Gary
GaryVasco
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Post by Gary on Mar 1, 2017 20:39:28 GMT
Maybe I didn't misunderstand you after all. In which case I would say that when the loop is traversed for a particular point $p$ we have to traverse the whole loop. So in diagram (d) for a point in the rightmost segment of the loop we cannot traverse just the segment of loop that surrounds it. Also, in figure 2 in the book on page 339, the values of $\nu$ in the little box on the right are derived from traversing the whole loop not just a segment of it for each of the sets $D_1,D_2,D_3,D_4$. If we applied what you are proposing to figure 2 the value of $\nu$ for each of the four sets $D_1,D_2,D_3,D_4$ would be $\pm1$. Also as you traversed the segment containing the set $D_3$ you would sometimes be going with the arrows and sometimes against them. Vasco Vasco, I agree entirely that you have to traverse the whole loop, and perhaps I was wrong to say that one could get the winding number by following just the segments of loop that border the $p$ in $\nu(L,p)$. But after several rechecks, it appears to me that the two methods result in the same $\nu(L,p)$, each shaded set with its own winding number. At least they match for this (d) loop. The most problematic set is the one that, starting from the bottom left corner, I would label p-set (2). If one were to draw in arrowheads on both the intersecting segments that form its boundary and trace always in the direction of the arrows, there is no need to reverse direction on the boundary. $\nu(L,p) = -1$ is the same by both methods. Tracing just the immediate boundary might not work for other complex loops. You are right to point out that the hand of the watch whose tip traces the whole loop would be sometimes going with the boundary trace and sometimes against, so that is a good argument for tracing the whole loop. Gary
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Post by Admin on Mar 1, 2017 21:17:52 GMT
Gary
Try checking figure 2 on p. 339.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 1, 2017 22:09:07 GMT
Gary Try checking figure 2 on p. 339. Vasco Vasco, Very interesting. It clearly proves that tracing the inner boundary of a region can give the wrong answer. Gary
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Post by Admin on Mar 2, 2017 6:25:57 GMT
Gary Try checking figure 2 on p. 339. Vasco Vasco, Very interesting. It clearly proves that tracing the inner boundary of a region can give the wrong answer. Gary Gary Also in loops (c) and (f) it would give the wrong answer for the inner loop. Your diagram (d) should now have all the values equal to 1 to be consistent with the others. I think that when we write an expression like $\nu(L,p)=2$ we need to be specific about what $L$ represents and what $p$ represents. Vasco
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Gary
GaryVasco
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Post by Gary on Mar 3, 2017 1:46:56 GMT
Vasco, Very interesting. It clearly proves that tracing the inner boundary of a region can give the wrong answer. Gary Gary Also in loops (c) and (f) it would give the wrong answer for the inner loop. Your diagram (d) should now have all the values equal to 1 to be consistent with the others. I think that when we write an expression like $\nu(L,p)=2$ we need to be specific about what $L$ represents and what $p$ represents. Vasco Vasco, The point about (c) and (f) is a good one. In diagram (d), p. 338 (and my diagram), I do not see how to score all the p-regions as 1. If we number the regions from 1 to 4 from the lower left, regions 2 and 4 seem to be differently oriented enclosures than regions 1 and 3. If we choose a point anywhere in region 2 and start anywhere on a segment bounding region 2 and follow the arrows all the way around the loop, I don't see how we avoid giving it a -1, as in (b). This may be a case where your third point would apply. Gary
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Post by Admin on Mar 3, 2017 5:25:41 GMT
Gary
Yes, I agree, you are right of course.
I see from the very first sentence of the section entitled, appropriately enough, "The Definition", on page 338, that the definition of $L$ includes its sense.
I will put the arrows back on my diagram.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 3, 2017 6:01:15 GMT
Vasco,
I had not noticed that before, "once in its given sense", right in the first sentence of the definition section! But it does resolve the problem. One has to think about each phrase.
Gary
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Post by Admin on Mar 3, 2017 6:25:20 GMT
Vasco, I had not noticed that before, "once in its given sense", right in the first sentence of the definition section! But it does resolve the problem. One has to think about each phrase. Gary Gary Dead right. Vasco
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Post by Admin on Mar 3, 2017 11:05:08 GMT
Gary
I have added another short commentary to my document just after my solution to the first suggested exercise. It has messed up the page numbering a bit as referred to in my previous posts, but as the document is short it shouldn't be to difficult for others to find what they are looking for. When we have finished the chapter I could add a contents page. I discovered a quantity called 'turning number' which is intrinsic to a loop, unlike the winding number.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 3, 2017 16:40:16 GMT
Gary I have added another short commentary to my document just after my solution to the first suggested exercise. It has messed up the page numbering a bit as referred to in my previous posts, but as the document is short it shouldn't be to difficult for others to find what they are looking for. When we have finished the chapter I could add a contents page. I discovered a quantity called 'turning number' which is intrinsic to a loop, unlike the winding number. Vasco Vasco, I have lost track of the location of the original document. Oh, I see that that the link occurs on page 1 of this thread. I'm not sure which commentary is the new one. Is it the one answering the question regarding the slope of the graph on p. in the last paragraph of subsection 2? I read it and it looked correct. I took away the lesson that when the term "speed" is introduced, one should introduce time as a parameter. I found the new material on turning number. Can we conclude that while the winding number depends on $R(\theta)$ and $\Phi(\theta)$ in a given sense of L, the turning number depends only on $\Phi(\theta)$ in a given sense of L? Gary
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Post by Admin on Mar 3, 2017 17:31:56 GMT
Gary I have added another short commentary to my document just after my solution to the first suggested exercise. It has messed up the page numbering a bit as referred to in my previous posts, but as the document is short it shouldn't be to difficult for others to find what they are looking for. When we have finished the chapter I could add a contents page. I discovered a quantity called 'turning number' which is intrinsic to a loop, unlike the winding number. Vasco Vasco, I have lost track of the location of the original document. Oh, I see that that the link occurs on page 1 of this thread. I'm not sure which commentary is the new one. Is it the one answering the question regarding the slope of the graph on p. in the last paragraph of subsection 2? I read it and it looked correct. I took away the lesson that when the term "speed" is introduced, one should introduce time as a parameter. I found the new material on turning number. Can we conclude that while the winding number depends on $R(\theta)$ and $\Phi(\theta)$ in a given sense of L, the turning number depends only on $\Phi(\theta)$ in a given sense of L? Gary Gary No, it comes immediately after my solution to the first suggested exercise - pages 1-2 of the document. Just seen your edit/question - I'll get back to you as soon as. Vasco
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Post by Admin on Mar 6, 2017 13:22:03 GMT
Gary
I'm not sure I understand what you mean here. If you walk round the loop while holding a stick so that it always points along the tangent to the curve of the loop then the turning number is the number of complete turns made by the stick when you arrive back at your starting point.
Vasco
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Gary
GaryVasco
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Post by Gary on Mar 14, 2017 20:24:51 GMT
Gary I'm not sure I understand what you mean here. If you walk round the loop while holding a stick so that it always points along the tangent to the curve of the loop then the turning number is the number of complete turns made by the stick when you arrive back at your starting point. Vasco Vasco, The question doesn't make sense. I withdraw it. Gary
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