Mixing
Finding the angle of a cone from a 3D point
$begingroup$
Given a point in $3$D space $(x,y,z)$ and a circular cone about the $x$ axis, I wish to find the angle of the cone such that the point is on the surface of the cone. For a given point, there is only one possible angle (I think). If the point lies in the plane defined by $z$, then the intersection between the plane and the cone is a hyperbola.
This image I found describes the problem fairly well - the plane is traveling along the $x$ axis:
The apex of the cone is in the origin $(0,0,0)$, and my point is in $(x,y,z)$, that is, somewhere on the ground. How do I find the angle of the cone such that the point lies on the hyperbola resulting from the intersection of the cone and the ground?
geometry
edited Feb 10 at 22:01
Glorfindel
3,41981830
asked May 21 '14 at 9:14
DhiDhi
62
$endgroup$
add a comment |
$begingroup$
Given a point in $3$D space $(x,y,z)$ and a circular cone about the $x$ axis, I wish to find the angle of the cone such that the point is on the surface of the cone. For a given point, there is only one possible angle (I think). If the point lies in the plane defined by $z$, then the intersection between the plane and the cone is a hyperbola.
This image I found describes the problem fairly well - the plane is traveling along the $x$ axis:
The apex of the cone is in the origin $(0,0,0)$, and my point is in $(x,y,z)$, that is, somewhere on the ground. How do I find the angle of the cone such that the point lies on the hyperbola resulting from the intersection of the cone and the ground?
geometry
edited Feb 10 at 22:01
Glorfindel
3,41981830
asked May 21 '14 at 9:14
DhiDhi
62
$endgroup$
add a comment |
$begingroup$
Given a point in $3$D space $(x,y,z)$ and a circular cone about the $x$ axis, I wish to find the angle of the cone such that the point is on the surface of the cone. For a given point, there is only one possible angle (I think). If the point lies in the plane defined by $z$, then the intersection between the plane and the cone is a hyperbola.
This image I found describes the problem fairly well - the plane is traveling along the $x$ axis:
The apex of the cone is in the origin $(0,0,0)$, and my point is in $(x,y,z)$, that is, somewhere on the ground. How do I find the angle of the cone such that the point lies on the hyperbola resulting from the intersection of the cone and the ground?
geometry
edited Feb 10 at 22:01
Glorfindel
3,41981830
asked May 21 '14 at 9:14
DhiDhi
62
$endgroup$
Given a point in $3$D space $(x,y,z)$ and a circular cone about the $x$ axis, I wish to find the angle of the cone such that the point is on the surface of the cone. For a given point, there is only one possible angle (I think). If the point lies in the plane defined by $z$, then the intersection between the plane and the cone is a hyperbola.
This image I found describes the problem fairly well - the plane is traveling along the $x$ axis:
The apex of the cone is in the origin $(0,0,0)$, and my point is in $(x,y,z)$, that is, somewhere on the ground. How do I find the angle of the cone such that the point lies on the hyperbola resulting from the intersection of the cone and the ground?
geometry
geometry
edited Feb 10 at 22:01
Glorfindel
3,41981830
asked May 21 '14 at 9:14
DhiDhi
62
edited Feb 10 at 22:01
Glorfindel
3,41981830
asked May 21 '14 at 9:14
DhiDhi
62
edited Feb 10 at 22:01
Glorfindel
3,41981830
edited Feb 10 at 22:01
Glorfindel
3,41981830
edited Feb 10 at 22:01
Glorfindel
3,41981830
3,41981830
asked May 21 '14 at 9:14
DhiDhi
62
asked May 21 '14 at 9:14
DhiDhi
62
asked May 21 '14 at 9:14
DhiDhi
62
62
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
I think I found a solution.
The point $(x,y,z)$ is not only on the hyperbola, but also on the circle that results from intersecting the cone with the plane defined by $x$ and that is given by the equation $y^2+z^2=r^2$, where $r=sqrt{y^2+z^2}$ is the radius of the circle, which has its center in $(x, 0, 0)$.
Now that I have the radius $r$ and length of the cone $x$, I can find the angle of the cone $alpha=arctan(r/x)$, which was what I was looking for.
Tl;dr: $alpha=arctan(sqrt{y^2+z^2}/x)$.
answered May 22 '14 at 7:18
DhiDhi
62
$endgroup$
add a comment |
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$begingroup$
I think I found a solution.
The point $(x,y,z)$ is not only on the hyperbola, but also on the circle that results from intersecting the cone with the plane defined by $x$ and that is given by the equation $y^2+z^2=r^2$, where $r=sqrt{y^2+z^2}$ is the radius of the circle, which has its center in $(x, 0, 0)$.
Now that I have the radius $r$ and length of the cone $x$, I can find the angle of the cone $alpha=arctan(r/x)$, which was what I was looking for.
Tl;dr: $alpha=arctan(sqrt{y^2+z^2}/x)$.
answered May 22 '14 at 7:18
DhiDhi
62
$endgroup$
add a comment |
$begingroup$
I think I found a solution.
The point $(x,y,z)$ is not only on the hyperbola, but also on the circle that results from intersecting the cone with the plane defined by $x$ and that is given by the equation $y^2+z^2=r^2$, where $r=sqrt{y^2+z^2}$ is the radius of the circle, which has its center in $(x, 0, 0)$.
Now that I have the radius $r$ and length of the cone $x$, I can find the angle of the cone $alpha=arctan(r/x)$, which was what I was looking for.
Tl;dr: $alpha=arctan(sqrt{y^2+z^2}/x)$.
answered May 22 '14 at 7:18
DhiDhi
62
$endgroup$
add a comment |
$begingroup$
I think I found a solution.
The point $(x,y,z)$ is not only on the hyperbola, but also on the circle that results from intersecting the cone with the plane defined by $x$ and that is given by the equation $y^2+z^2=r^2$, where $r=sqrt{y^2+z^2}$ is the radius of the circle, which has its center in $(x, 0, 0)$.
Now that I have the radius $r$ and length of the cone $x$, I can find the angle of the cone $alpha=arctan(r/x)$, which was what I was looking for.
Tl;dr: $alpha=arctan(sqrt{y^2+z^2}/x)$.
answered May 22 '14 at 7:18
DhiDhi
62
$endgroup$
I think I found a solution.
The point $(x,y,z)$ is not only on the hyperbola, but also on the circle that results from intersecting the cone with the plane defined by $x$ and that is given by the equation $y^2+z^2=r^2$, where $r=sqrt{y^2+z^2}$ is the radius of the circle, which has its center in $(x, 0, 0)$.
Now that I have the radius $r$ and length of the cone $x$, I can find the angle of the cone $alpha=arctan(r/x)$, which was what I was looking for.
Tl;dr: $alpha=arctan(sqrt{y^2+z^2}/x)$.
answered May 22 '14 at 7:18
DhiDhi
62
answered May 22 '14 at 7:18
DhiDhi
62
answered May 22 '14 at 7:18
DhiDhi
62
answered May 22 '14 at 7:18
DhiDhi
62
62
add a comment |
add a comment |
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