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Post by Admin on Nov 25, 2016 14:39:40 GMT
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Gary
GaryVasco
Posts: 3,352 
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Post by Gary on Dec 4, 2016 7:38:21 GMT
Vasco,
I've been working on problem 10 and getting basically the same results that you have up to the calculation of y. When I look at the similar triangles, whether on my plot or yours, it looks to me like $dy/d\sigma = X/R = X$ (unit sphere). Your version ($dy/d\sigma = 1/X$) has to be correct, because that's the only way to arrive at the proper equation for the Mercator map, but I don't see it. Can you suggest a reason why?
QUESTION WITHDRAWN. Triangle in my hemisphere figure was mislabeled.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 5, 2016 0:47:16 GMT
Vasco, Here is my attempt at Exercise 10. This seemed very tricky, so discussion would be welcome. nh.ch6.ex10.pdf (392.98 KB) Gary |
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Post by Admin on Dec 5, 2016 6:52:35 GMT
Gary
Your solution seems fine on a first look. What aspect in particular would you like to discuss?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 5, 2016 16:05:02 GMT
Gary Your solution seems fine on a first look. What aspect in particular would you like to discuss? Vasco Vasco, I was puzzled by the fact that the circles in the map overlap rather than touch. Gary |
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Post by Admin on Dec 5, 2016 16:29:11 GMT
Gary
Good point! How are you calculating the centres?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 5, 2016 18:04:48 GMT
Gary Good point! How are you calculating the centres? Vasco This is pseudocode for ease of reading, but very close to the actual: $\hspace{5em}DiskObject(\sigma,rDisc) =$ $\hspace{10em}\{$ $\hspace{10em}x = const.$ $\hspace{10em}X = |sin(\sigma)|;$ $\hspace{10em}y = -ln(cot(\sigma/2));$ $\hspace{10em}Disk((x, y), rDisc/X)$ $\hspace{10em}\}$ I think to get precise conformality, it might be necessary to plot the disc perimeters as points and apply $-ln(cot(\sigma/2))$ to every point of every disc. Gary |
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Post by Admin on Dec 7, 2016 6:23:58 GMT
Gary
I seem to remember that you drew figure 20 on page 297. How did the circles look then? It seems to me that the necessary maths is that used to do the suggested exercise on p. 309, but adapted to the sphere.
That is, calculate the top and bottom points of the vertical diameter and then combine them as in the exercise to get the radius and centre.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 7, 2016 7:11:04 GMT
Vasco,
The plot you mention contained only the small triangle. I didn't understand it well enough to plot the circles. Apparently, I still don't, but perhaps your suggestion will work. The equations on p. 309 appear to pertain only to the h-plane and the map, so as you say, they would have to be adapted to the sphere. I don't remember trying that exercise. I will have to look into that. Since the target is an Euclidean circle, it should be enough to calculate the radius.
Gary
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 7, 2016 19:51:48 GMT
Vasco, Here is the result of calculating the radius of the map discs with the following: $\hspace{5em}rMap = | | ln(cot((\sigma + r)/2)) | - | ln(cot(\sigma/2)) | |$ where r is the radius of a disc on the sphere. Instead of crowding the circles as happened with rDisc/X, it gives them too much space.  Gary |
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Post by Admin on Dec 7, 2016 21:05:39 GMT
Gary
What is your reason for not also calculating the position of the centre as in the suggested exercise?
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 7, 2016 22:16:06 GMT
Gary What is your reason for not also calculating the position of the centre as in the suggested exercise? Vasco Vasco, The centres were calculated as x = const. and y=−ln(cot(σ/2)) as above. I believe that conforms to the suggested exercise on p. 147. Gary |
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Post by Admin on Dec 8, 2016 7:06:58 GMT
Gary What is your reason for not also calculating the position of the centre as in the suggested exercise? Vasco Vasco, The centres were calculated as x = const. and y=−ln(cot(σ/2)) as above. I believe that conforms to the suggested exercise on p. 147. Gary Gary Sorry, I should have made it clear that I was referring to the suggested exercise on p. 309, where the centre is calculated by adding together the $y$ coordinates of the top and bottom of the vertical diameter and dividing by 2. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Dec 10, 2016 18:24:23 GMT
Vasco, The centres were calculated as x = const. and y=−ln(cot(σ/2)) as above. I believe that conforms to the suggested exercise on p. 147. Gary Gary Sorry, I should have made it clear that I was referring to the suggested exercise on p. 309, where the centre is calculated by adding together the $y$ coordinates of the top and bottom of the vertical diameter and dividing by 2. Vasco Vasco, I tried that, too, without success. Gary |
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Post by Admin on Dec 10, 2016 20:52:42 GMT
Gary
I will have a go at it myself with a spreadsheet, compasses and graph paper and get back to you.
Vasco
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