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How early could the radius of the Earth be calculated?


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How early could the radius of the Earth be calculated with better accuracy than Newton?

Objective is to calculate the radius of the Earth using methods available to the Greeks around the time of Thales of Miletus (c600BC) and Pythagoras (c500BC) and then by enhancing with modern science.

The numbers below, allowing for the effect of normal atmospheric refraction, were obtained from distance of sea horizon tables found in nautical tables/almanacs. In ancient times the same results could be obtained by practical observation.

With a height of eye of 2.1metres the distance of the sea horizon is found to be 5630.08 metres (3.04nm).

Using Pythagoras: A=90deg, b=5630.08metres, c=2.1metres giving C=0.02137deg

Using Thales’ Theorem: Angle at Earth centre C multiplied by 2, giving CE=0.04274deg.

Using Pythagoras: A=90deg, C=0.04274deg, c=5630.08metres, giving Earth radius =7547Km. This answer is incorrect because it was not corrected for refraction.

Amazingly, as the Sun has a disc size of 32.56 minutes of arc in January (when it’s closest) and 31.5’ in July (when it’s farthest), this means that because mean refraction is 34.5 minutes of arc at zero altitude, just before the Sun starts to dip below the horizon, in reality it is already completely out of sight due to refraction. Refraction needs to be corrected for. The ancients did not know about refraction or how to correct for it, meaning too big a radius figure, but we do, using the equation for terrestrial refraction correction:

 minutes of arc to be added to angle C = √ Height of Eye in metres

                                                                                    7

(Do not confuse the equation above with the one used by mariners: Dip = -1.76√ Height of eye in metres           where dip is the refraction correction in minutes of arc to be subtracted from the observed sextant altitude of the body being observed)      

Using the terrestrial refraction corr’n equation: minutes of arc to be added to C = √ 2.1

giving 0.21’arc, which is 3.45x10-3 of a degree.                                                             7

C + refraction = 0.02137 + (3.45x10-3)   giving corrected angle CC = 0.02482 degrees.

Using Thales’ Theorem: Multiply CC by 2 to give CE = 0.04964 degrees.

Using Pythagoras: A=90deg, C=0.04964deg, c= 5630.08metres, giving Earth radius 6498km. This figure is still too big and larger height of eye is needed for better accuracy.

Using a height of eye (a cliff) of 244 metres above the sea, the horizon tables give a horizon of 60190metres (32.5nm). See diagram. This may be achievable by observation.

Using Pythagoras: A=90deg, b=60190metres, c=244metres giving C=0.2322661deg

Using Thales Theorem: Angle at centre of Earth is C, multiplied by 2, giving CE=0.4645322deg.

Using Pythagoras: A=90deg, C=0.4645322deg, c=60190metres, giving Earth radius =7424Km. This answer is better using a bigger range, but it not corrected for refraction.

Using the refraction correction equation: minutes of arc to be added to C = √ 244         

giving 2.23’arc, which is 0.0371666 of a degree.                                                  7

C + refraction = 0.2322661 + (0.0371666)   giving corrected angle CC = 0.26943 degrees.

Using Thales Theorem: Multiply CC by 2 to give CE = 0.53887 degrees. Using Pythagoras: A=90deg, C=0.53887deg, c= 60190metres, giving Earth radius 6399km

I wish to give credit to the new book called ‘To Explain The World’ by Steven Weinberg (winner of the Nobel Prize) for inspiring this post.

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Very nice, but to answer your question ("How early could the radius of the Earth be calculated with better accuracy than Newton?") I note that you've used modern data for horizon distance and refraction correction. So I think the answer is "some time after Newton".

The first good estimate of the Earth's size was, I believe, that of Eratosthenes, though only his method is known, not the exact figure he arrived at. His method was to use the difference in the Sun's zenith angle at noon from two sites on the same meridian at known distance. His result was limited by the accuracy with which he could determine the variables, as is also the case in your presentation.

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I suspect the method to define the radius existed maybe with the designers of the pyramids or even before and the method is what is important the actual value is only ever going to be an approximation, what I find wierd is how anyone ever thought the Earth was flat I can stand on a hill and can clearly see the horizon is not a straight line.

Alan

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I suspect the method to define the radius existed maybe with the designers of the pyramids or even before and the method is what is important the actual value is only ever going to be an approximation, what I find wierd is how anyone ever thought the Earth was flat I can stand on a hill and can clearly see the horizon is not a straight line.

Alan

are you really seeing the curvature of the earth? this is a real question as I don't know. When you stand on a beach the horizon appears to be curved but how much of it is an illusion? the horizon is only 6 miles away at sea level (i think) so how much of an arc is scribed on a globe with a circumference of 24000 km with a circle just 20ish km ?

I seem to recall that pilots of high altitude aircraft  (and brian cox also saw it on one of his series) being blown away the first time they see the curvature of the earth and they are many miles above a hill

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I suspect the method to define the radius existed maybe with the designers of the pyramids or even before and the method is what is important the actual value is only ever going to be an approximation, what I find wierd is how anyone ever thought the Earth was flat I can stand on a hill and can clearly see the horizon is not a straight line.

Alan

The existence of the horizon was one of the proofs given by Aristotle to show that the Earth is round. But what do you mean by saying it's "not a straight line"? Do you mean you think you can see the curvature of the Earth? That's an optical illusion caused by the way we process visual signals. From a ship (or aircraft over sea) the horizon is a circle, with every point being equidistant from the observer. It should therefore appear "flat", if it weren't for our brain playing tricks.

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It may be an illusion but I remember seeing that Brian Cox episode and thinking how could you not see it, even on a normal flight it was clear as day to me and as said I can clearly see it with my feet planted firmly on the Earth but maybe I am weird.

Alan

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It may be an illusion but I remember seeing that Brian Cox episode and thinking how could you not see it, even on a normal flight it was clear as day to me and as said I can clearly see it with my feet planted firmly on the Earth but maybe I am weird.

Alan

I saw the curvature effect very clearly when I was on a plane a few days ago. I can certainly see why people are fooled by it. But just think of the geometry and you'll see why it can only be an illusion. We're not weird, just human.

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Just to add, or better, subtract from the "Flat Earth" idea, I believe there is very little evidence that anyone has actually believed the world is flat at almost any time of what may be referred to as human civilisation.

For some reason a lot of these come out of fairly recent idas that we are more "advanced" as previous people and the "proof" was to comment that they thought the world was flat. When they never did.

Along similar lines many of our ideas on "history" or "historical situations" are fictitious. A lot of this appears to stem from the Victorian era when a fictional book would somehow become "factual".

I believe the Chinese determined a radius or circumference for the Earth a few thousand years ago and I would also strongly suspect that the early Indian civilisations did also as they were seafaring.

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The illusion's interesting. If you visualize being on an imaginary flat Earth while looking at a very long, straight and evenly high wall some way off from you, won't the wall appear to be at its highest along the line of sight leading to its nearest point? The further away you looked to either side of this nearest point the lower the wall's height would appear to be because the further away it would be. In other words the top and bottom of the wall would seem to be curved and would eventually meet when the limit of your eye's resolution was met. Is this right? Erm...

Olly

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Whilst I understand the geometry and can follow your arithmetic, I don’t quite understand what is achieved by it.

You give a specific value for the distance to the horizon(DTH) of 60.190km, and suggest that this may be found by observation. 60km is a very long way to measure to an accuracy of 1 metre. I think anyone equipped with 2500 year old technology would struggle to achieve anywhere near that precision on land, let alone at sea. Any error in the DTH value would cause over a 100 fold error in R (approx. proportional to the cotangent of the small angle).

You take your DTH value from a nautical almanac. My understanding is that these tabulated DTH values are calculated, using a currently accepted (geodesic) value for the radius of the Earth. This would imply that you are using a DTH value based on a knowledge of the radius of the Earth, to calculate the radius of the Earth. In this situation, I’m not sure what is being achieved.

Refraction does affect the result, and your method does indeed show that it must be included. However, the 1/7 term in your function would be recognized by surveyors as a Refraction Coefficient, used to correct zenith angles. It is based on the ratio of the curve of a light path (due to atmospheric refraction), to the curve of the Earth’s surface. So this value is also based on a prior knowledge of the Earth’s radius.

So, the method works, but is dependent on an ability to make measurements. If we take those measurements from an almanac, we can only hope to find the radius value they used to compile the almanac.

There are several methods that can be used to find the radius of the Earth, but I believe it wasn’t until suitable technology came along in the 16th, 17th and 18th centuries (stop watches, theodolites, telescopes etc.), that the problem was suitably solved. Ancient civilizations may have understood the science, but lacked the technology to experimentally verify their theories. That some came close is outstanding.

Ray

I trust the OP understands that I only comment because I share their interest, and hope to be corrected if I’m wrong.  

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Whilst I understand the geometry and can follow your arithmetic, I don’t quite understand what is achieved by it.

You give a specific value for the distance to the horizon(DTH) of 60.190km, and suggest that this may be found by observation. 60km is a very long way to measure to an accuracy of 1 metre. I think anyone equipped with 2500 year old technology would struggle to achieve anywhere near that precision on land, let alone at sea. Any error in the DTH value would cause over a 100 fold error in R (approx. proportional to the cotangent of the small angle).

You take your DTH value from a nautical almanac. My understanding is that these tabulated DTH values are calculated, using a currently accepted (geodesic) value for the radius of the Earth. This would imply that you are using a DTH value based on a knowledge of the radius of the Earth, to calculate the radius of the Earth. In this situation, I’m not sure what is being achieved.

Refraction does affect the result, and your method does indeed show that it must be included. However, the 1/7 term in your function would be recognized by surveyors as a Refraction Coefficient, used to correct zenith angles. It is based on the ratio of the curve of a light path (due to atmospheric refraction), to the curve of the Earth’s surface. So this value is also based on a prior knowledge of the Earth’s radius.

So, the method works, but is dependent on an ability to make measurements. If we take those measurements from an almanac, we can only hope to find the radius value they used to compile the almanac.

There are several methods that can be used to find the radius of the Earth, but I believe it wasn’t until suitable technology came along in the 16th, 17th and 18th centuries (stop watches, theodolites, telescopes etc.), that the problem was suitably solved. Ancient civilizations may have understood the science, but lacked the technology to experimentally verify their theories. That some came close is outstanding.

Ray

I trust the OP understands that I only comment because I share their interest, and hope to be corrected if I’m wrong.  

Good question. I achieved exposing it’s an illusion that ancients thought the World’s flat.  

I was educated with the idea that sailors around the time of Columbus’s 1476 voyage across the Atlantic thought the Earth was flat and might fall off the edge. With this mindset I wanted to show how the ancients could have saved themselves the concern by using Thales’ Theorem and Pythagoras Theorem both available c500BC to calculate the radius of the Earth and establish it’s spherical nature. Thanks to the answers from stargazers lounge members to my post I only now realise it was my education at fault in giving me the illusion the ancients thought the world was flat!

Now, it needs to be said that the sea horizon tables used, although convenient and useful, are not essential for the success of the idea in the title and nothing magical (they use the equation: Distance in nautical miles = 2.08√Height of eye in metres). These tables give you the same horizon distance you can get from observation. So, the method I would use to find the horizon distance is to use a straight length of beach and have an assistant tend a lantern at the waterline while I marked the position the lantern appeared above the horizon while at the desired height of eye above the water at the beach. Trial and error would be employed. For a large (say 6metre) height of eye a floating platform moved up the beach by helpers would have to be employed. The length along the beach between the platform and the lantern could be measured using a rolling distance click wheel (5.09nautical miles for 6metre ht of eye). I acknowledge that a cliff top height of eye of 244metres in my post giving an eye watering 32.5 nm distance to measure is probably too difficult to achieve by observation, so I have used a more realistic maximum 6metre height of eye using simulated ancient method:

Re working the calculation using Height of eye of 6 metres the distance of the sea horizon is 9426.68 metres (5.09nm). Tables used, so distance rounded to 9430 metres to simulate the less accurate click wheel distance observed method.

Using Pythagoras: A=90deg, b=9430metres, c=6 metres giving C=0.0365(rounded)deg

Using Thales’ Theorem: Angle at Earth centre C multiplied by 2, giving CE=0.073deg.

Using Pythagoras: A=90deg, C=0.073deg, c=9430metres, giving Earth radius =7401km. This answer is incorrect because it was not corrected for refraction. Actual Earth radius is 6371km. The result is therefore16 per cent too high. Not bad for methods available 2500 years ago.

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