- This topic has 6,414 replies, 23 voices, and was last updated 1 month, 3 weeks ago by
.
- Posts
Before I reply to your post Mike, I want you to answer the following for me first which I have been asking for a while now.
Do you believe the sun move across the Australian sky at a faster speed than it does in the USA?
Three persons facing due south in the Southern Hemisphere, each on a different continent, why is it we see the same stars when they are experiencing night, but rotated? Why do they not see the same stars in the same rotation?
If you skip these questions, then I know that what matters more to you is you being right (pride) over truth winning (light).
I have no time to waste on people who can’t be polite enough to say “you’re mistaken”, or “what you said was inaccurate”. A decent person would have worded that like this…
Dig, instead of teaching and spreading inaccurate information…
Mike, there comes a point when a person repels all evidence to the contrary for their religion, doctrine, or science. They have in effect become brainwashed in a cult and are no longer serving the truth and are not useful for the Kingdom. Spreading inaccurate information stops when a person is debunked. That is, if they are truth seekers. Some abandon truth for other doctrines.
For those who are led by the Spirit of God are the children of God.
But when he, the Spirit of truth, comes, he will guide you into all the truth. He will not speak on his own; he will speak only what he hears, and he will tell you what is yet to come.
Now the Spirit expressly states that in later times some will abandon the faith to follow deceitful spirits and the teachings of demons, influenced by the hypocrisy of liars, whose consciences are seared with a hot iron.…
Jesus said to His disciples, “It is inevitable that stumbling blocks will come, but woe to the one through whom they come!
because they refused the love of the truth that would have saved them. For this reason, God will send them a powerful delusion so that they will believe the lie, in order that judgment will come upon all who have disbelieved the truth and delighted in wickedness.…
Whilst the immediate above verse appears to be talking about being saved, it is clear that delighting in anything but the truth is wickedness and God allows people into delusion if they pursue it in their heart.
Flat Earth and parabola
The graph below shows a basic plot of how the line that each approximation produces ‘falls away’ from a horizontal line at the starting point.
Here the distances are in miles, but you will notice that the horizontal and vertical scales are not at the same scale, so the curve seems steeper than it actually is. So the ‘linear rule’ is the blue line across the top, while the ‘parabolic rule’ is actually behind the purple ‘Circular 2’ line. The Circular 1 line is computed using distance along the surface of the Earth, i.e., going around the curve, while Circular 2 is the distance along the horizontal line from the starting point.If we plot out the entire circular cross-section of the Earth, and place the parabola on it at the same scale, we get the following picture:
where the red circle represents the cross-section of the Earth, and the blue line is the representation of ‘8 inches per mile squared’ at the same scale. This will give you an idea of how well the parabolic (binomial) approximation works. Distances are in miles, radius is 3,920 miles.
At 1,000 miles (along the straight line) the difference is about 0.722 miles, which is about 3,800 feet.
It should be stated clearly that 8” per mile is not a correct statement of the equation, but should be 8” × mi²/mi², since you are not dividing by distance, but multiplying 8 by the square of the distance in miles and dividing by mi² (implicitly, the 8”/mi² comes about through conversion from miles to inches). It should also be stated clearly that this equation is not the measure of the Earth’s curvature. Curvature is defined as the inverse of the radius of a curve. The curvature of the Earth’s surface is approximately 1/3960 mi⁻¹. Finally, it should be stated clearly that this is an approximate formula for the change in height over the ground of a straight line that begins parallel to and on or near the Earth’s surface and traveling some distance while not following the Earth’s curve.It is valid over short distances, depending on the level of accuracy you are looking for.
8”d² is a quick way to solve for h if you are a geodetic surveyor performing triangulation surveys and need to take into account errors that might accumulate when you are sighting distances. Over short distances, c, the actual distance over the ground, is very close to d (line of sight), but because the Earth is curving away, h will be at a different height than it appears to be if you confuse c and d. To calculate h, given d, you would use the pythagorean theorem to find a binomial equation which you would then solve using the quadratic formula.
From the picture,
R² + d² = (R + h)²
R² + d² = R² + 2Rh + h²
d² = 2Rh + h²
h² + 2Rh – d² = 0
Then by the quadratic equation,
h = (-2R ± √(4R² – (-4d²)))/2
and simplifying
= -R ± √(R² + d²)
Considering only positive distances,
h = √(R² – d²) – R
R = 3960
h = √(15,681,600 + d²) – 3960
This would be a cumbersome equation to have to solve by hand in the field in the years before pocket calculators, so a short cut was made at the 3rd step above:
d² = 2Rh + h²
d² = h(2R + h)
h = d²/(2R + h)
For h << R,
2R + h ≈ 2R
that is to say that for small heights h, much smaller than the radius of the Earth adding h to 2R is well within measurement error, and so for all intents and purposes, we can discard the addition of h.
So
h ≈ d²/2R,
Plugging in 3960 for R, we get
h ≈ d²/7920
Because typical line of sight distances will be small enough that h can be measured in inches, we can just convert from miles to inches using dimensional analysis:
h ≈ d² mi²/(7920 mi) × 5280ft/mi × 12 in/ft
h ≈ d² mi²/mi² × 8 in = 8 in × d²
So the most correct way to state the equation is 8 inches miles squared per miles squared.
The way this equation is misused is to confuse d (line of sight distance) and c (distance along the ground), and assume that the observation height doesn’t matter when sighting objects further than the horizon. Over long distances and varying vantage points, the relative errors can stack up and relying on the simplifyin 8”d² approximation will give erroneous answers.
To reiterate, the equation makes the assumption that viewer height is right at ground level and the horizon is 0 feet away. Because observation height is so much smaller that the Earth’s radius, it doesn’t matter too much whether you just add your observer height to the Earth’s radius R, or keep R as is.
However, over long distances, and/or high vantage points, errors in inches and feet do increase substantially and keeping with the approximate equation above, you can come to very erroneous conclusions about how much of an object should be hidden behind the horizon from a some distance and vantage point.
Take the following picture:
You will notice that as the vantage point h₁ increases, the drop in height h₂ decreases, and the line of sight distance changes, all despite the ground distance c remaining constant. Different heights at h₁ can cause significant changes in h₂.
To find h₂ given h₁, this equation (which assumes the Earth is a perfect sphere, refraction negligible, and heights given above sea level), you can use the better equation:
h₂ = R × sec(c/R – β) – R
Where,
h₂ is the maximum height an object can have and be hidden by the horizon;
c/R is the angle subtended between observer and object with the Earth’s center at the vertex;
β = cos⁻¹((R/(R + h₁))) is the angle subtended between the observer and the horizon with the Earth’s center at the vertex, and h₁ = the observer’s eye-height above the ground;
α = c/R – β is the angle subtended between the horizon and the object with the Earth’s surface at the vertex;
The length from the center of the Earth to the top of the object is R × sec(α), or R/cos(α). By subtracting this length by the Earth’s radius, we get the maximum height an object can have while still be below the horizon, or put another way, the amount an object is hidden behind the Earth’s curve from some vantagr point at h₁, neglecting effects of refraction, and terrain.
When h₁ = 0, then the above equation reduces to
h = R × sec(c/R) – R, and can be approximated, using a Taylor Series expansion, to h ≈ c²/2R.
When c ≈ d, h ≈ d²/2R, again giving us the original 8” × mi²/mi² equation.
A comparison of three equations for h:
Green curve: h = Rsec(c/R) – R
Blue curve: h = d²/2R
Red curve: h = √(R² + d²) – R
The graph shows observer height h along the vertical axis and distance along the horizontal. The green curve measures distance along a great circle, whereas blue and read measure distances along the line of sight. For distances up to about 46.5 miles, the difference between green and blue is less than an inch, and by about 53 miles, less than an inch between blue and red.
d is related to c by
c = R×tan⁻¹(d/R)
d = R×tan(c/R)
d is implicitly constrained by the geometry of the Earth, and by including this constraint in the equation for the red curve,
h = √(R² + d²) – R
= √(R² + R²tan²(c/R)) – R
= √(R² + R²(sec²(c/R)-1)) – R
= Rsec(c/R) – R
https://www.quora.com/Is-the-8-inch-per-mile-curvature-rule-valid-for-long-distances
Right, explain the math you just copied and pasted. And then explain why when it’s within 200 to 600 miles it is very accurate. And then explain why you lost this point.
But don’t forget this isn’t the only point about this topic! Remember there are other applications to these official documents from NASA, CIA, the Army and Russian agencies that assume a non-rotating flat earth.Still waiting for an explanation for this fact.
Right, explain the math you just copied and pasted.
I don’t have to explain it or understand it because I am not the one saying the earth is flat.
You are the one arguing that the earth is flat so you need to understand it to see if it debunks your point that this math claims to debunk.
Do the work yourself.
If you don’t understand it, then it shows why you believe the earth is flat because you cannot understand why you are wrong.
If you cannot be bothered to understand it, then you are not bothered about the truth.
If you know you cannot understand it, then you cannot claim to be right.
I claim the earth is a sphere because of the myriad of evidence that points to it coupled with all points raised for the Flat Earth being debunked so far
Victory for Team Globe once again.
In short Diggles, you should be trying to prove yourself wrong too.
A person who really wants to believe in the truth and be led by the Spirit of Truth will always question their own conclusions. A person who is prideful will only point out errors elsewhere and not with themselves. Paul gave Believers the following advice that would save a lot of deception and make us better.
Now if we judged ourselves properly, we would not come under judgment. But when we are judged by the Lord, we are being disciplined so that we will not be condemned with the world.…
Test yourselves to see if you are in the faith; examine yourselves! Or do you not recognize this about yourselves, that Jesus Christ is in you–unless indeed you fail the test?
But each one must examine his own work, and then he will have reason for boasting in regard to himself alone, and not in regard to another.
Game Over
Dig. Because you believe the sun in Australia moves at a faster speed than it does in the USA, this means you lost the debate because the sun moves at the same speed to the observer on the ground.
The flat earth model requires this faster sun, but reality contradicts this. So easy to test too and no one in history can or has shown the sun moving faster in Australia.
Another nail in coffin.
You’ve claimed this over and over but yet to quote my answer. So leave it alone or do the right thing and accurately state my response.
We’ll start with a basic question;
How could the extremely thin walls of the lunar lander withstand the extreme vacuum of space?
When as we know a vacuum chamber on earth needs 6’-8’ thick concrete walls with a steel insert in addition to a thick walled chamber itself. All of this is necessary for a fraction of the alleged “vacuum of space“.
When a container that has even a small amount of air or gas (pressure) is exposed to a vacuum, even an extremely weak vacuum, the container will expand or even explode. A “spacesuit” would look like the Michelin Man if not simply explode. We know this is true and many scientific experiments confirm this – every time. So what makes you think a few sheets of tin foil can withstand this strong vacuum?
Ed, I would love to get your opinion on this short video;
Hi Digger,
I already know the NASA stuff is fake.
Ed, it’s just a few pages back but I’ll paraphrase.
In my first response I said that the different speeds could be a possibility although I don’t think so. I posted a video where the program Stellarium automatically posted the different speeds for the different seasons. Now as you probably know this isn’t a flat earth program.
I then made a suggestion that the probable answer was that the firmament has the most important role in this phenomenon. This would be due to the reflective or refractive aspect of the firmament.
My last suggestion was somewhat tongue-n-cheek although may contain some truth. I said that the difference in speed was not noticeable as a take off on t8s answer that the drop in elevation of a flying plane was not enough to notice. Essentially, it’s not enough to matter or notice.
While the last answer may seem rather crazy keep in mind that it was inspired by t8.
Thise are the possibilities that I can think of now. Nowhere did I say assertively that the sun did in fact travel at different speeds. I only pointed out that some software did in fact change the speed of the sun automatically without having to be programmed in.
Yep, that video clearly shows the sun moving across Australia at a greater speed than the USA.
You lost.
Hi Digger,
That only explains a northern perspective, but that is NOT my question.
My question is: how do you get a longer than 12 hour sun-cycle in the SOUTH.
The model F-A-I-L-S. You need a model to explain a longer sun-cycle in the SOUTH.
That is the deal-breaker for me. No model – no believably
- You must be logged in to reply to this topic.
