Distances on RE and FE consistent thanks to bendy light.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #360 on: February 12, 2011, 04:01:47 PM »
Speed(not affected by bendy light) =/= distance(determined by bendy light)/time
Distance isn't determined by light. What on Earth are you talking about?
Ok, so you want length to be redefined due to bendy light...
Yes.
Yes. And?

Can we distill the confusion into the following?

RET: Speed = distance/time

FET: ?
The same equation will work. Of course, in both cases we're considering average speeds.
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Nolhekh

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #361 on: February 12, 2011, 04:13:05 PM »
PizzaPlanet, your model's contradictions are exposed.  I'm finished here.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #362 on: February 12, 2011, 09:12:18 PM »
PizzaPlanet, your model's contradictions are exposed.  I'm finished here.
Your insufficient reading skills have been exposed. I'm very glad you're finished here, though. We might have some actual discussion going on rather than "HERP DERP I THINK YOU SAID SOMETHING ELSE THAN WHAT YOU MEANT AND THEREFORE YOUR MODEL FAILS".
« Last Edit: February 13, 2011, 05:11:33 PM by PizzaPlanet »
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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #363 on: February 13, 2011, 01:35:32 PM »
well, if you said something else than you mean, you should go back and correct it
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Around And About

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #364 on: February 13, 2011, 02:00:58 PM »
Well PizzaPlanet, I for one appreciate you answering my questions (even though this is technically the Debate section, oops), thank you! :)
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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #365 on: February 13, 2011, 05:25:18 PM »
well, if you said something else than you mean, you should go back and correct it
I said exactly what I meant. Unfortunately, not everyone here (including you, sadly) can read.
For example, you have just misinterpreted this:
"HERP DERP I THINK YOU SAID SOMETHING ELSE THAN WHAT YOU MEANT AND THEREFORE YOUR MODEL FAILS".
Now, let's make the relevant part more visible, just for you:
"HERP DERP I THINK YOU SAID SOMETHING ELSE THAN WHAT YOU MEANT AND THEREFORE YOUR MODEL FAILS".

Well PizzaPlanet, I for one appreciate you answering my questions (even though this is technically the Debate section, oops), thank you! :)
You're welcome. And don't worry - questions are absolutely fine in the debate section, especially when they come from someone who genuinely wants to discuss rather than just stir things up. :)
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Nolhekh

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #366 on: February 13, 2011, 07:41:15 PM »
I'm afraid it is you who cannot read, if you cannot see the contradiction in your model.

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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #367 on: February 13, 2011, 09:06:11 PM »
Speed(not affected by bendy light) =/= distance(determined by bendy light)/time
Distance isn't determined by light. What on Earth are you talking about?

you still haven't answered to my request.
You forgot to make a request. Also, your attitude is not welcome in this thread. Please take your angry ranting to RM.

I'm not angry or ranting. I'm just saying you haven't fufilled my request, which is to show me how you would approximate the area of Colorado with the description from the FAQ and the coordinates from wikipedia.

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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #368 on: February 13, 2011, 09:24:10 PM »

i got too lazy to read the earlier posts, 17 pages is a lot

but, pizza, i want you to draw 3 parrellel lines on a orange peel the goes all the way around and meets itself again (aka, draw a circle, that are parellel.) one in the middle, and one equidistant from the top and the center, one equidistant from the bottom to the center
now i want you to mesure the length around the circles, do this to your best estimate

you should find the equator is longer than the ones near the pole

then cut it perpendicular to the lines with only one cut, that expands half of the whole circumfrence, in which it starts and ends at the poles

now, i want you to make that rectangular, without additional cuts

if you succeed, you will have 3 lines, that are of equal lenghts, but becuase the outer lines represent a smaller distance, due to a smaller original circumfrence, you must compensate for this

hence, if you were to divide the rectangle into fourths, a fourth of the equator, would not be the same length across as a fourth of the outer lines, hence your 1000 km to 200 km

google is makeing the round earth rectangular, but now, it must compensate for the increased length at the poles

hope you can understand this

do try this experiment!
seriously, try this experiment to clear your confusion with how round earth is presented on maps

and i see the contradiction

because light bends, the distances expressed from the original light source is distorted, and changed to what is measured

but as nohekl has made the point that becuase your speed isnt determined by light, you can measure speed by traveling a set distance, timing yourself, and knowing your rate

hence, you cannot use bendy light to explain the differences in what is observed and what isnt
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #369 on: February 13, 2011, 10:00:24 PM »
ITT:

PP lies
Noobs suck up to PP
PP theory having been shattered ignores everything

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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #370 on: February 13, 2011, 10:02:41 PM »
itt? what does that mean?

noobs dont all suck up to him, im a noob
if im not understanding what suck up means, pls correct me
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markjo

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #371 on: February 13, 2011, 10:05:44 PM »
itt? what does that mean?

ITT = In This Thread
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #372 on: February 13, 2011, 10:09:40 PM »
did i make a good point? in my above post?
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #373 on: February 13, 2011, 10:22:21 PM »
itt? what does that mean?

noobs dont all suck up to him, im a noob
if im not understanding what suck up means, pls correct me

In this thread,
and I mean "Around And About".
He's ignoring glaring holes in an attempt to remain non confrontational, a tactic I used for my first 300 or so posts a while back, but its all in vein.

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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #374 on: February 13, 2011, 10:23:27 PM »
so is it a re victory yet?
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markjo

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #375 on: February 13, 2011, 10:30:36 PM »
so is it a re victory yet?

It's not whether RE wins or loses.  It's how you present your evidence.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.

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Around And About

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #376 on: February 13, 2011, 11:28:00 PM »
itt? what does that mean?

noobs dont all suck up to him, im a noob
if im not understanding what suck up means, pls correct me

In this thread,
and I mean "Around And About".
He's ignoring glaring holes in an attempt to remain non confrontational, a tactic I used for my first 300 or so posts a while back, but its all in vein.

You made it to 300 posts that way? Good god, man.  :o


Also, you should probably just focus on your own behavior and not worry about mine. Here's some free advice, though: If you're getting increasingly frustrated by the endless debating, perhaps you should take a break? Gotta take care of those veins of yours.
« Last Edit: February 14, 2011, 12:01:57 AM by Around And About »
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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #377 on: February 13, 2011, 11:45:44 PM »
I'm afraid it is you who cannot read, if you cannot see the contradiction in your model.
Nope. I've already pointed out there's no contradiction. The distances appear different, but are not different. The definition would need to be adjusted to attribute for this fact, nothing else. You still don't understand the simple thing I've said a bajillion times here: mechanics != optics.

I'm not angry or ranting.
Incorrect and incorrect.

I'm just saying you haven't fufilled my request
Just because I liked your "native English speakers speak good English" argument: it's "fulfilled"; you have also misused a comma in that sentence.

which is to show me how you would approximate the area of Colorado with the description from the FAQ and the coordinates from wikipedia.
The following half-assed approximation assumes that Colorado is rectangular (hence the slightly imprecise result), and uses data from Wikipedia as requested:
« Last Edit: February 13, 2011, 11:51:26 PM by PizzaPlanet »
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Thevoiceofreason

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #378 on: February 14, 2011, 07:27:46 AM »
I'm afraid it is you who cannot read, if you cannot see the contradiction in your model.
Nope. I've already pointed out there's no contradiction. The distances appear different, but are not different. The definition would need to be adjusted to attribute for this fact, nothing else. You still don't understand the simple thing I've said a bajillion times here: mechanics != optics.

I'm not angry or ranting.
Incorrect and incorrect.

I'm just saying you haven't fufilled my request
Just because I liked your "native English speakers speak good English" argument: it's "fulfilled"; you have also misused a comma in that sentence.

which is to show me how you would approximate the area of Colorado with the description from the FAQ and the coordinates from wikipedia.
The following half-assed approximation assumes that Colorado is rectangular (hence the slightly imprecise result), and uses data from Wikipedia as requested:

How was I angry?
in this thread, I presented a logical proof of why light has nothing to do with physical distances, measured, real, or approximate. And since you're not American, I see how you could be confused about Colorado, but alas it was designed to be a perfect angular sector.
And the way you did your math is incorrect . That is not how you find surface area on a circle/disk. The margin of error turns out to be like>5% if done correctly


If you'd allow me to show you,

The upper bound is 41 N and the lower is 37 N, the rate is 12,450 miles /180 degrees or 69.166666 miles per degree. Therefore, it is between 3389.16 miles and 3665.83 miles radially. Since its Long. coordinates are 102.03 and 109.03, the length is 7 degrees, or 0.122173048 radians.

thus the integral for the angular sector is



or in other words, 119234 square miles, which is larger, as expected.

Thusly, either
A, space itself is twisted in such a way that the earth is curved, yet "flat" aka bent space
B, the data from wikipedia is a part of the Conspiracy to trick you into thinking that the earth is not flat.
C, The Earth is not flat.

A has been proven wrong, space has been found to be relatively flat.
B is still possible within the scope of this site.
C applying Occam's razor and common sense is most likely.
Note this does not prove the earth round, simply not flat and of the dimensions of the FAQ.
« Last Edit: February 14, 2011, 07:34:41 AM by Thevoiceofreason »

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vhu9644

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #379 on: February 14, 2011, 06:40:53 PM »
all i got from this

Hello,
The supposed inconsistency between the distances measured on the (supposedly round) Earth and the most commonly used model of FE has been a strikingly overused topic, especially lately. Need I remind anyone of TheJackel's "all I need is a time speed distance circular calculator you are pleading for credibility lol.. 8)"? If so, I just have. If not, let us proceed.
The map considered for this solution is:


The solution itself bases on the fact that, due to bendy light, rays hit the Earth at different angles, and thus the projections of the same length on the surface will differ. They will be longer as we approach the rim. Thus, what we currently consider a metre will appear considerably longer, leading us to an illusion of RE distances. The definition of our units of length is at fault - it is an application of optics (which already assume perfectly straight light rays, hitting the round Earth at approximately 90 degrees). An actual unit of length should take Bendy Light into account, and thus consider an apparently longer distance to be - more or less - the same as something smaller closer to the pole.

The following diagram explains the problem in detail:


Now, this would seem to be a baseless conclusion. And yes, you would be right, if it weren't for the fact that many RE'ers favourite map* - Google Maps - confirms it. Just have a look at these screens, both taken at the same objective zoom:



Oh, so 1000km is approximately the same as 200km in different places. How very peculiar. As you can see, it's the RE model that's inconsistent, not only with reality, but even with itself!
A round Earth is a geographical impossibility. That is all.
was that becuase of bendy light, distances from each other are optically different from higher altitudes, and re map doesnt work
am i wrong? if so, please correct me
but becuase something appears different from higher altitudes, doesnt mean it is actually different, so there, fe distances are not going to match the observed ones
and becuase a sphere is being projected on a flat surface, it will stretch

i got too lazy to read the earlier posts, 17 pages is a lot

but, pizza, i want you to draw 3 parrellel lines on a orange peel the goes all the way around and meets itself again (aka, draw a circle, that are parellel.) one in the middle, and one equidistant from the top and the center, one equidistant from the bottom to the center
now i want you to mesure the length around the circles, do this to your best estimate

you should find the equator is longer than the ones near the pole

then cut it perpendicular to the lines with only one cut, that expands half of the whole circumfrence, in which it starts and ends at the poles

now, i want you to make that rectangular, without additional cuts

if you succeed, you will have 3 lines, that are of equal lenghts, but becuase the outer lines represent a smaller distance, due to a smaller original circumfrence, you must compensate for this

hence, if you were to divide the rectangle into fourths, a fourth of the equator, would not be the same length across as a fourth of the outer lines, hence your 1000 km to 200 km

google is makeing the round earth rectangular, but now, it must compensate for the increased length at the poles

hope you can understand this

do try this experiment!

seriously try this
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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #380 on: February 15, 2011, 02:38:14 AM »
in this thread, I presented a logical proof of why light has nothing to do with physical distances, measured, real, or approximate.
"Logical" is a vast overstatement, as is "proof". However, light does have nothing to do with physical distances. I keep saying that all the time.
Mechanics != optics. Mechanics != optics. Mechanics != optics.

And since you're not American
Ah, again with the nationality wildcard. Are you intentionally trying to make your arguments less credible?

I see how you could be confused about Colorado, but alas it was designed to be a perfect angular sector.
Alas, that was done under RE assumptions. It is completely inapplicable to the FE model.

And the way you did your math is incorrect. That is not how you find surface area on a circle/disk.
Actually, it is very correct.

The margin of error turns out to be like>5% if done correctly
Did you check my margin of error? It's slightly more than 2%.
Since I don't trust your mathematical capabilities (as they keep giving you completely wrong results over and over again), I'll do the calculations for you.
For reference:
The area of Colorado as given on Wikipedia: 269837 km^2
My half-assed approximation: 276012 km^2
[(276012-269837)/269837]*100%=~2.288%
[(276012-269837)/276012]*100%=~2.237%
2.288<5 and 2.237<5


If you'd allow me to show you
Humour me.

The upper bound is 41 N and the lower is 37 N, the rate is 12,450 miles /180 degrees or 69.166666 miles per degree. Therefore, it is between 3389.16 miles and 3665.83 miles radially. Since its Long. coordinates are 102.03 and 109.03, the length is 7 degrees, or 0.122173048 radians.
And this is where you fail. You look like a map, which is an optical representation (and, as I stated before, gets more distorted as the scale diminishes), and treat it as if it could be used to make approximations. Sigh.
Mechanics != optics.

thus the integral for the angular sector is
something we don't care about.

Oh, come on. Surely you didn't think hotlinking a Wolfram output would work, did you?

or in other words, 119234 square miles, which is larger, as expected.
Or, in other words, you did some random inapplicable stuff and got a completely wrong result (14.44% margin of error!). How surprising. What's next? Proving that the Earth is round using the transfer function for a non-inverting operational amplifier?

Thusly, either
A, space itself is twisted in such a way that the earth is curved, yet "flat" aka bent space
B, the data from wikipedia is a part of the Conspiracy to trick you into thinking that the earth is not flat.
C, The Earth is not flat.
D, a^2+b^2=c^2 ergo camels are blue and the Earth is a spoon frequently used by Zeus. QED

A has been proven wrong, space has been found to be relatively flat.
B is still possible within the scope of this site.
C applying Occam's razor and common sense is most likely.
D is quite likely considering your use of inapplicable mathematics and your inability to understand one thing; and yes, I will say it again now; and yes, I will increase the font size again. Why? Because I want you to at least try to think about it.
Mechanics != optics.

Note this does not prove the earth round
Definitely, seeing how it doesn't prove anything whatsoever.

simply not flat and of the dimensions of the FAQ.
Definitely not, seeing how it doesn't prove anything whatsoever.
« Last Edit: February 15, 2011, 02:47:57 AM by PizzaPlanet »
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29silhouette

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #381 on: February 15, 2011, 08:18:40 PM »
"Logical" is a vast overstatement, as is "proof". However, light does have nothing to do with physical distances. I keep saying that all the time.
Mechanics != optics. Mechanics != optics. Mechanics != optics.
So a person could then have a laptop set up in a car to read speed and steering info to basically trace the vehicles path.  That vehicle could then travel the coastal highways to create an approximate outline of the continents (if a lot of people did this it would be much easier than one person obviously).  The data could be combined to create a flat map of a flat earth.

Or, use all the aerial photos available to start making a flat map of the earth.  Small sections at a time of the coastlines would also get the shapes of the continents.

Neither of these would be affected by bendy light or GPS.  They also wouldn't involve making a 1:1 scale map, but even so, a 1:1 scale map scaled down to 1:250,000, while maintaining it's original aspect ratio, would still present an accurate flat earth map. 

Distances across the oceans would still have to be measured, but at least you would have accurate maps of the continents.

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Around And About

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #382 on: February 16, 2011, 01:20:49 AM »
Is there any way to mathematically describe the surface of the FET model? For RET, we can very roughly approximate the surface area by using (4*pi*r²) where r is 6371km, right? So what is the FET equivalent?
« Last Edit: February 16, 2011, 01:23:29 AM by Around And About »
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silver

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #383 on: February 16, 2011, 04:20:44 AM »
Is there any way to mathematically describe the surface of the FET model? For RET, we can very roughly approximate the surface area by using (4*pi*rē) where r is 6371km, right? So what is the FET equivalent?

Well it can't be pi * r^2, because that's for describing a flat, circular object, which according to PizzaPlanet it's apparently not anyways.

Let's state a fact and see what comes out of it.

Two circles with a different radius (IE distance from the North Pole) cannot have the same circumference.

In reality they do. The only geomitrically viable explanation for this is a sphere.

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markjo

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #384 on: February 16, 2011, 06:04:03 AM »
Two circles with a different radius (IE distance from the North Pole) cannot have the same circumference.

In reality they do. The only geomitrically viable explanation for this is a sphere.

Or a cylinder.
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.


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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #386 on: February 16, 2011, 11:39:59 AM »
Is there any way to mathematically describe the surface of the FET model? For RET, we can very roughly approximate the surface area by using (4*pi*r?) where r is 6371km, right? So what is the FET equivalent?
A=πr23

A - area of the Earth
β - Bishop Constant, a very rough approximation of which being 3π/7=~1.35; has no unit.
r - meridional radius of the Earth, about 20004 km

Thus:
A=πr23
A=3.14*200042/1.353[km2]
A=510695503.83 km2

Two circles with a different radius (IE distance from the North Pole) cannot have the same circumference.

In reality they do. The only geometrically viable explanation for this is a sphere.
In Euclidean geometries, that would almost be the case. Also cylinder, egg, and many, many other arbitrary solids of revolution. However, I don't see how this is relevant to the topic at hand.
« Last Edit: February 16, 2011, 12:27:45 PM by PizzaPlanet »
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silver

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #387 on: February 16, 2011, 12:10:03 PM »
The physical circumference of the south and north pole is the same. That is why it's relevant.

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PizzaPlanet

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #388 on: February 16, 2011, 12:40:05 PM »
The physical circumference of the south and north pole is the same. That is why it's relevant.
The geometry is not Euclidean. That's why it isn't.
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markjo

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Re: Distances on RE and FE consistent thanks to bendy light.
« Reply #389 on: February 16, 2011, 12:48:52 PM »
A=πr23

How did you derive this formula?
Science is what happens when preconception meets verification.
Quote from: Robosteve
Besides, perhaps FET is a conspiracy too.
Quote from: bullhorn
It is just the way it is, you understanding it doesn't concern me.