# AI Throws Out Nonsense to Simple Questions About the Flat Earth

## Nothing new here, but I thought you’d be interested in reading this. This would be a typical response to someone who is completely brainwashed by the media. But at least AI doesn’t get nasty!

Using Spherical trigonometry, what is the drop-off of every mile at the equator?

Spherical trigonometry deals with the relationships between angles and distances in three-dimensional space. The drop-off of every mile at the equator can’t be determined solely using spherical trigonometry as it depends on the elevation of the starting point, the slope of the terrain, and other factors that aren’t related to spherical trigonometry.

Spherical trigonometry can be used to calculate angles and distances on the surface of a sphere, such as the Earth. For example, it can be used to determine the great-circle distance between two points on the Earth’s surface or to calculate the altitude of a celestial body above the horizon. However, it doesn’t provide information about changes in elevation or drop-off over a specific distance.

Say the starting point is at sea level, what is the drop-off?

If the starting point is at sea level, then there is no drop-off. At sea level, the elevation is constant and there is no change in height. The concept of drop-off implies a change in elevation or height over a certain distance, so a starting point at sea level would not result in a drop-off.

Say it’s at 1,000 feet altitude

I apologize, but the concept of “drop-off” is not applicable in this scenario. The Earth’s surface is curved, and the change in elevation over a given distance at a constant altitude of 1,000 feet would be a result of the curvature of the Earth’s surface, not a drop-off. To accurately calculate the change in elevation over a given distance, detailed topographical information would be needed, along with additional calculations taking into account the Earth’s curvature.

What is the curvature over a distance of 10 miles?

The Earth’s surface is a sphere, and its curvature can be calculated using mathematical formulas. The amount of curvature over a distance of 10 miles will depend on the latitude of the starting point. At the equator, the Earth’s circumference is greatest, and the curvature is greatest. The closer you are to the poles, the smaller the Earth’s circumference, and the smaller the curvature.

To give a general idea, the Earth’s circumference at the equator is approximately 24,901 miles. Therefore, over a distance of 10 miles, the drop in elevation due to the Earth’s curvature is very small and would not be noticeable to the naked eye. A more precise calculation of the Earth’s curvature over a specific distance would require more detailed information and a series of complex calculations.

How far does a boat at sea has to be before it’s behind the curve, assuming the boat is 10 feet high?

The distance at which a boat would be behind the curve of the Earth’s surface depends on the height of the boat above the water and the latitude of the boat. The concept of “behind the curve” is often used to describe the point at which the curvature of the Earth’s surface causes the bottom of an object to no longer be visible from a observer’s perspective.

In general, the farther the observer is from the object, the more the Earth’s curvature will come into play, causing the bottom of the object to disappear from view. However, determining the exact distance at which this would occur is not a simple task and would require a detailed calculation taking into account the height of the boat, the observer’s height above sea level, and the latitude of the boat.

As a rough estimate, the Earth’s curvature causes an object to disappear from view if it is about 8 inches lower in the horizon for every mile it is away from the observer. Based on this, a 10-foot high boat would disappear from view when it is approximately 5 miles away from an observer at sea level. However, this is just a rough estimate and the actual distance at which the boat would disappear from view could be different, depending on the observer’s height and the latitude of the boat.

To comment on what you said, “The ability of a telescope to see a distant object, such as a boat, is not affected by the water in between the observer and the object.” ANY object, including water, does effect what you see. If I’m looking at you at 10 feet away, I can see you, but if a wall was put in between us, then I would NOT be able to see you.

As for the second comment I would like to make, you said, “This increased size makes it possible to see object that would otherwise be obscured to the Earth’s curvature.” May I inform you that a telescope and only see in a STRAIGHT LINE, not behind a curve, no matter how powerful the telescope is!

Do you know that your answer does not make sense to a human. I think a more intelligent human needs to program you!

What this means is, that the earth is flat! For, with a powerful enough telescope on a clear day and calm sea, I can see a 100 foot tall building 20 miles away. 