Gary
GaryVasco
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Post by Gary on Jun 20, 2016 0:49:38 GMT
Vasco, I'm attaching a document with a few notes and a question regarding the Introduction. Rather than saving up all the notes for the section, I attach them now because they may help with Exercise 1. nh.ch6.notes.questions.pdf (17.16 KB) (An expanded version of these notes and questions was added in a new message on June 21, 12:30 am.) Gary |
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Post by Admin on Jun 20, 2016 15:14:03 GMT
Gary
I do not understand figure 4 on page 273:
1. In the left-hand part of the figure, is the piece of paper in the plane of the page? The designated right angle is not a right angle if you measure it and the little 'square' box is not square. The triangle is supposed to be a 60, 30, 90 triangle, but the sides are not the right length for that to be true (ie if the hypotenuse is 2 then the other two sides should be 1 and $\sqrt{3}$.
2. I don't see how the left-hand part of the figure can be bent into the right-hand part of the figure.
3. The middle figure looks like it would be a rectangle if it were laid out flat.
Maybe the left-hand part of the figure is supposed to be a perspective drawing of a flat rectangle. If so then it could be bent into the middle part of the diagram, but not the right-hand part. If not then it cannot be bent to resemble the middle or the right-hand part!
I'm ready to give up.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 20, 2016 16:11:44 GMT
Gary I do not understand figure 4 on page 273: 1. In the left-hand part of the figure, is the piece of paper in the plane of the page? The designated right angle is not a right angle if you measure it and the little 'square' box is not square. The triangle is supposed to be a 60, 30, 90 triangle, but the sides are not the right length for that to be true (ie if the hypotenuse is 2 then the other two sides should be 1 and $\sqrt{3}$. 2. I don't see how the left-hand part of the figure can be bent into the right-hand part of the figure. 3. The middle figure looks like it would be a rectangle if it were laid out flat. Maybe the left-hand part of the figure is supposed to be a perspective drawing of a flat rectangle. If so then it could be bent into the middle part of the diagram, but not the right-hand part. If not then it cannot be bent to resemble the middle or the right-hand part! I'm ready to give up. Vasco Vasco, I see what you mean. I don't expect these illustrative graphics to be precise, so I took Needham at his word that the angles are all as he described them. The diagram on the right is clearly conical, so the diagram on the left must be an isosceles trapezoid in order to get a cone without stretching the far right hand side, as Needham stipulated with "Clearly we can bend such a flat piece of paper into either of the two (extrinsically) curved surfaces on the right." Unfortunately, the right hand side of the left hand diagram should be the longest, but it does not appear to be so. And then the panel in the middle should also be an isosceles trapezoid, but, as you say, it appears to be a rectangle. The short sides are very close to the same length. So I agree with you, that the diagrams are not consistent with the text. I would resolve it this way: All the angles are as Needham specified and they don't change. All the diagrams are presented in three dimensions, so you have to imagine a bit of perspective for the diagram on the left. The perspective makes it look a bit smaller to the northeast. The center diagram is unproblematic. The diagram on the right hand side was mistakenly drawn in a conical shape. It should be a cylinder. One might get the conical appearance from perspective, but it appears too short for that and it would need to be rotated a bit more clockwise. Perhaps this one should go on the error list. Gary |
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Post by Admin on Jun 20, 2016 16:31:50 GMT
Gary I do not understand figure 4 on page 273: 1. In the left-hand part of the figure, is the piece of paper in the plane of the page? The designated right angle is not a right angle if you measure it and the little 'square' box is not square. The triangle is supposed to be a 60, 30, 90 triangle, but the sides are not the right length for that to be true (ie if the hypotenuse is 2 then the other two sides should be 1 and $\sqrt{3}$. 2. I don't see how the left-hand part of the figure can be bent into the right-hand part of the figure. 3. The middle figure looks like it would be a rectangle if it were laid out flat. Maybe the left-hand part of the figure is supposed to be a perspective drawing of a flat rectangle. If so then it could be bent into the middle part of the diagram, but not the right-hand part. If not then it cannot be bent to resemble the middle or the right-hand part! I'm ready to give up. Vasco Vasco, I see what you mean. I don't expect these illustrative graphics to be precise, so I took Needham at his word that the angles are all as he described them. The diagram on the right is clearly conical, so the diagram on the left must be an isosceles trapezoid in order to get a cone without stretching the far right hand side, as Needham stipulated with "Clearly we can bend such a flat piece of paper into either of the two (extrinsically) curved surfaces on the right." Unfortunately, the right hand side of the left hand diagram should be the longest, but it does not appear to be so. And then the panel in the middle should also be an isosceles trapezoid, but, as you say, it appears to be a rectangle. The short sides are very close to the same length. So I agree with you, that the diagrams are not consistent with the text. I would resolve it this way: All the angles are as Needham specified and they don't change. All the diagrams are presented in three dimensions, so you have to imagine a bit of perspective for the diagram on the left. The perspective makes it look a bit smaller to the northeast. The center diagram is unproblematic. The diagram on the right hand side was mistakenly drawn in a conical shape. It should be a cylinder. One might get the conical appearance from perspective, but it appears too short for that and it would need to be rotated a bit more clockwise. Perhaps this one should go on the error list. Gary Gary I think your resolution is good and the question posed in the footnote 3 can still be posed with respect to the right-hand diagram as shown in the book, to which the answer is then "No" as you say in your document. I feel better about it now. I agree it should go on the error list. Not a very auspicious start to chapter 6. All this is backed up by figure 6 on page 276 where the left-hand figure is obviously shown in perspective and has light and dark grey shading as does the figure under discussion. I will move on. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 20, 2016 16:56:50 GMT
Vasco, I see what you mean. I don't expect these illustrative graphics to be precise, so I took Needham at his word that the angles are all as he described them. The diagram on the right is clearly conical, so the diagram on the left must be an isosceles trapezoid in order to get a cone without stretching the far right hand side, as Needham stipulated with "Clearly we can bend such a flat piece of paper into either of the two (extrinsically) curved surfaces on the right." Unfortunately, the right hand side of the left hand diagram should be the longest, but it does not appear to be so. And then the panel in the middle should also be an isosceles trapezoid, but, as you say, it appears to be a rectangle. The short sides are very close to the same length. So I agree with you, that the diagrams are not consistent with the text. I would resolve it this way: All the angles are as Needham specified and they don't change. All the diagrams are presented in three dimensions, so you have to imagine a bit of perspective for the diagram on the left. The perspective makes it look a bit smaller to the northeast. The center diagram is unproblematic. The diagram on the right hand side was mistakenly drawn in a conical shape. It should be a cylinder. One might get the conical appearance from perspective, but it appears too short for that and it would need to be rotated a bit more clockwise. Perhaps this one should go on the error list. Gary Gary I think your resolution is good and the question posed in the footnote 3 can still be posed with respect to the right-hand diagram as shown in the book, to which the answer is then "No" as you say in your document. I feel better about it now. I agree it should go on the error list. Not a very auspicious start to chapter 6. All this is backed up by figure 6 on page 276 where the left-hand figure is obviously shown in perspective and has light and dark grey shading as does the figure under discussion. I will move on. Vasco Vasco, With some mental effort, I can perceive the right hand side diagram as an elliptical cylinder with roughly Pi/4 counter-clockwise rotation in the z axis. Perhaps that is what was intended. If so, it could be obtained by bending a rectangle. Gary |
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Post by Admin on Jun 21, 2016 6:54:14 GMT
Gary
In your document you write:
Since $\kappa$ is a number, we cannot think of it as rotating. It is the plane $\Pi$ that we think of as rotating, using the vector n (perpendicular to the surface at $p$), as the axis of rotation. As it rotates, $\Pi$ intersects the surface in a two dimensional curve which passes through $p$ and $\kappa$ is the curvature of this two dimensional curve at $p$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 21, 2016 7:15:18 GMT
Vasco,
I have expanded my notes and questions.
Revised and added to later reply, June 22, 2016.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 21, 2016 7:20:52 GMT
Gary In your document you write: Since $\kappa$ is a number, we cannot think of it as rotating. It is the plane $\Pi$ that we think of as rotating, using the vector n (perpendicular to the surface at $p$), as the axis of rotation. As it rotates, $\Pi$ intersects the surface in a two dimensional curve which passes through $p$ and $\kappa$ is the curvature of this two dimensional curve at $p$. Vasco Vasco, Thank you. I see it now. Careless reading. Gary |
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Post by Admin on Jun 21, 2016 14:28:45 GMT
Gary
In your document you write
On page 275 you will see that Needham is talking about the Gaussian curvature $k$ here, not $\kappa$ the curvature introduced in chapter 5 section IX. The way I see it, on the surfaces illustrated in figure 4 page 273, at least one of $\kappa_{min}$ and $\kappa_{max}$ must be zero at every point on the surface and so $k(p)\equiv\kappa_{min}\kappa_{max}\equiv 0$. None of these surfaces resembles the two left hand diagrams in figure 5 on page 274 at any point $p$.
In my opinion, there is a change in the principal curvatures if the surface is not a plane, but their product, the Gaussian curvature, is zero, because there is always a direction in which at least one of them is zero. At every point $p$ the surface is like the right hand diagram in figure 5.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 21, 2016 15:33:19 GMT
Vasco,
I'm in total agreement with your use of $k_{min}k_{max} ≡ 0$ as the intended way of thinking about this. I did not mean to imply that I was using the $\kappa$ curvature from Chapter 5. I just have difficulty distinguishing between $k$ and $\kappa$ in Needham, so I thought that Needham had written $\kappa$. While my comments may not be perfectly on target, it does seem to me that they are not entirely wrong. That is, if $k_{min}k_{max} = 0$, then $k = E(T)/A(T) = 0/A(T) = 0$. I think an illustration of how $k$ varies with rotation of the plane $\Pi$ would be helpful. I can't see how rotation of $\Pi$ would have anything to do with it unless one is observing $k$ along a particular axis on $\Pi$ as $\Pi$ rotates.
Aside from that, I'm not completely tranquil with the distinction between intrinsic and extrinsic curvature and the idea that one can bend a surface a little bit without stretching it, thereby leaving the intrinsic curvature undisturbed. How much is too much? The four examples we have all seem to illustrate flatness as the result of 0 curvature in at least one direction. For example, if we were to bend the $k(p) = 0$ example in [5] even an infinitesimal bit in direction of the $k(p) > 0$ example, would we not have $k(p) > 0$? What do you think about this?
Gary
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Post by Admin on Jun 21, 2016 15:47:50 GMT
Gary
I did not mean to imply that your comments were wrong, I was just worried about the use of $\kappa$ instead of $k$.
I'll have a think about your point concerning the distinction between intrinsic and extrinsic curvature and get back to you.
Vasco
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Post by Admin on Jun 22, 2016 11:04:54 GMT
Gary
On page 273 subsection 4 Needham is just pointing out that if we stretch the skin of the vegetable then $k(p)$ will change, and so in order to keep $k(p)$ the same what we do is to make sure we only bend it within the limits required to avoid stretching it. This could allow a large amount of bending or only a small amount of bending - it depends on the properties of the vegetable skin. If we make sure we do not stretch it then $k(p)$ will remain unchanged.
If we were to bend the $k(p)=0$ example in figure [5] even slightly, so that locally it was like a sphere, then we would have stretched it/compressed it and so, as you say, $k(p)$ would then be $>0$, but we could also bend it without stretching it and $k(p)$ would then still be zero.
The "small amount" phrase is nothing to do with infinitesimals, but just to do with the bending being small enough to avoid any stretching/compressing.
I think that the phrase "bend it a little" is misleading because, as you say, even a very, very small amount of bending could result in stretching/compressing. I think it would be better to say "bend it in such a way that there is no stretching/compressing".
I hope this is some help.
Vasco
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Post by Admin on Jun 22, 2016 15:54:00 GMT
Gary
In item 8 of your document you write
Since $\Lambda$ is the same whatever the direction of $d\widehat{s}$ then we can choose the $d\widehat{s}$ which is parallel to $\mathbb{C}$ and this enables us to derive the result $d\widehat{s}=\frac{2}{[Nz]^2}ds$ without using (6) on page 126.
From reading your other comments, I get the impression that you have misunderstood what Needham has written at the top of page 286.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 23, 2016 5:10:40 GMT
Gary In item 8 of your document you write Since $\Lambda$ is the same whatever the direction of $d\widehat{s}$ then we can choose the $d\widehat{s}$ which is parallel to $\mathbb{C}$ and this enables us to derive the result $d\widehat{s}=\frac{2}{[Nz]^2}ds$ without using (6) on page 126. From reading your other comments, I get the impression that you have misunderstood what Needham has written at the top of page 286. Vasco Vasco, I'm not sure I misunderstood it. I just didn't see the solution. I have revised. nh.ch6.notes.questions.pdf (271.75 KB) Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Jun 23, 2016 5:29:33 GMT
Vasco,
I agree, but what does that imply? Can we say that we can bend the plane without stretching or compressing so long as we leave the plane unbent in one direction?
By the way, did you notice my suggestion that the right hand side shape in Figure [4], p. 273 is a segment of an elliptic cylinder?
Gary
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