Recent Posts

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Flat Earth Theory / Re: Viewing Brighton Seafront from Worthing
« Last post by stack on Today at 10:02:28 PM »
I'm going back through the video to see if there's some modeling to be done around the landmarks, specifically the i360 tower and the Sussex Heights building.

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When I popped my numbers into an earth curve calc, I came out with almost 5.4 million feet hidden.
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Flat earth geometry is pretty simple:



Globe earth geometry is a little less straightforward:



For a globe of r=3959 miles, the tilt angle between my perpendicular and perpendicular at the location of solar noon (sun's location) at the time of my measurement would have been ≈33.67°.

(arc distance/earth circumference)*360°
or
(2329/24901)*360°

That plus the observed elevation angle above horizontal of 55.87° leaves "gap" of about 0.46° to perpendicular.

But that's not acute enough a sun supposedly 93,000,000 miles away. The "gap" angle between tilt and observed elevated angle should be more like 0.002°, which is too precise for the kind of measuring any of us can due with poles and measuring tape. In a more ideal measurement setting, I should have measured a shadow of around 46 1/4" or so. As it was, my measurement would put the sun only 290,000 miles away from a globe earth.

 
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Flat Earth Theory / Re: Viewing Mt Helix from Cabrillo Point 16.2 miles
« Last post by stack on Today at 09:35:47 PM »
Nice idea using Bislin's tool to model the difference between FE and GE calculations. Could add a 2nd distant target to visualize how the hill (Mt Helix) appears in the background relative to the hotel, and see how it compares.

I'm working on a pixel comparison now. But here's how Bilsins' calc lays out the Hilton to Mt Helix (no refraction):

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Flat Earth Investigations / Re: What Would You Do?
« Last post by stack on Today at 09:12:13 PM »
Go talk with or write any surveyor and ask them if the method I propose is or is not a legitimate method for measuring the height of objects.
There is nothing wrong with your method. But it does assume a flat earth.
If the earth isn’t flat (spoiler: it isn’t) then you will get a very different result. I don’t even know if it’s possible to calculate the distance to the sun with your method on a globe earth.

It definitely assumes a flat plane.

I put my calculations into an Earth Curvature Calculator. The result was that the target object (Sun), calculated at a distance of 2581 miles away and a height of 3673 feet would be hidden by the curvature of the earth by 5,383,000 feet. Yet it is above me at about a 60 degree angle.
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Flat Earth Investigations / Re: What Would You Do?
« Last post by AllAroundTheWorld on Today at 08:51:57 PM »
Go talk with or write any surveyor and ask them if the method I propose is or is not a legitimate method for measuring the height of objects.
There is nothing wrong with your method. But it does assume a flat earth.
If the earth isn’t flat (spoiler: it isn’t) then you will get a very different result. I don’t even know if it’s possible to calculate the distance to the sun with your method on a globe earth.
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If this experiment is so GLOBAL already, why not take sun's altitude (the vertical angle between horizon and sun) also from "timeanddate", instead of measuring this angle from the height of a pole and the length of it's shadow.

Advantage: You only need your PC and Internet connection, to "repeat" this experiment for any place and any time on earth.

Agreed all around. But I figured I'd just do the experiment as described. Here's mine:

Date: 9/21/2018
Time: 13:18 PDT (20:18 UTC)
Location: 37°46'09.5"N 122°27'06.8"W

Y - Object Height: 37"
X - Shadow Length: 26"
Ratio Y:X = 1.423:1

Sun location at 20:18 UTC: Latitude: 0° 29' North, Longitude: 126° 15' West
X' - Ground Distance: 2581 miles
Y' - Height of Sun:  3673 miles
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Date: 9/21/2018
Time: 1200 PDT (1900 UTC)
Location: 32°43'26.9"N 117°14'36.5"W

Y - Object Height: 67.5"
X - Shadow Length: 45.75"
Ratio Y:X = 1.4754:1

Sun location at 2000 UTC: 0° 30'N, 106° 45'W
X' - Ground Distance: 2329 miles
Y' - Height of Sun: 3436 miles ?

Alternate calculation
arctan (67.5/45.75) = 55.87°
(2329 miles)*tan(55.87°) = 3436 miles ?

Hi Bobby!
If you have searched the timeanddate site more thoroughly, you would also have found the "sun and moon today" calculator. This one gives, for the given time, as "altitude" for the sun at San Diego:  56°. Quite close to your 55.87°.

I cannot figure out, what this experiment should proof.
Advice is to take your position in lat/lon from a GLOBE earth map; to derive the position, where the sun is currently in zenith, from a "timeanddate" calculator based on GLOBE earth; derive the distance between these two points again from a GLOBE earth map.
If this experiment is so GLOBAL already, why not take sun's altitude (the vertical angle between horizon and sun) also from "timeanddate", instead of measuring this angle from the height of a pole and the length of it's shadow.

Advantage: You only need your PC and Internet connection, to "repeat" this experiment for any place and any time on earth.
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Date: 9/21/2018
Time: 1200 PDT (1900 UTC)
Location: 32°43'26.9"N 117°14'36.5"W

Y - Object Height: 67.5"
X - Shadow Length: 45.75"
Ratio Y:X = 1.4754:1

Sun location at 2000 UTC: 0° 30'N, 106° 45'W
X' - Ground Distance: 2329 miles
Y' - Height of Sun: 3436 miles ?

Alternate calculation
arctan (67.5/45.75) = 55.87°
(2329 miles)*tan(55.87°) = 3436 miles ?
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Flat Earth Investigations / Re: What Would You Do?
« Last post by Curious Squirrel on Today at 07:25:31 PM »
The angle of elevation of the sun is not required, that is correct. But you can use it to find the altitude of the sun in the exact same manner as you have described.
No, you cannot.

Surveyors have been using the method I describe for thousands of years and there is nothing required about the angles of the top of the object(s) being measured.

You have no clue about what you are writing.

Go talk with or write any surveyor and ask them if the method I propose is or is not a legitimate method for measuring the height of objects.
I have never said the method you posted isn't a way to measure the height. My statement as been there are other methods for doing so. But since you suggested it, I've sent an email to a surveying company to inquire as to what methods are currently being used for measuring heights and similar.

With the 90 degree angle from sun to Earth, and the distance to this point know, the angle of the sun up from the horizon will allow one to find the distance to the sun. Do you dispute this? If not I'd be more than happy to run some numbers, but I'm not going to waste my time if we can't agree on the base principles at work, and I have my doubts at present when you claim your experiment and the one done by Eratosthenes have 'nothing to do' with one another.
Yes, I do.

Surveyors do not use this method to determine the height of objects.

One does not need to know the angle of the hypotenuse.

One need only know the baseline distance from the object being observed to the vertex and the distance of the interceding pole to vertex.

This establishes a precise ratio of right triangles formed without any conjecture.
Not even sure where to begin with this. Are you claiming that with triangle ABC, you can't figure out the height of side a by knowing angles A and C, and the length side b?



If you are not, please explain what you are actually trying to say. If you are, well....not sure there's any point in continuing this.