The sun doesn't work as a spotlight

Quote from: Tom Bishop on January 14, 2012, 12:22:53 AM

Quote from: OrbisNonSufficit

You have never answered my question tom...  Why is the sun not visible from all parts of the earth while in a plane?  The atmosphere is significantly reduced, so distortion via the atmosphere is not an option.  Not to mention that in a plane the sun still does not shrink as it "moves away".

I've answered this already. If it's midnight and a plane is at an altitude of 5 miles (26400 feet), the sun is still 3000 miles in altitude above the earth and tens of thousands of miles away. The altitude of 5 miles is insignificant compared to the sun. The 5 mile altitude of the plane may result in the sun taking a bit longer to set into the horizon due to slightly more acute perspective lines at that altitude, but it still sets none-the-less, as the sun will eventually recede into the vanishing point of the plane.
to a tiny dot as it recedes from the observer.

Tom, Here is a side-on view of the flat earth, with an observer on the left and a midnight sun on the right.


Consider the 2D geometry of this atmospheric "section" and in paricular the photons leaving the sun: 

What path are the photons taking when they travel from the sun to the observer on the left?

Quote from: Tom Bishop on January 14, 2012, 12:22:53 AM

Quote from: OrbisNonSufficit

You have never answered my question tom...  Why is the sun not visible from all parts of the earth while in a plane?  The atmosphere is significantly reduced, so distortion via the atmosphere is not an option.  Not to mention that in a plane the sun still does not shrink as it "moves away".

I've answered this already. If it's midnight and a plane is at an altitude of 5 miles (26400 feet), the sun is still 3000 miles in altitude above the earth and tens of thousands of miles away. The altitude of 5 miles is insignificant compared to the sun. The 5 mile altitude of the plane may result in the sun taking a bit longer to set into the horizon due to slightly more acute perspective lines at that altitude, but it still sets none-the-less, as the sun will eventually recede into the vanishing point of the plane.
to a tiny dot as it recedes from the observer.

Tom, Here is a side-on view of the flat earth, with an observer on the left and a midnight sun on the right.


Consider the 2D geometry of this atmospheric "section" and in paricular the photons leaving the sun: 

What path are the photons taking when they travel from the sun to the observer on the left?

Quote from: Tom Bishop on January 14, 2012, 12:22:53 AM

Quote from: ClickTower

We've repeatedly documented the sun rise and sun set times for various parts of the world. We've documented that the FE perspective of the Sun is wrong from many places. You've admitted that your postings are wrong. I guess there's nothing more but conclude that FET is embarrassingly wrong.

http://www.timesunrisesunset.com/
http://www.almanac.com/sun/rise/NH/Dublin/2012-01-13

The values in both of those links are from calculators based on RET, not direct observations of reality.

From the bottom of your first link:

"Daylight Saving Time (DST) / Summer Time is taken into account for all time calculations on this site."

Tom, you're getting forgetful, or rather more so than usual.

Try refreshing your memory by reading this post: http://www.theflatearthsociety.org/forum/index.php?topic=52461.msg1286655#msg1286655

Personally, I think it's much more impressive that RET predicts these times for everywhere in the world. It's a piece of cake for you to get the time of the next events from these links and then observe RET's accuracy. Of course, you could always lie about this as well, like you do with your Monterey Bay experiment.

And you have done nothing to show that those calculations are inconsistent with real world observations.

Quote from: ClockTower on January 14, 2012, 01:01:56 AM

Personally, I think it's much more impressive that RET predicts these times for everywhere in the world. It's a piece of cake for you to get the time of the next events from these links and then observe RET's accuracy. Of course, you could always lie about this as well, like you do with your Monterey Bay experiment.

Those sun calculator lists are calculations, and not based on reality. Please stop posting those.

...your lies about your Monterey Bay experiment condemn you as dishonest; your outlandish unsupported claims condemn you as delusional.

Quote from: markjo

And you have done nothing to show that those calculations are inconsistent with real world observations.

A calculation by definition has no implied or guaranteed consistency with reality. It's a calculation, for what might be true, if certain variables were met.

As we know that the earth is not a globe, we know that the sun calculators are wrong, at least to some degree. I've seen inconstant suns calculations for myself, which I've described to you in the past.

Quote from: ClockTower on January 14, 2012, 12:12:14 PM

...your lies about your Monterey Bay experiment condemn you as dishonest; your outlandish unsupported claims condemn you as delusional.

How exactly have you determined this claim to be a lie?

Have you reviewed the whole of the science of atmospheric optics so that you know Tom's Monterey Bay experiment to be a lie?  How did you determine that it's a lie as opposed to mistake, misinterpretation, faulty instruments, etc?

Have you reviewed the whole of the science of atmospheric optics so that you know Tom's Monterey Bay experiment to be a lie?  How did you determine that it's a lie as opposed to mistake, misinterpretation, faulty instruments, etc?

Why don't you explain for me, in detail, what kind of margin of error occurs in my experiment across the Monterey Bay. Here it is again in case you've missed it:

I live along the California Monterey Bay. It is a relatively long bay that sits next to the Pacific Ocean. The exact distance between the extremes of the Monterey Bay, Lovers Point in Pacific Grove and Lighthouse State Beach in Santa Cruz, is 33.4 statute miles. See this map.

On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa. With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 33 miles away near the lighthouse. The entire beach is visible down to the water splashing upon the shore. Upon looking into the telescope I can see children running in and out of the water, splashing and playing. I can see people sun bathing at the shore and teenagers merrily throwing Frisbees to one another. I can see runners jogging along the water's edge with their dogs. From my vantage point the entire beach is visible.

Have you reviewed the whole of the science of atmospheric optics so that you know Tom's Monterey Bay experiment to be a lie?  How did you determine that it's a lie as opposed to mistake, misinterpretation, faulty instruments, etc?

Quote from: http://www.theflatearthsociety.org/forum/index.php?topic=17770.msg310005;topicseen#msg310005

On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa. With a good telescope, laying down on the stomach at the edge of the shore on the Lovers Point beach 20 inches above the sea level it is possible to see people at the waters edge on the adjacent beach 33 miles away near the lighthouse.
...
I can see people sun bathing at the shore...

Tom's descriptions are too specific to be misinterpretations. Faulty instruments don't take on superpowers.

It's either a lie or a hallucination.

Quote from: zarg on January 14, 2012, 07:20:15 PM

Tom's descriptions are too specific to be misinterpretations. Faulty instruments don't take on superpowers.

It's either a lie or a hallucination.

That is a false conclusion as true also remains a possibility.  Have you not looked for other possibilities regarding Monterey Bay's peculiar atmospheric optics or did you just leap in unprepared and cry "Liar"?

Quote from: zarg on January 14, 2012, 07:20:15 PM

Tom's descriptions are too specific to be misinterpretations. Faulty instruments don't take on superpowers.

It's either a lie or a hallucination.

That is a false conclusion as true also remains a possibility.  Have you not looked for other possibilities regarding Monterey Bay's peculiar atmospheric optics or did you just leap in unprepared and cry "Liar"?

The same result comes up over and over throughout the year under a plethora of different atmospheric conditions.

Really, if you start out with that list of possibilities disregarding 'true,' it's fatally flawed from the get-go as you're preconcluding the conclusion.

Really, if you start out with that list of possibilities disregarding 'true,' it's fatally flawed from the get-go as you're preconcluding the conclusion.

I think he said he had ruled out any 'chance' optical illusions. 

Really, if you start out with that list of possibilities disregarding 'true,' it's fatally flawed from the get-go as you're preconcluding the conclusion.

If Tom is making claims that defy the laws of optics (being able to see a beach ball at 30+ miles) then what possibilities should we consider?

Quote from: markjo on January 14, 2012, 09:09:17 PM

If Tom is making claims that defy the laws of optics (being able to see a beach ball at 30+ miles) then what possibilities should we consider?

I thought all this time it was a frisbee game.  Anyway, back to the preconcluded conclusion, I start out to determine if Markjo is a liar.  I have compiled this list of possibilities:

He is either (1.)telling a whopper or he is (2.) a perjurer.  Ergo he is a liar. 

Oh my, what is wrong with that?

I have compiled this list of possibilities:

He is either (1.)telling a whopper or he is (2.) a perjurer.  Ergo he is a liar. 

Everything viewed through the atmosphere is visually displaced and would not correspond to simple mathematical predictions if this displacement factor is left out of the equation.  If conditions are agreeable, this displacement is enough to enable Tom is see exactly what he claimed to see.

Everything viewed through the atmosphere is visually displaced and would not correspond to simple mathematical predictions if this displacement factor is left out of the equation.  If conditions are agreeable, this displacement is enough to enable Tom is see exactly what he claimed to see.

Displaced now means bigger.  Whoda thunk?