# How the climb rate of an airplane debunks the globe

I have watched and read FlatEarth (“FE”) content that talks about the issue with flight paths, times and other anomalies that are difficult or impossible on a GlobeEarth (“GE”).  However, many GE folks tend to write-off these anomalies by evoking gravity.  Though the gravity excuse can be rebuked, it gives the GE folks a free pass.  This pass is about to be taken away.

My previous work on curvature calculations has given me the idea to track the flight path of a 747-200 class aircraft during the ascent phase of the journey.  On a recent trip to France I recorded the angle of ascent using a bubble level and screen capture software.  This allowed me to compare (in real-time) the ascent angle over time with the flight data from Flight Aware.  The majority of the ascent was between 1-3°.  It lost about 1° of angle every minute.  During the initial take off the angle was much higher at around 12° but it quickly reduced to 2° in about 12 minutes. The final 13 minutes was at a nominal angle of 1°.  A quick calculation shows that TAN(2°) x 185 miles (6.46 miles) is close to the 6.62 miles achieved after 25 minutes.  In other words, the bubble level and the recorded values from flight aware seem to correlate.

If you have an account at flight aware, you can take a look at the actual flight taken.  I will add the data tables here for those who don’t wish to create an account.

The purpose of this post is to show that if there is curvature to the earth, the climb rate over time and the curvature of the earth must place any average international flight at more than double the recorded cruising altitude. This is because the amount of curvature during the ascent phase of the example flight below would be approximately 4.37 miles even without adding any climb at all.

All of the data is publicly available and anyone can do the same calculations I’m going to do.  In fact, any international flight can be used for this experiment.

If you take a look at this graph:

you can see the ascent is in the first 25 minutes and 25 seconds of the flight (01:59:02 – 02:24:27).  You can compare this to the table data below:

An examination of this data eliminates a potential objection by GE folks that plane is flying with the curvature of the earth since the plane cannot be tracking with the curvature and ascending at the same time.  By it’s very definition, the plane is moving away from the ground (ascending), not tracking with it as it should during cruising altitude.  This might seem obvious to most logically minded folks but we are dealing with hardened attitudes that will use any argument, no matter how illogical, to avoid the conclusions reached here.

Since the plane is ascending, and if we are living on a ball, we must take into account the ascending and curvature rates at the same time.  There is no getting away from this.  This also destroys the idea that gravity keeps the plane moving along the curve, since if this was the case, a plane could never ascend at all.  The ascent angle over time is the key.

For example, if a plane took off and ascended at a nominal climbing rate of 1 meter / hour at an average speed of 450 miles / hour, it would reach a cruising altitude of 4.7 miles in 26 minutes.  If it continued at this rate for 52 minutes (or an additional 26 minutes), it would be at a cruising altitude of 19.8 miles.  The plane can ascend because it’s under it’s own power.  It has already overcome the force of gravity or it would fall to the ground.  But I digress….

Using a real example, if we include curvature into the calculation with the recorded climb rate, the plane should reach a cruising altitude of 35,000 feet (or 6.62 miles) in approximately 13 minutes.  This was calculated using 51 specific data points.  Each point contains time, latitude, longitude, course, direction, kts (knots), km/h, meters, climb rate and reporting facility.  By calculating the speed by the time interval, I can calculate the total distance travelled during each time interval.

The first 12 data points are ignored for distance travelled since the plane took off in a westerly direction and then turned around to a northeast direction.  We still need to include the altitude reach during this time frame which is approximately 1.26 miles.  This altitude was reached at 2:02:17 pm.

To accurately calculate the distance travelled, I had to measure the time difference between each data point (between 15-60 seconds) and multiply by the speed of the plane at that interval.  This eliminated the use of an average speed which would most likely be used as an additional rebuttal against this proposal.

If you examine the table below, you can see each time interval along with the height and curvature reached.

The 7 columns of the far right show the curvature that would occur over the distance travelled.  The distance travelled was calculated by converting the speed to nautical miles/min (i.e 281 kts / 60 min = 4.68 nm/min).

I then divided the time frame for each data point by 60 seconds to get the fraction of a minute that the plane travelled during that time frame (ie. 25 seconds / 60 seconds = .41 min)

Once I had that value, I was able to multiply that by the plane’s current speed to get the distance travelled (i.e. 4.68 nm/min × .41 min = 1.95 nm).  I then converted the value from nautical miles to miles (i.e. 1.95 nm × 1.15 = 2.24 miles).

Once I had the distance at each interval I could calculate the amount of curvature in feet.  Whew !!!

Since curvature is accumulative, the total arc distance from the starting point to the specific time frame must be included not just the distance travelled in that particular time frame.

I finally converted the last 21 data points into miles of curvature.  Since the plane was already 1.26 miles in altitude when it started heading northeast, I had to add that value to the curvature accumulated from 02:02:17 pm onward.  I calculated both the altitude of the plane without curvature and one with curvature.

As you can see from the table, the cruising altitude should have been reached at approximately the 13 minute mark.  Since the plane does not stop ascending at the 13 minute mark but continues for an additional 12 minutes, we are left with the inescapable conclusion that there is no curvature.

I’ve included a link to a post that calculates the distance covered during take off and the time taken to reach cruising altitude for comparison purposes.

https://aviation.stackexchange.com/questions/14357/how-long-after-takeoff-for-a-boeing-747-400-to-reach-cruise-speed

# Distance to Horizon Equation Video

I wanted to create a short video that presents the ideas behind an equation that I came up with that properly presents what I call “Distance to Horizon”.  I supposed the point of the equation is to have an accurate way to measure distances on ball since certain inaccuracies in other equations have been used to dismiss the evidence presented by flat earth folk. Once you have an accurate way to measure, it makes even the most entrenched skeptic open to the evidence.

Another reason for this equation is to present what our experience should be like if we lived on a ball.  Of course, one equation can only go so far.  Empirical evidence needs to be collected and real scientific work need to be done to show the truth of our situation.  With certain work already accomplished by individuals in the flat earth community, any contribution towards greater truth is a good thing.

Hopefully this video can help explain this equation a bit better.  Please let me know if any other work needs to be done on this equation.

# Flat Earth – The Horizon, Curvature and Angles

Over the past few months I’ve been working on the curvature equation for a circle.  It doesn’t seem like very exciting stuff but it has enormous implications.  During this process I thought I had found this equation but it turned out to be incorrect.  I had fixed what I thought was the error but that turned out to be incorrect as well.

The nagging problem stems from the way one would experience curvature if they truly lived on a ball.  All the current methods of finding curvature don’t really have a good explanation and upon further study are shown to be calculating something other than curvature.  So what do we mean by curvature?  I will answer this question and provide a new and rational equation for curvature and show why the other equations do not work.

## Ball Earth Math

This famous document that has made the rounds in the flat earth community is not so much incorrect as it calculates an irrelevant number.  Though it does properly calculate the value for X (or the “drop” along the axis), X is not the value we are looking for.

As you can see, the line X is tilted so it is parallel to the axis.  This is where the calculation becomes misleading.  What we really need to do is extend the line in order to intersect the line of site projected from point 0 (in the Ball Earth Map document).  Please see my curvature image to see an example of this.

## Distance² x 8 / 12

The equation “distance² x 8 inches / 12” does not solve for curvature but only for the hypotenuse of the distance: “Height = R-R*(cos(θ))” and “Length = sin(θ)*R”.  Since these two values only represent the length and height of the distance travelled along the hypotenuse; and, since you cannot have a circle without height and length of equal value, you cannot have an arc without height and length of unequal value; therefore, curvature cannot simply be height (or the amount of “drop” along the axis) nor the value of the hypotenuse.

As an example, 1° of circumference is equal to 24901/360°=69.1 miles.  If we plug in that value to the equation we get “√((3959-3959*(cos(1°))² + (sin(1°)*3959))² = 69.09 miles.  This is not the value for curvature.  However, this does resolve the hypotenuse.  The equation was being used to measure curvature by plugging in the arc distance (the distance travelled along the ball) not the hypotenuse; therefore, the resultant value will only solve the value of the hypotenuse.  All we are calculating with this equation is the point at which two different lines of site intersect on a ball not the height required for an object to be visible from point A.

Two points would intersect if we “forced the line” and built along the ground from each point (A and G).  The two lines would have to be 3959 miles long.  This is equivalent to a building at point D being 1639 miles in height.  Since we build either along the surface or perpendicular to the Earth’s surface (i.e. buildings), we need to calculate something else.

## Line of Site, the Hypotenuse and θ

So what are we actually trying to calculate?  If the hypotenuse is not the distance nor the length along the axis, then what is?  The confounding problem is related to the line of site of the observer.  Once we establish the position of the observer, all the other pieces fall into place.  If you examine the image below, you will notice that a line of site moves away from point A towards infinity.  So we need to take the observer as being at point A and does not move.

Next, several dashed lines at various degrees have been drawn until they intersect with the line of site originating at point A.  The equation ((1/(COS(θ)/R))-R) calculates the hypotenuse from θ to the point at which it intersects with the line of site from point A minus the radius.  What we are calculating is the height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point A.  This is the key.

As an example (if the ball theory is to work), as a ship goes over the horizon, the mast (and the rest of the boat) will begin to tilt as per angle θ.  Though this is a tiny angle at first, it nonetheless must follow that angle.  The mast does not start tilting back to stay parallel with the axis of the ball, it stays fixed to the boat.   As the boat continues along the circumference, the mast would need to extend in height to remain visible to the observer at point A.  This is an effect of curvature.

In the diagram below, I put the original curvature calculation beside the equation I have proposed.  You can see that at smaller distances the two values are very similar.  However, as you move past 2° the values begin to diverge more rapidly.

A simple way to calculate the “distance to horizon” is to divide the distance travelled by 69.17 miles which equals θ and plug that into ((1/(COS(θ)/R))-R).

## Curvature

So what do we mean by curvature?  For example, if I travelled from point A to G, I would have experienced 6225 miles of the total circumference but there is not 6225 miles of curvature between point A and G.  If you look at line D1 – D0, you will see that there is only 1159 miles of arc height (or the maximum height of the arc between two points).  It also happens to be the value of X at 45° or 90°/2.  This works for any arc length.

If you know the distance travelled, you can calculate the equivalent θ travelled.  Using a variation of the Ball Earth Math equation, R-R(COS(θ/2)), we can solve for X which gives us the maximum height of the arc.

At a distance of 6225 miles (or ¼ of the globe) the maximum height of the arc has been shown to be 1159 miles.  In a similar fashion, the ship and the observer would have to lift off the surface at 45° and travel for 1159 miles to see each other.  However, we can see that all this is doing is altering the line of site for both the boat and the observer.

## Conclusion

The definition of curvature is the degree to which a curve deviates from a straight line, or a curved surface deviates from a plane.  The curvature of a circle is defined mathematically as the reciprocal of the radius (ie. κ = 1/r).  All this tells us is that as the circle becomes larger κ becomes smaller.  This does not help us figure out what the effect of curvature is to a person living on a ball.

However, by showing rational examples I have demonstrated the effect of curvature is the height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point X.  Without altering the line of site of the observer and keeping the object (in this case a boat) on the surface of the earth, we can measure the effect of curvature.

I would propose then, that what we are looking for is effective curvature and it is defined as “The height necessary for an object to be above the surface, at a particular arc distance away or at angle θ, in order for it to be visible to the observer at point X and X being a stationary position.”