41

Flat Earth Theory / Re: Around-the-World Sailing Races?

« on: May 19, 2020, 06:46:46 AM »

Not clever . It's a statement of the bleedin obvious.

The tropics mark the Northern and Southernmost limits of the sun's journey as it orbits around the plane . The equator is the midpoint on it's journey .

The globe model has the tropics at 66.6 degrees from the poles and the polar circles at 66.6 degrees from the equator . The apparent cause is the tilt of the earths axis being 23.4 degrees - which is 66.6 degrees if you take the angle from the horizontal.
And earth scoots around the sun at an average distance of 93,000,000 miles and completes it orbit in 365.25 days (hence our need for  leap years) which gives a velocity of 66,600 mph.
Interesting that.

42

Flat Earth Theory / Re: Around-the-World Sailing Races?

« on: May 18, 2020, 09:39:42 PM »

It doesn't.

43

Flat Earth Theory / Re: Around-the-World Sailing Races?

« on: May 18, 2020, 05:54:03 PM »

Okay, I gave the illustration of an equilateral triangle on a sphere with three 90 degree angles to contrast to the standard flat equilateral triangle of three 60 degree angles just to illustrate the difference and to make sure readers knew what I was talking about.  Never-the-less spherical trig is indeed different than flat trig, and celestial navigation is based on spherical trig.  It uses spherical trig to calculate the angles relative to where a celestial body is directly over the surface of the earth at the time you are taking the sight.  (In other words the place you would be if the celestial body you are using was straight up over your head.) If that were the case all the time then celestial would be absurdly easy. Since the celestial object (sun, moon stars, planets) is almost never directly over your head, you use a sextant to measure the angle that object appears above the horizon.  Celestial then uses that angle with spherical trig to determine your "line of position".  (It's a movie myth that a single sextant sight gives you an "X marks the spot" position.  Except for the famous "noon sight" you get a short line that you are somewhere on that line.  There are other ways to then upscale that to a more precise position.).  And in reality nobody does the math themselves.  You use either books of detailed tables or later computer/calculators to do that math.  BUT the reality is that the system is indeed based on spherical trig, not flat trig.  The respondent's statement that you don't need spherical trig to navigate, that it's "nonsense" is simply not true for celestial navigation.  it is true for navigating on a chart, within sight of land, to work out your bearing to a lighthouse for example.  Flat trig works fine for that, because the distances involved are so short that the "needed correction" for being on the surface of a sphere is trivial.  It's the same reason you can use a flat paper chart to navigate along the coast.  Technically that flat chart is slightly "wrong" since you cannot accurately depict a sphere on a flat paper, but again for the distances involved the "error" is trivial.  But out on the open ocean, trying to determine your position is an entirely different matter.  Prior to GPS, you needed celestial navigation to do that. And without the spherical trig behind celestial, it would fail. And it doesn't fail - it works.  It's worked for hundreds of years.  If the earth was flat, then the math for celestial would be standard flat trig, but it's not.  There would be no need to calculate spherical triangles to determine your position since you are on a alleged flat surface.  This is a simple non-disputable fact.

I feel I speak with some degree of real-world expertise about celestial navigation.  Anybody out there who has used celestial who disagrees with me?

Celestial navigation has been carried out for thousands of years. Mankind knows the celestial sphere rotates overhead and has always used this for navigation .The stars rise and set , and all reach zenith which has always given a good approximation of longitude .

The spherical calculations are pointless as shown , none of what you say points to earth being a globe .

A trip from N pole to the equator and back after travelling a quarter of the equators length is not a short journey so how can you say it only works on short trips?

44

Flat Earth Theory / Re: Around-the-World Sailing Races?

« on: May 18, 2020, 01:27:47 PM »

Yes, indeed you would end up back at the pole on a FE with your proposed trip, BUT the flaw in that is that you have not made an equilateral triangle on your flat earth since the equator part of the trip is a curved line, a radius around the North Pole.  The math would then fail.  On a globe my route from pole to equator and back to the pole is a perfectly fine 90-90-90 equilateral spherical triangle.  Yours is a pie wedge. 

And so far, no one has answered why celestial navigation, based on spherical trigonometry, works if the earth is truly flat.  Plus the matching issue as to why sea captains don’t believe in a flat earth.

They are the same journey with the same result . The math doesn't fail . In FE the angle between outward and return leg at the pole is 90 degrees . The angle at the equator is 90 E and when you reach the required distance you turn 90N . The difference is that you believe you are journeying over a curved surface and so are applying spherical trig.

This also answers your celestial navigation - latitudes are circular lines around the point on earth beneath the N pole.

45

Flat Earth Theory / Re: International Space Station

« on: May 18, 2020, 11:04:45 AM »

Dude, how do the satellites keep up with the earth? They’re moving at those speeds RELATIVE to the earth.
This is literally one of the first concepts you will learn in an introductory physics class.


[/quote]

Nah. One of the concepts of theoretical physics maybe - that's imaginary physics based on unproven assumption , physics based on thought experiment including postulates treated as reality law because some plagiarist patsy says so .

 Science might as well admit the earth is stationary ( no experiment has ever found rotation of earth) and get rid of that shite and use real science based on laws derived from repeatable scientific experiment and observation.
 

46

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 06:56:45 PM »

The solar panels absorb the suns radiation though although how much I don't know. Surely the panels if directed like mirrors wouldn't appear bright all over the earth ?

test for yourself by looking at any functional solar panels here on earth. can you see a reflection on the surface? Yea? Then it's directly reflecting a lot of light. Now look at a piece of limestone. No reflection in that rock? It's scattering light a lot more than a surface where you can clearly see a mirror reflection of some kind.

https://www.quora.com/What-percentage-of-sunlight-is-directly-reflected-by-a-solar-panel .

The solar panels in my garden are dull. They are designed to absorb light .

https://www.edn.com/international-space-station-iss-power-system/

47

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 06:05:42 PM »

The solar panels absorb the suns radiation though although how much I don't know. Surely the panels if directed like mirrors wouldn't appear bright all over the earth ?

48

Flat Earth Theory / Re: Around-the-World Sailing Races? - Part 2

« on: May 17, 2020, 05:59:25 PM »

AS an experienced sailor, and knowing my beloved sailboat, Serenity, very well, I can tell without looking at the knotmeter a one knot difference in speed through the water.  And as others who have replied to my original post, looking at the common FE models would require distances that are at least 2 times farther than on a RE, and likely closer to three times farther to circumnavigate.  Please believe me that even ignoring GPS data, anyone with enough sailing experience to be doing a round-the-world race would be aware of the massively increased distance involved.  And monohull sailboats have a "hull speed" past which they cannot reasonably go faster.  This is based on the well-known physics of water itself and wave generation.  The oversimplified explanation is that as you push a displacement boat hull faster and faster it starts sinking deeper into the water, which greatly increases the drag until you reach a limit = the hull speed.  You just cannot make up that degree of distance increase.

My second question/puzzle for FE's is that I have the honor of being old enough to have been trained as a celestial navigator - using a sextant and an accurate watch to determine position at sea.  I was lucky enough that when I was in graduate school at Washington University in St. Louis I was able to take a whole semester long course in Celestial taught by a retired Navy officer navigator.  Since it was a semester long course we got heavily into the basic math involved.  In practice there are books of tables and now calculators/computers that do the math for you, but if you want as a FE believer to assume those books and computers are somehow "doctored" by NASA, my instructor made us do the math ourselves.  My final project was to write a computer program that calculated the "great circle route" between two points on the globe.  When Lindbergh wanted to calculate the great circle course (the shortest distance) for his historic flight from New York to Paris he had to stretch a string taut on a large globe at the New York City library before the flight and take careful notes as to where he should be at each stage of his flight. 

The math required for Celestial is called "spherical trigonometry", and you need a higher level scientific calculator or computer.  The simplest example I can give is that on a flat plane an equilateral triangle (all three sides are on the same exact length) has three angles of 60 degrees each.  However on a sphere an equilateral triangle is quite different, with three angles of 90 degrees each, not 60.  Imagine traveling on a sphere.  You start at the north pole and travel due south to the equator.  Once you reach the equator you turn due east or west (doesn't matter which), that's a 90 degree turn, and then travel the exact same distance as you did from the pole to the equator.  At that point you turn north again (another 90 degree turn) and travel back to the pole.  Once back at the pole again you will discover the angle between your departing and returning path is also 90 degrees.  Spherical trig has no trouble with calculating all sorts of triangles on the surface of a sphere.  Regular flat trig that we all learned in school won't work.  Likewise if we really are on a flat earth, then the math behind sextants wouldn't work, not even close.  But it works fine.  Not as good, and certainly not as easy as GPS, but gets you within about a mile of where you want to be, and for most purposes that's good enough.  And it's been used since the invention of a reliable chronometer (accurate clock at sea) in the 1760s.  Long before space flight, NASA or any other alleged conspirators.  Or do you think all those tens of thousands of crusty old sea captains were all part of a massive conspiracy to hide the flat earth secret for 260 years?  And not one of them ever fessed up about the "secret" he/she was keeping?

A FE answer please?

Regarding your second question , am assuming use of a compass, on FE head due south from the N pole - reach the equator , turn 90 east travel along the equator , which is a circle ,  following your compass heading of due east for same distance ,turn 90 N travelling same distance again using compass . Wouldn't you end back up at the N pole or am I missing something ?

49

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 05:23:14 PM »

Should imagine intensity of light diminishes over distance . Objects tend to scatter light too , as does the air .

50

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 04:19:06 PM »

Apparently it's because the solar panels cover a big area and reflect sunlight. I always thought the point of solar panels was to absorb sunlight.

I didn’t want this to slip away, but you need to go out and understand things better before you come try to refute them. Solar panels are indeed meant to absorb light but 5ish percent still gets reflected. That’s enough to look a little bright when the reflection is of the sun.

Neither do I . Insolation at earth = 1370W/m^2 so 5% reflection = 70W/m^2 .  Area of solar panels = 2500m^2 so total reflected =18000W . Isnt that a 180kw lamp.

Those figures won't be accurate but it seems a tall order to reflect such a bright image from 250mls away.

51

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 03:03:31 PM »

Straight forward statement . Big G constant is not constant. https://www.sheldrake.org/essays/how-the-universal-gravitational-constant-varies .

     Your equation involves use of a variable constant Big G and little r which I presume is the average value given for radius of earth 6371km . The earth is either pear shaped or oblate depending on which theoretic shape you want so the value 6371km cannot be used as an accurate value for little r.

The  equation therefore cannot give you a value for v which equates to reality according to the globe model.

How do satellites orbit and follow an object which is constantly accelerating at 18.5 miles per second around the sun ?

You quoted Rupert Sheldrake who studies ghosts and telepathy and is most assuredly not a physicist.

You will need to find an actual scientific paper, not a blog from someone who thinks he can see the future.   

You also do not understand scale.  Please refer to the Wiki.

https://en.wikipedia.org/wiki/Spatial_scale

https://physicsworld.com/a/gravitational-constant-mystery-deepens-with-new-precision-measurements/

https://phys.org/news/2015-04-gravitational-constant-vary.html

https://www.scientificamerican.com/article/puzzling-measurement-of-big-g-gravitational-constant-ignites-debate-slide-show/

You need to read a few o these

52

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 01:40:49 PM »

Straight forward statement . Big G constant is not constant. https://www.sheldrake.org/essays/how-the-universal-gravitational-constant-varies .

     Your equation involves use of a variable constant Big G and little r which I presume is the average value given for radius of earth 6371km . The earth is either pear shaped or oblate depending on which theoretic shape you want so the value 6371km cannot be used as an accurate value for little r.

The  equation therefore cannot give you a value for v which equates to reality according to the globe model.

How do satellites orbit and follow an object which is constantly accelerating at 18.5 miles per second around the sun ?

53

Flat Earth Theory / Re: International Space Station

« on: May 17, 2020, 07:30:31 AM »

Gravity will pull it to earth

It will try to.
Throw a stone horizontally and it will land some distance in front of you.
Fire a bullet and it will land much further from you.
If two people did those things at the same time then which would hit the ground first, the stone or the bullet?
It’s counter intuitive but they’d hit the ground at the same time (ignoring air resistance and assuming the bullet doesn’t go far enough that the earth’s curve is a significant factor.)
They hit at the same time because gravity pulls both of them to earth at the same rate, the bullet goes further because it is going faster and so can travel horizontally further in the time.

What if you had a more powerful gun which shot the bullet faster? It would go further still. Now, if you imagine we live on a globe (I know, but humour me) then the ground would slope away from you. Let’s ignore hills and mountains. You should be able to see that if you shoot the bullet fast enough then the bullet would never land, it would fall but as it falls the ground slopes away. Get the speed right and it would go all the way around the earth (assuming it maintains a constant speed so ignoring air resistance). That is how orbit works.

Gravitational constant , oxymoronic name, changes with altitude according to the inverse square law and pulls to the centre of mass or so the theory goes.

Here you have quite succinctly shown you don’t understand English or science.
An oxymoron is two adjacent words which contradict one another. These do not.
The Gravitational constant is, as the name suggests, constant. The gravitational force the earth exerts in a body is not constant.

It 's a plane, balloon satellite, or whatever lighter than air craft they want to wow you with . It's a hologram maybe. It isn't what OP thinks it to be.

And where is your evidence for that?

The gravitational constant ,big G , is not constant . Do some research. Start here if you wish . Article from New Scientist. https://www.newscientist.com/article/dn24180-strength-of-gravity-shifts-and-this-time-its-serious/

There's loads more on tinterweb .

Watched the ISS plane fly over last night 11.40pm. Ridiculously bright for an object 250mls or so away . Apparently it's because the solar panels cover a big area and reflect sunlight. I always thought the point of solar panels was to absorb sunlight.

Are there orbital parameters for iss that allow for the alledged spin of the earth ? Surely it must be constantly accelerating ?

Earth travels around the sun , according to theory , at 66,600mph, devilish number that . Now that is 18.5 miles per second or 30kms .Iss travels at 7kms we are led to believe . How does it keep up ? How does it maintain its orbit? How do the geostationary satellites ,thousands of miles away , cope with earths motions ? There is no wonder it took a scifi writer to dream up such nonsense or nonscience.

There is no magic velocity ,in globe theory , that allows a satellite to orbit a planet , it either escapes or is pulled back to earth .
If there is ,where is the magic formula?

54

Flat Earth Theory / Re: International Space Station

« on: May 16, 2020, 06:50:32 PM »

I think somerled may be correct.  Its a plane.  Its been travelling at over 7 km per second since 1998.  Need to get me some of that fuel. 

Or a balloon.  A really aerodynamic, pointy, balloon.

Yeah you watched it all the way eh . Angular velocity tells you nothing about distance .

55

Flat Earth Theory / Re: International Space Station

« on: May 16, 2020, 06:03:06 PM »

How does ISS stay in orbit ? How do you know how far away it is when you see it pass by ?

Several questions were just asked of FE with respect to the ISS. Your response is then to ask questions of RE. This implies:

a) you do not have answers to the questions, and so you are trying to

b) shift the burden of proof

Fine with me! I accept the implication that you lack these answers. But feel free to correct me

Answers to YOUR questions:

By gravity.

One complete orbit every 90 mins tells you how far away it is.

Gravity will pull it to earth . Gravity does not provide centripetal acceleration. Gravitational constant , oxymoronic name, changes with altitude according to the inverse square law and pulls to the centre of mass or so the theory goes.

You have no  independent way of proving what you say . Angular velocity doesn't give you distance. It 's a plane, balloon satellite, or whatever lighter than air craft they want to wow you with . It's a hologram maybe. It isn't what OP thinks it to be.

Even Kepler couldn't explain where he derived his "laws" from. They are not natural laws.

56

Flat Earth Theory / Re: International Space Station

« on: May 16, 2020, 04:10:45 PM »

How does ISS stay in orbit ? How do you know how far away it is when you see it pass by ?

57

Flat Earth Community / Re: Rocket Propulsion

« on: May 14, 2020, 03:12:51 PM »

No rebuttal there.
 
Any change in momentum requires application of a force - always . Defined by the equation ,  F = ma as stated by the laws of physics .

The Principle of Conservation of Momentum applied to rockets stems from theoretical physics , Neorems theory applied to Newton's laws , which allows calculation of velocities from change of momentum and other variables , but the starting point is always that F must be greater than zero , no force = no change in momentum .

Your first link generalizes this but omits to point out the fact that expansion of hot gas into a vacuum produces no work or force and ignores the fact that thrust is reactive force that requires pressure .

Nasa ignores the laws of science and this neat math trickery which leads to such statements as " rockets work better in a vacuum",
which anyone carrying out an actual experiment in vacuum chamber can see is wrong.

58

Flat Earth Theory / Re: Submarine cable distances

« on: May 13, 2020, 04:23:41 PM »

All globe worshipers require calculations because they think the pole star is at some fantastic distance and its light reaches us in parallel rays . FE does not require calculations

Sextant measures angles only from the position you take the measurement. You interpret whether the 45 degree angle is due to a spherical earth and you are angled towards the parallel rays or you are on a flat plane and the star is not at a ludicrous distance

1. Depends on the length of degrees.

2. It's difficult to see any star at less than about 3 degrees .

One can draw a diagram of anything one wants if you've a pen and paper.

59

Flat Earth Theory / Re: Submarine cable distances

« on: May 13, 2020, 03:17:56 PM »

No - you misunderstand the sextant . It measures angles .

If you measure an angle of 45 degrees to the polestar then that tells you you are at 45 north on FE or RE .

FE requires no calculation for mapping . RE does for mapping onto a sphere.

60

Flat Earth Theory / Re: Submarine cable distances

« on: May 13, 2020, 02:45:47 PM »

https://en.m.wikipedia.org/wiki/Sextant

Professional sextants use a click-stop degree measure and a worm adjustment that reads to a minute, 1/60 of a degree. Most sextants also include a vernier on the worm dial that reads to 0.1 minute. Since 1 minute of error is about a nautical mile, the best possible accuracy of celestial navigation is about 0.1 nautical miles (200 m). At sea, results within several nautical miles, well within visual range, are acceptable. A highly skilled and experienced navigator can determine position to an accuracy of about 0.25-nautical-mile (460 m).[4]

That doesn't sound too inaccurate to me. It sounds more like you guys are just making things up.

I just have to quote this twice, as I have to ask.

Are you really arguing that sextants, which depend on the Earth being a sphere and rotating, are highly accurate and dependable to locate your position on a globe?

After a sight is taken, it is reduced to a position by looking at several mathematical procedures. The simplest sight reduction is to draw the equal-altitude circle of the sighted celestial object on a globe. The intersection of that circle with a dead-reckoning track, or another sighting, gives a more precise location.

The fact that sextants work is alone very good evidence that the earth is indeed a globe.

Good to know you accept how well they work.

Sextants measure the angle between two objects . Used in marine navigation to measure between horizon and sun or other celestial body. Nothing to do with the shape of the earth .