totallackey

I encourage everyone to perform this measurement on their own.
« on: September 21, 2018, 12:53:31 PM »
Step One: Establish your latitude and longitude.
Step Two: Find a pole (with a base so the pole can stand erect without additional support) of known height.
Step Three: Measure the length of the shadow cast by the erect pole.
Step Five: Go to timeanddate.com and find the distance between you and the solar noon of the Sun at the time of the measurement.
Step Six: Do the math. The distance from the base of the pole to the end of the shadow (vertex of the angle) establishes a precise relationship with the distance between the Sun and vertex of the angle.

I think you will find the height of the Sun over the flat earth lane to be approximately 5600 miles.

Re: I encourage everyone to perform this measurement on their own.
« Reply #1 on: September 21, 2018, 12:57:06 PM »
Can you explain Step 5 in more detail?
I don't understand how to use timeanddate.com to establish the distance to the solar noon.
"On a very clear and chilly day it is possible to see Lighthouse Beach from Lovers Point and vice versa...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
- An excerpt from the account of the Bishop Experiment. My emphasis

totallackey

Re: I encourage everyone to perform this measurement on their own.
« Reply #2 on: September 21, 2018, 01:14:58 PM »
Can you explain Step 5 in more detail?
I don't understand how to use timeanddate.com to establish the distance to the solar noon.
https://www.timeanddate.com/worldclock/sunearth.html

Latitude and longitude of the Sun at any time of day can be found here.

Re: I encourage everyone to perform this measurement on their own.
« Reply #3 on: September 21, 2018, 05:48:04 PM »
How do you find the distance to an object from one measurement?

totallackey

Re: I encourage everyone to perform this measurement on their own.
« Reply #4 on: September 21, 2018, 06:29:11 PM »
How do you find the distance to an object from one measurement?
Of what do you write?

Why would I need more than one measurement?

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Offline Bobby Shafto

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Re: I encourage everyone to perform this measurement on their own.
« Reply #5 on: September 21, 2018, 07:53:51 PM »
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" 68.375" (68 3/8")
X - Shadow Length: 45.75"
Ratio Y:X = 1.4754:1 1.4945:1

Sun location at 2000 UTC: 0° 30'N, 106° 45'W
X' - Ground Distance: 2329 miles
Y' - Height of Sun: 3436 3481 miles ?

Alternate calculation
arctan (67.5 68.375/45.75) = 55.87 56.21°
(2329 miles)*tan(55.87 56.21°) = 3436 3480miles ?
« Last Edit: September 21, 2018, 11:51:28 PM by Bobby Shafto »

HorstFue

Re: I encourage everyone to perform this measurement on their own.
« Reply #6 on: September 21, 2018, 08:31:39 PM »
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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Offline stack

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Re: I encourage everyone to perform this measurement on their own.
« Reply #7 on: September 21, 2018, 08:49:20 PM »
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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Offline Bobby Shafto

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Re: I encourage everyone to perform this measurement on their own.
« Reply #8 on: September 21, 2018, 09:39:41 PM »
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.

------
Edit: applying corrected pole length
Recalculated elevation angle = 56.21°
Recalculated "gap" = 0.12°
Recalculated distance to sun ≈1,100,000 (still well off from 93,000,000, but better)
« Last Edit: September 21, 2018, 11:57:07 PM by Bobby Shafto »

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Offline stack

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Re: I encourage everyone to perform this measurement on their own.
« Reply #9 on: September 21, 2018, 09:54:06 PM »
When I popped my numbers into an earth curve calc, I came out with almost 5.4 million feet hidden.

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Offline Bobby Shafto

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Re: I encourage everyone to perform this measurement on their own.
« Reply #10 on: September 21, 2018, 11:42:22 PM »
I mis-measured the pole.

It's 58.375"   Crap!  Typo. Correct figure: 68.375"

I made another observation at 1607 PDT (same location). Took the most precise measurement of shadow length I could, which was 108.75" (tough without a sharp edge on the shadow).

I'll do the math on this one and update my previous post with the correct pole height later. Feel free to do the calculations yourself, if you want. See my earlier post for location. Again, time of measurement was 1607 local, which was 2307 UTC.
« Last Edit: September 21, 2018, 11:57:52 PM by Bobby Shafto »

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Re: I encourage everyone to perform this measurement on their own.
« Reply #11 on: September 22, 2018, 12:37:04 AM »
I might of done something wrong, but recalculating your latest I came up with:

Date: 9/21/2018
Time: 1607 PDT (2307 UTC)
Location: 32°43'26.9"N 117°14'36.5"W

Y - Object Height: 68.375"
X - Shadow Length: 108.75"
Ratio Y:X = .6287:1

Sun location at 2307 UTC: 0° 26' North, Longitude: 168° 16' West
X' - Ground Distance: 3983 miles
Y' - Height of Sun: 2504 miles

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Offline Bobby Shafto

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Re: I encourage everyone to perform this measurement on their own.
« Reply #12 on: September 22, 2018, 12:42:33 AM »
I have sun zenith location at that time slightly different.

Date: 9/21/2018
Time: 1607 PDT (2307UTC)
Location: 32°43'26.9"N 117°14'36.5"W

Y - Object Height:68.375" (68 3/8")
X - Shadow Length: 108.75"
Ratio Y:X = 0.6287:1

Sun location at 2307 UTC: 0° 25' 59"N, 168° 31'W
X' - Ground Distance: 4005 miles
Y' - Height of Sun: ~2518 miles?

Alternate calculation
arctan (68.375/108.75) = 32.158°
(2518 miles)*tan(32.158°) = ~2521 miles?

TimeandDate.com's elevation figure:
rounded to nearest degree: 32° (thanks, HorstFue...never saw that before)


Globe Earth circumference ~24901 miles
Ground distance to sun zenith=4005
Tilt angle: 360°*(4005/24901)= 57.9°

57.901°+32.158°=90.059°
No "gap"; angles overlap, so lines are beyond parallel.
Just not enough precision to be able to calculate distance to sun in the globe earth model with these coarse tools.

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Offline Bobby Shafto

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Re: I encourage everyone to perform this measurement on their own.
« Reply #13 on: September 22, 2018, 12:49:11 AM »
Globe earth calculation notwithstanding, totallackey's approach produced markedly different sun heights over a 4 hour span.

1900 UTC: 3481 miles (San Diego vantage point)
1918 UTC: 3673 miles (San Francisco vantage point)
2307 UTC: 2518 miles (San Diego vantage point)

None of which is in the ballpark of ~5600 miles.

Re: I encourage everyone to perform this measurement on their own.
« Reply #14 on: September 22, 2018, 02:27:36 AM »
How do you find the distance to an object from one measurement?
Of what do you write?

Why would I need more than one measurement?
Yea I sorted it out. Full disclosure I don't buy into flat earth but I'm not here to troll. I'm here because I'm curious. Also I'm new. So it took me a bit to work out how the measurement would work on a disk. My bad.

totallackey

Re: I encourage everyone to perform this measurement on their own.
« Reply #15 on: September 22, 2018, 09:24:46 AM »
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" 68.375" (68 3/8")
X - Shadow Length: 45.75"
Ratio Y:X = 1.4754:1 1.4945:1

Sun location at 2000 UTC: 0° 30'N, 106° 45'W
X' - Ground Distance: 2329 miles
Y' - Height of Sun: 3436 3481 miles ?

Alternate calculation
arctan (67.5 68.375/45.75) = 55.87 56.21°
(2329 miles)*tan(55.87 56.21°) = 3436 3480miles ?
Yeah, your calculation is skewed due to the difference in times listed for your measurements.

Quit behaving like such a disingenuous jerk.

And do not post if your intention is jet to muddy the issue and troll.

I swear...

totallackey

Re: I encourage everyone to perform this measurement on their own.
« Reply #16 on: September 22, 2018, 09:27:27 AM »
Every single post here is nothing but obvious disingenuous falsehoods.

"Mismeasured a pole."

LMMFAO!!!

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Offline junker

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Re: I encourage everyone to perform this measurement on their own.
« Reply #17 on: September 22, 2018, 11:08:23 AM »
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" 68.375" (68 3/8")
X - Shadow Length: 45.75"
Ratio Y:X = 1.4754:1 1.4945:1

Sun location at 2000 UTC: 0° 30'N, 106° 45'W
X' - Ground Distance: 2329 miles
Y' - Height of Sun: 3436 3481 miles ?

Alternate calculation
arctan (67.5 68.375/45.75) = 55.87 56.21°
(2329 miles)*tan(55.87 56.21°) = 3436 3480miles ?
Yeah, your calculation is skewed due to the difference in times listed for your measurements.

Quit behaving like such a disingenuous jerk.

And do not post if your intention is jet to muddy the issue and troll.

I swear...

Stick to the arguments and lay off the personal attacks, please. You let people get you riled up too easily.

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Offline Bobby Shafto

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Re: I encourage everyone to perform this measurement on their own.
« Reply #18 on: September 22, 2018, 01:47:01 PM »
Yeah, your calculation is skewed due to the difference in times listed for your measurements.

Quit behaving like such a disingenuous jerk.

And do not post if your intention is jet to muddy the issue and troll.

I swear...
What I would have done if I saw someone say they took measurements at one time and listed position of the sun an hour later is look up the correct info and correct the post to show the error instead of just leaping to insult.

If you had done that you would have found my listed time for sun position was a typo, but the location was correct for 1900 UTC, leaving my calculations unaffected:



I took this seriously. I spent time out of my day, twice, trying your suggestion. If you want to discuss the conduct and results of the experiments with civility, I'm in. If not, cya.

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Offline stack

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Re: I encourage everyone to perform this measurement on their own.
« Reply #19 on: September 22, 2018, 06:17:57 PM »
Yeah, your calculation is skewed due to the difference in times listed for your measurements.

Quit behaving like such a disingenuous jerk.

And do not post if your intention is jet to muddy the issue and troll.

I swear...

Let’s look at the facts, shall we:

- Presumably, based upon the experiment you defined, you came up with a Sun distance of 5600 miles.
- You have not shown your calculations, so it is unclear how you arrived at that number.
- Two people performed your experiment as you defined, three times.
- All calculations for the three efforts were presented.
- There was an error that was noticed by an experimenter and self acknowledged and corrected.
- The post-corrected results for the Sun distance in the three efforts were:
   3481 miles
   3673 miles
   2518 miles


And from a logical standpoint, why would we try and sully your experiment when we believe that however you came up with 5600 miles, in our belief, is off from reality by about 93,000,000 miles, give or take?