Flat earth debunked.
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"Therefore I will make the heavens tremble, and the earth will be shaken from its place at the wrath of the LORD of Hosts on the day of His burning anger." Haggai 2:6. I guess the earth can be shaken after all
"Therefore I will make the heavens tremble, and the earth will be shaken from its place at the wrath of the LORD of Hosts on the day of His burning anger." Isaiah 13:13
Go to New Zealand, aka "The Shaky Isles." Earthquakes are real. The problem is not with God's word or with the rotating globe. It's with your interpretation.
"Therefore I will make the heavens tremble, and the earth will be shaken from its place at the wrath of the LORD of Hosts on the day of His burning anger." Isaiah 13:13
Go to New Zealand, aka "The Shaky Isles." Earthquakes are real. The problem is not with God's word or with the rotating globe. It's with your interpretation.
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1 Chronicles 16:
30 Fear before him, all the earth: the world also shall be stable, that it be not moved.
Psalms 93:
1 The LORD reigneth, he is clothed with majesty; the LORD is clothed with strength, wherewith he hath girded himself: the world also is stablished, that it cannot be moved.
Psalms 96:
10 Say among the heathen that the LORD reigneth: the world also shall be established that it shall not be moved: he shall judge the people righteously.
Psalms 104:
5 Who laid the foundations of the earth, that it should not be removed for ever.
Isaiah 45:
18 For thus saith the LORD that created the heavens; God himself that formed the earth and made it; he hath established it, he created it not in vain, he formed it to be inhabited: I am the LORD; and there is none else.
We know that God will eventually destroy the 'old' heaven and earth. The wording in these verses is not intended to indicate a "never will" meaning. Rather, they indicate the intent of God to maintain the earth as His will dictates until such a time that His will allows it to change.
In a way, there is a cause-and-effect relationship here. It is like saying that you built something a certain way and positioned it a certain way so that it should/will never fall over. That does not mean that it absolutely never will fall over. It means that you intend for it to stay in that position for as long as necessary to carry out the purpose that it was built for.
God is saying that He made it and positioned it like He wants it and that only He Himself will change it.
God 'laid the foundations' in such a way that - if He Himself never changed it - it literally would 'not be removed for ever'.
This discussion is to explain the "sense" of the meaning of the words/phrases. God can change it. No one else will be able to.
The wording in the verses still indicate very plainly that [the earth] is physically stationary and does not move. It is fixed in place. It does not spin, wobble, or fly through space at a high rate of speed.
And, I will suggest that your interpretation is 'on shaky ground'.
Nowhere in the context of these verses is it talking about earthquakes in a localized context.
We all know earthquakes are real. Your example suggestion conflicts with the context of scripture.
All of the verses - his and yours - are true - each in its own particular context. There is no contradiction.
The earth-as-a-whole will not move or be shaken until God Himself does so according to His own will.
30 Fear before him, all the earth: the world also shall be stable, that it be not moved.
Psalms 93:
1 The LORD reigneth, he is clothed with majesty; the LORD is clothed with strength, wherewith he hath girded himself: the world also is stablished, that it cannot be moved.
Psalms 96:
10 Say among the heathen that the LORD reigneth: the world also shall be established that it shall not be moved: he shall judge the people righteously.
Psalms 104:
5 Who laid the foundations of the earth, that it should not be removed for ever.
Isaiah 45:
18 For thus saith the LORD that created the heavens; God himself that formed the earth and made it; he hath established it, he created it not in vain, he formed it to be inhabited: I am the LORD; and there is none else.
We know that God will eventually destroy the 'old' heaven and earth. The wording in these verses is not intended to indicate a "never will" meaning. Rather, they indicate the intent of God to maintain the earth as His will dictates until such a time that His will allows it to change.
In a way, there is a cause-and-effect relationship here. It is like saying that you built something a certain way and positioned it a certain way so that it should/will never fall over. That does not mean that it absolutely never will fall over. It means that you intend for it to stay in that position for as long as necessary to carry out the purpose that it was built for.
God is saying that He made it and positioned it like He wants it and that only He Himself will change it.
God 'laid the foundations' in such a way that - if He Himself never changed it - it literally would 'not be removed for ever'.
This discussion is to explain the "sense" of the meaning of the words/phrases. God can change it. No one else will be able to.
The wording in the verses still indicate very plainly that [the earth] is physically stationary and does not move. It is fixed in place. It does not spin, wobble, or fly through space at a high rate of speed.
And, I will suggest that your interpretation is 'on shaky ground'.
Nowhere in the context of these verses is it talking about earthquakes in a localized context.
We all know earthquakes are real. Your example suggestion conflicts with the context of scripture.
All of the verses - his and yours - are true - each in its own particular context. There is no contradiction.
The earth-as-a-whole will not move or be shaken until God Himself does so according to His own will.
Yeah, not buying the distance of the stars in the Heliocentric model, but I did take a picture of a star with my telescope. It looked like a moon and wondered why I never seen a picture like this before.
The video of the star is far superior than mine, they probably have better gear, than my beginner's telescope, and iphone mount.
Lots of distortion in the video, but that is a really good video!
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I believe I managed to display your complete ignorance (or willful deception), given your failure to refute the post that contradicts the falsity you espouse. I post it again below, either to aid your memory or your honesty. whichever seems to be the cause of your error. I note also the logical fallacy in your assertion, that somehow given you've worked for four years with microwave devices, this somehow is evidence that you know they are evidence for ball-Earth. Some people are slow learners - be thankful I'm here to correct you after 4 years, and not 40! If radar are evidence for ball-Earth, explain the below.
https://christianchat.com/threads/flat-earth-debunked.209031/post-5007887
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I made it *really* simple. Not so much for you, because I think you are shilling for ball-Earth, rather than genuinely trying to understand. But for anyone honest that reads your posts and for whatever reason believes you might be right.
https://www.quora.com/What-is-the-range-of-an-average-marine-radar-on-US-Navy-ships?share=1
"Almost every one of the 50+ US Navy ships I sailed on had an installed “average marine radar” from a manufacturer like Raytheon or Furuno. These tend to be “fishing boat” radars with ranges of 25 to 75 nm, depending on how high above the sea surface they were installed."
Note that 25 nautical miles (the lower estimate given) is approximately 46km.
● Let A, B and E represent 3 ships, of height H1, H2, H3, on a ball-Earth of circumference C and radius r.
● Let ships A and B be located on the ball-Earth at such a distance that a straight line between the highest point on A and the highest point on B is tangent with ball-Earth. [This sets ship B at the maximum possible detectable distance, by straight line, from A].
● Let T represent the point where the tangent line AB intersects the ball-Earth.
● Let O represent the center of ball-Earth.
● Let ϴ1 represent the angle between OA and OT.
● Let ϴ2 represent the angle between OT and OB.
● Let ϴT represent the sum of the angles ϴ1 and ϴ2.
● Let D represent the distance of the arc along the circumference of ball-Earth, between the ships A and B.
Also, let:
● H1 = 10m
● H2 = 10m
● C = 40,075,017m
● r = 6,378,000m
Then:
● From basic trigonometry, we know that cos(ϴ) = Adjacent/Hypotenuse
cos (ϴ1) = r / (r + H1)
cos (ϴ1) = 0.999998432112838
cos (ϴ2) = r / (r + H2)
cos (ϴ2) = 0.999998432112838
ϴ1 = acos(0.999998432112838)
ϴ1 = 0.00177081200133473 radians
ϴ2 = acos(0.999998432112838)
ϴ2 = 0.00177081200133473 radians
● We also know that 360º = 2π radians
ϴ1 = 0.001770812 * 360 / (2.π)
ϴ1 = 0.101460053987595º
ϴ2 = 0.001770812 * 360 / (2.π)
ϴ2 = 0.101460053987595º
ϴT = ϴ1 + ϴ2
ϴT = 0.10146º + 0.10146º
ϴT = 0.20292º
● There are 360º in a circle.
D = ϴT / 360 * C
D = 0.20292º / 360º * 40,075,017m
D = 22588.951249m
D = 22.59km
● Therefore, if we lived on a ball-Earth of circumference C = 40,075km, the maximum range that a straight‑beam radar mounted 10m above sea-level could detect a 10m-high ship, is 22.59km.
● In reality, radar can detect ships at much greater distances than this (e.g. easily at 50km).
● The trigonometric diagram of ball-Earth, together with radar ranges which greatly exceed the maximum range predicted by the ball-Earth model, therefore demonstrate that we do not live on a ball‑Earth at all, as some would have us imagine.
"
https://christianchat.com/threads/flat-earth-debunked.209031/post-5007887
"
I made it *really* simple. Not so much for you, because I think you are shilling for ball-Earth, rather than genuinely trying to understand. But for anyone honest that reads your posts and for whatever reason believes you might be right.
https://www.quora.com/What-is-the-range-of-an-average-marine-radar-on-US-Navy-ships?share=1
"Almost every one of the 50+ US Navy ships I sailed on had an installed “average marine radar” from a manufacturer like Raytheon or Furuno. These tend to be “fishing boat” radars with ranges of 25 to 75 nm, depending on how high above the sea surface they were installed."
Note that 25 nautical miles (the lower estimate given) is approximately 46km.
● Let A, B and E represent 3 ships, of height H1, H2, H3, on a ball-Earth of circumference C and radius r.
● Let ships A and B be located on the ball-Earth at such a distance that a straight line between the highest point on A and the highest point on B is tangent with ball-Earth. [This sets ship B at the maximum possible detectable distance, by straight line, from A].
● Let T represent the point where the tangent line AB intersects the ball-Earth.
● Let O represent the center of ball-Earth.
● Let ϴ1 represent the angle between OA and OT.
● Let ϴ2 represent the angle between OT and OB.
● Let ϴT represent the sum of the angles ϴ1 and ϴ2.
● Let D represent the distance of the arc along the circumference of ball-Earth, between the ships A and B.
Also, let:
● H1 = 10m
● H2 = 10m
● C = 40,075,017m
● r = 6,378,000m
Then:
● From basic trigonometry, we know that cos(ϴ) = Adjacent/Hypotenuse
cos (ϴ1) = r / (r + H1)
cos (ϴ1) = 0.999998432112838
cos (ϴ2) = r / (r + H2)
cos (ϴ2) = 0.999998432112838
ϴ1 = acos(0.999998432112838)
ϴ1 = 0.00177081200133473 radians
ϴ2 = acos(0.999998432112838)
ϴ2 = 0.00177081200133473 radians
● We also know that 360º = 2π radians
ϴ1 = 0.001770812 * 360 / (2.π)
ϴ1 = 0.101460053987595º
ϴ2 = 0.001770812 * 360 / (2.π)
ϴ2 = 0.101460053987595º
ϴT = ϴ1 + ϴ2
ϴT = 0.10146º + 0.10146º
ϴT = 0.20292º
● There are 360º in a circle.
D = ϴT / 360 * C
D = 0.20292º / 360º * 40,075,017m
D = 22588.951249m
D = 22.59km
● Therefore, if we lived on a ball-Earth of circumference C = 40,075km, the maximum range that a straight‑beam radar mounted 10m above sea-level could detect a 10m-high ship, is 22.59km.
● In reality, radar can detect ships at much greater distances than this (e.g. easily at 50km).
● The trigonometric diagram of ball-Earth, together with radar ranges which greatly exceed the maximum range predicted by the ball-Earth model, therefore demonstrate that we do not live on a ball‑Earth at all, as some would have us imagine.
"
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