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Flat Earth Theory / Re: Flat Earth Map Should Be Easy

« on: October 04, 2017, 08:05:11 PM »

Setting cable lengths and our abilities to search the internet aside....

Does anyone have anything else to add to the flight time map discussion?

• We have established that care needs to be taken when comparing short flights to long flights because a larger percentage of the flight is take up in ascending and descending from cruising altitude and speed.
• We have established that the purpose of the project is not to calculate distances, but to layout the general size and relative location of continents
• We have established that the margin of error is not great enough to change a flat earth into a globe or to change a globe into a flat earth.

You should certainly avoid using short flights - but for longer flights, the additional takeoff/landing times don't affect the result by much.  You're not going to need short-distance flights (under a couple of hours of flight time) anyway - you're constructing a map of the world, not of a single country.

I'm quite sure that even with HUGE errors (like 20%) you'd be able to show that there is no way to flatten out the results.

But I'm sorry to say that the Flat Earthers simply won't accept your results - they're just going to carp on and on about how we don't know DISTANCES - and their poor little brains aren't sharp enough to understand that what you're doing is sufficient proof.

The only things they trust for distances are things like car odometers (and I bet they'd discount those if we found a proof involving them!).

Well not so much- If you are accepting errors as high as they have stated- you could flatten anything out.

Its why you can draw North America on a cartesian projection and basically get away with it. 

If you allow 20-40% error- well thats well above a cartesian error except at the very very highest latitudes.  The math was tongue in cheek to show the absurdity of trying to make a valid measurement when you make assumptions on 95% of the dependent variables.  Especially when you start with fundamentally faulty beliefs.  I am not even getting into the shape of the world.  I am talking about basic assumptions that all planes are the same, they fly the exact same speed, along the exact same path, at the same altitude, in perfectly straight lines, and that error, if it did exist, could just get averaged away (instead of magnified, like it would here- by giving a false sense of accuracy).  When even 30 seconds on google would prove it otherwise.  But then again, you wouldn't expect much math or science here...  Which is why this is so fun.  Fact free zone!

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Flat Earth Theory / Re: Flat Earth Map Should Be Easy

« on: October 04, 2017, 07:55:23 PM »

-math snipped for brevity of reply-

I get where you're coming from here, but once again: Distance is not being calculated here. We're not looking for an incredibly accurate map.
These are the parameters/assumptions:

1) Flight times between two locations are the same within a reasonable margin of error so long as it's the same model plane.
3) Planes will always attempt to fly the shortest distance between two points.
4) Flight times in one direction can be combined with flight times in the other direction for an average flight time within a reasonable margin of error.
5) Flight times can thus be used as a reasonable approximation for distance, and should reveal the shape of the Earth if plotted out.
6) Flight times should not be skewed enough to make a flat Earth appear round, nor a round Earth appear flat.

Now, we'll see how well 1 and 4 turn out for margins of error, but I see nothing wrong with the rest of these otherwise. Once again, no distance in miles or kilometers is being calculated, nor is there an attempt to calculate said distances being made. This is about plotting air flight times onto a map, and seeing what shape the flight times dictate the map to be.

Point 1- We don't have the model of plane.
Point 3- This is flat out wrong.  Like the assumption that planes ALWAYS fly their cruising speeds.  This is wrong because planes can't fly the shortest distance between 2 points.  There are federally mandated rules on flight paths.  This is so planes don't hit each other.

Also, planes fly curved paths.  Even if you pretend the earth is flat, its still a curved path.

Point 4- I know from experience 4 is untrue.  I fly back and forth between CLT and DFW very often.  I know that the flight from CLT to DFW is always 20 minutes shorter than the flight there.  I have taken this trip many many times.

Point 5- If you have your stated acceptable error of 40% you can't make a map of shit.  Period.

Point 6- people on this site always lose sight of scale.  They think that a spherical earth must mean a small earth and a flat earth means the same.  An 8 hour flight is ~1500 -2400 miles.  Diameter of a spherical earth is listed as 8,000 miles- or 24,800 miles in circumference.  This means that the distance if the earth was flat, vs. if the earth was round would be only about  7%, WAY below the error you "allow".

This is besides the fact that any distance this way is still the same. This is perhaps the very MOST IMPORTANT fact. Imagine an ant on a beach ball.  If the Ant walks 8 inches on the beach ball or 8 inches on the sidewalk, he still walks.... wait for it... 8 inches!

So this is pointless anyway.

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Flat Earth Theory / Re: Flat Earth Map Should Be Easy

« on: October 03, 2017, 08:44:59 PM »

there should be closed parentheses after the h term before /I in the adjustment term.  It gets hard to read with all the markup it took to document that semi-correctly and it won't let me edit.

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Flat Earth Theory / Re: Flat Earth Map Should Be Easy

« on: October 03, 2017, 08:35:51 PM »

This won't work.

Lets look at this rationally.  Anyone who has ridden a plane knows that it takes off and then goes to a certain altitude, and during that time accelerates to speed v^c. This takes time t.  During time t, v is increasing from v⁰ to v^c at some rate, which we don't know.  But lets say for the sake of argument- that it's always 30 min to get to altitude and its always 30 min to land, and during that hour speed is always less than v^c.  Lets say this distance traveled during this approximately and hour is constant (which its not, but for ease of calculation) and call it X.  Otherwise we'll have to deal with integrals, and I don't want to do that here.

So in a 10 hour flight, 9 hours are at cruising speed v^c, but in a 4 hour flight only 3 hours are at v^c.  This is 90% and 75% of the flight respectively.  Which means when you compare 2 non-similar lengths, the ratio will be severely skewed.

Next, lets take short flights vs. long flights.  Flights of different lengths usually have different planes, which have different cruising speeds.

So if you want to make a map using only this information and not the speed themselves, then you have to have an adjustment factor, I, to adjust the time by the ratio of speeds between Plane 1 and Plane 2 (v¹/v²).  It's easiest to pick a single base plane and set that as I=1 and compare all the rest to that. So now your times will be comprable.

So, a barely reasonable formula might be ((t-1)*v^c)*I+X. 

Now you also have to account for prevailing wind speed, which will change the speed of the plane- +w for tailwinds and -w for headwinds.  You have to adjust the speed w to account for the fact that it might not be perfectly parallel to the plane, which we can find by taking the angle between the direction of travel and the direction of wind-Theta.

This gives ((t-1)*(v^c)+w cos (theta))*I+X

 Then, you also have to account for general human elements, that might change a flight- usually the planes will go a little faster or slower to make sure that they arrive in a particular time window so that there is a gate available. Your presumption that planes always travel the perfect speed is wrong.  My car has a listed cruising speed of 60mph, but I'll push it to 80mph if I'm late, and slow down to 40mph if the weather is bad.  Planes also have different cruising speeds at different altitudes.  This is because the oxygen content in the air changes and the engine has to maintain sufficient flow through it, etc. etc.   So all of these things together add an error of lets say- extrememly conservatively 10%.

So to find the distance between two seperate line segments since we don't believe in maps or modern technology such as GPS-you want the ratio between two distances which would be sufficient to draw an unscaled map- you can calculate:
((t₁-1*h)*(v₁^c)+w₁ cos (theta₁))*I₁+X/ ((t₂-1h)*(v₂^c)+w₂ cos (theta₂))*I₂+X.

It might seem that X would cancel, but it won't. We also need to adjust that by I because I is the term that adjusts based on v^c.  We can adjust X by dividing by I (since we will cover distance X faster if v^c is faster.  But we also need to adjust X by an additional factor-h that takes into account 2 things- the altitude we are at before we hit v^c, and the time it takes to get there-
 
((t₁-1h)*(v₁^c)+w₁ cos (theta₁))*I₁+(X*(h₂/h₁/I) / ((t₂-1h)*(v₂^c)+w₂ cos (theta₂))*I₂+(X*(h₁/h₂/I)   +/- 10%.

This would give you a minimimally accurate ratio that you can calculate using the information you have at hand.  You can  not do any less, because you are talking about making a map afterall.  Not just guesstimating how far it is.  You could do this all in Excel.  Only from flight aware you also need to grab the type of plane, and maybe the windspeed from there or somewhere else.  The +/- 10% is also critical. 

You also have to remove the additional distance that the planes fly because they think they are travelling along a curved earth.  I've got to get back to work, but you can find the math here- https://math.stackexchange.com/questions/830413/calculating-the-arc-length-of-a-circle-segment

Basically if you fly from Paris to NYC you have to figure out what that distance would be on a round earth- and divide the arc length by the line segment length to come up with a ratio.  This would be a correction factor.  This correction factor, A, grows in significance the longer the distance- to a maximum of pi/2*d where d is the diameter of a supposedly round earth.  Since you are using flights that go across most the world, or at least across the country, this becomes pretty important.  You can find it by multiplying the change in Latitude squared + the change in latitude squared to the 1/2 power (pythagoras' theorom). (ΔLat² + ΔLong²)^(1/2)- thats your round Earth distance in Lat/Long- convert that to miles (since our speeds are in miles)- and then divide that by d.  Thats your Round Earth distance traveled.  To Calculate the Flat Earth Equivalent is tougher.  I'd have to think about that. but it would be:
FEE/RDE * (  ((t₁-1h)*(v₁^c)+w₁ cos (theta₁))*I₁+(X*(h₂/h₁/I) / ((t₂-1h)*(v₂^c)+w₂ cos (theta₂))*I₂+(X*(h₁/h₂/I)  )   +/- 10%.

So, most of the eq. is:

  Flat Earth Eq. / ((ΔLat² + ΔLong²)^(1/2)) * (  ((t₁-1h)*(v₁^c)+w₁ cos (theta₁))*I₁+(X*(h₂/h₁/I) / ((t₂-1h)*(v₂^c)+w₂ cos (theta₂))*I₂+(X*(h₁/h₂/I)  )   +/- 10%.

Just leaving you one piece to find yourself.  You should be able to do it from there.