« **Reply #38 on:** October 18, 2022, 10:34:12 AM »

you don't demand anything from me, first of all. I told you there's questions i have about your model and that i don't make claims with no evidence. Now, the bullshit number you propose is based on several fallacies that I'm asking you to clarify. Going to Wikipedia for their "r" for radius and their presumed distances is an appeal to authority fallacy. The first thing i stated was i want the fundamentals. According to your calculations there's supposed to be curvature of the earth. When we actually measure geometrically(to measure the earth) we don't see your proported 8"per mile squared. So all those equations you did don't describe observed reality.

Do you think the sun is a physical thing?

3,306.5 miles high. Too high for either of us to actually measure. So i would like to stick with things that are verifiable. Looking at the ceiling does not tell you the shape of the floor.

this is from wiki. Seems like they still go based on erestothenes estimates and guess work. But never the less the radius they propose still doesn't match what we observe. Could it be because erestothenes never witnessed parallel light rays from the sun, which is where our conversation was at before you detoured into assumptive mathematics.

upon further review i find that the tropic of cancer is at 23.4⁰north, while Santa Barbra is at 34.4⁰north. That said you will never get the sun at 90⁰over santa barbra. So not only have you assumed the radius and the distances but you're also assuming the degrees of angles. I feel you're being disingenuous in our conversation.

➽ Too high for either of us to actually measure.

That is not correct. Such measuring is done with trigonometry. And it is NOT just you or I with an interest in measuring such things.

➽ 3,306.5 miles high.

This concurs with my result on a flat earth.

Now I will examine the "perspective" claim of FE.

Earlier I told you I'm not interested in the perspective theory for apparent elevation.

I am now going to engage the theory via the trig math.

In running these numbers, the math basically supports the "plausible" claim of a flat earth.

This does NOT prove a flat earth.

I'm handing you a Sorry chip. Ⓢ

Apparent Elevation = Ground Truth * tangent (50.6°)

Apparent Elevation 3,306.51 = 2716 * 1.2174199245579775791481703519979

Working the problem in reverse shows the differences because of rounding.

ArcTangent (Elevation / Ground Truth) = angle.

ArcTangent (1.2174153166421207658321060382916) = angle.

ArcTangent (Elevation / Ground Truth) = 50.599978623973347305787204622484°

This is close enough to confirm the 50.6° angle.

With the perspective theory, as the baseline doubles, the apparent angle of declination halves.

ArcTangent (3306 / 2 * Ground Truth) = 31.325346542640755203814710186387°

ArcTangent (3306 / 4 * Ground Truth) = 16.92541442526817730978967653951°

ArcTangent (3306 / 8 * Ground Truth) = 8.6514212337380432987587103592594°

I just showed you the math I expected from you to support the perspective theory.

Now gimme back my sorry chip.

As celestial navigating sailors have known for over a thousand years, nothing below 20°.

So I discard the apparent angles less than 20°.

Which means I will need to do the calc's with decreasing ground truth distances.

ArcTangent (3306 / 9083) = 20.00032508386290913198252757956°

ArcTangent (3306 / 2 * Ground Truth) = 31.325346542640755203814710186387°

ArcTangent (3306 / .5 * Ground Truth) = 67.668695575603243959058067598244°

ArcTangent (3306 / .25 * Ground Truth) = 78.393752870033447190350441427844°

ArcTangent (3306 / .125 * Ground Truth) = 84.136730555754599482739892766326°

The line of sight angle from the sighting location to the sun will be the longest distance.

This is the hypotenuse of a right triangle.

These 5 lines show the steps used to determine the LOS distance with the original ground truth and the FE elevation.

LOS distance = ²√(3306² + 9083²)

LOS distance = ²√(3306² + 2716²)

LOS distance = ²√(10,929,636 + 7,376,656)

LOS distance = ²√(18,306,292)

LOS distance = 4,278.58

LOS distance calculated with the FE elevation and the baseline of the FE distance to the directly under the sun location.

(<20°)

21,978.07= ²√(3306² + 21,728²)

(<20°)

11,183.80 = ²√(3306² + 10,864²)

Those two values are discarded because the are less than 20°.

The following six values are not less than 20°.

(20°)

9,665.94 = ²√(3306² + 9083²)

(31.32°)

6,358.95 = ²√(3306² + 5,432²)

(50.6°)

4,278.58 = ²√(3306² + 2716²)

(67.66°)

3,574.04 = ²√(3306² + 1,358²)

(78.39°)

3,306.00 = ²√(3306² + 679²)

(84.13°)

3,323.38 = ²√(3306² + 339.5²)

I am not using the top values of LOS distance because of <20° declination.

The sun's distance changes by a factor of:

9,665.94 / 3,323.38 = 2.908

This means the angular measure of the sun's circle would change from 0.5° at overhead distance to 0.172° at the 20° distance.

If I used the longest <20° distance above, the angular size of the sun would change by a factor of 6. The sun's angular measure would be 0.083°.

Think on that. And please, do not reply until tomorrow. I have things to do today.

« *Last Edit: October 19, 2022, 09:45:49 AM by Dale Eastman* »

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