Sunrise and sunset is in the wrong place

I live at roughly 52° north (about 2660 miles south of the north pole), and at the summer solstice the sun rises at approximately 4.43am and sets in the evening at 9.33pm. That’s nearly 17 hours of daylight.

Now in FET with the sun essentially orbiting around the north pole that means that the sun is visible for 245° of its 360° but more importantly it means that the sun should rise in the north east and set in the north west.

I can go into the maths and show where FET says the sun should appear and disappear but I suspect it’s pointless as I’m sure the magical refractive properties of the atmosphere will explain why the sun appears to be in the wrong place.

I live at roughly 52° north (about 2660 miles south of the north pole), and at the summer solstice the sun rises at approximately 4.43am and sets in the evening at 9.33pm. That’s nearly 17 hours of daylight.

Now in FET with the sun essentially orbiting around the north pole that means that the sun is visible for 245° of its 360° but more importantly it means that the sun should rise in the north east and set in the north west.

I can go into the maths and show where FET says the sun should appear and disappear but I suspect it’s pointless as I’m sure the magical refractive properties of the atmosphere will explain why the sun appears to be in the wrong place.

Yes, the reflective qualities of the atmosphere in the northern section, and a host of magical angles with reflecting parasols, that adjust the direction of the suns rays. Its all very logical really.

Is there no way in FET to explain how about 200 miles of atmosphere can make the sun appear to set approximately 90^o further south than the theory would say.

I've had a look through the archives but wasn't able to find this coming up before, I found a couple about the equinox but for the most part they were more concerned with the path the sun takes. If I missed something, probably by searching on the wrong terms then I'm happy to read the previous explanation if someone can direct me to it.

Please note that I haven't said the earth is a sphere, merely that the FET model as it stands doesn't account for where the sun rises and sets.

Right I'm going to have to debunk myself then am I?

Here's a link to where the sun rises and sets for London

http://www.suncalc.net/#/51.5073,-0.1277,7/2012.06.20/21:45

the sun does rise in the north east and set in the north west, not exactly where FE predicts but not too far off.

However here's a link to the mirror image (basically) of it seen from South America in winter

http://www.suncalc.net/#/-51.6248,-69.2358,5/2012.12.20/21:45

if you lived in the southern hemisphere and looked at that you'd swear the sun was going round the south pole.

The FE model is biased in favour of the Northern Hemisphere, if you're a flat earther in the southern hemisphere then it looks like the south pole is the centre.

Who said that the sun should rise in NE at 52 degrees north?  No one is answering you because you did not provide any sources to this claim and we don't know what you are talking about. 

Who said that the sun should rise in NE at 52 degrees north?  No one is answering you because you did not provide any sources to this claim and we don't know what you are talking about.

But I thought that's how things work here!

Before I write out the math and maybe even draw a picture can you confirm the distance from the north pole to the tropic of cancer, I've had to use the RET distances so have been using a distance of 4595 miles.

Let's go with your numbers.  However, no FEer, as far as I know, has ever said that the sun will rise in the NE from that far north.  I still do not know where you are even getting your information from.

Ok, I'll be basing this on the model I've seen quoted most often where the sun rotates around the north pole and is 3000 miles above the surface of the earth.

I'll go with Berlin, it's at approx 52.5^o north. It's 2590 miles south of the north pole.
At the summer solstice sunrise is at 03:43 and sunset is at 20:33, that means that a viewer in Berlin can observe the sun for 16 hours and 50 minutes of it's 24 hour orbit around the north pole, put another way it is visible  for 252^o of it's 360^o orbit.

I forgot to label the circle as the suns path.

If we just concentrate on sun rise for now we can draw 2 triangles as follows


We know the hypotenuse and the internal angles for the top triangle so we know how long the other sides are. As we now know the length of 2 of the 3 sides of the larger triangle we can work out the internal angles etc.

So in FET the sun is 35^o from north and 25^o from the horizon when it can be seen rising above the horizon. I tried to draw this but it didn't come out very well.



Now if you're actually at 52.5^o north then the sun rises 47.72⁰ and sets at 312.28⁰ which is pretty close by FE standards requiring only a 12^o shift horizontally and a 25^o vertically. This leads me to think that who ever came up with this model lived just a few degrees further south.

I was going to go on and show where FET predicts the sun will be on June 20th for different locations, Havana as it's pretty close to the tropic and Cape Town for the southern hemisphere.

I didn't want to do the math and draw pictures though if the response I get is "no that's not what the FE model predicts, number x is wrong"

I'm happy to be corrected about it, it's your model not mine. I'm just doing the best I can with the limited details on the website.

Considering this is a valid question, why hasn't anybody responded since last year? Perhaps another, more glaring example will be in order.

Having lived a short while in the Falkland Islands, I've seen the sun rise in the south-east (about midnight) and set south-west about fifteen hours later, which means the sun was visible for 225 degrees of it's 360 orbit above the north pole. Since the south pole doesn't exist, and the sun orbits the north pole, let's calculate my distance from the north pole. Stanley's about 5.7 Mm* from the equator, which is about 10 Mm away from the north pole, so I was, roughly 15.65 Mm from the point around which the sun orbits.

The Tropic of Cancer is roughly 7.4 Mm from the north pole, with the Tropic of Capricorn about 5.2 Mm from that. Since I was there in November, close to the summer solstice, let's say the sun's orbital diameter was the Tropic of Capricorn, at 12.6 Mm with me 3.1 Mm south of its path.

Plugging these numbers into the triangles so nicely drawn by Manarq**, I find that Flat Earth predicts the sun should have risen about 26 degrees east of north, about 90 degrees different from my observations.

_{*Mm is a 'mega-meter', or 1,000 kilometers. A kilo-kilometer, if you will. 1 Mm is roughly 621,000 miles
**I'm happy to spend the time to draw up my equations and diagrams, if you wish, though don't expect anything pretty. I'm not an artist, and I don't have fancy geometry software--just a pen, paper, and simple calculator.}

Considering this is a valid question, why hasn't anybody responded since last year? Perhaps another, more glaring example will be in order.

Having lived a short while in the Falkland Islands, I've seen the sun rise in the south-east (about midnight) and set south-west about fifteen hours later, which means the sun was visible for 225 degrees of it's 360 orbit above the north pole. Since the south pole doesn't exist, and the sun orbits the north pole, let's calculate my distance from the north pole. Stanley's about 5.7 Mm* from the equator, which is about 10 Mm away from the north pole, so I was, roughly 15.65 Mm from the point around which the sun orbits.

The Tropic of Cancer is roughly 7.4 Mm from the north pole, with the Tropic of Capricorn about 5.2 Mm from that. Since I was there in November, close to the summer solstice, let's say the sun's orbital diameter was the Tropic of Capricorn, at 12.6 Mm with me 3.1 Mm south of its path.

Plugging these numbers into the triangles so nicely drawn by Manarq**, I find that Flat Earth predicts the sun should have risen about 26 degrees east of north, about 90 degrees different from my observations.

_{*Mm is a 'mega-meter', or 1,000 kilometers. A kilo-kilometer, if you will. 1 Mm is roughly 621,000 miles
**I'm happy to spend the time to draw up my equations and diagrams, if you wish, though don't expect anything pretty. I'm not an artist, and I don't have fancy geometry software--just a pen, paper, and simple calculator.}

Hi Alex,

thanks for the post. Unfortunately stuff like this where you can make repeatable observations and measurements tend to be ignored, Flat Earthers seem to take a more philosophical approach in their world view as opposed to one based on pesky observations and measurements.

Unfortunately stuff like this where you can make repeatable observations and measurements tend to be ignored, Flat Earthers seem to take a more philosophical approach in their world view as opposed to one based on pesky observations and measurements.

So they're not even going to acknowledge that this thread exists?

Reminding everyone that this concern is left unanswered. Hmm, maybe I should add an actual question at the end.... I'll even make it bold so you can skip right to the question if you want.

Having lived a short while in the Falkland Islands, I've seen the sun rise in the south-east (about midnight) and set south-west about fifteen hours later, which means the sun was visible for 225 degrees of it's 360 orbit above the north pole. Since the south pole doesn't exist, and the sun orbits the north pole, let's calculate my distance from the north pole. Stanley's about 5.7 Mm^[1] from the equator, which is about 10 Mm away from the north pole, so I was, roughly 15.65 Mm from the point around which the sun orbits.

The Tropic of Cancer is roughly 7.4 Mm from the north pole, with the Tropic of Capricorn about 5.2 Mm from that. Since I was there in November, close to the summer solstice, let's say the sun's orbital diameter was the Tropic of Capricorn, at 12.6 Mm with me 3.1 Mm south of its path.

Plugging these numbers into the triangles so nicely drawn by Manarq^[2], I find that Flat Earth predicts the sun should have risen about 26 degrees east of north, about 90 degrees different from my observations.

How can the flat-earth hypothesis explain the sun rising a full 90 degrees away from where it should rise, according to the flat-earth hypothesis?

[1]Mm is a 'mega-meter', or 1,000 kilometers. A kilo-kilometer, if you will. 1 Mm is roughly 621,000 miles
[2]I'm happy to spend the time to draw up my equations and diagrams, if you wish, though don't expect anything pretty. I'm not an artist, and I don't have fancy geometry software--just a pen, paper, and simple calculator.

Still unanswered, and further research has added to the problem, so I'll, once again, summarize for ease of counter-argument:

Time spent in the Falkland Islands allowed me to observe sunrise and sunset a full 90 degrees away from what the mono-poled FE model predicts I should have seen it.

The bi-polar model more accurately predicts the observed locations, but this model falls apart now that I'm in the Pacific Northwest, and see sunrise and sunset about 40 degrees off what the model predicts. So neither model covers both hemispheres.

Again I ask how can either model be true if it not only fails to agree with observations, but fails quite dramatically depending on where one is observing?

Once again going to bring this to the top of the threads, so it might be available for quick-access by anyone wishing to explain these observations in an FE model.

Time spent in the Falkland Islands allowed me to observe sunrise and sunset a full 90 degrees away from what the mono-poled FE model predicts I should have seen it.

The bi-polar model more accurately predicts the observed locations, but this model falls apart now that I'm in the Pacific Northwest, and see sunrise and sunset about 40 degrees off what the model predicts. So neither model covers both hemispheres.

Again I ask how can either model be true if it not only fails to agree with observations, but fails quite dramatically depending on where one is observing?

Hey alex, check out videos on the mignight sun, its a sun that never sets near the arctic, it proves that the sun makes circles around the earth.

Hey alex, check out videos on the mignight sun, its a sun that never sets near the arctic, it proves that the sun makes circles around the earth.

No, it doesn't.

http://en.wikipedia.org/wiki/Midnight_sun

http://answers.yahoo.com/question/index?qid=20091209191240AAcmvNq

Hey alex, check out videos on the mignight sun, its a sun that never sets near the arctic, it proves that the sun makes circles around the earth.

It does set, it takes a day to do so however.  The same thing which causes seasons results in the midnight sun over both the arctic and antarctic for a full day during the respective hemisphere's summers.

Hello? Guys? We're getting derailed.

This topic isn't about midnight suns or seasons. This is about how my observations of sunrise and sunset in the Falkland Islands differed from the mono-polar FE model by about 90 degrees and current observations of sunrise and sunset in Cascadia differs from the bi-polar FE model by about 40 degrees.

That is, in Stanley, the sun was rising and setting 90 degrees farther south than FE says it should, and here the sun is rising and setting 40 degrees farther south than FE says it should.

Neither model correctly predicts observations from both hemispheres, so neither model can be correct. Please either explain how the sun can appear to be up to 90 degrees farther south than it really is, or provide a model that fits observations.

Bump.

Still hasn't been answered.

Once again, a summary: Time spent in the Falkland Islands allowed me to regularly observe sunrises and sunsets during the southern summer. These observations disagreed from the predictions by a north-pole centered FE model by as much as 90 degrees, the observations always farther south than the predictions.

Currently living in the northern hemisphere, I regularly observe sunrises and sunsets during the northern winter (same time period as above). These observations disagree from the predictions by a dual-poled FE model (Antarctica as a Continent) by as much as 40 degrees, the observations always farther south than the predictions.

Is there an explanation as to how the sun can appear much farther south than it really is? If not, is there a model that correctly predicts observations of both hemispheres in all seasons?

Bump.

Still hasn't been answered.

Once again, a summary: Time spent in the Falkland Islands allowed me to regularly observe sunrises and sunsets during the southern summer. These observations disagreed from the predictions by a north-pole centered FE model by as much as 90 degrees, the observations always farther south than the predictions.

Currently living in the northern hemisphere, I regularly observe sunrises and sunsets during the northern winter (same time period as above). These observations disagree from the predictions by a dual-poled FE model (Antarctica as a Continent) by as much as 40 degrees, the observations always farther south than the predictions.

Is there an explanation as to how the sun can appear much farther south than it really is? If not, is there a model that correctly predicts observations of both hemispheres in all seasons?

I'm sorry, but I don't have full knowledge of what your saying. What exactly are these predictions presented on the dual-poled model?

Quote from: Alex Tomasovich on August 26, 2013, 10:50:03 AM

Bump.

Still hasn't been answered.

Once again, a summary: Time spent in the Falkland Islands allowed me to regularly observe sunrises and sunsets during the southern summer. These observations disagreed from the predictions by a north-pole centered FE model by as much as 90 degrees, the observations always farther south than the predictions.

Currently living in the northern hemisphere, I regularly observe sunrises and sunsets during the northern winter (same time period as above). These observations disagree from the predictions by a dual-poled FE model (Antarctica as a Continent) by as much as 40 degrees, the observations always farther south than the predictions.

Is there an explanation as to how the sun can appear much farther south than it really is? If not, is there a model that correctly predicts observations of both hemispheres in all seasons?

I'm sorry, but I don't have full knowledge of what your saying. What exactly are these predictions presented on the dual-poled model?

How about you do a bit of work and work out where the dual-poled model predicts the sun will be at specific times for where you are and see if it matches up with what you observe.

In fact how about working it out and putting the predictions in the thread for some relatively close future date, oh and don't forget to include how you worked it out.

Formula for calculating sunrise location predictions from any FE model. For the bi-polar model, between the Northern Hemisphere's fall and spring equinoxes, the pole in question is the South Pole. All other times both models predict the same thing.


Deriving the Equation
Givens
1° lat = 111 km

Knowns - can be found with a bit of research
lat_sun
lat_observer
hours_daylight

Calculations - we're trying to find ε
First, let's solve triangle ESP:

Now we can solve triangle OSE:

ε gives us the angle away from the pole the sun would be during sunrise or sunset. If this is the south pole, sunrise will be 180° - ε. If it's the north pole, it will just be ε.

Full equation in terms of givens and knowns:

---

Note that although Figure 1 shows the observer outside the sun's path, this equation still works for an observer inside of it.

Definition of terms in FIGURE 1:
^{Let d be the distance from the observer to the pole the sun is orbiting
Let t be the distance between the sun's orbit and the pole around which the sun orbits
Let S be the point over which the sun appears to be on the horizon for the observer
Let O be the point upon which the observer is standing
Let P be the pole around which the sun is orbiting
Let y be the distance between S and d
Let E be the point at the intersection of y and d
Let z be the distance between O and E
Let x be the distance between E and P
Let α be the angle of arc through which the sun is visible
Let β be the angle OPS
Let ε be the angle EOS}

Formal experiment: Sunrise 31 August 2013 from Portland, Oregon.

FE Prediction: 54.06° east of north
RE Prediction: 77.38° east of north

Actual sunrise location: [Will be measured tomorrow if weather permits]

Planned methods of measurement:

• With a compass measure the direction of the shadow of a vertical spirit level. Subtract 180° from the result
• Point a yardstick at the rising sun observing from a low angle--this should make the yardstick accurate to within a few degrees. With a compass measure the direction of the yardstick

Are you FE or RE? Either way I like the fact your conducting actual, measurable, and replicable experiments to validate your claim.

Okay, results are in. I've quoted the original post and put in bold the things I've changed. Some notes: some low mountains existed on the eastern horizon, making sunset about 10 minutes later than it should. This shouldn't affect the readings too much, but I think a 5° uncertainty can be expected in this case (aside from the compass measurement itself, which seems to be 1-2 ° uncertainty).

Formal experiment: Sunrise 31 August 2013 from Portland, Oregon.

FE Prediction: 54.06° east of north
RE Prediction: 77.38° east of north

Actual sunrise location: 85° east of north ± 7°

Deviation from FE Prediction: 31° ± 7° (31% - 41% error)
Deviation from RE Prediction: 08° ± 7° (01% - 16% error)

Planned methods of measurement:

• With a compass measure the direction of the shadow of a vertical spirit level. Subtract 180° from the result
• Point a yardstick compass at the rising sun observing from a low angle--this should make the yardstick accurate to within a few degrees. With a compass measure the direction of the yardstick Record the compass reading

You forgot that Samuel Birley Rowbotham has established that perspective does not allow one to see for extremely long distances, making your FE predictions unfaithful to the model.

You forgot that Samuel Birley Rowbotham has established that perspective does not allow one to see for extremely long distances, making your FE predictions unfaithful to the model.

So you're saying I shouldn't have been able to see the sun during sunrise? Sounds like your model has a lot more problems than being a few dozen degrees off for sunrise.

For discussions on how Rowbotham's ideas affect perspective, please use the "Rowbotham's Perspective" thread.

You are also making the assumption that 1 lat = 111km. But who proved this and how? I doubt anyone laid down a tape measurer to every latitude on earth to make sure they were all consistent distances from each other.

The only way to measure your latitude is to look at number of angles the pole star is from the horizon. But the distance of the pole star from the horizon means different things under a Round Earth and a Flat Earth, making one's presumed distance from the North Pole incorrect.