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 August 4th, 2015, 05:32 AM #11 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 I am still worried about both the original formula and the third diagram (bottom right). There is the issue of radians v degrees and the question of is there a bit missing from the formula? That is should the RHS be arccot (cot^-1) or should the left hand side contain a trigonometric function? In the third diagram you show a (straight) line from the centre passing through a point on the surface of the sphere and then projected on to apparantly intersect the surface again at another point. This does not make sense unless the thid diagram is some sort of perspective view. Please explain.
August 4th, 2015, 05:52 AM   #12
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Quote:
 Originally Posted by steveupson We KNOW that when your point B is at the North pole, then the tangent of this point is congruent with a line of longitude passing through this point. E=90° gives α=0°
How do you know that E=90° gives α=0°? I don't see anything wrong with E=90° gives α=-90°. Basically, the plane is vertical, cuts the circumference of the circle at the North pole and the tangent to the circle is a horizontal line, running along the plane. Since every longitudinal line extending out from the North pole goes South, the particular "South" line of interest is the one which runs through the North pole over the spherical surface to A, which is the opposite side of the circle at the equator.

Quote:
 These are not the same. There is a method to use three discrete rotations about the x, y, and z in order to get a straight-on look at the angle α.
Yeah, we have probably calculated different things. However, I don't understand what angle $\displaystyle \alpha$ is in your case now. What plane is your $\displaystyle \alpha$ in? In my case it is in the plane of the circle.

In any case, I cannot reproduce the values you posted. Are you sure the equation is as you stated in the first post? Blue line is my function, green line is yours.

Here's the plotting script (Python). No cot function was available in numpy, so I calculated 1/tan instead. Let me know if there's a mistake.

Code:
import matplotlib.pyplot as mpl
import numpy as np

def f(x):
"""
Benit13's function
This functions returns alpha for a given E (in degrees).
"""
return np.arcsin((1.0 - np.tan(x))/(1.0 + np.tan(x)))*180.0/np.pi

def g(x):
"""
SteveUpson's function
This functions returns alpha for a given E (in degrees).
"""
return (1.0/(np.tan((1.0 - np.sin(x))/(np.sin(x)))))*180.0/np.pi

x = []
y1 = []
y2 = []

# Iterate over E between 0 and pi/2 radians
for angle in np.arange(0.0, np.pi/2.0, 0.0001):
x.append(angle/np.pi*180.0)
y1.append(f(angle))
y2.append(g(angle))

mpl.plot(x,y1, color = "blue", label="f")
mpl.plot(x,y2, color = "green", label="g")
mpl.xlabel(r"E (degrees)")
mpl.ylabel(r"$\alpha$ (degrees)")
mpl.ylim(-90.0, 90.0)
mpl.xlim(0.0, 90.0)

mpl.show()
Attached Images
 alpha_Eplot2.jpg (11.5 KB, 3 views)

August 4th, 2015, 07:34 AM   #13
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Quote:
 Originally Posted by studiot I am still worried about both the original formula and the third diagram (bottom right). There is the issue of radians v degrees and the question of is there a bit missing from the formula? That is should the RHS be arccot (cot^-1) or should the left hand side contain a trigonometric function? In the third diagram you show a (straight) line from the centre passing through a point on the surface of the sphere and then projected on to apparantly intersect the surface again at another point. This does not make sense unless the thid diagram is some sort of perspective view. Please explain.

All three views are in 3D. The line from the center passes through the point which is on the surface. Therefore, the angle shown for E is in perspective, yes. To see the actual angle in the plane in which it is described, the sphere should be rotated until the point on the surface is coincident with the outline of the sphere, thereby placing the point on the surface in that view. (Another way is to draw a longitude line that intersects the point and then extend a line from the center to the point where ths latttude contacts the circle representing the sphercal surface.)

As for degrees vs. radians, I don't quite follow what you are asking. The function is exactly the same as all other known trigonometric functions. I don't understand the question.

August 4th, 2015, 08:55 AM   #14
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Quote:
 Originally Posted by Benit13 How do you know that E=90° gives α=0°? I don't see anything wrong with E=90° gives α=-90°. Basically, the plane is vertical, cuts the circumference of the circle at the North pole and the tangent to the circle is a horizontal line, running along the plane. Since every longitudinal line extending out from the North pole goes South, the particular "South" line of interest is the one which runs through the North pole over the spherical surface to A, which is the opposite side of the circle at the equator.

It's a 3D explanation, and this is why an animation would be much better than trying to draw this out. If we have basically two separate planes, the one you describe that contains the longitude and passes through the center of the sphere and the one that contains the tangent line and passes through the center of the sphere.

These planes intersect each other at a defined angle. The function describes this angle.

Quote:
 Yeah, we have probably calculated different things. However, I don't understand what angle $\displaystyle \alpha$ is in your case now. What plane is your $\displaystyle \alpha$ in? In my case it is in the plane of the circle.
I only speak pidgin math so my guess is that my expression is not stated correctly. Perhaps you can help me sort it out. My ability to do algebra isn't very useful.

For the example E=60° I get:

1/tan(8.79°) = sin(60°)/(1-sin(60°)

If this is something else, then that's my bad, and I'll apologize in advance.

August 4th, 2015, 09:06 AM   #15
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Quote:
 α = cot ((1 – sin E) / sin E) = cot (1/sin E – 1)
Quote:
 1/tan(8.79°) = sin(60°)/(1-sin(60°)
are you trying to say that

tan (alpha) = 1/sinE - 1?

August 4th, 2015, 09:22 AM   #16
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Math Focus: non-euclidean geometry
Quote:
 Originally Posted by studiot are you trying to say that tan (alpha) = 1/sinE - 1?
That very well could be. I honestly don't know.

edited to add: I'll get my notes on this scanned and post them later on.

Last edited by steveupson; August 4th, 2015 at 09:24 AM.

 August 4th, 2015, 10:11 AM #17 Senior Member   Joined: Jun 2015 From: England Posts: 915 Thanks: 271 Thank you steve.
 August 4th, 2015, 05:52 PM #18 Senior Member     Joined: Jul 2015 From: Florida Posts: 154 Thanks: 3 Math Focus: non-euclidean geometry Except for the handwritten formula at the very bottom of the first page, the entire write-up is from more than ten years ago. So, even though I personally wrote it, my memory about it has faded. All of the illustrations were done in with a word processor, so nothing has been actually graphed or drawn to scale. None of this has been looked at or checked for errors or anything. Feel free to pick it apart mercilessly. It is fairly difficult to follow. Step 3.pdf -- 1.5 meg
 August 5th, 2015, 02:27 AM #19 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,157 Thanks: 732 Math Focus: Physics, mathematical modelling, numerical and computational solutions Let's call the point where the red line intersects the surface of the sphere C. The horizontal line in Fig. 2 then represents a horizontal plane. What determines the green point in Figure 2? Is it the intersection of the horizontal plane with the radius of the circle (that is, the radius pointing towards the North pole) or is it the intersection of the horizontal plane with the circumference of the circle? It is important because in the first case, alpha will always lie in the vertical plane (together with E). In the second case, the plane of alpha will change with E. Last edited by Benit13; August 5th, 2015 at 02:51 AM.
August 5th, 2015, 07:42 AM   #20
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Math Focus: non-euclidean geometry
Quote:
 Originally Posted by Benit13 Let's call the point where the red line intersects the surface of the sphere C. The horizontal line in Fig. 2 then represents a horizontal plane. What determines the green point in Figure 2? Is it the intersection of the horizontal plane with the radius of the circle (that is, the radius pointing towards the North pole) or is it the intersection of the horizontal plane with the circumference of the circle? It is important because in the first case, alpha will always lie in the vertical plane (together with E). In the second case, the plane of alpha will change with E.
It's the second case, where the intersection lies on the circumference. The plane of alpha rotates only 90°, though, when moving from the bottom of the small circle (equator) to the top (pole).

When we say "plane of alpha" we are talking about the plane passing through the center of the sphere that the tangent of the intersection lies in. This plane rotates 90° (relative to a line of longitude) from the equator to the pole.

 Tags function, non-euclidean, spherical, trig

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