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Flat Earth?


gerrydandridge

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I hope this sketch will help to work things out.

 

The radius of the Earth is R.

 

We are looking for the angle X.

 

They are at latitude 50 degrees.

 

Does that make sense?

 

Spot any errors?

 

So, how do you work out X? Any ideas?

 

attachicon.gifSketch.png

not something that immediately makes sense to me china no. but i see how well you have demonstrated your point in order to help me understand it correctly. i can't spot any errors as i'm not sure what they pertain to? and i don't know how to work out x without looking it up. x looks about 38 by eye! but i'm here to learn china and i'm sure if you explained it to me one to one i could understand what you are showing me perfectly. and i'm sure you would be able to verify my understanding and say well done paul good lad. i appreciate your kindness x

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It's a bit late so I'll leave it to tomorrow to try to explain, but can you see the connection between my drawing and Antonio's globe?

 

Antonio insisted that drawings aren't 3D - which is true, but not really relevant because you can still build up the maths for a 3D object from a 2D sketch.

 

I hope you can see how the drawing is like one of the pieces of Antonio's spaghetti going up to Polaris.

 

I've not included the other two - they will have the same view as the first one - they are all on the same latitude.

 

I really hope you can see the sketch represents the same situation as the globe. If you disagree it means you won't be able to accept the conclusions you can draw from the mathematics, so this issue is the clincher.

 

Do you get what the drawing represents, and see how it is the same situation as in the Youtube video?

 

One way to think about it is that we've cut the globe exactly in two where one of the spaghetti pieces is stuck to the surface at a latitude of 50 degrees. I hope that makes sense to you.

 

If it doesn't, try to say why and we'll try again, but in all these things unless we can agree the conceptual model it is very hard to get some mutual understanding.

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yes very good china i see exactly how well you have demonstrated your drawing to fit the model in the video. i think with a cup of tea, pencil and paper, you could sort it out with me in 20 mins or so. and i would then understand exactly what you are showing me and the points you are making. it's excessively laborious, especially for you i mean, post by post rather than face to face. so thank you and that's an invite by the way if you feel it would benefit us both, as i do x

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OK ... so ... we have a diagram where we are assuming Polaris is directly above the North Pole and is 4 Earth Radiuses above it.

 

post-1364-0-83718400-1450785325_thumb.png

 

Now, from this we can extract a triangle with one vertex having an angle of 40 degrees, one side length R (the radius of the Earth) and one side length 5R.

 

post-1364-0-35066900-1450785404_thumb.png

 

We are wanting to know the angle X. X + 90 degrees is the angle a second vertex of the triangle subtends.

 

To work out the angles around a triangle you need to know the cosine and sine rules.

 

post-1364-0-25822300-1450785532_thumb.png

 

The cosine rule is a2 = b2 + c2 - 2 x b x c x cos A

 

The sine rule is Sin(A)/a = Sin(B)/b = Sin©/c

 

So, looking at the triangle up to Polaris we've got a triangle with one side of length R, another side of length 5R and one we don't know how long it is - let's say that side is length a.

 

The angle A - opposite the side of length a has an angle of 40 degrees = 90 - the latitude.

 

So a2 = R2 + (5R)2 - 2 x R x 5R x cos(90-latitude)

 

So a = (26R2 - 10xR2 x cos(90-latitude))0.5

 

You can simplify this by taking R2 out.

 

a = (26- 10x cos(90-latitude))0.5 x R

 

Knowing a - we can now use the sine rule.

 

Sin(A)/a = Sin(B)/b = Sin©/c

 

We've just worked out what a is, and A = (90-latitude)

 

We can say B = 90+x what we are trying to find, and if that is the case then b = 5R.

 

So:

 

Sin(90-Latitude)/((26- 10x cos(90-latitude))0.5 x R) = Sin(90+x)/5R

 

Simplifying

 

Sin(90+x) = Sin(90-latitude) x 5R / ((26- 10x cos(90-latitude))0.5 x R)

 

Notice we can cancel out the Rs

 

Sin(90+x) = Sin(90-latitude) x 5 / ((26- 10x cos(90-latitude))0.5 )

 

Taking the inverse sine*

 

90+x = asin(Sin(90-latitude) x 5 / ((26- 10x cos(90-latitude))0.5 )

 

So x = asin(Sin(90-latitude) x 5 / ((26- 10x cos(90-latitude))0.5 ) - 90

So if the latitude is 50 degrees and Polaris was 5 earth radiuses away it would subtend an angle of 41.3 degrees.

 

Less than the 50 degrees of latitude.

 

So Antonio was incorrect - he tried to measure his spaghetti with a protractor and thought the angle was 50 degrees - it isn't, it is a bit less than that - though it isn't that surprising he was in error - getting the protractor at the right angle and measuring the angle accurately isn't easy with a model built like that - maths is a much better way to go!

 

Now this maths is all much of a muchness.

 

The first thing to ask is do you get what I've done, and the second thing is to ask how you can actually do some science with this result, but that is probably enough for the moment.

 

PGW do you get what I've done?

 

 

 

*There is a little bit of a complication here as the inverse sine function, asin, will only give you a number between -90 and 90 degrees even when the angle is between -360 and 360 - you need to know the form of the sine wave and transform it when dealing with an obtuse angle (which we have: the angle is 90+x). If you didn't know this you'd get an answer of -41.3 degrees as the answer not 41.3 - as ever in maths once you've got the answer you have to think, is the domain of your answer correct!

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sorry china, yes i am with you so far, although i would be asking lots of quick questions if we were together doing this, just to verify the accuracy of my understanding with you. like the cos sin values, rules, properties etc. i am not used to practicing maths nor have i furthered my understanding of the classic curriculum since school. i do enjoy it though and can follow along well enough. i'm very good at learning and remembering what i have learned, when i put my mind to it. i like studying and asking questions of what i learn about.

 

so just to clarify, how and why are we assuming the distance to polaris. from the lattitude and convergence of the lines?

 

out of interest would you like to assess these videos from the ball earth skeptics jon le bon and brian mullins. i enjoyed them as a novice.

 

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i'd just like to reiterate china and all that my position here has not changed and there has been a genuine resurgence in the topic of the shape of the earth. i know some people can't believe that or like china, see so many flaws in the whole subject. but i have remained open minded throughout and continue to post various things that interest me or just make me laugh like the spaghetti science.

 

it's really not as simple as some might think, to actually have a good understanding of the history and milestones, and events/research which is required, in order to know/believe the status quo regarding the earth's shape and size. i still don't know or believe one way or the other and i don't care. but if it's so obvious and there's so much evidence, then it makes it all the more peculiar how it has now become subject so many are making videos about. i'm not bein trite by sayin that either. i respect the giants who've gone before me but i would have to research for the rest of my life in every field just to cover the bases of scientific history. but watching this unfold across a broad spectrum of people has been fascinating and entertaining.

 

i don't think anyone should be ridiculed for looking into the subject for themselves and relearning, as i am, so much of what we were taught at school. it's just that my understanding is improving now all the time and that can only be a good thing. believe me china, if i knew for a fact and was absolutely sure of my belief i would just say so. i am genuinely intrigued as to the why and how of all things flat/ball/concave! earth. yes seriously there are concavers as well. lord steven christ none the less! thanks again china, looking forward to the next step x

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... i have remained open minded throughout ...

 

So open-minded your brains've fallen out ...

no quilp that's just your self inflated, programmed, cliched ego tellin you porkies mate. ye can't ridicule me, you're just hiding from a reflection you don't want to see. pretending faces never change a thing fella.

 

the mind can be said to be like a parachute or an umbrella, it needs to be open to work properly. good luck finding your ripcords x

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So ... PGW ... let's try and continue with this.

 

We've developed a mathematical model to examine the elevation Polaris that simply uses the how far away Polaris is, measured in radiuses of the earth, and the latitude of the observer on the earth.

 

This model could have been built out of spaghetti and a protractor, but by doing it with maths we can get an exact mathematical calculation.

 

When Polaris is 4 earth radiuses away from the Earth - like in Antonio's spaghetti model - the result is:

 

The Elevation of Polaris = x = asin(Sin(90-latitude) x 5 / ((26- 10x cos(90-latitude))0.5 ) - 90 = 41.3 degrees.

 

Now we can generalise this equation - instead of Polaris being 4 Radiuses from the pole - which is 5 radiuses from the centre of the earth.

 

Lets say Polaris is N radiuses from the centre of the Earth.

 

If you do that you will get the following equation

 

The Elevation of Polaris = x = asin(Sin(90-latitude) x N / ((N2+1 - 2 x N x cos(90-latitude))0.5 ) - 90

 

So here is a graph looking at what Polaris would look like if it was 5, 10 and 1,000,000 Earth Radiuses away from the centre of the earth for latitudes 0 to 90 N. IE from the equator to the North Pole.
post-1364-0-29641400-1450879999_thumb.png

 

Now at this stage we don't know how far Polaris is actually away, but we have created the lines on the graph IF IT WAS 5 or 10 or 1,000,000 earth radiuses away.

 

In all of them if you stand on the North Pole (latitude 90 degrees), Polaris would be directly over your head - an elevation of 90 degrees.

 

But the elevation you would measure at latitudes other than 90 degrees is different depending how faraway Polaris is.

 

The closer it is to the earth the more quickly the star would drop down towards the horizon as you moved away from the North Pole.

 

Are you understanding?

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