Distance and math

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AmateurAstronomer

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Re: Distance and math
« Reply #30 on: September 14, 2008, 08:41:58 AM »
Hint: Flights are curved in both models.

The curve in RET is necessary to not hit the ground. Hitting the ground would cause problems. The curve in FET would have added time to the trip. Both RET and FET curve is irrelevant to this question. The question itself exists because the sub-equatorial stretch makes flights of that distance at that sub-equatorial longitude improbable in that time frame.

Your post adds the question of why it's impossible to fly in a straight line between a number of locations on FE maps. The "tendencies" of given craft aside, it should be possible to make a craft that could, you know, make a quarter degree turn right/left every few dozen/hundred miles, and cut 5-10% or more off it's flight path.
« Last Edit: September 14, 2008, 08:47:27 AM by AmateurAstronomer »
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Re: Distance and math
« Reply #31 on: September 14, 2008, 08:50:06 AM »
The curve in FET would have added time to the trip.

What are you basing this on? The curves would be identical; adding no time and exhibiting the exact same flight times.

Your post adds the question of why it's impossible to fly in a straight line between a number of locations on FE maps.

I never said it was impossible.
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Re: Distance and math
« Reply #32 on: September 14, 2008, 08:57:10 AM »
Hint: Flights are curved in both models.

Yes, but that doesn't explain how I can then get  from Sydney, Australia to Buenes Aires, Argentina (EZE) in 13 hours and 0 minutes and then from Sao Paulo, Brazil (GRU) in

5190 miles from JNB to PER
7240 miles from SYD to EZE
4646 miles from GRU back to JNB

Given this I will have circumnavigated the southern hemisphere in just a bit over 17,000 miles of travel.  There have been absolutely zero suggestions on how this would be possible with a flat map.  So please spare me the condescension, it's not my fault that you can't do simple math.

It seems that you should be able to put together some of relative map given this, but no such luck.

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AmateurAstronomer

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Re: Distance and math
« Reply #33 on: September 14, 2008, 09:05:27 AM »
The curve in FET would have added time to the trip.

What are you basing this on? The curves would be identical; adding no time and exhibiting the exact same flight times.
These would have the same flight times?

Care to explain why you'd think that?

If you're saying FET and RET would have the same flight time for a sub-equatorial flight you must be joking. Go back a page and look at my reference materials, and look at the OP's assertions. On our world it gets smaller again after the equator.

Your post adds the question of why it's impossible to fly in a straight line between a number of locations on FE maps.

I never said it was impossible.

Fair enough. Why do you feel ships and aircraft are inclined to move in curves though? Did they catch the north pole?
« Last Edit: September 14, 2008, 09:20:48 AM by AmateurAstronomer »
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Re: Distance and math
« Reply #34 on: September 14, 2008, 10:02:36 AM »
If the earth is flat, how come you need to use spherical geometry in order to accurately calculate the distance between two objects?  For instance, many "distance by zip code" applications are based on using the coordinates of the center of a zip code area and plugging them into something like the Haversine formula:

R = earth’s radius (mean radius = 6,371km)
Δlat = lat2− lat1
Δlong = long2− long1
a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2)
c = 2.atan2(√a, √(1−a))
d = R.c

Since GPS data has been opened up to civilians and GPS data sources (by their very nature) overlap in area, it would be impossible to fake this data in a way that would produce the illusion of a sphere where there wasn't one.  This is compounded by the fact that you can verify the above data simply by walking between two coordinates and doing the math.

So you know, how does this fit into the FE way of life?



Your formula is incorrect. On a RE, the distance between two points with geodisic coordinates (latitude, longitude) (λ1, ϕ1) and (λ2, ϕ2) is:

d = R * cos-1(sinλ1*sinλ2 + cosλ1*cosλ2*cos(ϕ2-ϕ1))

The problem with FE is that we only know that points with constant lattitude correspond to concentric circles with center at the North Pole, but we cannot exactly determine their radius. In any case, the radius depends on the latitude only r = f(λ).The only thing we know for sure is that f(+90°) = 0, because this point corresponds to the North Pole and that is in the center of the circle. Then, we can find the shortest distnace between two points as:

d = √(r12+ r22 - 2*r1*r2*cos(ϕ2 - ϕ1))

However, one can prove that no choice of the function f(λ) can give the same result for all possible points. The proof is simple. First, notice that both formulas depend only on the difference in longitudes ϕ2 - ϕ1, as it should be, because we are free to choose the meridian of zero longitude (which, by convention today is through Greenwich). That's why we can choose in our formula ϕ1 = 0 and ϕ2 = ϕ in general. Furthermore, let us choose λ1 = 90°, because we know the value of the function f for this latitude, and λ2 = λ in general. Then, the two formulas give:

d = R*(π/2 - λ) - on RE
d = f(λ) - on FE
from which we conclude that

f(λ) = R*(π/2 - λ)
where the angle is measured in radians now.
Now, for points on the Equator (λ1 = λ2 = 0), the two equations give:

d = R*|ϕ2 - ϕ1| - in RE
d = R*π*|sin(ϕ2 - ϕ1)/2)| - in FE
These two estimates are obviously not the same. If they were, we would have the function

|sin(x)/x| = 2/π, 0 ≤ x ≤ 2π
which is certainly not correct. Hence, we cannot find f(λ) making the two distance estimates equal for any two points on the Earth.

However, this does not mean that FE model is not true. One has to give an empirical definition of longitude and latitude before one can use the above formulas. You provided the GPS data as a source for longitude and latitude. Until you provide us with the method with which this system gets the latitude and longitude of any point on Earth, without inverting the RE distance formula I stated above, you don't have a proof of anything.
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Jack

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Re: Distance and math
« Reply #35 on: September 14, 2008, 10:41:09 AM »
These would have the same flight times?

Care to explain why you'd think that?

If you're saying FET and RET would have the same flight time for a sub-equatorial flight you must be joking
In FE, our flight paths are curved in the FE's frame of reference. The reason behind this is that our compass always point to the North, which is the center of the FE. Going East or West is basically circling around the FE.

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Re: Distance and math
« Reply #36 on: September 14, 2008, 10:44:05 AM »
You need to meet our reasoning half way. That is only fair.

All right!  We'll call it a draw then.
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Re: Distance and math
« Reply #37 on: September 14, 2008, 11:09:27 AM »

Your formula is incorrect. On a RE, the distance between two points with geodisic coordinates (latitude, longitude)

No, my formula is correct.  I suspect that you've memorized one method for calculating distance on a globe while neglecting the others.

More importantly, it doesn't matter what method the GPS uses to calculate lat/long.  The fact that it does so consistently on all parts of the globe means that there is no distortion in its calculations.  If the earth was flat, GPS devices would only be accurate on certain parts of the planet.

Re: Distance and math
« Reply #38 on: September 14, 2008, 11:36:01 AM »

Your formula is incorrect. On a RE, the distance between two points with geodisic coordinates (latitude, longitude)

No, my formula is correct.  I suspect that you've memorized one method for calculating distance on a globe while neglecting the others.

More importantly, it doesn't matter what method the GPS uses to calculate lat/long.  The fact that it does so consistently on all parts of the globe means that there is no distortion in its calculations.  If the earth was flat, GPS devices would only be accurate on certain parts of the planet.

If the earth is flat, how come you need to use spherical geometry in order to accurately calculate the distance between two objects?  For instance, many "distance by zip code" applications are based on using the coordinates of the center of a zip code area and plugging them into something like the Haversine formula:

R = earth’s radius (mean radius = 6,371km)
Δlat = lat2− lat1
Δlong = long2− long1
a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2)
c = 2.atan2(√a, √(1−a))
d = R.c

I am sorry, I did not see the function atan2(y,x), which I interpreted as the regular atan(x). By your formula, we have:

sin(c/2) = √a, cos(c/2) = √(1 - a)

Therefore, by the double angle formula:

cos(c) = cos2(c/s) - sin2(c/2) = (1 - a) - a = 1 - 2*a

Using what you have as a, we get:

cos(c) = -2*sin2(∆lat/2) - 2*cos(lat1)*cos(lat2)*sin2(∆long/2) + 1

Grouping the first and the third term and using the double angle formula again, we have:

cos(c) = -2*cos(lat1)*cos(lat2)*sin2(∆long/2) + cos(lat2 - lat1)

Using the addition theorem for the cosine, we can write:

cos(c) = sin(lat1)*sin(lat2) + cos(lat1)*cos(lat2)*(1 - 2*sin2(∆long/2))

Finally, using the double angle formula once more, the term in the parentheses can be reduced to cos(long2 - long1).

We have:

cos(c) = sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(long2 - long1)

If you take the inverse cosine you will get:

c = cos-1[sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(long2 - long1)]

So, the distance is:

d = R*cos-1[sin(lat1)*sin(lat2)+cos(lat1)*cos(lat2)*cos(long2 - long1)]

This is equivalent to the formula I presented. However, you did not tell us how GPS finds the longitude and latitude of different points on the Earth.
« Last Edit: September 14, 2008, 11:52:22 AM by John Jackson »
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Re: Distance and math
« Reply #39 on: September 14, 2008, 12:16:54 PM »
This is equivalent to the formula I presented. However, you did not tell us how GPS finds the longitude and latitude of different points on the Earth.

You're right, I didn't.  I told you that it was immaterial since GPS is accurate in both the northern and southern hemispheres.  Oddly enough, nobody has offered even a wild guess as to how this could happen.

Re: Distance and math
« Reply #40 on: September 14, 2008, 12:20:44 PM »
You're right, I didn't.  I told you that it was immaterial since GPS is accurate in both the northern and southern hemispheres.  Oddly enough, nobody has offered even a wild guess as to how this could happen.

There is another problem: Measuring the distances on the Earth itself. How do you do that? If you don't measure these distances, there is no way to prove that GPS is accurate.
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Re: Distance and math
« Reply #41 on: September 14, 2008, 12:36:47 PM »
You're right, I didn't.  I told you that it was immaterial since GPS is accurate in both the northern and southern hemispheres.  Oddly enough, nobody has offered even a wild guess as to how this could happen.

There is another problem: Measuring the distances on the Earth itself. How do you do that? If you don't measure these distances, there is no way to prove that GPS is accurate.

Well, so far, we've used RADAR, SONAR, LIDAR, and the odometers in our cars. The only way you could debunk these would be to somehow assert that the length of a mile/foot/kilometer/meter is relative to how far from the north pole you are at any given time, at which point I bet men in Australia probably aren't the target demographic of penis enlargement spam.

Re: Distance and math
« Reply #42 on: September 14, 2008, 12:39:41 PM »
Well, so far, we've used RADAR, SONAR, LIDAR, and the odometers in our cars. The only way you could debunk these would be to somehow assert that the length of a mile/foot/kilometer/meter is relative to how far from the north pole you are at any given time, at which point I bet men in Australia probably aren't the target demographic of penis enlargement spam.

Ok, then, give us the data where distances between two points along the surface of the Earth were measured by using a RADAR.
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Re: Distance and math
« Reply #43 on: September 14, 2008, 01:34:00 PM »
Ok, then, give us the data where distances between two points along the surface of the Earth were measured by using a RADAR.

Try google.  I'm not your research assistant.  I'll give you a hint:  weather RADAR also measures ground distances.

Re: Distance and math
« Reply #44 on: September 14, 2008, 02:56:06 PM »
Ok, then, give us the data where distances between two points along the surface of the Earth were measured by using a RADAR.

Try google.  I'm not your research assistant.  I'll give you a hint:  weather RADAR also measures ground distances.


Somehow I think this is more a proof for a Flat Earth than a Round Earth. As far as I know, shortwave EM radiation travels in straight lines. How do these follow the supposed curvature of a Round Earth?
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Re: Distance and math
« Reply #45 on: September 14, 2008, 02:58:14 PM »
Somehow I think this is more a proof for a Flat Earth than a Round Earth. As far as I know, shortwave EM radiation travels in straight lines. How do these follow the supposed curvature of a Round Earth?

They don't.  That's why they had to go off and invent OTH RADAR for long ranges.

Re: Distance and math
« Reply #46 on: September 14, 2008, 03:02:29 PM »
Somehow I think this is more a proof for a Flat Earth than a Round Earth. As far as I know, shortwave EM radiation travels in straight lines. How do these follow the supposed curvature of a Round Earth?

They don't.  That's why they had to go off and invent OTH RADAR for long ranges.

Or, they just built a more powerfull transmitter, and they could detect reflections from further away because the Earth is flat. Then, they had to come up with the funny theory that light bends.
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Re: Distance and math
« Reply #47 on: September 14, 2008, 03:06:20 PM »
Or, they just built a more powerfull transmitter, and they could detect reflections from further away because the Earth is flat. Then, they had to come up with the funny theory that light bends.

Or they just bounce RF off of the atmosphere, which conveniently doesn't require anything like unicorns, magic, flat earths or bendy light.

Re: Distance and math
« Reply #48 on: September 14, 2008, 03:10:53 PM »
Or, they just built a more powerfull transmitter, and they could detect reflections from further away because the Earth is flat. Then, they had to come up with the funny theory that light bends.

Or they just bounce RF off of the atmosphere, which conveniently doesn't require anything like unicorns, magic, flat earths or bendy light.

Yes, but once you bounce off the atmosphere, you must use the model to calculate the distance from the point you are looking. Bouncing off atmosphere is perfectly possible in a FE, too.
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Re: Distance and math
« Reply #49 on: September 14, 2008, 03:19:12 PM »
Yes, but once you bounce off the atmosphere, you must use the model to calculate the distance from the point you are looking. Bouncing off atmosphere is perfectly possible in a FE, too.

So how was I wrong here?

Re: Distance and math
« Reply #50 on: September 14, 2008, 03:31:51 PM »
Yes, but once you bounce off the atmosphere, you must use the model to calculate the distance from the point you are looking. Bouncing off atmosphere is perfectly possible in a FE, too.

So how was I wrong here?

Circular logic.
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Re: Distance and math
« Reply #51 on: September 14, 2008, 06:09:11 PM »
I still don't see any evidence that spherical math is accurate for distances.[/size]

At the risk of being accused by Tom of being an idiot again.

I have used spherical trig to calculate courses and distances for transoceanic voyages numerous times; both in the northern and southern hemispheres.  Each time, the courses and distances that were calculated were the same as what was measured during the transit.  The longest of these was the transit between the western entrance to the Malacca Strait and Cape Town, South Africa.
« Last Edit: September 14, 2008, 06:17:49 PM by Rig Navigator »

Re: Distance and math
« Reply #52 on: September 14, 2008, 06:11:35 PM »
At the risk of being accused by Tom of being an idiot again.

I have used spherical trig to calculate courses and distances for transoceanic voyages numerous times; both in the northern and southern hemispheres.  Each time, the courses and distances that were calculated were the same as what was measured during the transit.  The longest of these was the transit between the western entrance to the Malacca Strait and Cape Town, South Africa.

Is this a new computer game that I am not aware of?
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Re: Distance and math
« Reply #53 on: September 14, 2008, 06:17:30 PM »
Is this a new computer game that I am not aware of?

No, just my my job.  Like most jobs, it probably isn't considered exciting enough to make into a video game.

Re: Distance and math
« Reply #54 on: September 14, 2008, 06:19:54 PM »
What are you basing this on? The curves would be identical; adding no time and exhibiting the exact same flight times.

No they aren't identical.  The curved course on a RE the apex of the curve is toward the poles, but on FE it would be toward the equator.

Re: Distance and math
« Reply #55 on: September 14, 2008, 06:25:45 PM »
Ok, then, give us the data where distances between two points along the surface of the Earth were measured by using a RADAR.

Well, you can measure the distance between the two sides of the Strait of Gibraltar, Dover Strait or any other narrow channel using a surface search radar.  In fact any body of water less than 48 nautical miles across can be measured by most ship's radar.

As for weather radar, you aren't detecting surface weather out to the maximum distance of the set.  Rain (which is what you are detecting) is detected above sea level which extends the radar horizon significantly.

Re: Distance and math
« Reply #56 on: September 14, 2008, 06:27:44 PM »
Ok, then, give us the data where distances between two points along the surface of the Earth were measured by using a RADAR.
Well, you can measure the distance between the two sides of the Strait of Gibraltar, Dover Strait or any other narrow channel using a surface search radar.  In fact any body of water less than 48 nautical miles across can be measured by most ship's radar.

This means that 48 miles are visible, which is only possible if the Earth is flat.
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Re: Distance and math
« Reply #57 on: September 14, 2008, 06:30:44 PM »
This means that 48 miles are visible, which is only possible if the Earth is flat.

Um, how is this only possible on FE?  A ship located in the middle of the strait will easily have 24 miles of radar search.  Especially when you consider the fact that the antenna is located over 50 feet above sea level.

Re: Distance and math
« Reply #58 on: September 14, 2008, 06:38:45 PM »
This means that 48 miles are visible, which is only possible if the Earth is flat.

Um, how is this only possible on FE?  A ship located in the middle of the strait will easily have 24 miles of radar search.  Especially when you consider the fact that the antenna is located over 50 feet above sea level.

According to RE model, an object that is h feet high can be seen from at most s miles on the surface of the Earth, such that:
h = 2*s2/3
so, a 50 ft high antenna can have a range of only √(50 × 3 / 2) = 8.7 miles or 7.6 nautical miles. That's quite smaller than what you stated as evidence. I'm afraid your data is in favor of Flat Earth Theory.
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spacemanjones

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Re: Distance and math
« Reply #59 on: September 14, 2008, 07:14:13 PM »