trigonometry

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Re: trigonometry
« Reply #30 on: December 08, 2017, 09:04:56 AM »
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.
This is not having the distance of the sun changing depending upon time of year.
This is having it change depending on the location you are at, so at the same time, it has a different height. That is impossible.

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect.
So that's your excuse? Just ignore the data that contradicts your view.

How about you address the massive lie in the image you provided?
So there is no data or sources backing up your little picture? I ask you to supply them, and you come back and retort 'you are ignoring them!'

How can I ignore what has not been given?

You are correct, there are two misspellings on that poster. Its unfortunate I had the plates made before recognizing that.
All data can be found on any site such as timeanddate that offers elevations. Checking those locations shows elevations to be correct on that day at the listed time. A search also shows the distances shown to be accurate using any map of the globe. The proposed sun heights are easy to determine using simple trig once these two facts are known. So what exactly is your problem with this image, beyond showing the impossibility of a single sun for a FE?

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Re: trigonometry
« Reply #31 on: December 08, 2017, 10:08:15 AM »
So you can't supply the data? My problem is that the data is incorrect and you refuse to source it so we can find out why.
Quantum Ab Hoc

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Re: trigonometry
« Reply #32 on: December 08, 2017, 10:19:03 AM »
So you can't supply the data? My problem is that the data is incorrect and you refuse to source it so we can find out why.
You claim it's incorrect. The angles are correct as checked at https://www.suncalc.org and the distances to the equator are correct as checked using https://sciencing.com/distance-city-equator-7484864.html the calculation for said distance. I do not know the original source, but both of those agree (within acceptable margin of error) with the data given in his diagram. As well a look at them upon a map shows them to indeed be upon the same line of longitude as their coordinates indicate.

So please, explain what is incorrect.

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Re: trigonometry
« Reply #33 on: December 08, 2017, 10:25:51 AM »
I have already supplied you with a list.

Are you saying you came up with your data points by calculating them assuming a round earth?

Why would you be measuring based off the distance to the equator? This would be wrong.
Quantum Ab Hoc

1 + 1 = 2
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Re: trigonometry
« Reply #34 on: December 08, 2017, 10:45:10 AM »
So you can't supply the data? My problem is that the data is incorrect and you refuse to source it so we can find out why.
The data is in the picture.  Do you object to existence of the locations?  Do you object to their latitude and longitude?  Do you object to the distances given between them?  Do you object to the angle of elevation of the Sun at each location on 20 March 2016 16:48 UTC? 

There is nothing there that requires a source.  That is all data that is either correct or incorrect, regardless of where it came from.  If you believe something is in error, specify what it is and we can all peer review the picture together.

FWIW, I just used google myself to look up the coordinates, so the locations appear to exist.  The coordinates in the picture appear to be accurate to 3 decimal places.   I used google to check the distance and I got a 40 km variance from Punta Arenas to Taraira and another 40 km variance from Taraira to Kimmirut, but that seems negligible given the distances involved.  I used the NOAA Solar Calculator (https://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html) to verify the Sun's angle of elevation at each coordinate on March 20, 2016 at 16:48 UTC, and found that the picture was accurate within a decimal place.  In other words, everything in the picture appears to be correct upon my, admittedly cursory, independent analysis.  If you think something is in error, explain exactly what and we can look into it with more scrutiny.


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Re: trigonometry
« Reply #35 on: December 08, 2017, 10:47:41 AM »
I have already supplied you with what you asked for. However, presumably the list of ridiculous errors were too long for you to hold in your short term memory. Here, let's start with the first issue:

Why would you be measuring based off the distance to the equator?
Quantum Ab Hoc

1 + 1 = 2
"The above proposition is occasionally useful." - Bertrand Russell

Re: trigonometry
« Reply #36 on: December 08, 2017, 10:49:12 AM »
I have already supplied you with a list.

Are you saying you came up with your data points by calculating them assuming a round earth?

Why would you be measuring based off the distance to the equator? This would be wrong.
I'm saying calculating the distance to the equator using the normal distribution of lines of longitude agrees with the distances given. If you disagree, please tell me how a flat km differs from a round km so I may adjust the calculations.

What would be wrong about using ASA trigonometry to find the height of the sun using the side as the distance from the point to the equator, which would have an angle of 90 degrees up to the sun on the equinox?

Also, are you referring to this?
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.

What is the source of this data?

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect. What is this trash? Did you find it on the back of a box of cereal? Also, these locations do not lie along a straight line as you seem to claim they do.
Correct, the suns elevation cannot be correct because each locations angle demands it be in a different location.
The angles are correct according to every source I can locate, and all sources are correct when checked against my location. If you believe the angles to be incorrect, supply the correct ones and your source.
The distances are correct based upon simple lines of latitude and the regular distance between them. Once again, describe the difference between a km upon a round Earth and one upon a flat Earth and I will happily run things again using those corrected distances.
Those locations lie upon a nearly straight line, as the longitude of them suggest. This is easily verified by plotting them out upon a map. The slight difference does not introduce enough error to account for the issues in the final numbers.

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Re: trigonometry
« Reply #37 on: December 08, 2017, 10:56:16 AM »
I am asking you why you think the equator is the appropriate location to measure from? You are assuming the location of the sun as above this to calculate its height. This is incorrect and discordant with flat earth theory.

Further, you are not using actual data, but calculating the angles based on the round earth model. Are you not worried about circularly justifying your own claim? Wouldn't real data be the correct choice here? Why don't you supply us with the sources you used to verify, or am I just to guess? I'm not going to source your arguments for you, you have to support your own argument with more than "Well, that's the angle that the round earth model says it should be, so its right."
Quantum Ab Hoc

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Re: trigonometry
« Reply #38 on: December 08, 2017, 11:00:20 AM »
I am asking you why you think the equator is the appropriate location to measure from? You are assuming the location of the sun as above this to calculate its height. This is incorrect and discordant with flat earth theory.
Did you miss the "@ equinox" note on the drawing?  I think the sun is over the equator on the equinox, isn't it?

Re: trigonometry
« Reply #39 on: December 08, 2017, 11:14:14 AM »
I am asking you why you think the equator is the appropriate location to measure from? You are assuming the location of the sun as above this to calculate its height. This is incorrect and discordant with flat earth theory.
Taraira is the actual point being measured from.  It is only a half degree south of the equator, so measuring to the equator would only result in about 55 km of error. 

We could replace this point entirely though if you want, because it has nothing to do with the Sun's position, it is just to a data point chosen close to the 70 degree longitude line in common with each of the others.  The distance is only necessary to provide one known side of a triangle that we can use with the angle of elevation to calculate the Sun's position.

Further, you are not using actual data, but calculating the angles based on the round earth model. Are you not worried about circularly justifying your own claim? Wouldn't real data be the correct choice here? Why don't you supply us with the sources you used to verify, or am I just to guess? I'm not going to source your arguments for you, you have to support your own argument with more than "Well, that's the angle that the round earth model says it should be, so its right."
The NOAA solar calculator I linked above is based on actual observational data.  If you question the accuracy of the solar calculator, you could start by making observations of the solar elevation at any point convenient to you at any time and check those observations against the calculator.  If you don't trust the agreement because it's only one location, you could choose several locations and check them. 

Using the angles you observe, the distance between the points, and trigonometry, you can calculate the height of the Sun.  The more points you use, the more positions the Sun will have to be in simultaneously in any flat Earth model.  There will always be one answer if the surface is curved into a sphere of approximately 4000 miles in radius. 

Re: trigonometry
« Reply #40 on: December 08, 2017, 11:43:12 AM »
I am asking you why you think the equator is the appropriate location to measure from? You are assuming the location of the sun as above this to calculate its height. This is incorrect and discordant with flat earth theory.

Further, you are not using actual data, but calculating the angles based on the round earth model. Are you not worried about circularly justifying your own claim? Wouldn't real data be the correct choice here? Why don't you supply us with the sources you used to verify, or am I just to guess? I'm not going to source your arguments for you, you have to support your own argument with more than "Well, that's the angle that the round earth model says it should be, so its right."
So the sun ISN'T 90 degrees above the equator in the FE hypothesis on the equinox? So we can just throw it out since it doesn't represent reality at this point can we not? That's the point of using the equator on the equinox after all, is the handy 90 degree elevation on that date when viewed from the equator. A well known fact.

ItsRoundIPromise has noted the data correlates with observational data. So the angles are correct. If you wish to dispute the distances I'm all ears, but first you must show the difference that would exist between a round Earth km and a flat Earth km. To my knowledge the differences would be well within normal margins of error and thus irrelevant.

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Re: trigonometry
« Reply #41 on: December 08, 2017, 12:40:29 PM »
So there is no data or sources backing up your little picture?
The data can be obtained from numerous sources, the majority of which you will just ignore. Would you trust time and date with their sun's position?
What about suncalc.org?
Regardless, the picture has the data on it.
Before even dealing with which is correct, there is the bigger issue to discuss, which is what was raised in the OP.

You are correct, there are two misspellings on that poster. Its unfortunate I had the plates made before recognizing that.
This is not the lie I am referring to.
I am referring to the claim that without assumptions and using simple trig you show Earth is flat, when in fact it assumes Earth is flat to arrive at its result.

Care to address that?

I have already supplied you with a list.
Where is your list?

Are you saying you came up with your data points by calculating them assuming a round earth?
How about the typical FE BS:
By using "strange, unjustified formulas" (i.e. those based upon a round Earth) which match patterns observed in reality to allow future predictions of the azimuth and angle of elevation of the sun.
Every test of these angles has been correct, at least to the nearest degree.

Why would you be measuring based off the distance to the equator? This would be wrong.
Because at the time the sun was over the equator?
Yes, it will have a small error due to the cities not being in a perfectly straight line, but that is irrelevant.

But like I said, first I want you to address the lie in your picture.
Using just those 2 points (I am ignoring the third due to symmetry), with the angle of the ground and the distance between those points, can you determine the shape of Earth and the distance to the sun? Yes or no?

Re: trigonometry
« Reply #42 on: December 09, 2017, 12:36:24 AM »
For those in the northern latitudes, we could extend the same problem to the altitude angle of polaris. In this comparison, you remove longitude as a factor and everyone in the northern latitudes can compare answers.

How does flat earth explain this:
at 10N, polaris is 10 deg above the horizon
20N, 20 deg
45N, 45 deg
etc.
up to the north pole, where it's directly overhead.

If the earth were flat, then polaris must be closer. If polaris is closer, and at a fixed altitude (it can only be in one place), then you could use the same trig to calculate the relative distances between latitudes given the observation angle above the horizon.

I did the calculation for 4 latitudes: 5N, 10N, 80N, and 85N and here's what I found:

The distance between latitudes 5N and 10N would need to be 64.7 x the distance between 80N and 85N, regardless of the altitude of polaris. Seems a bit unreasonable to me.

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rabinoz

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Re: trigonometry
« Reply #43 on: December 09, 2017, 02:47:55 AM »
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.

What is the source of this data?

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect. What is this trash? Did you find it on the back of a box of cereal? Also, these locations do not lie along a straight line as you seem to claim they do.
Incorrect!

Let's go right back here and get to where those distances and angles came from. If you don't like the distances coming from Google Earth and the angles from Sun Earth Tools., I can later post similar results in miles and with all the measurements from your own Wiki. The only significant if that the latter, not having any refraction or measurement error fit the Globe even better.
Note that the equator does not come into it, other than the sun being over the equator at solar noon on an equinox.

So sorry about the length, but you seemed to want it all.
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The sun's elevation is a physical measurement we can all easily make, but I'm lazy and will use values from Sun Earth Tools..
If you don't like such a "Globe Earth" source, I'll work out the angles from "the Wiki" - they look better!
There is more detail in Flat Earth Debate / So you think the sun is about 5,000 km high? « Message by rabinoz on August 24, 2016, 01:22:33 PM ».

The usual "a bit over 3000 miles" or 5000 km for the FE sun height comes from a "non-measurement" in about 1899 by Thomas Winship, author of Zetetic Cosmogony. He provides a calculation demonstrating that the sun can be computed to be relatively close to the earth's surface if one assumes that the earth is flat:
Quote
On March 21-22 the sun is directly overhead at the equator and appears 45 degrees above the horizon at 45 degrees north and south latitude. As the angle of sun above the earth at the equator is 90 degrees while it is 45 degrees at 45 degrees north or south latitude, it follows that the angle at the sun between the vertical from the horizon and the line from the observers at 45 degrees north and south must also be 45 degrees. The result is two right angled triangles with legs of equal length. The distance between the equator and the points at 45 degrees north or south is approximately 3,000 miles. Ergo, the sun would be an equal distance above the equator.
This is illustrated in this diagram from Modern Mechanics - Oct, 1931:

Voliva's Flat Earth Sun Distance.
This is also shown in the Wiki under Distance to the Sun in the section Sun's Distance - Modern Mechanics.
He "calculates" the height from only ONE location, Latitude 45°.

It would seem that we would get a more accurate result by taking measurements from a number of different latitudes and comparing the results.


So this time, I will present the sun elevations and azimuth from five locations all close to longitude 70°W.


These locations are shown on the Google Earth map on the right.


The sun azimuth and elevations have been found from: Sun Earth Tools.

If you have any doubts as the accuracy of this site, I suggest that a good test would be to check its accuracy where you live. I think if it is accurate at a lot of random locations is could be relied on for these locations.


   

Locations for Sun Height Calculations

The following table gives the data for each location. All sun elevation was obtained from Sun Earth Tools as close as possible to the local midday on the last equinox. The time was UTC 20/Mar/2016  16:48.


Location   

Latitude   

Longitude   

Sun Elev   
Dist from   
Vaupes   

Flat Sun Ht   
Lat Diff from   
Vaupes   
Calc
Circum
Kimmirut, Canada   
62.847°   
-69.869°   
27.36°   
7,034 km   
3,609 km   
63.58°   
39,828 km
Santo Domingo   
18.486°   
-69.931°   
71.72°   
2,107 km   
6,077 km   
19.22°   
39,465 km
Vaupes, Colombia   
-0.565°   
-69.634°   
89.06°   
0 km   
------   
   
   
Chupa District, Peru   
-15.109°   
-69.998°   
74.69°   
1,610 km   
6,256 km   
14.37°   
40,334 km
Punta Arenas, Chile   
-53.164°   
-70.917°   
36.63°   
5,830 km   
4,388 km   
52.43°   
40,031 km

These locations and the directions to the sun on a flat earth are shown in the left hand  diagram below:
Once we have the angles from two sites the height of the sun can be calculated from: h = d/(1/tan(A1) + 1/tan(A2)).


Sun Height on Flat Earth along 70°W Long
   

Sun Height on Globe Earth along 70°W Long

Using this method to find the height of the sun on the Flat earth gives measurements from 3609 km (for Kimmirut and Vaupes) to 6256 km (for Chupa District to Vaupes) depending on the spacing of the measurement sites. Using locations closer to the poles gives more extreme values still.

In other words, claiming that the Flat Earth sun is at about 5,000 km altitude has no foundation whatever.

It is very telling when we note that when we plot these angles on a spherical earth the directions to the sun are all parallel.
Explain that!

Now, if instead of using these measurements to determine the Flat Earth sun height, we use them as Eratosthenes did, assuming a distant sun and use this data to calculate the circumference of the earth.

The circumference can be calculated from (distance from Vaupes) * 360°/(angle difference of sun from Vaupes)

This time, we get far better consistency.
              The estimated figures for the circumference of the earth range from 39,465 km to 40,334 km.
Not too bad, when the accepted polar circumference of the Globe is 40,008 km.

Certainly these figures would indicate that the earth is a globe with a distant sun.

As I said before, some might doubt Sun Earth Tools., but using the elevations from the method in "the Wiki" gives "better results" for the Globe!

So, for me, that is one physical mathematical proof that the earth is a sphere.

But there's still more for a later post!

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rabinoz

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Re: trigonometry
« Reply #44 on: December 09, 2017, 03:07:18 AM »
Here is the sun height using data calculated from the Flat Earth Wiki.

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I will use the same locations (as in the previous post), so the latitudes will be the same, but even that doesn't matter.
Quote from: the Flat Earth Society Wiki
Finding your Latitude and Longitude
Latitude
To locate your latitude on the Flat Earth, it's important to know the following fact: The degrees of the Earth's Latitude are based upon the angle of the sun in the sky at noon equinox.

That's why 0˚ N/S sits on the equator where the sun is directly overhead, and why 90˚ N/S sits at the poles where the sun is at a right angle to the observer. At 45˚ North or South from the equator, the sun will sit at an angle 45˚ in the sky. The angle of the sun past zenith is our latitude.

Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.

From: Finding your Latitude and Longitude

Here is the basic method, often called "Voliva's Method".
<< same as previous post >>
I will calculate the sun's elevation as just 90˚ at the equator (Lat = 0˚), decreases by 1˚ for we move North of South of the equator. The distance from the equator will be calculated as (degrees of lat) x 69.5 miles as in the Wiki.

Also, I will make it more like the "Voliva Method" and just take all distances from the equator, not Vaupes.
 
The following table gives the data for each location. As for Voliva the date and time are for the overhead sun at an equinox, here UTC 20/Mar/2016 16:48 - the time the sun was overhead at long 70˚ West.


Location   

Latitude   

Sun Elev   
Dist from   
Equator   

Flat Sun Ht   
Calc
Circum
Kimmirut, Canada   
62.847°   
27.15°   
4,368 miles   
2,240 miles   
25,020 miles
Bingham, Maine, USA   
45.059°   
44.94°   
3,132 miles   
3,125 miles   
25,020 miles
Santo Domingo   
18.486°   
71.51°   
1,285 miles   
3,843 miles   
25,020 miles
Equator   
0°   
90°   
0 miles   
- - - -   
- - - -
Chupa District, Peru   
-15.109°   
74.89°   
1,050 miles   
3,889 miles   
25,020 miles
Los Tamariscos, Argentina   
-45.033°   
44.97°   
3,130 miles   
3,126 miles   
25,020 miles
Punta Arenas, Chile   
-53.164°   
36.63°   
3,695 miles   
2,747miles   
25,020 miles

Once we have the elevation angle of the sun from the site the height of the sun can be calculated from: h = d x tan(ElevAng) .
Also, if the earth is taken as a Globe the calculated circumference is found from (distance from equator) x 360°/(latitude difference from equator).

This time, I must stress, the sun elevations and distances are calculated exactly as in "the Wiki".
So, please no excuses like blaming Google earth or "Globalist" figures - they are Flat Earth distances from the Society's Wiki!

Using this method to find the height of the sun on the Flat earth gives measurements from 2,240 miles (for Kimmirut to the Equator) to 3,889 miles (for Chupa District to the Equator) depending on the spacing of the measurement sites.
Also for at locations close to 45° from the equator we get the usual Flat Earth sun height of "a bit over 3,000 miles, vis 3,125 miles and 3,126 miles.

In other words, claiming that the Flat Earth sun is at about 3,000 miles altitude has no foundation whatever.

It is very telling when we use the same data to calculate the earth's circumference, as Eratosthenes did, we get an absolutely consistent 25,020 miles circumference.
  Explain that!
This consistency is simply because we used "the Wiki's" figure of exactly 69.5 miles/degree, whereas the actual sun elevation figures from SunEarthTools would have little errors.

Certainly these figures would indicate that the earth is a globe with a distant sun.

<< typo, swapped let for lat >>
« Last Edit: December 27, 2017, 12:19:45 AM by rabinoz »

Re: trigonometry
« Reply #45 on: December 09, 2017, 09:38:41 AM »
I am asking you why you think the equator is the appropriate location to measure from? You are assuming the location of the sun as above this to calculate its height. This is incorrect and discordant with flat earth theory.

Ok, let’s do the math for a day when the sun isn’t over the equator.  Hourly sun elevation and azimuth data for eighteen cities on April 21 was provided by one of your own, a user who could never be accused of being a round earth guy, but instead must surely be considered among the flattest of flatties, the artist formerly known as İntikam.  He presented his data as a series of two city pairs, which pairs he used to calculate sun height using trigonometry.  As any round earther would expect (and most of us have stated outright at some point or other) he got widely varying results.  How “widely” you ask?  His lowest calculated sun altitude was 2289 km using the city pair of Athens and İzmir (3rd most populous city in Turkey); the highest was more than twice that altitude, 4652 km, coming from using the city pair of Pituffik, Greenland and Hamilton, Bermuda.  He calculated the average, 3352 km, and discarded it (appropriately, I might add: the sample set was too small and too clustered geographically to make averaging a valid method).  Instead he chose ONE of these results, the smallest, ignored all the large numbers and confidently declared

 ”We can easily say that the altitude of the sun should be about 2200 kms +-%10 error limit.”

Never mind that by his own calculations, all but two of his city-pair results are outside his own stated “+-10% error limit”.

Maths not being his strong suit, I suspected some mistakes were made and so I took his data and did the calculations myself.  Instead of going city-to-city as he did, I calculated where the sub solar point was (which would be north of the equator on April 21) and did the math from there.  My calcs agreed with İntikam: different solar heights from different cities.  My conclusions did not agree with İntikam: we can easily say the earth is not flat.



And here's a graph of the resulting sun heights, as viewed from each city in İntikam's data set.  The horizontal axis represents latitude of the cities on his list, the vertical axis is the sun elevation.


Re: trigonometry
« Reply #46 on: December 15, 2017, 11:18:12 AM »
I took his sun angles and plotted the straight-line visual location of the sun from each city.  The intersections of lines from various city pairs not only give different heights for the sun, but also different distances from the equator!  There are two conclusions one can draw from this:
  • The initial assumption of "the earth is flat" is not supported by the observations
  • The initial assumption of "light travels in straight lines" is not supported by the observations

In case #1, we can go on to use the observations to calculate the shape of the earth under varying assumptions of solar distance until we find consistent results.
In case #2, we can go no further, because with light not traveling in straight lines even the "3000 miles away" number has no basis, since it was calculated under the assumption that the sun's light DOES travel in straight lines (at least at the 45° north and south latitudes).


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rabinoz

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Re: trigonometry
« Reply #47 on: December 27, 2017, 12:51:25 AM »
Everybody else seems to be resurrecting old threads, so why shouldn't I?

I am asking you why you think the equator is the appropriate location to measure from? You are assuming the location of the sun as above this to calculate its height. This is incorrect and discordant with flat earth theory.

Further, you are not using actual data, but calculating the angles based on the round earth model. Are you not worried about circularly justifying your own claim? Wouldn't real data be the correct choice here? Why don't you supply us with the sources you used to verify, or am I just to guess? I'm not going to source your arguments for you, you have to support your own argument with more than "Well, that's the angle that the round earth model says it should be, so its right."
Well John, you seem to have gone very quiet after I posted similar calculations for the sun height using elevations and distances calculated only from the Flat Earth Wiki.

Here have another look! Re: trigonometry « Reply #44 on: December 09, 2017, 09:07:18 PM ».
This is the result:

Location   

Latitude   

Sun Elev   
Dist from   
Equator   

Flat Sun Ht   
Calc
Circum
Kimmirut, Canada   
62.847°   
27.15°   
4,368 miles   
2,240 miles   
25,020 miles
Bingham, Maine, USA   
45.059°   
44.94°   
3,132 miles   
3,125 miles   
25,020 miles
Santo Domingo   
18.486°   
71.51°   
1,285 miles   
3,843 miles   
25,020 miles
Equator   
0°   
90°   
0 miles   
- - - -   
- - - -
Chupa District, Peru   
-15.109°   
74.89°   
1,050 miles   
3,889 miles   
25,020 miles
Los Tamariscos, Argentina   
-45.033°   
44.97°   
3,130 miles   
3,126 miles   
25,020 miles
Punta Arenas, Chile   
-53.164°   
36.63°   
3,695 miles   
2,747miles   
25,020 miles

As before, over a flat earth, the sun's height shows no consistency, but on the Globe the circumference comes out with exactly the same values.

Any thoughts?
« Last Edit: December 27, 2017, 12:58:50 AM by rabinoz »

Re: trigonometry
« Reply #48 on: December 28, 2017, 10:29:55 AM »


As before, over a flat earth, the sun's height shows no consistency, but on the Globe the circumference comes out with exactly the same values.

Any thoughts?

You have seen trigonometry that shows the sun to have zero variance in height on a flat earth. My thought: why do you keep bringing this up? 

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EvolvedMantisShrimp

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Re: trigonometry
« Reply #49 on: December 28, 2017, 11:11:30 AM »
Here is a trigonometric problem to work out: On December 21st 2017(the solstice) in Ushuaia, Argentina (Lat 54° 48' S, Long 68° 18' W) at 09:00 UTC, the sun is at an azimuth of 121.59(to the southeast of one of the southernmost cities in the world) and an elevation of 7.0 degrees. At that same time, in Perth, Australia(31°57′8″S 115°51′32″E) the same sun is at an azimuth of 258.17(almost due west) an an elevation of 27.39 degrees.

Now, where is the sun on a Flat Earth map?
Nullius in Verba

Re: trigonometry
« Reply #50 on: December 28, 2017, 11:33:41 AM »
Here is a trigonometric problem to work out: On December 21st 2017(the solstice) in Ushuaia, Argentina (Lat 54° 48' S, Long 68° 18' W) at 09:00 UTC, the sun is at an azimuth of 121.59(to the southeast of one of the southernmost cities in the world) and an elevation of 7.0 degrees. At that same time, in Perth, Australia(31°57′8″S 115°51′32″E) the same sun is at an azimuth of 258.17(almost due west) an an elevation of 27.39 degrees.

Now, where is the sun on a Flat Earth map?

Please provide info on how you measured these values, especially at the same time.

Since all of Flat Earth theory relies on massive global conspiracies throughout every portion of government, industry, and academia, and since Zetetic Astronomy is based on observation (which most FE take to mean personal observation), then sourcing this information becomes a crucial question.

*

EvolvedMantisShrimp

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  • Physical Comedian
Re: trigonometry
« Reply #51 on: December 28, 2017, 11:49:42 AM »
Here is a trigonometric problem to work out: On December 21st 2017(the solstice) in Ushuaia, Argentina (Lat 54° 48' S, Long 68° 18' W) at 09:00 UTC, the sun is at an azimuth of 121.59(to the southeast of one of the southernmost cities in the world) and an elevation of 7.0 degrees. At that same time, in Perth, Australia(31°57′8″S 115°51′32″E) the same sun is at an azimuth of 258.17(almost due west) an an elevation of 27.39 degrees.

Now, where is the sun on a Flat Earth map?

Please provide info on how you measured these values, especially at the same time.

Since all of Flat Earth theory relies on massive global conspiracies throughout every portion of government, industry, and academia, and since Zetetic Astronomy is based on observation (which most FE take to mean personal observation), then sourcing this information becomes a crucial question.

My data was sourced from https://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html based on latitude and longitude data provide by Wikipedia.

If you believe that data is faulty, I welcome you to use your zetetic skills to provide more accurate data as long as you submit it for peer review.
Nullius in Verba

Re: trigonometry
« Reply #52 on: December 28, 2017, 12:53:38 PM »

My data was sourced from https://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html based on latitude and longitude data provide by Wikipedia.

If you believe that data is faulty, I welcome you to use your zetetic skills to provide more accurate data as long as you submit it for peer review.

Calculations done by a government web site which already accepts round earth theory, based on data provided by a web-site where anyone can update data.

'Nuff said.

If I believe the data sources to be suspect, I can do so without providing more accurate data, as I am not attempting to do a calculation. I was simply asking about your sources before you continued any farther in this train of thought.

*

rabinoz

  • 26528
  • Real Earth Believer
Re: trigonometry
« Reply #53 on: December 28, 2017, 12:58:24 PM »
Here is a trigonometric problem to work out: On December 21st 2017(the solstice) in Ushuaia, Argentina (Lat 54° 48' S, Long 68° 18' W) at 09:00 UTC, the sun is at an azimuth of 121.59(to the southeast of one of the southernmost cities in the world) and an elevation of 7.0 degrees. At that same time, in Perth, Australia(31°57′8″S 115°51′32″E) the same sun is at an azimuth of 258.17(almost due west) an an elevation of 27.39 degrees.

Now, where is the sun on a Flat Earth map?

Please provide info on how you measured these values, especially at the same time.

Since all of Flat Earth theory relies on massive global conspiracies throughout every portion of government, industry, and academia, and since Zetetic Astronomy is based on observation (which most FE take to mean personal observation), then sourcing this information becomes a crucial question.

My data was sourced from https://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html based on latitude and longitude data provide by Wikipedia.

If you believe that data is faulty, I welcome you to use your zetetic skills to provide more accurate data as long as you submit it for peer review.
I don't believe that Curiouser and Curiouser, Alice in future, is arguing about the correctness of your data.
Alice is claiming that most flat earthers will not believe data from a "globularist source" such as  NOAA ESRL, Solar Position Calculator.

Most flat-earthers will doubt the data unless they themselves take the measurements and since that is impossible, you case is unproven.

Of course "the Wiki" has essentially the same information,
Quote from: The Flat Earth Society Wiki
Finding your Latitude and Longitude
Latitude
To locate your latitude on the Flat Earth, it's important to know the following fact: The degrees of the Earth's Latitude are based upon the angle of the sun in the sky at noon equinox.

That's why 0˚ N/S sits on the equator where the sun is directly overhead, and why 90˚ N/S sits at the poles where the sun is at a right angle to the observer. At 45˚ North or South from the equator, the sun will sit at an angle 45˚ in the sky. The angle of the sun past zenith is our latitude.

Knowing that as you recede North or South from the equator at equinox, the sun will descend at a pace of one degree per 69.5 miles, we can even derive our distance from the equator based upon the position of the sun in the sky.
From: Finding your Latitude and Longitude

But even using that data some will come up with any old excuse like "perspective", "refraction" or simply fudge the maths and
others will claim the the Wiki is not accurate anyway.

Best of luck!

PS It might help if you read this:
      How can I debunk the arguments used by Flat Earth advocates? You can’t. Or rather, you can, but it won’t do any good.

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EvolvedMantisShrimp

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  • Physical Comedian
Re: trigonometry
« Reply #54 on: December 28, 2017, 01:02:49 PM »

My data was sourced from https://www.esrl.noaa.gov/gmd/grad/solcalc/azel.html based on latitude and longitude data provide by Wikipedia.

If you believe that data is faulty, I welcome you to use your zetetic skills to provide more accurate data as long as you submit it for peer review.

Calculations done by a government web site which already accepts round earth theory, based on data provided by a web-site where anyone can update data.

'Nuff said.

If I believe the data sources to be suspect, I can do so without providing more accurate data, as I am not attempting to do a calculation. I was simply asking about your sources before you continued any farther in this train of thought.

So, you are saying the data is wrong?
Nullius in Verba

Re: trigonometry
« Reply #55 on: December 28, 2017, 01:08:56 PM »

So, you are saying the data is wrong?

No, I simply asked you how you got this "data".

You've answered the question, and that is all I need to know about the thoroughness of your understanding of the issues.

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EvolvedMantisShrimp

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  • Physical Comedian
Re: trigonometry
« Reply #56 on: December 28, 2017, 01:13:29 PM »

So, you are saying the data is wrong?

No, I simply asked you how you got this "data".

You've answered the question, and that is all I need to know about the thoroughness of your understanding of the issues.

What issues? It's a trigonometric question. Either the data I gave is right, or wrong. Pick one. Then if you think it's wrong, provide right data so I can correct the problem. If it's right, choose whether you want to try the problem or not.

This is math, not philosophy.
Nullius in Verba

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JackBlack

  • 18132
Re: trigonometry
« Reply #57 on: December 28, 2017, 01:13:47 PM »
You have seen trigonometry that shows the sun to have zero variance in height on a flat earth. My thought: why do you keep bringing this up?
No, we saw your pathetic excuse of a formula which was completely unjustified and still had significant variation.
We also say my made up "correction" based upon a round Earth which would have "allowed" a "flat Earth" to have the sun at whatever height you wish, but the math was based on a round Earth not a flat Earth.

Meanwhile, the trig for a flat Earth without any unjustified BS, shows massive variation in the height of the sun.

That is why we keep bringing it up; because a FE has not adequately explained the apparent angle to the sun.

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JackBlack

  • 18132
Re: trigonometry
« Reply #58 on: December 28, 2017, 01:16:13 PM »
Calculations done by a government web site which already accepts round earth theory, based on data provided by a web-site where anyone can update data.

'Nuff said.

If I believe the data sources to be suspect, I can do so without providing more accurate data, as I am not attempting to do a calculation. I was simply asking about your sources before you continued any farther in this train of thought.
A simpler example is the equinox.
For everywhere on Earth at the equinox, the sun appears to rise from ~ due east and set ~ due west, while remaining above some point on the equator.
So where is the sun?

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EvolvedMantisShrimp

  • 927
  • Physical Comedian
Re: trigonometry
« Reply #59 on: December 28, 2017, 01:19:13 PM »
Calculations done by a government web site which already accepts round earth theory, based on data provided by a web-site where anyone can update data.

'Nuff said.

If I believe the data sources to be suspect, I can do so without providing more accurate data, as I am not attempting to do a calculation. I was simply asking about your sources before you continued any farther in this train of thought.
A simpler example is the equinox.
For everywhere on Earth at the equinox, the sun appears to rise from ~ due east and set ~ due west, while remaining above some point on the equator.
So where is the sun?

I prefer examples using the winter solstice from the Southern Hemisphere. They are far more interesting.
Nullius in Verba