Scripture Based Flat Earth Proposition

[Sculpt]

How did they do this while telling us they couldn't figure out how to get past the Van Allen belts? This distance is almost four times the distance to the moon.

Sorry, Rom, that's my typo. As the website notes, the DSCOVR satellite is 901,613 miles from earth (not 93,606,069 miles).

You wrote, "Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object."

My source is the distance between earth and the satellites and the circumference of the earth at the equator (probably on an hundred difference websites, but I believe I used Nasa's), pi (3.1415927), and a little math (22232 satellite distance / 7926 earth diameter = 2.8).

I meant what's your source that "Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object." What's your source about cameras being able to do this? (I wasn't questioning the circumference of earth or satellite distance.)

According to this NOAA website, "Since 1975, NOAA's Geostationary Operational Environmental Satellites (GOES) have provided continuous imagery and data on atmospheric conditions and solar activity (space weather)." I wouldn't make sense NOT to put a camera on a $200 million project. Besides, they could shut up all the Flat Earthers . . . if they would only take the shot and publish it, instead of making up all the excuses why they cannot. Only, there's one problem . . . they really cannot do it (because nothing that far away exists to do it)!

You may not have read my post right. I was saying even though there wasn't a good enough reason to put up a satellite that could take pics of the whole earth, they did it anyway. The DSCOVR satellite takes multiple pics of the whole earth every day. You can see those photos here https://epic.gsfc.nasa.gov/

 

[Romans34]

Sorry, Rom, that's my typo. As the website notes, the DSCOVR satellite is 901,613 miles from earth (not 93,606,069 miles).

901,631 miles is almost four times the distance to the moon (238,855 miles).

You wrote, "Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object."

I meant what's your source that "Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object." What's your source about cameras being able to do this? (I wasn't questioning the circumference of earth or satellite distance.)

I simply used my Cannon PowerShot SD1200 IS (not the most powerful camera on the market). Viewer pulls in more than enough to get the whole object.

You may not have read my post right. I was saying even though there wasn't a good enough reason to put up a satellite that could take pics of the whole earth, they did it anyway. The DSCOVR satellite takes multiple pics of the whole earth every day. You can see those photos here https://epic.gsfc.nasa.gov/

Then I say, if "they did it anyway", let's see the pictures (not composites). The pics on the website are not photographs.

 

[Sculpt]

901,631 miles is almost four times the distance to the moon (238,855 miles).

Yes, that is correct.

I simply used my Cannon PowerShot SD1200 IS (not the most powerful camera on the market). Viewer pulls in more than enough to get the whole object.

That's cool. But where did you get the calculation of 2.8 when you said, "cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object."? Did you do some measurements with a tape measure with your camera distance and the object length? Or did you get 2.8 off some FE meme or website?

Then I say, if "they did it anyway", let's see the pictures (not composites). The pics on the website are not photographs.

Yes, they are photographs. A photograph is defined as "an image created by light falling on a photosensitive surface, usually photographic film or an electronic image sensor, such as a CCD or a CMOS chip."

This is the info about DSCOVR's camera:

"The Earth Polychromatic Imaging Camera (EPIC) takes images of the sunlit side of Earth for various Earth science monitoring purposes in ten different channels from ultraviolet to near-infrared. Ozone and aerosol levels are monitored along with cloud dynamics, properties of the land, and vegetation.[29]

EPIC has an aperture diameter of 30.5 cm (12.0 in), a focal ratio of 9.38, a field of view of 0.61°, and an angular sampling resolution of 1.07 arcseconds. Earth's apparent diameter varies from 0.45° to 0.53° full width. Exposure time for each of the 10 narrowband channels (317, 325, 340, 388, 443, 552, 680, 688, 764, and 779 nm) is about 40 ms. The camera produces 2048 × 2048 pixel images, but to increase the number of downloadable images to ten per hour the resolution is averaged to 1024 × 1024 on board. The final resolution is 25 km/pixel (16 mi/pixel)."

 

[Romans34]

That's cool. But where did you get the calculation of 2.8 when you said, "cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object."? Did you do some measurements with a tape measure with your camera distance and the object length? Or did you get 2.8 off some FE meme or website?

I simply looked through the viewer standing the appropriate distance from the object based upon the calculations of the distance of satellites from earth (which is 2.8 times the diameter of the object as already shown).
See Post#639.

 

[Sculpt]

I simply looked through the viewer standing the appropriate distance from the object based upon the calculations of the distance of satellites from earth (which is 2.8 times the diameter of the object as already shown).
See Post#639.

So you're eye-balling it? Making a guestimation? Not measuring. And you get 2.8?

 

[Romans34]

So you're eye-balling it? Making a guestimation? Not measuring. And you get 2.8?

No. The 2.8 is calculated based upon the ratio of the satellite distance from earth and the diameter of the earth. That simple.

 

[Gideon300]

That giant beach ball would only be about 4.2 inches in diameter at one foot away. Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object. The geosynchronous satellites are said to be at least 35780 km (22232 miles) from the earth's surface. The earth's equator is estimated to be 40,075 km (24,901 mi). If you do the simple math, they (satellites) are in orbit above the earth at about 2.8 times the diameter of the earth. So it shouldn't be difficult to take a picture of the entire earth from a geostationary satellite.

Earth's diameter is about 12,750 km. Since the earth is a globe, you will never be able to get a complete picture. So a composite is the only way. Get your beach ball and see for yourself.

 

[Sculpt]

No. The 2.8 is calculated based upon the ratio of the satellite distance from earth and the diameter of the earth. That simple.

You wrote, "Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object." I keep asking you how you know modern cameras can do this? What a camera can do is based on the camera's specifications. What do you know about modern camera's specifications?

Also, your math is incorrect. You wrote:

The geosynchronous satellites are said to be at least 35780 km (22232 miles) from the earth's surface. The earth's equator is estimated to be 40,075 km (24,901 mi). If you do the simple math, they (satellites) are in orbit above the earth at about 2.8 times the diameter of the earth. So it shouldn't be difficult to take a picture of the entire earth from a geostationary satellite.

If earth equator is 24,901 miles in diameter, and you times that by 2.8 you get 69,722.8 miles. But you said geosynchronous satellites are 22,232 miles from the earth's surface. The satellite's distance from earth is not 2.8 times the diameter of the earth... it's only 0.89 times the diameter of the earth.

And of course Gary liked your post. Not sure what that says about Gary.

 

[Romans34]

You wrote, "Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object." I keep asking you how you know modern cameras can do this? What a camera can do is based on the camera's specifications. What do you know about modern camera's specifications?

Also, your math is incorrect. You wrote:


If earth equator is 24,901 miles in diameter, and you times that by 2.8 you get 69,722.8 miles. But you said geosynchronous satellites are 22,232 miles from the earth's surface. The satellite's distance from earth is not 2.8 times the diameter of the earth... it's only 0.89 times the diameter of the earth.

And of course Gary liked your post. Not sure what that says about Gary.

Please, this is not that complicated. Equator (24901 miles) divided by pi (3.1415927) is equal to the diameter of the globe (7926.23 miles). The diameter (7926.23 miles) times 2.8 equals the distance between the globe and camera (22193.44), or approximately the 22232 miles that the satellites are from the surface.

Again, 22232 miles (distance geosynchronous satellites are from the surface) divided by 7926 miles (diameter of globe) equals 2.80. A camera positioned at 22232 miles from earth should be able to snap a shot of the whole thing. Using a 24" diameter beach ball, camera positioned 5' 7.2" (24" x 2.8) from the ball would take in the whole thing. Same ratio as the satellites to globe. Using this ratio, is there any object that you cannot get into the viewer of modern cameras (even without the zoom)? Get a camera and go try it out and you will see that there's no excuse as to why we don't have thousands of actual photographs of the whole globe. Period.

 

[Romans34]

Earth's diameter is about 12,750 km. Since the earth is a globe, you will never be able to get a complete picture. So a composite is the only way. Get your beach ball and see for yourself.

So you agree that the pics on https://epic.gsfc.nasa.gov/ are composites and not real photographs?

 

[GaryA]

And of course Gary liked your post. Not sure what that says about Gary.

My guess would be that it says that Gary liked his post...

That giant beach ball would only be about 4.2 inches in diameter at one foot away. Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object. The geosynchronous satellites are said to be at least 35780 km (22232 miles) from the earth's surface. The earth's equator is estimated to be 40,075 km (24,901 mi). If you do the simple math, they (satellites) are in orbit above the earth at about 2.8 times the diameter of the earth. So it shouldn't be difficult to take a picture of the entire earth from a geostationary satellite.

Gary understands what he is saying.

 

[Sculpt]

Please, this is not that complicated. Equator (24901 miles) divided by pi (3.1415927) is equal to the diameter of the globe (7926.23 miles). The diameter (7926.23 miles) times 2.8 equals the distance between the globe and camera (22193.44), or approximately the 22232 miles that the satellites are from the surface.

Again, 22232 miles (distance geosynchronous satellites are from the surface) divided by 7926 miles (diameter of globe) equals 2.80. A camera positioned at 22232 miles from earth should be able to snap a shot of the whole thing. Using a 24" diameter beach ball, camera positioned 5' 7.2" (24" x 2.8) from the ball would take in the whole thing. Same ratio as the satellites to globe. Using this ratio, is there any object that you cannot get into the viewer of modern cameras (even without the zoom)? Get a camera and go try it out and you will see that there's no excuse as to why we don't have thousands of actual photographs of the whole globe. Period.

OK, I see what you're doing. You're using a circle measurement. I would have just taken the circumference of the earth at the equator (24,901.5 miles) and divided that by two: 12,450.8 miles. I was confused by the way you wrote it in #363, as I assumed you meant the diameter was 24,901.5 (I was just taking your numbers, not looking them up myself).

The math question is beside the point because there is a satellite that takes real, non-composite photos of the entire earth. It's called DSCOVR. You can see those photos here https://epic.gsfc.nasa.gov/.

Let's move to an area that you actually base your beliefs on. In the quote below, this what you said in response to the video presentation you watched from Answers in Genesis fellow Dr. Danny Faulkner "Does The Bible Describe the Earth as Flat?" (video at bottom). You said Faulkner countered the weakest arguments that some flatearthers use. What are the best arguments? I assume you might mean biblical verses. What are the best biblical verses and/or best arguments?

I watched the entire video and heard his argument for a spherical earth, though he really spent most of his time trying to disprove the flat earth. I'm afraid I have to say that what I noticed is that for the most part his explanations and defense against flat earth were based on the weakest arguments that some flat earthers use to bolster their argument (some of which I would NEVER use because they are so weak). And his arguments were weak as well. Many things he mentioned, explaining what the flat earther believes, verbally refuted it, while not really proving anything to the contrary. I want to propose for those that may be willing to reason with me, to take a particular train of thought one step at the time until we come to some conclusion that makes sense, but I will present this in another post as soon as I am able.

 

[Susanna]

That giant beach ball would only be about 4.2 inches in diameter at one foot away. Modern cameras can take pictures of most any object at a distance of 2.8 times the diameter of the object. The geosynchronous satellites are said to be at least 35780 km (22232 miles) from the earth's surface. The earth's equator is estimated to be 40,075 km (24,901 mi). If you do the simple math, they (satellites) are in orbit above the earth at about 2.8 times the diameter of the earth. So it shouldn't be difficult to take a picture of the entire earth from a geostationary satellite.

Are you for real?

 

[Romans34]

Let's move to an area that you actually base your beliefs on. In the quote below, this what you said in response to the video presentation you watched from Answers in Genesis fellow Dr. Danny Faulkner "Does The Bible Describe the Earth as Flat?" (video at bottom). You said Faulkner countered the weakest arguments that some flatearthers use. What are the best arguments? I assume you might mean biblical verses. What are the best biblical verses and/or best arguments?

I've already stated them many times. How about someone (and I mean ANYONE who is not a shill) who is truly interested in addressing the subject of this proposition with intelligence, self-discipline (mature enough to avoid arrogant judgemental and derogatory remarks), and sincerity review posts 432 & 436 and let us move forward from there.

 

[GaryA]

The math question is beside the point because there is a satellite that takes real, non-composite photos of the entire earth.

It is not beside the point - a much closer satellite still far enough away to take a photo of the entire earth would be able to capture much more detail.

Compare taking a photo of the entire earth both at ~22,000 miles away and at ~901,000 miles away.

Consider hue/tint/tone/shade - which would make a better truer picture? The closer one, of course.

If you could take a photo of the entire earth at ~22,000 miles away, why would you need to go out to ~901,000 miles away?

Do you REALLY believe that there is a satellite in orbit around the earth that is so far beyond the moon? It is obvious from the pictures that it has an inordinately long slow orbit - what keeps it in orbit? The gravitational force of the moon would easily throw it out of orbit.

Do you realize that the "sunspot" in the center of those pictures is impossible considering the necessary orbit path position based on the spin direction? The only possible legitimate scenario for it is at the equinox times - it cannot exist right now in July. Therefore, it is false - it is edited into the pictures.

Do you realize that any metallic satellite in "outer space" would be destroyed (burn up, melt down, ...) from absorbing energy from sunlight and not being able to dissipate it into a vacuum? Electronic circuits certainly would not be able to handle the heat and function properly.

BTW - I took a picture of a 12" diameter desk globe - entirely in the frame - with the camera lens only 7" away from the globe.

That's right - less than one diameter away!

 

[GaryA]

There is no reason whatsoever why any-and-every satellite in "outer space" should not be able to take pictures of the entire earth.

We should have thousands/millions of actual original unedited single photographs of the entire earth by now...

 

[GaryA]

I've already stated them many times. How about someone (and I mean ANYONE who is not a shill) who is truly interested in addressing the subject of this proposition with intelligence, self-discipline (mature enough to avoid arrogant judgemental and derogatory remarks), and sincerity review posts 432 & 436 and let us move forward from there.

We should get the thread back on the rails...

 

[GaryA]

For what it may be worth - here are quick links to posts #432 and #436:

https://christianchat.com/conspirac...earth-proposition.213032/page-22#post-5296467

https://christianchat.com/conspirac...earth-proposition.213032/page-22#post-5297443

 

[Gideon300]

So you agree that the pics on https://epic.gsfc.nasa.gov/ are composites and not real photographs?

NASA states that they are composites. It's simply impossible otherwise. A composite is a a real photo. Stand on a high point and look to the horizon. Can you see the whole of the view? No. Your field of view is insufficient. Likewise with cameras. And no camera can see the other side of a sphere, no more than the human eye can.

Whatever happened to logic and reason? I don't know why the saying is "common sense". There's not much of that around any more.

 

[GaryA]

They claim to receive “ribbons of imagery” from satellites ...

These "ribbons of imagery" do not come from satellites; rather, they come from airplane fly-overs.