What are you talking about? What situation? What data?
I am asking for a detailed picture, along with any supporting graphs, equations, and explanations needed, which can demonstrate how the following situation can be possible according to Rowbotham's claims.
The following four individuals can, simultaneously, take pictures of a crescent moon at night and share it online. The individuals took the pictures from the following locations:
San Francisco, California
Manhattan, New York
Lima, Peru
Buenos Aires, Argentina
I am asking for a detailed picture, along with any supporting graphs, equations, and explanations needed, which can demonstrate how the following situation can be possible according to Rowbotham's claims.
The following four individuals can, simultaneously, take pictures of a crescent moon at night and share it online. The individuals took the pictures from the following locations:
San Francisco, California
Manhattan, New York
Lima, Peru
Buenos Aires, Argentina
This is how four individuals in SF, NYC, Lima, and Buenos Aries can do what you described:Step 1: Get a job and make some money
Step 2: Use said money that you made to buy a camera, a computer, and internet service
Step 3: Take a picture of the crescent moon
Step 4: Upload it online
Here is the equation for that:
Consumer Goods and Services = Labor + Good Work Ethic + Time
Are there any further questions?
Unless you have data for exactly what is seen of the crescent moon and its orientation from various points on earth simultaneously, I don't see that there is anything that needs to be explained.
Here is one for you:
Get a shovel and start digging. You fall out the other side and fall to your death.
How do you explain that?
What is there that we need to answer for? You asked "how is it possible for people to take a picture of the moon from different locations" without showing what the result is from such an experiment, or even really explaining what you think the expected result will be.
That is not my problem, Thomas. You have proposed a FET as valid. The burden on proof lies with you. I have proposed a scenario, and asked for an explanation in the FE model.
That is not my problem, Thomas. You have proposed a FET as valid. The burden on proof lies with you. I have proposed a scenario, and asked for an explanation in the FE model.
Burden of proof to explain what? You provided no data for the result of your hypothetical experiment would be.
What data do you imagine might follow from a hypothetical experiment? You do understand what these words mean, yes?
What data do you imagine might follow from a hypothetical experiment? You do understand what these words mean, yes?
Yes. You want us to explain the results of an experiment that happened in your head.
We've looked at those calculators. They are based on patterns of the moon's presumed movement and occurrences. They even still use Ptolmy's famous lunar perturbations from his pattern-predicting methods used in his geocentric earth calculations.
If you want to compare the results from various points on earth, be my guest. I'm not going to do your experiments for you.
You're not interested in what others do at their points, so I suggest you do it for yours.
.... neither of us will demand that the other does work for the other. Does that sound agreeable?
What data do you imagine might follow from a hypothetical experiment? You do understand what these words mean, yes?
Yes. You want us to explain the results of an experiment that happened in your head.
If you have nothing to actually contribute... we know where this thread is going to end up.
What data do you imagine might follow from a hypothetical experiment? You do understand what these words mean, yes?
Yes. You want us to explain the results of an experiment that happened in your head.
If you have nothing to actually contribute... we know where this thread is going to end up.
No, Thomas. A theory that describes reality means that it is a theory that can predict what will happen...in reality. I used to live in San Francisco, where I remember seeing a crescent moon one night. What does your theory predict would been seen in the other following locations that I specified? This IS the forum for FET, yes? Where we investigate FET?
This is what scientists do, Thomas. They continuously apply their model to different scenarios to test it.
The more you hide from actually attempting to test your model, the less anyone will ever believe it to be true.
Scientists who believe in their models do not hide like this. It is evident that you do not believe in the FET.
What data do you imagine might follow from a hypothetical experiment? You do understand what these words mean, yes?
Yes. You want us to explain the results of an experiment that happened in your head.
If you have nothing to actually contribute... we know where this thread is going to end up.
No, Thomas. A theory that describes reality means that it is a theory that can predict what will happen...in reality. I used to live in San Francisco, where I remember seeing a crescent moon one night. What does your theory predict would been seen in the other following locations that I specified? This IS the forum for FET, yes? Where we investigate FET?
This is what scientists do, Thomas. They continuously apply their model to different scenarios to test it.
The more you hide from actually attempting to test your model, the less anyone will ever believe it to be true.
Scientists who believe in their models do not hide like this. It is evident that you do not believe in the FET.
What reality? Test against what? You have not posted any data for what will be seen.
Does FET need data before it can provide an answer to a simple question?
Convince me or us that this is "inexplicable" ...
Convince me or us that this is "inexplicable" ...
Real photographs would be better.
I took one of your times and went down the Greenwich Longitude (0 degrees E), at 45 degrees North, 0 degrees North, and 45 degrees South. Here are the results:
I took one of your times and went down the Greenwich Longitude (0 degrees E), at 45 degrees North, 0 degrees North, and 45 degrees South. Here are the results:
What have you 'demonstrated'?
It is easy to see how perspective plays very little part in the Round Earth system.
In RET the distance to the moon is 238,900 miles.
Imagining that distance as a radius of a circle, with observers positioned all around the moon, we can get the circumference of that circle with C=2*pi*R.
The circumference is 1501052.96989 miles
We divide it by 360 to get 4169.59158301 miles per degree
The diameter of the earth is 7,917.5 miles
7,917.5 / 4169.59158301 = a ratio of 1.89, or a little less than 2 degrees.
If we place the earth on the circumference we created above, with two observers standing on direct opposite sides of the earth, about 7917.5 miles apart, looking at the moon, should see a difference in shift of less than 2 degrees.
Here is an image which may provide some clarification:
Here is an image which may provide some clarification:
The differences between the tips of the crescents are 2 x 45 degrees, or 90 degrees in total.
The differences in observer latitude are two sets of 45 degrees, making 90 degrees in total.
What does this indicate to you?
lat·i·tude
ˈladəˌt(y)o͞od/
noun
noun: latitude; plural noun: latitudes
the angular distance of a place north or south of the earth's equator, or of a celestial object north or south of the celestial equator, usually expressed in degrees and minutes.
Your graphics are faulty.
In the second one, all you've done is draw lines away from the surface. What's the point of view here? Are we behind the observers? To the side of them?
You've taken no account of the position of the Moon in the sky for each observer, and not shown this on your diagram.
It's a side view.
Here is an example, showing that from the North Pole to South Pole (90o N to 90o S), that the RET difference is about 2 degrees.
Angles are measured in degrees. Where do you see this angle of 2 degrees? It must be placed somewhere.It is easy to see how perspective plays very little part in the Round Earth system.
In RET the distance to the moon is 238,900 miles.
Imagining that distance as a radius of a circle, with observers positioned all around the moon, we can get the circumference of that circle with C=2*pi*R.
Observers on the Moon will not be on that circle. You're calculating the circumference of the Moon's orbit with that formula. A number of observers on the Moon at any one time would, at max, be upon a circle of (2*3.14*1079 =) 6780 miles.
The circumference is 1501052.96989 miles
That's the length of the Moon's orbit, not the circumference of the Moon
We divide it by 360 to get 4169.59158301 miles per degree
So the Moon moves 4170 miles for each degree of its orbit.
The diameter of the earth is 7,917.5 miles
7,917.5 / 4169.59158301 = a ratio of 1.89, or a little less than 2 degrees.
If we place the earth on the circumference we created above, with two observers standing on direct opposite sides of the earth, about 7917.5 miles apart, looking at the moon, should see a difference in shift of less than 2 degrees.
Does FET need data before it can provide an answer to a simple question?
The fact is that the orientation that will be seen will be inexplicable in the Round Earth Theory. I have been waiting for you to post some examples to point it out.
The degrees of Latitude were originally defined on the angles celestial bodies at the horizon come out of the horizon. At 0 degrees N the Sun comes straight out of the horizon when it is over the equator. At 45 degrees N and S the sun comes out of the horizon at a 45 degree angle N or S.
Degrees of latitude only have meaning as angles. Where would you draw the angle, if not at the centre of a circle or globe?
The fact that the moon shifts by that much is not unexpected, since that the longitude lines are based on the changing angles of celestial bodies that come out of the horizon.
Here is an example, showing that from the North Pole to South Pole (90o N to 90o S), that the RET difference is about 2 degrees.
Angles are measured in degrees. Where do you see this angle of 2 degrees? It must be placed somewhere.It is easy to see how perspective plays very little part in the Round Earth system.
In RET the distance to the moon is 238,900 miles.
Imagining that distance as a radius of a circle, with observers positioned all around the moon, we can get the circumference of that circle with C=2*pi*R.
Observers on the Moon will not be on that circle. You're calculating the circumference of the Moon's orbit with that formula. A number of observers on the Moon at any one time would, at max, be upon a circle of (2*3.14*1079 =) 6780 miles.
The circumference is 1501052.96989 miles
That's the length of the Moon's orbit, not the circumference of the Moon
We divide it by 360 to get 4169.59158301 miles per degree
So the Moon moves 4170 miles for each degree of its orbit.
The diameter of the earth is 7,917.5 miles
7,917.5 / 4169.59158301 = a ratio of 1.89, or a little less than 2 degrees.
If we place the earth on the circumference we created above, with two observers standing on direct opposite sides of the earth, about 7917.5 miles apart, looking at the moon, should see a difference in shift of less than 2 degrees.
Moon circumference? I'm talking about the distance around the moon. Here is an illustration:
IMG
If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.
(https://i.imgur.com/lOa8DUh.jpg)
If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.
IMG
I don't see how that makes any sense at all. That doesn't use Round Earth Geometry. In fact, it is assuming that the moon is very close to the earth.
The difference between 45o N and 45o S would be very slight under the geometry of the Earth-Moon System.
IMG
The difference between 45o N and 45o S would be very slight under the geometry of the Earth-Moon System.
(https://i.imgur.com/CTOCE1I.png)
The difference between 45o N and 45o S would be very slight under the geometry of the Earth-Moon System.
https://i.imgur.com/CTOCE1I.png
This is just my side view flipped horizontally, isn't it? You've shown Earth left, Moon right, I showed Moon left, Earth right.
Surely the easy way to do this is to calculate a triangle with two sides of 240k miles, and one of the distance between the two observers at 45N and 45S, and solve for the angle opposite the side which connects them? That would take away your totally artificial circle of the Earth going around the Moon.
The short side will be the length of chord BX, where the angle at M is 90 degrees, assuming two observers at 45N and 45S.
(https://upload.wikimedia.org/wikipedia/commons/thumb/e/ea/Chord_in_mathematics.svg/200px-Chord_in_mathematics.svg.png)
https://en.wikipedia.org/wiki/Chord_(geometry)
I genuinely thought you were drawing the orbit of the Moon around the Earth. Way to make things more confusing
Thing is Tom, on your "close moon - flat Earth" diagram, completely different faces of the moon would be seen. This is not observed.
Between the different latitudes on a globe however, the face of the moon rotates due to the different angles between different latitudes. This is what is observed.
there are not many good sources of pictures of the moon taken simultaneously, to say what exactly happens.Doesn't need to be simultaneously. If the moon were close to a flat plain, different faces would be seen as it moved overhead. Has this ever been observed?
Your Round Earth model does not explain it at all.It does actually. Anyone with a ball, camera, and an object with surface features can try it. At 45N, when viewing the moon set, there is a 90 degree difference in viewing angle from viewing it when it rises.
It does actually. Anyone with a ball, camera, and an object with surface features can try it. At 45N, when viewing the moon set, there is a 90 degree difference in viewing angle from viewing it when it rises.
If you have ever seen a desktop globe in real life, or perhaps any ball, have a think about it.
(https://i.imgur.com/iJ2Mmy8.png)
While researching this, I found several other inexplicable Round Earth items.
- The Moon is above the horizon for observers on the Prime Meridian and the Antimeridian simultaneously, when it should not be
- The Moon is pointing in the wrong direction on the horizons of the PM and AM. While the Moon will never "tilt" or "rotate" in the Round Earth model, it is possible to look at it upside-down when standing on opposite sides of the earth.
- The Moon is above the Western Horizon for observers on the Prime Meridian and the Antimeridian, when it should be on the Western Horizon for the Prime Meridian and the Eastern Horizon for the Antimeridian
Here is what the Mooncalc shows for the Prime Meridian and the Antimeridian:
IMGs
While researching this, I found several other inexplicable Round Earth items.
- The Moon is above the horizon for observers on the Prime Meridian and the Antimeridian simultaneously, when it should not be
- The Moon is pointing in the wrong direction on the horizons of the PM and AM. While the Moon will never "tilt" or "rotate" in the Round Earth model, it is possible to look at it upside-down when standing on opposite sides of the earth.
- The Moon is above the Western Horizon for observers on the Prime Meridian and the Antimeridian, when it should be on the Western Horizon for the Prime Meridian and the Eastern Horizon for the Antimeridian
Explanatory image:
(https://i.imgur.com/E8Ogpn6.png)
(https://i.imgur.com/itm8Kfk.png)
Here is what the Mooncalc shows for the Prime Meridian and the Antimeridian:
Prime Meridian
(https://i.imgur.com/pa7oQA7.jpg)
Link: https://www.mooncalc.org/#/0,0,3/2018.08.12/17:58/1/2
Antimeridian
(https://i.imgur.com/IoQQE5r.jpg)
Link: https://www.mooncalc.org/#/0,180,3/2018.08.12/17:58/1/2
Do you know what you have not taken into account in your cartoon?
The Sun.
Create a proper scale-model figure which includes the thing that is illuminating the Moon in the first place (that would be the Sun, Thomas), and re-submit.
Thomas, you have the same time entered for both locations. You do recognize that this is actually comparing them at different times, yes?
Do you know what you have not taken into account in your cartoon?
The Sun.
Create a proper scale-model figure which includes the thing that is illuminating the Moon in the first place (that would be the Sun, Thomas), and re-submit.
In addition, whilst he may have indicated the POSITION of the observers, he hasn't made any allowance for their orientation ...
If you change location you need to change the time. Why would it do that for you? That's not a buggy program, that's GIGO. 16:50 in CST is not the same time as 16:50 in GST. You need to change the time manually when you move timezones.Thomas, you have the same time entered for both locations. You do recognize that this is actually comparing them at different times, yes?
I didn't change the time, I just changed the location. If Mooncalc changed the time, then I wouldn't be surprised. It is an incredibly buggy program.
I didn't change the time, I just changed the location. If Mooncalc changed the time, then I wouldn't be surprised. It is an incredibly buggy web application.
By all means, list all the other bugs that you have found, IF you have found any.
It expects you to enter the LOCAL time for the place you're interested in. Why would anyone want to look at the Moon in Perth, Australia, for a time in London?
I've shown that it cannot be the moon that is rotating.
Did anyone claim that it was?
The only out is that it is the earth that is shifting.
Or it's the orientation of the observers ....
It doesn't seem that the Mooncalc can reproduce the Moon Tilt Illusion.
Model it. Use the Round Earth Distances to explain it.What is there to model? If you tilt your head/camera 45 degrees to the right while looking at something (insert distance of your choosing here) away, and then tilt your head/camera 90 degrees to the left while looking at the same object, it will appear to tilt, or rotate if you prefer, in your field of view.
Model it. Use the Round Earth Distances to explain it.What is there to model? If you tilt your head/camera 45 degrees to the right while looking at something (insert distance of your choosing here) away, and then tilt your head/camera 90 degrees to the left while looking at the same object, it will appear to tilt, or rotate if you prefer, in your field of view.
If you were to the rear of all the observers, looking past the Earth toward the Moon, their views would vary, because of the difference in their LATITUDE. The observer at the equator will have the crescent toward the bottom of the Moon, and the observers at 45N or 45S will have it inclined to this aspect by 45 degrees each.
(https://i.imgur.com/lOa8DUh.jpg)
I don't see how that makes much sense. That doesn't use Round Earth Geometry. In fact, it seems to be assuming that the moon is very close to the earth.
I'm Australian and haven't fallen off yet, so what do you say to that
Which effect are you talking about? I'm talking about two observers, one at 45N, one at 45S, looking at the Moon at the same time, with each seeing the crescent in different positions, separated by approx 90 degrees. You seem to understand what I'm getting at. You seemed to confirm it with your red arrows on the moon graphics in Reply #23 ... - 90 degrees difference in latitude, 90 degrees difference in the crescent
I'm not talking about the passage of the Moon over time, over the course of a night, or any other period. If both observers look at the same time, we have already seen, from mooncalc, how their views would differ. You agree that if a camera is held in one position, looking at a crescent moon, then moved around its axis, the crescent is shifted by 90 degrees or so. With two observers separated by 90 degrees of latitude, can you see how this shift of the camera applies, assuming both observers keep their cameras upright?
|
|
V
---------
west |
west | <---------
|
Think about what I described above. You are at the equator looking at the Western Horizon and the moon is passing by from behind overhead.
No, that's not the situation that is being discussed. We're not talking about any movement in the Moon, only observations from different places AT THE SAME TIME. The Moon is at a static point in the sky at this time. Where it has been. or where it is going don't enter into it.
Now you rotate the camera by 90 degrees to simulate what would happen at the North Pole.
Again, not the situation being discussed. We were talking about two or three observers, spanning 45S, equator, and 45N
Now the moon seems to be passing over the North Pole...
We're not concerned with its path, only how its crescent appears to different observers at the same time.
Rotating the camera is not enough to simulate the curvature of the earth.
But you agree that if you have two cameras which are inclined separately to the Moon with a difference of 90 degrees, that the two pictures will have the Moon's crescent differing by 90 degrees?
I do think that the Mooncalc is trying to simulate the tilt of the earth, however. But I no longer think it is accurate due to ignoring the Moon Tilt Illusion which takes place at all times; so this point of what the calculator shows is moot with me.
Again, not the situation being discussed. We were talking about two or three observers, spanning 45S, equator, and 45N
Tom, will you accept the above as a starting point, and that I was genuinely looking at the Moon, with my baseball held at arm's length, in the daytime, and in the sunlight?
QuoteTom, will you accept the above as a starting point, and that I was genuinely looking at the Moon, with my baseball held at arm's length, in the daytime, and in the sunlight?
Sure. I don't have any reason to doubt you. If you want to talk about the Moon Tilt Illusion, lets take it to the Moon Tilt Illusion thread.