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July 11th, 2015, 05:37 AM   #1
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Math Focus: non-euclidean geometry
New Spherical Trig Function

I have recently begun sorting through some of our old writings and I came across what looks to me like a new spherical trigonometric function.

It goes like this:

The relationship of the angle α, formed between the tangent of a point along a 45° small circle laid on the surface of a sphere and tilted to 45° (such that the small circle intersects both the pole and the equator) and a line of longitude passing through that same point, and the elevation angle E of this point (defined as the angle between the equator and tangent point as measured from the center of the sphere) is:

α = cot ((1 – sin E) / sin E) = cot (1/sin E – 1)

At this time, there are no hits at all on google for this equation:

https://www.facebook.com/photo.php?f...3424896&type=1

https://www.facebook.com/photo.php?f...3424896&type=1

This was posted at Wolfram a couple days ago and there has been no discussion over there. Geometry Forum—Wolfram Community

It would be ok if anyone has any questions.
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July 28th, 2015, 09:24 AM   #2
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Could you draw a diagram please?
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July 28th, 2015, 10:26 AM   #3
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An animation would be a lot better.
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File Type: jpg An animation would be a lot better.JPG (31.2 KB, 20 views)
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July 29th, 2015, 01:29 AM   #4
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Have you a derivation of this formula?

It seems strange since you are working in degress yet you have an angle allegedly equal to the cotangent of an angle that can only be measured in radians.

I can confirm that the small circle has to be the 45 degree parallel of altitude to fit the between the equator and pole as shown.
It has to be translated as well as rotated to fit into the assigned position.


Quote:
have recently begun sorting through some of our old writings
The question appears to me to be the sort of thing a 1920s teacher might have set his class to keep them quiet.

Last edited by studiot; July 29th, 2015 at 01:57 AM.
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July 29th, 2015, 08:15 AM   #5
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I actually managed to obtain the following:

$\displaystyle \tan E = \frac{1-\sin\alpha}{1+\sin\alpha}$

which makes sense to me. When $\displaystyle \alpha$ = 90 degrees, E = 0 (equator). When $\displaystyle \alpha$ = 0 degrees, E = 45 degrees.

Here's the derivation (sorry, it's a bit wordy... I should draw a diagram and add points to it).

Draw a line from the centre of the small, tilted circle (C) to the point where the circle touches the equator (A). This has length r (radius of circle).

Now subtend an angle E upwards from the equatorial plane from the centre of the sphere and draw a plane. The plane will cut the radius of the circle at a given position X. Let's assign a length x to the line XA. Let's also name the point (actually two points, but we only care about one...) where the plane intersects the circumference of the circle B.

So, there is a triangle BCX in the plane of the circle.

The angle $\displaystyle \alpha$ is the angle between the South longitude line and the tangent of the circle at the point on the circle's circumference where the plane intersects the circle. This angle is the same as angle CBX.

Therefore:
$\displaystyle \sin \alpha = \frac{r-x}{r} = 1 - \frac{x}{r}$

$\displaystyle x = r(1-\sin\alpha)$

Now let's draw a line vertically downwards (from the persepective of someone sitting at the North pole) from X. The line hits the equatorial plane somewhere within the sphere at a point D. Let's assign the length h = DX.
If the sphere has radius R, then

$\displaystyle \tan E = \frac{h}{R-h}$ (since the small circle is tilted at 45 degrees)

So far, we should have everything in accordance with the diagram posted by SteveUpson, albeit with names to some of the points and some algebraic letters assigned to various lengths.

We also know $\displaystyle h = \sqrt{2} x$ because of the 45 degree isosceles triangle with hypoteneuse AX. Therefore:

$\displaystyle \tan E = \frac{\frac{x}{\sqrt{2}}}{R - \frac{x}{\sqrt{2}}} = \frac{x}{\sqrt{2}R - x}$

We also know that $\displaystyle r = \frac{diameter}{2} = \frac{\sqrt{2} R}{2} = \frac{R}{\sqrt{2}}$

So $\displaystyle \tan E = \frac{x}{\sqrt{2}R-x} = \frac{x}{2r - x}$

Substituting for $\displaystyle x$:

$\displaystyle \tan E = \frac{r(1-\sin\alpha)}{2r - r(1-\sin\alpha)} = \frac{1-\sin\alpha}{1 + \sin\alpha}$

I don't know if it's equivalent to your result or not.
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July 30th, 2015, 03:58 AM   #6
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I did a bit more investigating...

The relation in my last post can be rearranged to form

$\displaystyle \sin \alpha = \frac{1-\tan E}{1 + \tan E}$

I attached a plot of the function and this seems like a sensible curve. I think this is the correct result.

Therefore, to answer your question, I think you cannot find your result in the literature because it is incorrect.
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July 30th, 2015, 05:06 AM   #7
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Attachment. The y-axis limits are 90 and -90 degrees, but for some reason the numbers didn't get displayed by my plotting tool.

Also, here's an error correction from my first post: The line

"We also know $\displaystyle h=\sqrt{2}x$ because of the 45 degree isosceles triangle with hypoteneuse AX. Therefore:"

should read

"We also know $\displaystyle h=\frac{x}{\sqrt{2}}$ because of the 45 degree isosceles triangle with hypoteneuse AX. Therefore:"

The rest of the derivation is unaffected by this error.
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File Type: jpg alpha_Eplot.jpg (8.4 KB, 8 views)

Last edited by Benit13; July 30th, 2015 at 05:11 AM.
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August 2nd, 2015, 07:29 PM   #8
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Quote:
Originally Posted by Benit13 View Post

Therefore, to answer your question, I think you cannot find your result in the literature because it is incorrect.

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°


You started off correct with E=0° gives α=90°


Quote:
The angle α is the angle between the South longitude line and the tangent of the circle at the point on the circle's circumference where the plane intersects the circle. This angle is the same as angle CBX.



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.





also, if we graph the new function, we get this:

E=0° α=90°

E=30° α=45°

E=45° α=22.5°

E=60° α=8.79°

E=90° α=0°
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August 2nd, 2015, 07:45 PM   #9
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transform for viewing angle in a plane tangent to the point

transform for viewing angle in a plane tangent to the point (shear plane)
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File Type: pdf pg transfrm.pdf (16.3 KB, 14 views)
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August 2nd, 2015, 08:58 PM   #10
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Quote:
Originally Posted by studiot View Post
Have you a derivation of this form"cot((1-sin(x))/sin(x))"ula?

My notes claim that we found a proof for this by "solving a rather awkward right spherical triangle (Napier)" but I haven't tried to go back and figure out what we were talking about.


Edited to add> My best guess is that it's the right triangle formed by the equator, longitude line, and tangent.


Edited again to add> It's awkward because the right angle is on the opposite side of the sphere.


Nope, it's got to be the great circle formed by the plane that passes through the center of small circle. It's been years.

Last edited by steveupson; August 2nd, 2015 at 09:25 PM.
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