $$begin{bmatrix}cosphi&sinphi&tauend{bmatrix} begin{bmatrix}a^2&0&0\0&b^2&0\0&0&-1end{bmatrix} begin{bmatrix}cosphi\sinphi\tauend{bmatrix} = a^2cos^2phi+b^2sin^2phi-tau^2=0,$$ from which the distance between the two tangents is simply $$2sqrt{a^2cos^2phi+b^2sin^2phi}.$$
answered Feb 3 at 1:11
amdamd
31.7k21052
$endgroup$
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
add a comment |
$begingroup$
This is quite easy to work out if you use homogeneous coordinates of points and lines and a matrix to represent the ellipse, and then work with the dual conic. Setting $mathbf x=(x,y,1)^T$, we can write the equation of your standard-position ellipse as $mathbf x^TCmathbf x=0$, where $C=operatorname{diag}(1/a^2,1/b^2,-1)$. (Multiply it out for yourself to verify.) At the same time, by setting $mathbf l=(a,b,c)^T$, the general equation $ax+by+c=0$ of a line can be written $mathbf l^Tmathbf x=0$, i.e., lines can also be represented as vectors of real numbers. A major advantage of this representation in my mind is that there are no special cases to worry about. Vertical lines, which are usually problematic if you’re basing things on slope, are accommodated in this framework with no fuss.
Now, if $C$ is the matrix of a nondegenerate conic, then tangent lines to the conic satisfy the dual equation $mathbf l^TC^{-1}mathbf l=0$. Also, all lines that have a normal that makes an angle of $phi$ with the positive $x$-axis are of the form $(cosphi,sinphi,tau)^T$, where the signed distance of the line from the origin is given by $tau$. By symmetry, then, the distance between two parallel tangents is $2|tau|$. Expanding the dual equation, we have $$begin{bmatrix}cosphi&sinphi&tauend{bmatrix} begin{bmatrix}a^2&0&0\0&b^2&0\0&0&-1end{bmatrix} begin{bmatrix}cosphi\sinphi\tauend{bmatrix} = a^2cos^2phi+b^2sin^2phi-tau^2=0,$$ from which the distance between the two tangents is simply $$2sqrt{a^2cos^2phi+b^2sin^2phi}.$$
answered Feb 3 at 1:11
amdamd
31.7k21052
$endgroup$
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
add a comment |
$begingroup$
This is quite easy to work out if you use homogeneous coordinates of points and lines and a matrix to represent the ellipse, and then work with the dual conic. Setting $mathbf x=(x,y,1)^T$, we can write the equation of your standard-position ellipse as $mathbf x^TCmathbf x=0$, where $C=operatorname{diag}(1/a^2,1/b^2,-1)$. (Multiply it out for yourself to verify.) At the same time, by setting $mathbf l=(a,b,c)^T$, the general equation $ax+by+c=0$ of a line can be written $mathbf l^Tmathbf x=0$, i.e., lines can also be represented as vectors of real numbers. A major advantage of this representation in my mind is that there are no special cases to worry about. Vertical lines, which are usually problematic if you’re basing things on slope, are accommodated in this framework with no fuss.
Now, if $C$ is the matrix of a nondegenerate conic, then tangent lines to the conic satisfy the dual equation $mathbf l^TC^{-1}mathbf l=0$. Also, all lines that have a normal that makes an angle of $phi$ with the positive $x$-axis are of the form $(cosphi,sinphi,tau)^T$, where the signed distance of the line from the origin is given by $tau$. By symmetry, then, the distance between two parallel tangents is $2|tau|$. Expanding the dual equation, we have $$begin{bmatrix}cosphi&sinphi&tauend{bmatrix} begin{bmatrix}a^2&0&0\0&b^2&0\0&0&-1end{bmatrix} begin{bmatrix}cosphi\sinphi\tauend{bmatrix} = a^2cos^2phi+b^2sin^2phi-tau^2=0,$$ from which the distance between the two tangents is simply $$2sqrt{a^2cos^2phi+b^2sin^2phi}.$$
answered Feb 3 at 1:11
amdamd
31.7k21052
$endgroup$
This is quite easy to work out if you use homogeneous coordinates of points and lines and a matrix to represent the ellipse, and then work with the dual conic. Setting $mathbf x=(x,y,1)^T$, we can write the equation of your standard-position ellipse as $mathbf x^TCmathbf x=0$, where $C=operatorname{diag}(1/a^2,1/b^2,-1)$. (Multiply it out for yourself to verify.) At the same time, by setting $mathbf l=(a,b,c)^T$, the general equation $ax+by+c=0$ of a line can be written $mathbf l^Tmathbf x=0$, i.e., lines can also be represented as vectors of real numbers. A major advantage of this representation in my mind is that there are no special cases to worry about. Vertical lines, which are usually problematic if you’re basing things on slope, are accommodated in this framework with no fuss.
Now, if $C$ is the matrix of a nondegenerate conic, then tangent lines to the conic satisfy the dual equation $mathbf l^TC^{-1}mathbf l=0$. Also, all lines that have a normal that makes an angle of $phi$ with the positive $x$-axis are of the form $(cosphi,sinphi,tau)^T$, where the signed distance of the line from the origin is given by $tau$. By symmetry, then, the distance between two parallel tangents is $2|tau|$. Expanding the dual equation, we have $$begin{bmatrix}cosphi&sinphi&tauend{bmatrix} begin{bmatrix}a^2&0&0\0&b^2&0\0&0&-1end{bmatrix} begin{bmatrix}cosphi\sinphi\tauend{bmatrix} = a^2cos^2phi+b^2sin^2phi-tau^2=0,$$ from which the distance between the two tangents is simply $$2sqrt{a^2cos^2phi+b^2sin^2phi}.$$
answered Feb 3 at 1:11
amdamd
31.7k21052
answered Feb 3 at 1:11
amdamd
31.7k21052
answered Feb 3 at 1:11
amdamd
31.7k21052
answered Feb 3 at 1:11
amdamd
31.7k21052
31.7k21052
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
add a comment |
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
Thanks for this identical result to my second attempt, but via a more elegant route. Amazing the answers were posted within two minutes of each other. Well, almost identical: your $phi$ is at right angles to mine.
$endgroup$
– James Ashton
Feb 3 at 1:17
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
@JamesAshton In your question, you ask about tangents “perpendicular to $phi$.” That’s what I computed. Your solution is for tangents parallel to $phi$, which is not what you asked for originally.
$endgroup$
– amd
Feb 3 at 1:25
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
$begingroup$
Another way to approach this is to simply perform the projection onto a line through the origin perpendicular to $phi$. The outline of the shadow of the ellipse will be a pair of points given by the same quadratic equation as above.
$endgroup$
– amd
Feb 3 at 1:27
add a comment |
$begingroup$
The equation for an eclipse in polar coordinates
$r(theta) = frac {ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
When $theta = 0, r=a$, and when $theta = fracpi 2, r = b$
So the formula you’re looking for (where $d = 2r$) is
$r(theta) = frac {2ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
EDIT: misunderstood the question
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
$endgroup$
2
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
add a comment |
$begingroup$
The equation for an eclipse in polar coordinates
$r(theta) = frac {ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
When $theta = 0, r=a$, and when $theta = fracpi 2, r = b$
So the formula you’re looking for (where $d = 2r$) is
$r(theta) = frac {2ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
EDIT: misunderstood the question
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
$endgroup$
2
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
add a comment |
$begingroup$
The equation for an eclipse in polar coordinates
$r(theta) = frac {ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
When $theta = 0, r=a$, and when $theta = fracpi 2, r = b$
So the formula you’re looking for (where $d = 2r$) is
$r(theta) = frac {2ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
EDIT: misunderstood the question
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
$endgroup$
The equation for an eclipse in polar coordinates
$r(theta) = frac {ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
When $theta = 0, r=a$, and when $theta = fracpi 2, r = b$
So the formula you’re looking for (where $d = 2r$) is
$r(theta) = frac {2ab} {sqrt {(b costheta)^2 + (a sintheta)^2}}$
EDIT: misunderstood the question
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
edited Feb 2 at 4:23
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
answered Feb 2 at 3:31
Jonathan ViccaryJonathan Viccary
1866
1866
2
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
add a comment |
2
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
2
2
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Note that's not what is asked in the question. You answered what I would consider the "diameter", but what he question implies is the projection, not the diameter. There is a confusion in the terms.
$endgroup$
– Andrei
Feb 2 at 3:35
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
$begingroup$
Thanks, but it does answer something different since the tangent at a point on an ellipse is not in general perpendicular to the line from the point to the centre.
$endgroup$
– James Ashton
Feb 2 at 3:42
add a comment |
$begingroup$
My previous answer was wrong, because I used the distance from the tangent point to the origin. What I should have used was the closest approach of the tangent line to the origin. Per distance from a point to a line, the line
$$Ax+By+C=0$$
approaches within
$$frac{C}{sqrt{A^2+B^2}}$$
of the origin. Per ellipse tangent slope as parameter, the ellipse
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
has a tangent line at slope $m$ of
$$y=mxpmsqrt{m^2a^2+b^2}quadimpliesquad mx-ypmsqrt{m^2a^2+b^2}=0,.$$
Substituting gives the origin distance of the ellipse tangent of
$$sqrt{frac{m^2a^2+b^2}{m^2+1}},.$$
We simply double the above to find the width of the ellipse projected at angle $phi$ such that the slope $m=tanphi$:
$$2sqrt{frac{a^2tan^2phi+b^2}{tan^2phi+1}},.$$
Noting that $tanphicosphi=sinphi$ we can multiply denominator and numerator by $cos^2phi$ to give:
$$2sqrt{frac{a^2sin^2phi+b^2cos^2phi}{sin^2phi+cos^2phi}} \= 2sqrt{a^2sin^2phi+b^2cos^2phi}$$
and so remove the singularities at angles like $phi=fracpi2$ which were an artefact of the parameterisation.
answered Feb 3 at 1:09
James AshtonJames Ashton
61
$endgroup$
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
add a comment |
$begingroup$
My previous answer was wrong, because I used the distance from the tangent point to the origin. What I should have used was the closest approach of the tangent line to the origin. Per distance from a point to a line, the line
$$Ax+By+C=0$$
approaches within
$$frac{C}{sqrt{A^2+B^2}}$$
of the origin. Per ellipse tangent slope as parameter, the ellipse
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
has a tangent line at slope $m$ of
$$y=mxpmsqrt{m^2a^2+b^2}quadimpliesquad mx-ypmsqrt{m^2a^2+b^2}=0,.$$
Substituting gives the origin distance of the ellipse tangent of
$$sqrt{frac{m^2a^2+b^2}{m^2+1}},.$$
We simply double the above to find the width of the ellipse projected at angle $phi$ such that the slope $m=tanphi$:
$$2sqrt{frac{a^2tan^2phi+b^2}{tan^2phi+1}},.$$
Noting that $tanphicosphi=sinphi$ we can multiply denominator and numerator by $cos^2phi$ to give:
$$2sqrt{frac{a^2sin^2phi+b^2cos^2phi}{sin^2phi+cos^2phi}} \= 2sqrt{a^2sin^2phi+b^2cos^2phi}$$
and so remove the singularities at angles like $phi=fracpi2$ which were an artefact of the parameterisation.
answered Feb 3 at 1:09
James AshtonJames Ashton
61
$endgroup$
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
add a comment |
$begingroup$
My previous answer was wrong, because I used the distance from the tangent point to the origin. What I should have used was the closest approach of the tangent line to the origin. Per distance from a point to a line, the line
$$Ax+By+C=0$$
approaches within
$$frac{C}{sqrt{A^2+B^2}}$$
of the origin. Per ellipse tangent slope as parameter, the ellipse
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
has a tangent line at slope $m$ of
$$y=mxpmsqrt{m^2a^2+b^2}quadimpliesquad mx-ypmsqrt{m^2a^2+b^2}=0,.$$
Substituting gives the origin distance of the ellipse tangent of
$$sqrt{frac{m^2a^2+b^2}{m^2+1}},.$$
We simply double the above to find the width of the ellipse projected at angle $phi$ such that the slope $m=tanphi$:
$$2sqrt{frac{a^2tan^2phi+b^2}{tan^2phi+1}},.$$
Noting that $tanphicosphi=sinphi$ we can multiply denominator and numerator by $cos^2phi$ to give:
$$2sqrt{frac{a^2sin^2phi+b^2cos^2phi}{sin^2phi+cos^2phi}} \= 2sqrt{a^2sin^2phi+b^2cos^2phi}$$
and so remove the singularities at angles like $phi=fracpi2$ which were an artefact of the parameterisation.
answered Feb 3 at 1:09
James AshtonJames Ashton
61
$endgroup$
My previous answer was wrong, because I used the distance from the tangent point to the origin. What I should have used was the closest approach of the tangent line to the origin. Per distance from a point to a line, the line
$$Ax+By+C=0$$
approaches within
$$frac{C}{sqrt{A^2+B^2}}$$
of the origin. Per ellipse tangent slope as parameter, the ellipse
$$frac{x^2}{a^2}+frac{y^2}{b^2}=1$$
has a tangent line at slope $m$ of
$$y=mxpmsqrt{m^2a^2+b^2}quadimpliesquad mx-ypmsqrt{m^2a^2+b^2}=0,.$$
Substituting gives the origin distance of the ellipse tangent of
$$sqrt{frac{m^2a^2+b^2}{m^2+1}},.$$
We simply double the above to find the width of the ellipse projected at angle $phi$ such that the slope $m=tanphi$:
$$2sqrt{frac{a^2tan^2phi+b^2}{tan^2phi+1}},.$$
Noting that $tanphicosphi=sinphi$ we can multiply denominator and numerator by $cos^2phi$ to give:
$$2sqrt{frac{a^2sin^2phi+b^2cos^2phi}{sin^2phi+cos^2phi}} \= 2sqrt{a^2sin^2phi+b^2cos^2phi}$$
and so remove the singularities at angles like $phi=fracpi2$ which were an artefact of the parameterisation.
answered Feb 3 at 1:09
James AshtonJames Ashton
61
answered Feb 3 at 1:09
James AshtonJames Ashton
61
answered Feb 3 at 1:09
James AshtonJames Ashton
61
answered Feb 3 at 1:09
James AshtonJames Ashton
61
61
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
add a comment |
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
$begingroup$
This solution is for tangents parallel to $phi$, not perpendicular to it as you asked originally. You need to use $m=-cotphi$ instead of $tanphi$.
$endgroup$
– amd
Feb 3 at 1:30
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3096989%2fdiameter-of-an-ellipse-at-an-angle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Regarding the word “diameter”, there are objects known as shapes of constant diameter, where “diameter” means exactly the same thing you asked about. So I think “diameter” is the word you want, but you do need to define it carefully.
$endgroup$
– David K
Feb 3 at 4:34