relative travel times in hemispheres

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odes

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relative travel times in hemispheres
« on: April 23, 2013, 12:33:13 PM »
Hi, just discovering this site. Very interesting.

I thought it might be a good question to compare distances in respective north/south latitudes. Looking at a globe, I found approximately 35 degrees north and south to be fairly useful:

Montevideo to Cape Town (about 35 degrees south latitude)
Raleigh NC to Sevilla, Spain (about 35 degrees north latitude)

The distances, according to http://www.mapcrow.info/ , are 4143 and 3923 miles, respectively.

But, looking at the Johnson map linked from the FES home page, here, it seems that Montevideo would be roughly twice as far from Cape Town, as Raleigh would be from Sevilla. How is this explained?
Quote from: Rushy
No bawwing is necessary.

Re: relative travel times in hemispheres
« Reply #1 on: May 02, 2013, 02:28:19 AM »
Has anyone actually traveled directly from between NC and Spain?  No?  You are making the classic mistake of assuming what you are trying to prove.  The website you used presumed the earth is globular and the same shape as the globe.  Big deal if they agree with each other.  Like two old-time Russians saying the USSR will soon crush the USA.  I suggest you ACTUALLY TRAVEL from one place to the other and then post again.  No offence, but your life experience is a little like that Jim Carey movie where he was trapped in a director's set. 

My apologies in advance if you have ACTUALLY TRAVELED between the four points.   

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Rama Set

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Re: relative travel times in hemispheres
« Reply #2 on: May 02, 2013, 05:16:05 AM »
You are making the classic mistake of saying that one cannot know anything unless it is experienced first hand. If you look at your life you will no doubt find hundreds of examples where you rely on accurately published information and it works. I have tracked flights online from Hawaii to Sydney, Australia and the flight time was impossible on a FE by the flight path they claimed to take or any other flight path for that matter. Air travel to the Southern Hemisphere is largely nonsensical on the FE map and yet we accurately navigated hundreds of times daily.
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Re: relative travel times in hemispheres
« Reply #3 on: May 02, 2013, 06:12:19 AM »
Sydney to Perth, Australia.

Sydney 33.8683S 151.2086E
Perth    31.9554S 115.8585E

I have personally been on this flight, in both directions, and can confirm that it takes approximately 5 hours, takeoff to landing, in a Boeing 737. The mapped distance between these two points is about 3300km by air.

The calculated distance on the north pole centred (NPC) FE map would be about 8300km (they are about 35 degrees apart in longitude, and about 13,700km South of the North pole, you can confirm the maths yourself). This is straight line distance too, not following any kind of curve (which would go further). I'm ignoring the bipolar FE map, as it is utterly useless, and has been debunked.

With a maximum speed of 876km/h for the Next Generation 737's, it would have taken about 10 hours to do this flight if the earth looked like the NPC map, not to mention there would have been a need to refuel as well.

Incidentally, as this flight follows a Great Circle to make it as short as possible, it passes over Adelaide (again, personally confirmed by me: I watched Adelaide pass under the nose of the aeroplane), which is at about 35 degrees South latitude. Further South than either of the other two cities, which would make the trip a great deal longer on a NPC map.

Of course, the earth could be a disc centered on the South pole, but then all the Northern hemisphere (sorry, hemiplane  ::) ) flights would be longer...
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

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odes

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Re: relative travel times in hemispheres
« Reply #4 on: May 02, 2013, 09:23:39 AM »
The website you used presumed the earth is globular and the same shape as the globe.

Exactly. The web site might 'prove' a point, while simply calculating in accordance with a predetermined assumption. I see that now, but I didn't when I posted originally.
Quote from: Rushy
No bawwing is necessary.

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Rama Set

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Re: relative travel times in hemispheres
« Reply #5 on: May 02, 2013, 09:28:02 AM »
The website you used presumed the earth is globular and the same shape as the globe.

Exactly. The web site might 'prove' a point, while simply calculating in accordance with a predetermined assumption. I see that now, but I didn't when I posted originally.

The website has predictive power.  This is a good indication that it is working under a correct hypothesis.  In other words that the theory is true.
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Re: relative travel times in hemispheres
« Reply #6 on: May 02, 2013, 02:04:55 PM »
It should be pretty easy to find a flight in the Northern Hemishpere similar to the one I posted earlier for the Southern Hemisphere. If the times are even remotely similar, I'd say its pretty sound proof.
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

?

odes

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Re: relative travel times in hemispheres
« Reply #7 on: May 02, 2013, 02:12:03 PM »
Happy hunting for south-south direct flights. For a variety of social reasons, they seem to be rare. For example, I don't think there's a listed Montevideo - Cape Town flight. But a study of flight charts might yield something.

Flight times may also be affected by wind currents, which may vary from area to area.
Quote from: Rushy
No bawwing is necessary.

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Scintific Method

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Re: relative travel times in hemispheres
« Reply #8 on: May 02, 2013, 02:26:35 PM »
Happy hunting for south-south direct flights. For a variety of social reasons, they seem to be rare. For example, I don't think there's a listed Montevideo - Cape Town flight. But a study of flight charts might yield something.

Flight times may also be affected by wind currents, which may vary from area to area.

Plenty of south-south direct flights to be found my friend. Within Australia, and between Australia and New Zealand being examples I am familiar with.

Also, if flights were effected by winds, there would be a difference between the east-west time, and the west-east time. I can confirm that this difference is minimal, if any.
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

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Rama Set

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Re: relative travel times in hemispheres
« Reply #9 on: May 02, 2013, 02:27:47 PM »
There are also regular flights from Johannesburg to Sydney.
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markjo

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Re: relative travel times in hemispheres
« Reply #10 on: May 02, 2013, 04:53:51 PM »
Happy hunting for south-south direct flights.
Maybe this will help: http://www.qantas.com.au/travel/airlines/home/au/en
Science is what happens when preconception meets verification.
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odes

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Re: relative travel times in hemispheres
« Reply #11 on: May 02, 2013, 09:11:41 PM »
Maybe this will help: http://www.qantas.com.au/travel/airlines/home/au/en

Hm, funny results so far, maybe you can try it.

Christchurch NZ (45 degrees) - Sydney AUS (35 degrees) = 1326 miles according to mapcrow.info
Flight either 3:10 or 3:00

Raleigh NC (35 degrees) = Minneapolis MN (45 degrees) = 995 miles according to mapcrow.info
Flight either 2:45 or 2:25

On a cardboard globe, Christchurch-Sydney is a titch further than Raleigh-Minneapolis. All measurements are approximate. The trick is to find a pair of cities between which direct flights are possible, which have equivalent latitudes, and which seem to be equally far apart on a globe artifact.
Quote from: Rushy
No bawwing is necessary.

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Scintific Method

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Re: relative travel times in hemispheres
« Reply #12 on: May 02, 2013, 11:22:21 PM »
The latitude and longitude are both important.

Christchurch 43.5S 172.6E
Sydney 33.9S 151.2E

A difference of 9.6 degrees in latitude, and 21.4 degrees in longitude.

Raleigh NC 35.8N 78.6W
Minneapolis MN 45.0N 93.3W

A difference of 9.2 degrees in latitude, and almost identical latitudes to the other pair (good find!), but only 14.7 degrees in longitude. This is almost certainly the reason for the difference in distances. A little maths should give us a better idea:

Using the difference in degrees, and a bit of Pythagoras, to give a rough ratio of distances gives us 23.45 / 17.34 = 1.35

Using your distances: 1326 / 995 = 1.33

That's not too bad really, pretty close to proportional. Bloody lot closer to proportional than they would be on a FE map, that's for sure!
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

?

darknavyseal

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Re: relative travel times in hemispheres
« Reply #13 on: May 03, 2013, 01:21:03 AM »
If you want a FE explanation, here is one:

There are winds that propel the airplane in order to speed it up in the "southern hemisphere". These winds work in both directions, somehow, and are undetectable by any human instrument.

FE: 1
RE: 0

Alternatively, the actual orientation of the continents are not correct on the map.

Cheers.
« Last Edit: May 03, 2013, 10:31:14 AM by darknavyseal »

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odes

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Re: relative travel times in hemispheres
« Reply #14 on: May 03, 2013, 05:40:45 PM »
The latitude and longitude are both important.

Christchurch 43.5S 172.6E
Sydney 33.9S 151.2E

A difference of 9.6 degrees in latitude, and 21.4 degrees in longitude.

Raleigh NC 35.8N 78.6W
Minneapolis MN 45.0N 93.3W

A difference of 9.2 degrees in latitude, and almost identical latitudes to the other pair (good find!), but only 14.7 degrees in longitude. This is almost certainly the reason for the difference in distances. A little maths should give us a better idea:

Using the difference in degrees, and a bit of Pythagoras, to give a rough ratio of distances gives us 23.45 / 17.34 = 1.35

Using your distances: 1326 / 995 = 1.33

That's not too bad really, pretty close to proportional. Bloody lot closer to proportional than they would be on a FE map, that's for sure!

I'm not in a good position right now to quibble with your math, because I need to review my high school trig and so on. But in my somewhat childlike way of estimating, i.e. by using the edge of an index card, a corner on Christchurch and the edge marked at Sydney, and then the same corner on Raleigh and the edge marked again at Minneapolis, I find that the Raleigh-Minneapolis is only about 10% shorter as the airplane would likely go (no doubt making 'whrrrrr' sounds!). :)

The other comment I have for now is that this is an experiment in other ways. I don't know that mapcrow data and methods are accurate. I think I can trust the flight information, though, because commercial interests attempt accuracy. But as for what's actually happening on the ground, I don't know what we know for sure. Maybe on my globe the circles representing cities are in the wrong places.
Quote from: Rushy
No bawwing is necessary.

Re: relative travel times in hemispheres
« Reply #15 on: May 11, 2013, 06:42:56 AM »
"The calculated distance on the north pole centred (NPC) FE map would be about 8300km (they are about 35 degrees apart in longitude, and about 13,700km South of the North pole, you can confirm the maths yourself). This is straight line distance too, not following any kind of curve (which would go further). I'm ignoring the bipolar FE map, as it is utterly useless, and has been debunked."

We have your word for that.  I am sure you are being truthful.  Please post confirmatory images of your calculations.

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Scintific Method

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Re: relative travel times in hemispheres
« Reply #16 on: May 11, 2013, 08:28:16 AM »
"The calculated distance on the north pole centred (NPC) FE map would be about 8300km (they are about 35 degrees apart in longitude, and about 13,700km South of the North pole, you can confirm the maths yourself). This is straight line distance too, not following any kind of curve (which would go further). I'm ignoring the bipolar FE map, as it is utterly useless, and has been debunked."

We have your word for that.  I am sure you are being truthful.  Please post confirmatory images of your calculations.


Why? Are you incapable of basic maths and reasoning?
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

Re: relative travel times in hemispheres
« Reply #17 on: May 22, 2013, 01:13:58 PM »
Sydney to Perth, Australia.

Sydney 33.8683S 151.2086E
Perth    31.9554S 115.8585E

I have personally been on this flight, in both directions, and can confirm that it takes approximately 5 hours, takeoff to landing, in a Boeing 737. The mapped distance between these two points is about 3300km by air.

The calculated distance on the north pole centred (NPC) FE map would be about 8300km (they are about 35 degrees apart in longitude, and about 13,700km South of the North pole, you can confirm the maths yourself). This is straight line distance too, not following any kind of curve (which would go further). I'm ignoring the bipolar FE map, as it is utterly useless, and has been debunked.

With a maximum speed of 876km/h for the Next Generation 737's, it would have taken about 10 hours to do this flight if the earth looked like the NPC map, not to mention there would have been a need to refuel as well.

Incidentally, as this flight follows a Great Circle to make it as short as possible, it passes over Adelaide (again, personally confirmed by me: I watched Adelaide pass under the nose of the aeroplane), which is at about 35 degrees South latitude. Further South than either of the other two cities, which would make the trip a great deal longer on a NPC map.

Of course, the earth could be a disc centered on the South pole, but then all the Northern hemisphere (sorry, hemiplane  ::) ) flights would be longer...

Good job. You've used evidence and experiment to completely disprove the current FE theory. I'm sure the FE people will see this as proof and change their beliefs (as they say they will once RE is proven) and we can all go home!.. if only.

Re: relative travel times in hemispheres
« Reply #18 on: May 22, 2013, 02:01:01 PM »
Sydney to Perth, Australia.

Sydney 33.8683S 151.2086E
Perth    31.9554S 115.8585E

I have personally been on this flight, in both directions, and can confirm that it takes approximately 5 hours, takeoff to landing, in a Boeing 737. The mapped distance between these two points is about 3300km by air.

The calculated distance on the north pole centred (NPC) FE map would be about 8300km (they are about 35 degrees apart in longitude, and about 13,700km South of the North pole, you can confirm the maths yourself). This is straight line distance too, not following any kind of curve (which would go further). I'm ignoring the bipolar FE map, as it is utterly useless, and has been debunked.

With a maximum speed of 876km/h for the Next Generation 737's, it would have taken about 10 hours to do this flight if the earth looked like the NPC map, not to mention there would have been a need to refuel as well.

Incidentally, as this flight follows a Great Circle to make it as short as possible, it passes over Adelaide (again, personally confirmed by me: I watched Adelaide pass under the nose of the aeroplane), which is at about 35 degrees South latitude. Further South than either of the other two cities, which would make the trip a great deal longer on a NPC map.

Of course, the earth could be a disc centered on the South pole, but then all the Northern hemisphere (sorry, hemiplane  ::) ) flights would be longer...

Good job. You've used evidence and experiment to completely disprove the current FE theory. I'm sure the FE people will see this as proof and change their beliefs (as they say they will once RE is proven) and we can all go home!.. if only.

Not that I don't get the sarcasm, but I'm sure that will never happen. Not even when cheap commercial space flights become available. Not even when humans populate other planets.

Re: relative travel times in hemispheres
« Reply #19 on: May 22, 2013, 07:10:54 PM »
Yeah, I don't consider that evidence.  There are too many factors to consider and using "about" (a certain number of miles) and "about" (a certain speed/altitude) is not reliable data.

Re: relative travel times in hemispheres
« Reply #20 on: May 23, 2013, 12:20:31 AM »
Yeah, I don't consider that evidence.  There are too many factors to consider and using "about" (a certain number of miles) and "about" (a certain speed/altitude) is not reliable data.

It's not reliable data for an exact calculation, but it's more than enough to show that the distances in the southern hemisphere are definitely not doubled. Which is an automatic fail on FET's part.

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Scintific Method

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Re: relative travel times in hemispheres
« Reply #21 on: May 23, 2013, 02:22:18 AM »
Yeah, I don't consider that evidence.  There are too many factors to consider and using "about" (a certain number of miles) and "about" (a certain speed/altitude) is not reliable data.

Did you want it to 3 decimal places? I could provide that, but it wouldn't change anything, there's still no way that this fits with the NPC flat earth model.
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

Re: relative travel times in hemispheres
« Reply #22 on: May 24, 2013, 12:51:34 PM »
While we do not have an accurate map. If you take the conceptual map from the website and shear it along the proper axis, it will account for the differences in travel. The Great Ice Wall is not a perfect circle, but the edges of the disc are probably so.

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DuckDodgers

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Re: relative travel times in hemispheres
« Reply #23 on: May 24, 2013, 12:59:30 PM »
It seems pretty major that a flat earth cannot be properly protected onto a flat map.
markjo, what force can not pass through a solid or liquid?
Magnetism for one and electric is the other.

Re: relative travel times in hemispheres
« Reply #24 on: May 24, 2013, 01:43:58 PM »
It seems pretty major that a flat earth cannot be properly protected onto a flat map.

That's precisely because you can't project a globe's surface onto a two-dimensional surface, continuously (without breaking it somewhere), while also maintaining the distance mapping everywhere.

This is just another of the countless conclusive proofs that the Earth cannot be flat... a concept even a primary school pupil would understand. And there's no conspiracy of false distances in the southern hemisphere, anyone with any vehicle and a watch can verify.

Re: relative travel times in hemispheres
« Reply #25 on: May 24, 2013, 01:45:50 PM »
It seems pretty major that a flat earth cannot be properly protected onto a flat map.
It is simple (though not easy) to do. Take accurate measurements and reduce to the proper scale.

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DuckDodgers

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Re: relative travel times in hemispheres
« Reply #26 on: May 24, 2013, 01:52:20 PM »
It seems pretty major that a flat earth cannot be properly protected onto a flat map.
It is simple (though not easy) to do. Take accurate measurements and reduce to the proper scale.

Surely this would have been done in the few centuries that surveying has been around.   The only distances that would cause trouble in this regard is transoceanic, but I'm sure there is some surveying technique to gather these distances.   Long story short,  if it could be done,  it should have by now.   By the way,  flat maps which are a projection of a globe show accurate scale, when the distortion is removed by " cutting" the surface to spread it flat. 
markjo, what force can not pass through a solid or liquid?
Magnetism for one and electric is the other.

Re: relative travel times in hemispheres
« Reply #27 on: May 24, 2013, 01:54:43 PM »
It seems pretty major that a flat earth cannot be properly protected onto a flat map.
It is simple (though not easy) to do. Take accurate measurements and reduce to the proper scale.

The problem is you can't make it proper scale without skewing it or using a third dimension to account for the distances. Like I said, there is no way (as in geometrically impossible) a globe can be continuously mapped to a two-dimensional surface, equidistantly. You either have to cut it in a few places, which wouldn't work since it'd mean huge areas of emptiness in the middle of our world, or you have to stretch and skew distances, which isn't equidistant anymore, and once again doesn't work without some very weird space-time distortion, which is invalidated by real time communication between any two points on Earth.

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Scintific Method

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Re: relative travel times in hemispheres
« Reply #28 on: May 24, 2013, 02:49:41 PM »
It seems pretty major that a flat earth cannot be properly protected onto a flat map.
It is simple (though not easy) to do. Take accurate measurements and reduce to the proper scale.

Surely this would have been done in the few centuries that surveying has been around.   The only distances that would cause trouble in this regard is transoceanic, but I'm sure there is some surveying technique to gather these distances.   Long story short,  if it could be done,  it should have by now.   By the way,  flat maps which are a projection of a globe show accurate scale, when the distortion is removed by " cutting" the surface to spread it flat.

Duck, I think transoceanic distances may have been calculated from latitude and longitude measurements. There is a page in the flat earth wiki about how to find lat and long, and distances measured across a Southern landmass could have been used to confirm how many miles there were per degree of longitude at any given Southern latitude.
Quote from: jtelroy
...the FE'ers still found a way to deny it. Not with counter arguments. Not with proof of any kind. By simply denying it.

"Better to keep your mouth shut and be thought a fool, than to open it and remove all doubt."

Flat Earth Map
« Reply #29 on: June 03, 2013, 05:00:30 PM »
« Last Edit: June 03, 2013, 05:04:06 PM by EarthIsASpaceship »