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Post by Admin on Oct 2, 2016 15:05:23 GMT
Gary I have published my solution to exercise 3. There are two diagrams which I found easier to draw freehand and so I took photos of them with my phone and included them in the pdf document. You can make them larger if you find them hard to read, by clicking on the zoom. My graphics system is not very good for drawing spheres. Vasco |
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Gary
GaryVasco
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Post by Gary on Oct 3, 2016 21:36:46 GMT
Gary I have published my solution to exercise 3. There are two diagrams which I found easier to draw freehand and so I took photos of them with my phone and included them in the pdf document. You can make them larger if you find them hard to read, by clicking on the zoom. My graphics system is not very good for drawing spheres. Vasco Vasco, This seems a strange question. I have gone round and round with it. I think $v$ must be taken to be one pole of a stereographic projection. The axis of which $v$ is a pole must then be orthogonal to the plane of $\widehat{C}$. Otherwise, I can make no sense of "$\mathcal{I}_{C}(z)$ becomes projection from the vertex v of the cone that touches $\Sigma$ along $\widehat{C}$." If this sentence were to be interpreted literally as applying to some vertex off the cone, the projection could not be conformal. If the projection were not conformal, we could not guarantee that $C$, $G$, or $J$ would be circles or that angles would be preserved, a point that was necessary to the proof of (18). With v at the pole, projections from the inversions fall into place as you have indicated. Then (18) is just the limiting case of reflection across a great circle. I would suggest a change of $\mathcal{P}_{v}(\widehat{z})$ in Figure 1 to $\widehat{w}$. Also, it doesn't seem to matter whether we let $C$ or $\widehat{C}$ be an arbitrary circle. If one is arbitrary, then so is its stereographic projection in the sense that $\widehat{C}$ may not be a great circle. If you think I'm wrong about this (not unlikely), then how does one place $v$ prior knowing the planes of $\widehat{G}$ and $\widehat{J}$, so one can say that extension of the line through $v$ and $\widehat{z}$ will intersect $\Sigma$ again at $\mathcal{P}_{v}(\widehat{z})$? Gary |
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Post by Admin on Oct 4, 2016 5:31:24 GMT
Gary I have published my solution to exercise 3. There are two diagrams which I found easier to draw freehand and so I took photos of them with my phone and included them in the pdf document. You can make them larger if you find them hard to read, by clicking on the zoom. My graphics system is not very good for drawing spheres. Vasco Vasco, This seems a strange question. I have gone round and round with it. I think $v$ must be taken to be one pole of a stereographic projection. The axis of which $v$ is a pole must then be orthogonal to the plane of $\widehat{C}$. Otherwise, I can make no sense of "$\mathcal{I}_{C}(z)$ becomes projection from the vertex v of the cone that touches $\Sigma$ along $\widehat{C}$." If this sentence were to be interpreted literally as applying to some vertex off the cone, the projection could not be conformal. If the projection were not conformal, we could not guarantee that $C$, $G$, or $J$ would be circles or that angles would be preserved, a point that was necessary to the proof of (18). With v at the pole, projections from the inversions fall into place as you have indicated. Then (18) is just the limiting case of reflection across a great circle. I would suggest a change of $\mathcal{P}_{v}(\widehat{z})$ in Figure 1 to $\widehat{w}$. Also, it doesn't seem to matter whether we let $C$ or $\widehat{C}$ be an arbitrary circle. If one is arbitrary, then so is its stereographic projection in the sense that $\widehat{C}$ may not be a great circle. If you think I'm wrong about this (not unlikely), then how does one place $v$ prior knowing the planes of $\widehat{G}$ and $\widehat{J}$, so one can say that extension of the line through $v$ and $\widehat{z}$ will intersect $\Sigma$ again at $\mathcal{P}_{v}(\widehat{z})$? Gary Gary Thanks for the feedback. I don't see how $v$ can be a pole of $\Sigma$ because then the cone could not "touch $\Sigma$ along $\widehat{C}$". I take "touch" here to mean that the cone is tangential to (ie does not intersect) $\Sigma$ on $\widehat{C}$. I don't think you need worry about this projection being conformal in the sense that the stereographic projection of $\Sigma$ to $\mathbb{C}$ is conformal. It is just a rule for obtaining the second intersection point, given the first, $\widehat{z}$, and then applying stereographic projection to map $\Sigma$ to $\mathbb{C}$ in the usual (stereographic) way. Because of the above interpretation of "touches", the vertex $v$ becomes the centre of a sphere $S$ which intersects $\Sigma$ orthogonally in $\widehat{C}$. Vasco |
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Gary
GaryVasco
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Post by Gary on Oct 4, 2016 15:04:52 GMT
Vasco, This seems a strange question. I have gone round and round with it. I think $v$ must be taken to be one pole of a stereographic projection. The axis of which $v$ is a pole must then be orthogonal to the plane of $\widehat{C}$. Otherwise, I can make no sense of "$\mathcal{I}_{C}(z)$ becomes projection from the vertex v of the cone that touches $\Sigma$ along $\widehat{C}$." If this sentence were to be interpreted literally as applying to some vertex off the cone, the projection could not be conformal. If the projection were not conformal, we could not guarantee that $C$, $G$, or $J$ would be circles or that angles would be preserved, a point that was necessary to the proof of (18). With v at the pole, projections from the inversions fall into place as you have indicated. Then (18) is just the limiting case of reflection across a great circle. I would suggest a change of $\mathcal{P}_{v}(\widehat{z})$ in Figure 1 to $\widehat{w}$. Also, it doesn't seem to matter whether we let $C$ or $\widehat{C}$ be an arbitrary circle. If one is arbitrary, then so is its stereographic projection in the sense that $\widehat{C}$ may not be a great circle. If you think I'm wrong about this (not unlikely), then how does one place $v$ prior knowing the planes of $\widehat{G}$ and $\widehat{J}$, so one can say that extension of the line through $v$ and $\widehat{z}$ will intersect $\Sigma$ again at $\mathcal{P}_{v}(\widehat{z})$? Gary Gary Thanks for the feedback. I don't see how $v$ can be a pole of $\Sigma$ because then the cone could not "touch $\Sigma$ along $\widehat{C}$". I take "touch" here to mean that the cone is tangential to (ie does not intersect) $\Sigma$ on $\widehat{C}$. I don't think you need worry about this projection being conformal in the sense that the stereographic projection of $\Sigma$ to $\mathbb{C}$ is conformal. It is just a rule for obtaining the second intersection point, given the first, $\widehat{z}$, and then applying stereographic projection to map $\Sigma$ to $\mathbb{C}$ in the usual (stereographic) way. Because of the above interpretation of "touches", the vertex $v$ becomes the centre of a sphere $S$ which intersects $\Sigma$ orthogonally in $\widehat{C}$. Vasco Vasco, My selected quotes are dropping large segments, so I have to quote the whole message with previous quotes. I see now what you are doing. The problem is really "How can one construct two orthogonal circles on $\widehat{C}$." I was under the impression that this was unproblematic---just a matter of placing the centers of $G$ and $J$ on $C$ and making sure that they intersect---but I see now that this method would probably work only for straight lines on $\Sigma$ or $\mathbb{C}$, and so it would just be an approximation. It's a fair approximation, as one can see from my most recent plot, but we'll leave that for engineers to decide. I will replot. Gary |
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Post by Admin on Oct 5, 2016 10:49:04 GMT
Gary
As a result of reading your solution I have decided to add a couple of lines of extra explanation to my solution, at the end of the fourth paragraph. I have posted the amended solution.
Vasco
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Post by Admin on Oct 5, 2016 15:18:24 GMT
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 5, 2016 17:50:36 GMT
Vasco,
I would like to suggest a new take on the problem without using orthogonal circles. There is no mention of orthogonal circles in Needham's problem and introducing them creates the problem of establishing their orthogonality.
We project the line from v (vertex of cone with base in $\widehat{C}$) to create the puncture points $\widehat{z}$ and $\widehat{w}$. These are just two points on $\Sigma$ and on the line v to $\widehat{w}$. They are not necessarily stereographic projections of anything. Draw a line (great circle segment) from $\widehat{z}$ to $\widehat{w}$. Stereo-project $\widehat{C}$ and $\widehat{z}$ to $C$ and $z$. Find $w = \mathcal{I}_{C}(z)$. When $w$ is projected back to $\Sigma$, it must lie on a line (great circle) that connects it with $\widehat{z}$! (because the line from z to w also projects back to $\Sigma$). In fact, it must be the same great circle that was created by the original transect from v. Therefore, the stereo-projection of w must also lie on the straight line from v to $\widehat{w}$ and $\widehat{w}$ must be induced by inversion in C.
Gary
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Post by Admin on Oct 5, 2016 18:20:58 GMT
Vasco, I would like to suggest a new take on the problem without using orthogonal circles. There is no mention of orthogonal circles in Needham's problem and introducing them creates the problem of establishing their orthogonality. We project the line from v (vertex of cone with base in $\widehat{C}$) to create the puncture points $\widehat{z}$ and $\widehat{w}$. These are just two points on $\Sigma$ and on the line v to $\widehat{w}$. They are not necessarily stereographic projections of anything. Draw a line (great circle segment) from $\widehat{z}$ to $\widehat{w}$. Stereo-project $\widehat{C}$ and $\widehat{z}$ to $C$ and $z$. Find $w = \mathcal{I}_{C}(z)$. When $w$ is projected back to $\Sigma$, it must lie on a line (great circle) that connects it with $\widehat{z}$! (because the line from z to w also projects back to $\Sigma$). In fact, it must be the same great circle that was created by the original transect from v. Therefore, the stereo-projection of w must also lie on the straight line from v to $\widehat{w}$ and $\widehat{w}$ must be induced by inversion in C. Gary Gary It is easy to see that the circle of intersection $C$ with $\Sigma$, of any plane through $v$ must be orthogonal to $\widehat{C}$, because the line from $v$ to the points of intersection of $C$ with $\widehat{C}$ is tangent to $C$ and orthogonal to $\widehat{C}$.Vasco PS I will look at your argument above and get back to you. |
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Gary
GaryVasco
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Post by Gary on Oct 5, 2016 18:59:48 GMT
Vasco, I would like to suggest a new take on the problem without using orthogonal circles. There is no mention of orthogonal circles in Needham's problem and introducing them creates the problem of establishing their orthogonality. We project the line from v (vertex of cone with base in $\widehat{C}$) to create the puncture points $\widehat{z}$ and $\widehat{w}$. These are just two points on $\Sigma$ and on the line v to $\widehat{w}$. They are not necessarily stereographic projections of anything. Draw a line (great circle segment) from $\widehat{z}$ to $\widehat{w}$. Stereo-project $\widehat{C}$ and $\widehat{z}$ to $C$ and $z$. Find $w = \mathcal{I}_{C}(z)$. When $w$ is projected back to $\Sigma$, it must lie on a line (great circle) that connects it with $\widehat{z}$! (because the line from z to w also projects back to $\Sigma$). In fact, it must be the same great circle that was created by the original transect from v. Therefore, the stereo-projection of w must also lie on the straight line from v to $\widehat{w}$ and $\widehat{w}$ must be induced by inversion in C. Gary Gary It is easy to see that the circle of intersection $C$ with $\Sigma$, of any plane through $v$ must be orthogonal to $\widehat{C}$, because the line from $v$ to the points of intersection of $C$ with $\widehat{C}$ is tangent to $C$ and orthogonal to $\widehat{C}$.Vasco PS I will look at your argument above and get back to you. Vasco, My argument might work, but I retract the bit about orthogonal circles not being mentioned in the problem, because he says to generalize the argument in figure [14], which is based on orthogonal circles. So unless I can figure out a way to show that my proposal is a generalization of the argument of [14], it's back to orthogonal circles. Regarding the point about the orthogonality of the tangents to $\Sigma$ in the lines from v touching $\Sigma$ at $\widehat{C}$, I see that they are orthogonal to $\widehat{C}$, but it is not so clear that circles can be easily constructed in $\Sigma$ to be orthogonal to $\widehat{C}$, since $\widehat{C}$ is not a great circle. We can't just place their centers on $\widehat{C}$. But assuming that they are orthogonal to $\widehat{C}$, can we also say that they are orthogonal to each other? And perhaps it doesn't matter for the proof, but we can't easily say that the two points in $\Sigma$ are inversions. Gary |
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Post by Admin on Oct 5, 2016 19:09:59 GMT
Gary
What is S? Do you mean $\Sigma$?
If so it's easy to draw the circles. Any two points lie on a great circle. Just draw a line perpendicular to this great circle and draw a circle centred anywhere on this line that passes through the two points.
The circles do not have to be orthogonal to each other and, as I wrote before, the two points in $\Sigma$ are definitely not inverses and are not meant to be.
Vasco
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Gary
GaryVasco
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Post by Gary on Oct 5, 2016 20:26:39 GMT
S is $\Sigma$ induced by a habit in Mathematica in which $\Sigma$ is typed with esc-S-esc.
Thanks for the note on how to draw the circles. I take it that by "line perpendicular to this great circle" you mean another great circle. But I have rewritten my proof and replaced it in the other thread. It uses elements of my proposed proof and orthogonal circles in $\Sigma$, which I obtained by another route. I also noted your response to the limiting case question and tried to write something similar.
Gary
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