trigonometry

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trigonometry
« on: December 05, 2017, 02:30:29 PM »
Somewhere on the earth right now, the sun is directly overhead. Some far distance away from there, the sun is 5 degrees above the horizon. If you go halfway between those two points on a flat earth, the sun should be roughly 10 degrees above the horizon. Instead, you'll find it to be much higher in the sky than 10 degrees. In my specific case: this time of year, the sun is about 5 degrees above my horizon when it is directly over Madagascar. Somewhere around halfway is the Egypt Sudan border, where the sun is 40 degrees above the horizon at the same time. Furthermore, my measurement would indicate the sun is 500 miles up, but an observer in the halfway point would calculate the sun at 2400 miles up. Check out suncalc.org or mooncalc.org and find that they accurately state the sun and moon positions for all locations and times.

Bottom line:
Learn basic trigonometry, and realize that the earth isn't flat.

Re: trigonometry
« Reply #1 on: December 05, 2017, 05:10:53 PM »
Bendy light, refraction, perspective, have you been to those places to verify distances lalalalalalala I've got my fingers in my ears...

Sorry, thought I was a FE'er there for a second.

Re: trigonometry
« Reply #2 on: December 06, 2017, 01:47:18 PM »
Somewhere on the earth right now, the sun is directly overhead. Some far distance away from there, the sun is 5 degrees above the horizon. If you go halfway between those two points on a flat earth, the sun should be roughly 10 degrees above the horizon. Instead, you'll find it to be much higher in the sky than 10 degrees. In my specific case: this time of year, the sun is about 5 degrees above my horizon when it is directly over Madagascar. Somewhere around halfway is the Egypt Sudan border, where the sun is 40 degrees above the horizon at the same time. Furthermore, my measurement would indicate the sun is 500 miles up, but an observer in the halfway point would calculate the sun at 2400 miles up. Check out suncalc.org or mooncalc.org and find that they accurately state the sun and moon positions for all locations and times.

Bottom line:
Learn basic trigonometry, and realize that the earth isn't flat.
I think you should learn the difference between a point directly above you and a point somewhere else that is 5 degrees above the surface you are also standing on.

1) Where do you come up with 10 degrees at a halfway point?

2) What effect would the Sun taking a circular pattern or elliptical pattern of travel over the flat plane have on your calculation?

Re: trigonometry
« Reply #3 on: December 06, 2017, 02:42:44 PM »
Somewhere on the earth right now, the sun is directly overhead. Some far distance away from there, the sun is 5 degrees above the horizon. If you go halfway between those two points on a flat earth, the sun should be roughly 10 degrees above the horizon. Instead, you'll find it to be much higher in the sky than 10 degrees. In my specific case: this time of year, the sun is about 5 degrees above my horizon when it is directly over Madagascar. Somewhere around halfway is the Egypt Sudan border, where the sun is 40 degrees above the horizon at the same time. Furthermore, my measurement would indicate the sun is 500 miles up, but an observer in the halfway point would calculate the sun at 2400 miles up. Check out suncalc.org or mooncalc.org and find that they accurately state the sun and moon positions for all locations and times.

Bottom line:
Learn basic trigonometry, and realize that the earth isn't flat.
I think you should learn the difference between a point directly above you and a point somewhere else that is 5 degrees above the surface you are also standing on.

1) Where do you come up with 10 degrees at a halfway point?

2) What effect would the Sun taking a circular pattern or elliptical pattern of travel over the flat plane have on your calculation?

I'm talking about the position of one object from two locations at the same time, not the position of two objects from one location.

tan(Θ)=opposite/adjacent, where the right triangle is defined as:
opposite: the altitude of the object
adjacent: the distance from the observation point to the point directly below the object.

so the equations for locations 1 (5 deg observation point) and 2 (halfway point) are:
tan(Θ_1) = opposite_1 / adjacent_1 and tan(Θ_2) = opposite_2 / adjacent_2

but opposite_1 and opposite_2 are describing the altitude of the object, so they are the same. this is how to relate the equations:
adjacent_1*tan(Θ_1) = adjacent_2*tan(Θ_2)

if adjacent_2 is half the distance to the point below the object, it becomes 1/2*adjacent_1
so:
adjacent_1*tan(Θ_1) = 1/2*adjacent_1*tan(Θ_2)

solving for Θ_2 simplifies to:
Θ_2 = atan(2*tan(Θ_1))

since Θ_1 is 5 degrees:
Θ_2 is 9.9 degrees, but you could've also just used small angle approximations for this.

The motion of the sun has no effect on this calculation, because it is a snapshot in time. You can also see that the altitude of the object is also not a factor.

Re: trigonometry
« Reply #4 on: December 06, 2017, 02:50:24 PM »
Somewhere on the earth right now, the sun is directly overhead. Some far distance away from there, the sun is 5 degrees above the horizon. If you go halfway between those two points on a flat earth, the sun should be roughly 10 degrees above the horizon. Instead, you'll find it to be much higher in the sky than 10 degrees. In my specific case: this time of year, the sun is about 5 degrees above my horizon when it is directly over Madagascar. Somewhere around halfway is the Egypt Sudan border, where the sun is 40 degrees above the horizon at the same time. Furthermore, my measurement would indicate the sun is 500 miles up, but an observer in the halfway point would calculate the sun at 2400 miles up. Check out suncalc.org or mooncalc.org and find that they accurately state the sun and moon positions for all locations and times.

Bottom line:
Learn basic trigonometry, and realize that the earth isn't flat.
I think you should learn the difference between a point directly above you and a point somewhere else that is 5 degrees above the surface you are also standing on.

1) Where do you come up with 10 degrees at a halfway point?

2) What effect would the Sun taking a circular pattern or elliptical pattern of travel over the flat plane have on your calculation?

I'm talking about the position of one object from two locations at the same time, not the position of two objects from one location.

tan(Θ)=opposite/adjacent, where the right triangle is defined as:
opposite: the altitude of the object
adjacent: the distance from the observation point to the point directly below the object.

so the equations for locations 1 (5 deg observation point) and 2 (halfway point) are:
tan(Θ_1) = opposite_1 / adjacent_1 and tan(Θ_2) = opposite_2 / adjacent_2

but opposite_1 and opposite_2 are describing the altitude of the object, so they are the same. this is how to relate the equations:
adjacent_1*tan(Θ_1) = adjacent_2*tan(Θ_2)

if adjacent_2 is half the distance to the point below the object, it becomes 1/2*adjacent_1
so:
adjacent_1*tan(Θ_1) = 1/2*adjacent_1*tan(Θ_2)

solving for Θ_2 simplifies to:
Θ_2 = atan(2*tan(Θ_1))

since Θ_1 is 5 degrees:
Θ_2 is 9.9 degrees, but you could've also just used small angle approximations for this.

The motion of the sun has no effect on this calculation, because it is a snapshot in time. You can also see that the altitude of the object is also not a factor.
But why do I need trigonometry to tell me the Sun would be higher in the sky than 10 degrees at a halfway point?

And how do you know altitude of the object above or movement shape/path of the object above has no affect?

I think your whole math issue is likened to the joke how three guys go into a hotel and get back the wrong change from a hundred dollar bill...
« Last Edit: December 06, 2017, 03:07:08 PM by totallackey »

Re: trigonometry
« Reply #5 on: December 06, 2017, 03:26:43 PM »
Somewhere on the earth right now, the sun is directly overhead. Some far distance away from there, the sun is 5 degrees above the horizon. If you go halfway between those two points on a flat earth, the sun should be roughly 10 degrees above the horizon. Instead, you'll find it to be much higher in the sky than 10 degrees. In my specific case: this time of year, the sun is about 5 degrees above my horizon when it is directly over Madagascar. Somewhere around halfway is the Egypt Sudan border, where the sun is 40 degrees above the horizon at the same time. Furthermore, my measurement would indicate the sun is 500 miles up, but an observer in the halfway point would calculate the sun at 2400 miles up. Check out suncalc.org or mooncalc.org and find that they accurately state the sun and moon positions for all locations and times.

Bottom line:
Learn basic trigonometry, and realize that the earth isn't flat.
I think you should learn the difference between a point directly above you and a point somewhere else that is 5 degrees above the surface you are also standing on.

1) Where do you come up with 10 degrees at a halfway point?

2) What effect would the Sun taking a circular pattern or elliptical pattern of travel over the flat plane have on your calculation?

I'm talking about the position of one object from two locations at the same time, not the position of two objects from one location.

tan(Θ)=opposite/adjacent, where the right triangle is defined as:
opposite: the altitude of the object
adjacent: the distance from the observation point to the point directly below the object.

so the equations for locations 1 (5 deg observation point) and 2 (halfway point) are:
tan(Θ_1) = opposite_1 / adjacent_1 and tan(Θ_2) = opposite_2 / adjacent_2

but opposite_1 and opposite_2 are describing the altitude of the object, so they are the same. this is how to relate the equations:
adjacent_1*tan(Θ_1) = adjacent_2*tan(Θ_2)

if adjacent_2 is half the distance to the point below the object, it becomes 1/2*adjacent_1
so:
adjacent_1*tan(Θ_1) = 1/2*adjacent_1*tan(Θ_2)

solving for Θ_2 simplifies to:
Θ_2 = atan(2*tan(Θ_1))

since Θ_1 is 5 degrees:
Θ_2 is 9.9 degrees, but you could've also just used small angle approximations for this.

The motion of the sun has no effect on this calculation, because it is a snapshot in time. You can also see that the altitude of the object is also not a factor.
But why do I need trigonometry to tell me the Sun would be higher in the sky than 10 degrees at a halfway point?

And how do you know altitude of the object above or movement shape/path of the object above has no affect?

I think your whole math issue is likened to the joke how three guys go into a hotel and get back the wrong change from a hundred dollar bill...

No, trig tells you it would be 10 degrees up at the halfway point IF the earth were flat. The angle there is more like 40 degrees, so the earth isn't flat.

Its the relationship between the horizontal distance and the angle, and is not dependent on the height of the object.

Snapshot in time means motion is irrelevant.

Go get a cup of coffee, take your time, and re-read the posts. It's clearly stated.

Re: trigonometry
« Reply #6 on: December 06, 2017, 09:12:31 PM »
My Calculations came up with 1603 miles above my town using the suncalc.org

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JackBlack

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Re: trigonometry
« Reply #7 on: December 06, 2017, 11:34:58 PM »
But why do I need trigonometry to tell me the Sun would be higher in the sky than 10 degrees at a halfway point?
You don't.
The trig tells you it would be 10 degrees (slightly less). Reality says otherwise.
You can use this trig (along with observations from reality) to show Earth isn't flat.

And how do you know altitude of the object above or movement shape/path of the object above has no affect?
Can you see height or time in the final equation at all? No. So it is quite clear that neither affect it.

I think your whole math issue is likened to the joke how three guys go into a hotel and get back the wrong change from a hundred dollar bill...
Do you mean the one asking about the missing dollar from a $30 note?

Re: trigonometry
« Reply #8 on: December 07, 2017, 03:01:10 AM »
My Calculations came up with 1603 miles above my town using the suncalc.org
And what do your calculations come up with above your town at a different time, and above a different place at the exact same time?

Please share the results, we could all use the laugh, I'm sure!

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Re: trigonometry
« Reply #9 on: December 07, 2017, 12:36:28 PM »
Somewhere on the earth right now, the sun is directly overhead. Some far distance away from there, the sun is 5 degrees above the horizon. If you go halfway between those two points on a flat earth, the sun should be roughly 10 degrees above the horizon. Instead, you'll find it to be much higher in the sky than 10 degrees. In my specific case: this time of year, the sun is about 5 degrees above my horizon when it is directly over Madagascar. Somewhere around halfway is the Egypt Sudan border, where the sun is 40 degrees above the horizon at the same time. Furthermore, my measurement would indicate the sun is 500 miles up, but an observer in the halfway point would calculate the sun at 2400 miles up. Check out suncalc.org or mooncalc.org and find that they accurately state the sun and moon positions for all locations and times.

Bottom line:
Learn basic trigonometry, and realize that the earth isn't flat.
If you'd care to pick up a book, this experiment has been done twice - once by Taoists and again by Rowbotham era researchers. Both came to the rightful conclusion that the Earth is flat. I find nothing so ridiculous as to state that the beams of light from a radial source would be parallel simply to have the luxury of assuming the Earth is round! Talk about circular logic.
Quantum Ab Hoc

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"The above proposition is occasionally useful." - Bertrand Russell

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JackBlack

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Re: trigonometry
« Reply #10 on: December 07, 2017, 01:00:12 PM »
If you'd care to pick up a book, this experiment has been done twice - once by Taoists and again by Rowbotham era researchers. Both came to the rightful conclusion that the Earth is flat.
No, this specific experiment was not done by Rowbotham and co. He did a much simpler one, and came to a completely different height of the sun than others.

I find nothing so ridiculous as to state that the beams of light from a radial source would be parallel simply to have the luxury of assuming the Earth is round! Talk about circular logic.
Not parallel, mostly parallel.
They come from 150 000 000 km away.
This fact (not the specific distance but them being mostly parallel due to the massive distance of the sun) is firmly established by observations on the equinox, where at roughly sunrise (but the same time for everyone along the line) the sun is due east for everyone along a line of longitude.
This shows the sun to be very far away.
« Last Edit: December 07, 2017, 01:37:31 PM by JackBlack »

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Re: trigonometry
« Reply #11 on: December 07, 2017, 01:13:56 PM »
It takes a special kind of person to believe this kinda malarkey. With any shadow in all of existence, except that one supposedly created by the sun, you would simply draw a line between the end of the shadow and the top of the object to locate the light source. But oh no! not with the sun. To sustain the unsustainable belief in a globe they blow our universe up to ridiculous scales and claim the beams from a radial source are 'almost parallel.'

Truth be told, the further away you are from a radial source, the distance between any two of its rays grows.


Quantum Ab Hoc

1 + 1 = 2
"The above proposition is occasionally useful." - Bertrand Russell

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JackBlack

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Re: trigonometry
« Reply #12 on: December 07, 2017, 01:37:05 PM »
It takes a special kind of person to believe this kinda malarkey.
Yes, it does take a special kind of person to believe the malarkey you are spouting, to believe you can have a close sun that somehow appears due east along a line of longitude all at once.

(yes, I noticed an error in my prior post but will fix it).

It takes a special kind of person to believe the sun can magically set while remaining above you (i.e. above a flat Earth).

With any shadow in all of existence, except that one supposedly created by the sun, you would simply draw a line between the end of the shadow and the top of the object to locate the light source.
No, you wouldn't.
That is because that only gives 2 dimensions, not 3. The sun could be anywhere along that line.
There is also a certain degree of error in this. So to do it properly, instead of having a line you would have a cone, growing larger as it gets further away. The light source would be somewhere in that cone, a region of 3D space.

In order to determine the location in 3D space, you need 2 such lines, which again due to error means 2 cones and instead of a point you get a small region.
But that requires knowing the relationship between those 2 lines/cones, i.e. you need to know where a point on each line/cone is and you need to know what angle they are in relation to each other.

Otherwise you can only make assumptions and see if they hold.

With close objects you can typically get it fairly accurate.
With very distant objects, like the sun, you can't.

But oh no! not with the sun. To sustain the unsustainable belief in a globe they blow our universe up to ridiculous scales and claim the beams from a radial source are 'almost parallel.'
Again, you dismiss this as ridiculous without any basis, even though it is what all the evidence shows.
Again, along a line of longitude, at a particular time on the equinox, the sun is due east. This shows these beams of light are roughly parallel.

Truth be told, the further away you are from a radial source, the distance between any two of its rays grows.
Truth be told, the further away you are from a  radial source, the slower the proportional rate of divergence.

For example, if you are 1 m away from a radial source, and looking at 2 beams which are 1 m apart, that subtends an angle of 60 degrees.
If you follow them for an additional 1 m, they end up 2 m apart.

If instead it is 1 km away, and you look at 2 beams which are 1 m apart, they now subtend an angle of roughly 1 milliradian. If you follow it for another m, it will go to roughly 1.001 m

Similar relations to those above can be used.
You start at a distance d1, with 2 beams separated by 2*s1. This forms a right angle triangle with the relation tan(a)=s1/d1.
If you then go to some further distance, d2, the separation grows to 2*s2 and the following relation holds: tan(a)=s2/d2

Thus s1/d1=s2/d2.
Thus s2/s1=d2/d1.

If you let d2=d1+dd, you get:
s2/s1=(d1+dd)/d1=d1/d1+dd/d1=1+dd/d1
Thus the proportional rate of divergence is dd/d1.

So the further away you go, the less they diverge.

So if you are at 150 000 000 km away from a light source, then travelling the entire width of Earth (roughly 13 000 km, which I will over estimate as 15 000 km) would only give a divergence of 15 000/150 000 000 = 1/10 000=0.01%. I would say that is nearly parallel.

Now do you have a rational objection, or just emotional appeals of dismissing it as ridiculous?

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Re: trigonometry
« Reply #13 on: December 07, 2017, 01:38:58 PM »
My rational objection is that the Sun's light acts differently than every single other light source in human experience. If you'd care to address the actual point, instead of going off on unrelated nonsense, that would be fantastic.
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Re: trigonometry
« Reply #14 on: December 07, 2017, 01:42:12 PM »
Locating the Source of the Sun:


Locating any other light source in existence:



And again:
« Last Edit: December 07, 2017, 01:43:54 PM by John Davis »
Quantum Ab Hoc

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JackBlack

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Re: trigonometry
« Reply #15 on: December 07, 2017, 01:52:36 PM »
My rational objection is that the Sun's light acts differently than every single other light source in human experience. If you'd care to address the actual point, instead of going off on unrelated nonsense, that would be fantastic.
I did address the actual point, but you have ignored mine.

The sun's light does not act differently than every other light source.

How about you address the fact that the sun needs to be very far away?

Locating any other light source in existence:

Now go put a tiny ball in that light source and see how it acts?

Or better still, how about we "locate the sun" using your simple math (or at least the start):
At sunset, the sun is observed to drop below the horizon, that is go to a negative angle of elevation.
That means it must be below you. This happens even when right near Earth's surface.
This shows the sun must go below the surface of Earth.

It will deal with your lying image in the other thread.

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realNarcberry

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Re: trigonometry
« Reply #16 on: December 07, 2017, 01:58:05 PM »
At sunset, the sun is observed to drop below the horizon, that is go to a negative angle of elevation.

This isn't the case

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Re: trigonometry
« Reply #17 on: December 07, 2017, 02:02:24 PM »
I have never seen a sun 'drop below the horizon' nor have I been able to measure the shadow created by the sun after sunset. What kind of nonsense are you on about? I'm not sure if you are aware, but the sun does not create shadows during the night. You are thinking of the moon, friend. I suggest you stay away from it, for your own safety.

The sun does not need to be far away, except if one wishes to support ludicrous ideas like the Earth being a Globe. You have failed to show it does need to be this far away.

We already have located the sun using "my" simple math. Simply review Earth Not A Globe, the FAQ for this website, HUAI NAN TZU, or any of the thousands of other places this experiment has taken place and has been verified.
Quantum Ab Hoc

1 + 1 = 2
"The above proposition is occasionally useful." - Bertrand Russell

Re: trigonometry
« Reply #18 on: December 07, 2017, 02:12:56 PM »
I have never seen a sun 'drop below the horizon'
Then you've clearly never watched a sunset where you can see the horizon while the sun sets.

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realNarcberry

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Re: trigonometry
« Reply #19 on: December 07, 2017, 02:17:22 PM »
I have never seen a sun 'drop below the horizon'
Then you've clearly never watched a sunset where you can see the horizon while the sun sets.

Yeah, if there are mountains in the way. Sheesh

Re: trigonometry
« Reply #20 on: December 07, 2017, 02:18:00 PM »
It takes a special kind of person to believe this kinda malarkey. With any shadow in all of existence, except that one supposedly created by the sun, you would simply draw a line between the end of the shadow and the top of the object to locate the light source. But oh no! not with the sun. To sustain the unsustainable belief in a globe they blow our universe up to ridiculous scales and claim the beams from a radial source are 'almost parallel.'

Truth be told, the further away you are from a radial source, the distance between any two of its rays grows.

It is difficult to communicate with a person who has taken a seat in Plato's cave where the flat Earth is playing. When presented with evidence, that is contrary to flat Earth, it is automatically labeled as false, lies, and malarkey. Things like the sunrise, are seen through different eyes. Things like satellites, and the ISS, are false and do not exist. Then there is gravity, that is misunderstood by the flat earthling. I have yet to find away to drag that person out in the daylight, where the rear world exists. At least I know when I am in the Plato's cave, but the people seated do not. They will claim that I am the one that is lost, and filled with falsehoods. And that is the a state, of affairs, of the flat earthlings.
The the universe has no obligation to makes sense to you.
The earth is a globe.

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JackBlack

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Re: trigonometry
« Reply #21 on: December 07, 2017, 02:27:40 PM »
I have never seen a sun 'drop below the horizon'
So you have never seen a sunset.
I recommend trying to watch one at some point.

nor have I been able to measure the shadow created by the sun after sunset.
Not sure if it was before or after, but how about this one:

You can even do it yourself, but it works best with the stick horizontal, right near sunset.

So something like this:
|
|
|
|-----
|
|
|


Have that setup and observe the shadow around sunset and sunrise.

The sun does not need to be far away, except if one wishes to support ludicrous ideas like the Earth being a Globe. You have failed to show it does need to be this far away.
No, i showed that it must be far away. If you wish to disagree, explain how a close sun can appear due east along an entire line of longitude.

We already have located the sun using "my" simple math. Simply review Earth Not A Globe, the FAQ for this website, HUAI NAN TZU, or any of the thousands of other places this experiment has taken place and has been verified.
Your "simple" math, which must ignore numerous points and is based upon the false assumption that Earth is flat.

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realNarcberry

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Re: trigonometry
« Reply #22 on: December 07, 2017, 02:48:35 PM »
Did you just post an image of one of the worlds largest volcanos to prove the sun rises/sets behind the earth? Of course the sun can be obscured by mountains, that doesn't mean it drops below ocean level.

Re: trigonometry
« Reply #23 on: December 07, 2017, 02:52:54 PM »
Locating the Source of the Sun:


Locating any other light source in existence:



And again:

Take the Globe and turn it 90 clockwise, one Peg is still at mid afternoon, the second is at sunset, The third is in the night.
On the flat example move the sun 90 degrees in its orbit and all that happens is the shadows change length, no sunsets, no nighttime.
The the universe has no obligation to makes sense to you.
The earth is a globe.

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rabinoz

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Re: trigonometry
« Reply #24 on: December 07, 2017, 03:09:18 PM »
Locating the Source of the Sun:


Locating any other light source in existence:



And again:

Take the Globe and turn it 90 clockwise, one Peg is still at mid afternoon, the second is at sunset, The third is in the night.
On the flat example move the sun 90 degrees in its orbit and all that happens is the shadows change length, no sunsets, no nighttime.
Then add a few more points and:
         
And you clearly see that FE does not match, but a RE with a distant sun does.

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Re: trigonometry
« Reply #25 on: December 07, 2017, 03:17:56 PM »
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.

What is the source of this data?

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect. What is this trash? Did you find it on the back of a box of cereal? Also, these locations do not lie along a straight line as you seem to claim they do.
« Last Edit: December 07, 2017, 03:22:12 PM by John Davis »
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JackBlack

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Re: trigonometry
« Reply #26 on: December 08, 2017, 01:51:07 AM »
Did you just post an image of one of the worlds largest volcanos to prove the sun rises/sets behind the earth? Of course the sun can be obscured by mountains, that doesn't mean it drops below ocean level.
No, I posted an image of the sun casting a shadow upwards, to show the sun with a negative angle of elevation.
Or do you think the sun is below the height of the mountain?

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JackBlack

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Re: trigonometry
« Reply #27 on: December 08, 2017, 01:52:56 AM »
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.
This is not having the distance of the sun changing depending upon time of year.
This is having it change depending on the location you are at, so at the same time, it has a different height. That is impossible.

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect.
So that's your excuse? Just ignore the data that contradicts your view.

How about you address the massive lie in the image you provided?

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Re: trigonometry
« Reply #28 on: December 08, 2017, 08:37:36 AM »
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.
This is not having the distance of the sun changing depending upon time of year.
This is having it change depending on the location you are at, so at the same time, it has a different height. That is impossible.

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect.
So that's your excuse? Just ignore the data that contradicts your view.

How about you address the massive lie in the image you provided?
So there is no data or sources backing up your little picture? I ask you to supply them, and you come back and retort 'you are ignoring them!'

How can I ignore what has not been given?

You are correct, there are two misspellings on that poster. Its unfortunate I had the plates made before recognizing that.
Quantum Ab Hoc

1 + 1 = 2
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Re: trigonometry
« Reply #29 on: December 08, 2017, 08:52:32 AM »
Yes, the distance of the sun changes depending on the time of year. This is in our FAQ.
This is not having the distance of the sun changing depending upon time of year.
This is having it change depending on the location you are at, so at the same time, it has a different height. That is impossible.

The suns elevation, as labelled is incorrect. The angles as reported are incorrect. The distances as labelled are incorrect.
So that's your excuse? Just ignore the data that contradicts your view.

How about you address the massive lie in the image you provided?
So there is no data or sources backing up your little picture? I ask you to supply them, and you come back and retort 'you are ignoring them!'

How can I ignore what has not been given?
The data is IN the picture, and it doesn't need a source because the math speaks for itself.  All the numbers are provided so that anyone who believes it is inaccurate can check the calculations independently.  It's transparent to anyone with a high school level of trigonometry.