[Tumeni]

If the observer is at 800-850 feet, with an intervening hill between him and the target buildings of 650 feet, and we know the distance between them, then simple school-level trig will tell us the downward angle from observer to intervening hill, and by extension from that, the height H at which that sightline should meet the buildings - IF the underlying plane and reference level for all the heights is perfectly flat.

Like this;



Storm, I invite you do the calcs for us ...

[model 29]

Yet, in the southern U.S., Venus is visible at 45 degrees -- WAY up in the night sky, after sundown for at least two months STRAIGHT. Tonight, Venus is visible for as long FOUR HOURS after sunset. That means I can see it til ONE O'CLOCK in the morning.

Can you tell us what latitude you are at in the mid-northern latitudes, and what time sunset is for that particular latitude in March on a particular day that you claim to see Venus at 1am?

[CJO]

I know this is an old topic but I just thought I would throw it out there the first time I read this post I looked up how to distance of curve of earth in a mile.  I thin it was only 8 inch fall making 30 miles distans only 240 inches not 600 ft. Or 240 ft

[JSS]

I know this is an old topic but I just thought I would throw it out there the first time I read this post I looked up how to distance of curve of earth in a mile.  I thin it was only 8 inch fall making 30 miles distans only 240 inches not 600 ft. Or 240 ft

The correct answer is for 30 miles the drop will be 600.18 feet. 

The formula for estimating the curvature of the Earth you are trying to remember is "Eight Inches Per Mile Squared" that is also used in the 1800's FE bible by Samuel Rowbotham so is used almost exclusively instead of the correct math by most of the Flat Earth community.

The problem is this is wrong.  That 8 inches per mile squared function plots a parabola, not a circle. It's obvious when you note that the FE formula doesn't include PI, which is essential for calculating anything involving circles.

It gives close results for some distances but with very large values it becomes completely inaccurate. It's used a lot in back of the envelope engineering calculations because it's good enough for a few miles and works well as a rough estimate for more, bu eventually it stops becoming useful.

See the image below for how that formula works.  It fits pretty well for a while, but then starts to go wrong and gets worse very fast.

[CJO]

Yeah I apologise 2 all. My math was wrong.  Still in using the calculated drop for 30 miles at the distance of the photo that drop is imperceptible being it falls equally in both directions.  Perception of what ur eyes are seeing is up to ur brain.  That should explain it. Better.

[Storm]

I'd like to issue you a challenge.

Fire up your CAD diagram machine and create a diagram of your ROUND EARTH depiction of that photo, from the observer to the target city--using ACCURATE dimensions, showing ACCURATE curvature/drop for that distance (1350 feet), with the observer on the left and the target on the right - just like you've been showing with a FLAT ground line and make this one with a ROUND ground line. Wow us with your results.

Don't have CAD, but ... why should I? ...

Exactly. Why should you make it THAT obvious that you're completely wrong.

The fact is ... you can't. Plain and simple.

If you were to post the diagram that you've been challenged to create, everyone who saw it would see immediately that the very photograph this thread is all about 100% proves the Flat Earth.

The formula for estimating the curvature of the Earth you are trying to remember is "Eight Inches Per Mile Squared" that is also used in the 1800's FE bible by Samuel Rowbotham so is used almost exclusively instead of the correct math by most of the Flat Earth community.

The problem is this is wrong.  That 8 inches per mile squared function plots a parabola, not a circle. It's obvious when you note that the FE formula doesn't include PI, which is essential for calculating anything involving circles.

It gives close results for some distances but with very large values it becomes completely inaccurate. It's used a lot in back of the envelope engineering calculations because it's good enough for a few miles and works well as a rough estimate for more, bu eventually it stops becoming useful.

See the image below for how that formula works.  It fits pretty well for a while, but then starts to go wrong and gets worse very fast.

None of this makes any difference at all. It's all completely irrelevant because NEITHER formula works one single bit on the Earth that we all live on.

Not once has either curvature formula ever been proven to reflect the reality of the physical plane that we all inhabit.

Furthermore:

Though there were a handful of responses, not one person has tried to explain why Venus can be seen every single night in a row, for going on THREE MONTHS now, at 45 degrees in the sky after dark.

Not ONE diagram or explanation has even come close to explaining this tom-foolery.

Now, somebody will try to claim that Earth and Venus are completely in-sync with each other despite the fact that I've already posted their 'far-from-close' orbit speeds and orbit circumference distances.

So, ....... what's the explanation?

[Tom Bishop]

I know this is an old topic but I just thought I would throw it out there the first time I read this post I looked up how to distance of curve of earth in a mile.  I thin it was only 8 inch fall making 30 miles distans only 240 inches not 600 ft. Or 240 ft

The correct answer is for 30 miles the drop will be 600.18 feet. 

The formula for estimating the curvature of the Earth you are trying to remember is "Eight Inches Per Mile Squared" that is also used in the 1800's FE bible by Samuel Rowbotham so is used almost exclusively instead of the correct math by most of the Flat Earth community.

The problem is this is wrong.  That 8 inches per mile squared function plots a parabola, not a circle. It's obvious when you note that the FE formula doesn't include PI, which is essential for calculating anything involving circles.

It gives close results for some distances but with very large values it becomes completely inaccurate. It's used a lot in back of the envelope engineering calculations because it's good enough for a few miles and works well as a rough estimate for more, bu eventually it stops becoming useful.

See the image below for how that formula works.  It fits pretty well for a while, but then starts to go wrong and gets worse very fast.

Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?

[JSS]

Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?

What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.

Can you show me where I claimed he used it to measure over thousands of miles? I never said that.

[JSS]

None of this makes any difference at all. It's all completely irrelevant because NEITHER formula works one single bit on the Earth that we all live on.

Not once has either curvature formula ever been proven to reflect the reality of the physical plane that we all inhabit.

The "curvature formula" is just basic geometry about the curvature of a sphere. Nothing special about it. I'm not sure how you can claim it's never been proven. The thousands of 'sinking ship' style observations match the expected curve. Bridges long enough have to have their supports at slight angles to account for it. Millions of pictures from space match the "curvature formula" by showing a sphere. All the thousands of satellites, several space stations and hundreds of people, and a few monkeys and dogs that were launched into orbit depend on the Earth being a sphere.

It's basic geometry.

Could you please show me your examples of where the "curvature formula" does not work?

Thanks.

[JSS]

Though there were a handful of responses, not one person has tried to explain why Venus can be seen every single night in a row, for going on THREE MONTHS now, at 45 degrees in the sky after dark.

Not ONE diagram or explanation has even come close to explaining this tom-foolery.

Now, somebody will try to claim that Earth and Venus are completely in-sync with each other despite the fact that I've already posted their 'far-from-close' orbit speeds and orbit circumference distances.

So, ....... what's the explanation?

You can see Venus for about 9 months out of the year before becoming too hard to see due to appearing too close to the sun from our perspective, so seeing it for 3 months in a row is hardly surprising.

I'm not sure what you mean by 45 degrees. Venus will be at 0 degrees when it rises on the horizon, go up and then back down to 0 when it sets. Not knowing your location I can't give any specifics on how high that might currently be, and it will change nightly.

You asked for an explanation. This is a pretty good animation of the phases and why you see them.

https://earthsky.org/astronomy-essentials/tracking-venus-in-late-2019-and-early-20

[Storm]

Could you please show me your examples of where the "curvature formula" does not work?

Thanks.

Oh, if you insist...



...it does a smash-up job of working here.

The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.

I'd ... say it's not workin there too well.

I'm not sure what you mean by 45 degrees. ... and it will change nightly.

Well, it means when somebody in the southern U.S. is perpendicular to the flat ground, the very bright star Venus is at a 45 degree angle from their person. If they look straight up at a 90 degree angle, the star is at 45 degrees.

And, no, it doesn't change each night. Not in any perceptible way. It's been at the same angle (roughly) for months now; every single night after dark. Brightest star in the sky; hard to miss.

Many of the explanations would only make sense if it were way down right on top of the horizon line at sundown. Anybody observing it will confirm that is not the case.

And it remains visible in the night sky for 3-4 hours after sundown.

The only way that is possible is for Venus to be somewhere on the opposite side of the Earth from the sun; it would need to have an orbit outside the Earth's orbit; meaning a further orbit from the sun than the Earth.

And, regardless of that particular detail, the FACT that it has been visible so high in the night sky, at roughly the SAME ANGLE, for months now, demands that it be in some sort of sync with the Earth. Which has already been proven to be impossible per their widely differing orbit speeds and orbit circumferences.

[stack]

Could you please show me your examples of where the "curvature formula" does not work?

Thanks.

Oh, if you insist...



...it does a smash-up job of working here.

The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.

I'd ... say it's not workin there too well.

We went through this already. Unless the data you provided was incorrect, here's the deal in reality. If you don't like the answer, then formulate with facts why that is the case, don't just stomp your feet.

This, here, is absolute NONSENSE.

This is exactly the kind of computer generated fantasy abstract cartoon foolery that'd get laughed out of a court of law.

Without the magical math theorems and computer-aided Disney nonsense, there doesn't seem to be any other way to refute the Plane TRUTH of the Flat Earth we ALL inhabit.

BUSTED!!

Actually, it's your data. Is your data incorrect? And it seems you don't have a rebuttal of any sort. Just stomping your feet and exclaiming the world is flat - Well done. I'm sure that would play nicely in a court of law.

Why don't you think about the data you provided and come up with a cogent explanation and actually defend your position rather than just whining.

[JSS]

The city to the left of the image is 45 miles from the observer. Should be 1,350 feet below the horizon based on Rowbotham's curvature formula that he took directly from the scientific-stated (claimed) circumference of the Earth.

I'd ... say it's not workin there too well.

This has been debunked multiple times, I'm not going to duplicate it. Please re-read all the times it's been explained.

I'm not sure what you mean by 45 degrees. ... and it will change nightly.

Well, it means when somebody in the southern U.S. is perpendicular to the flat ground, the very bright star Venus is at a 45 degree angle from their person. If they look straight up at a 90 degree angle, the star is at 45 degrees.

And, no, it doesn't change each night. Not in any perceptible way. It's been at the same angle (roughly) for months now; every single night after dark. Brightest star in the sky; hard to miss.

Many of the explanations would only make sense if it were way down right on top of the horizon line at sundown. Anybody observing it will confirm that is not the case.

And it remains visible in the night sky for 3-4 hours after sundown.

The only way that is possible is for Venus to be somewhere on the opposite side of the Earth from the sun; it would need to have an orbit outside the Earth's orbit; meaning a further orbit from the sun than the Earth.

And, regardless of that particular detail, the FACT that it has been visible so high in the night sky, at roughly the SAME ANGLE, for months now, demands that it be in some sort of sync with the Earth. Which has already been proven to be impossible per their widely differing orbit speeds and orbit circumferences.

Do you have any proof that it reaches exactly 45% every night and doesn't change? Have you been taking a series photos all these months? All I see is you claiming a fact with zero evidence, just you stating it is. What date did you first observe Venus at 45 degrees? From what latitude? What is it at now?

You are making claims based on a misunderstanding of how angles and orbits work, I'm providing some more information to help below.

I also know you're wrong from my own observations, I do amateur astrophotography and have never once pointed my telescope where science predicted a planet to be and have it not be there. I've taken pictures of Venus, even got shots of it crossing the sun.

Please read the link I posted, and read the new one below.  Here is another that explains it. It has a diagram of why we see Venus, and from what angles.  It also explains why it will vary in height from 19% to 46%, but that also changes based on your position on the Earth.

https://in-the-sky.org/article.php?term=venus

Some quotes and images from that page for reference:



The journey that Venus makes from the evening sky to the morning sky is much quicker than the reverse journey. This is because Venus's orbit is so close to our own, so the points of greatest elongation are much closer to inferior conjunction than to superior conjunction.



The inclination of the ecliptic to the horizon changes over the course of the year, affecting how high planets close to the Sun appear in the sky.

[Tom Bishop]

Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?

What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.

Can you show me where I claimed he used it to measure over thousands of miles? I never said that.

Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine for measuring the distances in the book.

[JSS]

Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?

What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.

Can you show me where I claimed he used it to measure over thousands of miles? I never said that.

Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine to measure the distances in the book.

Again, you are making up things I said and then arguing against them. Classic straw man. You again completely ignored my question so I will repeat.

Where did I state that Rowbotham was measuring thousands of miles.  Quote it. <-- I'm asking you a question right here.

Otherwise, please admit that I never said that and stop arguing like I did.

I'm waiting for that quote.

[Tom Bishop]

Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?

What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.

Can you show me where I claimed he used it to measure over thousands of miles? I never said that.

Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine to measure the distances in the book.

Again, you are making up things I said and then arguing against them. Classic straw man. You again completely ignored my question so I will repeat.

Where did I state that Rowbotham was measuring thousands of miles.  Quote it. <-- I'm asking you a question right here.

Otherwise, please admit that I never said that and stop arguing like I did.

I'm waiting for that quote.

That's what I asked you. You said the equation is wrong because it becomes inaccurate over thousands of miles. Thousands of miles were not tested, however. Therefore the equation is correct.

Are you assuming that math did not exist in the 1800's and that this was not known?

[JSS]

Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?

What does that have to do with anything? I said it works fine for short distances, but is only an estimate that gets worse the further you go.

Can you show me where I claimed he used it to measure over thousands of miles? I never said that.

Rowbotham says that it's the surveyor's rule for the drop of the Earth. Not many surveyors are measuring over thousands of miles. The equation is just fine to measure the distances in the book.

Again, you are making up things I said and then arguing against them. Classic straw man. You again completely ignored my question so I will repeat.

Where did I state that Rowbotham was measuring thousands of miles.  Quote it. <-- I'm asking you a question right here.

Otherwise, please admit that I never said that and stop arguing like I did.

I'm waiting for that quote.

That's what I asked you. You said the equation is wrong because it becomes inaccurate over thousands of miles. Thousands of miles were not tested, however. Therefore the equation is correct.

Are you assuming that math did not exist in the 1800's and that this was not known?

I'm really sorry you are having so much trouble understanding this. I'm not sure how simpler I can make this, but will try, again.

I never mentioned Rowbotham's tests. I mentioned his use of an approximation. A parabola is not a circle. These are all indisputable facts.

You were the one that brought up his experiments. Nobody else mentioned it before that. Only you. It was NEVER mentioned before your comment. Can I make this any clearer?

Please, take a little time to read and understand this before replying. If you need something clarified, ask.

But if you insist on demanding references to things, you must quote where I said such things. Quote me, or please quit claiming I said it. Understand that you can only argue about things I said, not thing you imagined I said.

[Tom Bishop]

I didn't say that you said anything. I asked: "Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?"

The answer is "No" and gives us everything we need to know about the applicability of the math used in this work.

[JSS]

I didn't say that you said anything. I asked: "Can you show us the part in ENAG where Rowbotham is using it to measure over thousands of miles?"

The answer is "No" and gives us everything we need to know about the applicability of the math used in this work.

That is the very definition of a straw man argument.  You ask a question that has nothing at all to do with the discussion, answer it, and claim victory.

Can you not see what you did here?

[Storm]

This photograph of Dallas Fort Worth...



...has not been "gone thru", it has not been dealt with and it most certainly has not been "DEBUNKED."

Not one single person in this entire thread has explained away why the majority of the city of Dallas can be seen CLEARLY from 45 miles away, from just north of Fort Worth, which is 30 miles away from Dallas; when it should be 1,350 feet below the horizon.

Nobody has argued that math.

These are facts.

There is no more proof NEEDED, and there is no satisfactory proof that can be presented online to convince those who refuse to accept what their eyes plainly see.

The computer generated images from Stack of this...



...are absolutely worthless, and have zero basis in reality. Especially the last one.

JSS demands photos of all my observations over time, as well as further evidence, while JSS provides nothing concrete and expects this 'computer generated' image to serve as undebatable evidence of Venus' dynamics in relation to Earth.



Tumeni refuses to create an image, very similar to this one...



..., of the photograph in question, but on a curved plane -- which would prove the entire position of the OP and lay this entire thread to rest.

So, all that is going on here is people talking in circles.

The photograph that started this entire thread is YET to be explained, from a round earth perspective, simply because it cannot be explained away.

Once again, in a court of law, the ONLY piece of evidence that would hold up to legal scrutiny is the photograph that this thread is based upon.

Nothing else that has been presented to the contrary holds one single drop of water.