Flat Earth

[BADecker]

@notbatman

You can end most of the disagreement and misunderstanding by simply showing us the math/trig that allows us to find the distance and size using only the angle. If you do this, 90% of all the rest of the talk will disappear.

Just spit it out right here. You know, like 1+1=2. Or whatever it is. And if you use unconventional math, show us why your math stands over standard math.

   The angular resolution limit of the human eye determines how far the human eye can see (source: Ophthalmology 3rd Edition, ISBN 978-0444511416). Tell us why you think the angular resolution limit of the human eye isn't a factor in determining the distance the human eye can see?

Why are you asking me to provide a formula for calculating the distance to an object without including the angular resolution limit of the eye? If you don't include the angular resolution limit of the eye in your calculation, then the distance to an object can not be calculated.

If there's an error in my formulas for calculating size and distance then show us!

I ask for this formula because of what you said.

If we get rid of the limited human eye in the equation altogether, we also get rid of eye limits.

Trigonometry doesn't use human eye limitations. Rather, trig calculates the answer accurately no matter what the eye thinks it sees... and especially if we use calculus along with the trig.

If there is an error in your calc, the error is using your formulas at all, because there is trig and calculus that will give the answer easily and accurately, and (as you said) the eye has limitations.

What is the basic answer that trig and calculus gives? It gives the answer that says that you can't find distance and size with only the angle. You need another measurement along with the angle to show distance or size.

Let me say it another way. Go inside a building with no windows, let someone give you the angle (32 degrees), and calculate the size/distance of anything accurately without using trig or calculus. You can do it accurately if you have the distance or size (and use trig), but not without one of these in addition to the angle (and trig).

Or do you have a way without non-accurate, hazy, limited observations of the eye? Show us if you do.

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[notbatman]

^^^ You get the same response as in the Moon thread:

^^^ You're a clown using pipul to win an argument with intellectual dishonestly. Every eye has an angular resolution limit or distance and I'm using that distance to measure other objects. You response (you fucking clown) is to conflate the maximum distance the eye (any eye or camera) can see with with the concept that the human eye has limited abilities thus can't be used.

This like telling somebody you're having guests for dinner then killing and eating them when they arrive for supper. You had "guests for dinner" and everybody you told about having "guests for dinner" believes that you met with some people and shared a meal together.

You have to have a distance and I provide a distance. You provide semantics and pipul, go curl up inside a gas oven.

[BADecker]

^^^ You get the same response as the moon thread... https://bitcointalk.org/index.php?topic=5150220.msg51577028#msg51577028.

Besides, you don't show us how you get the distance. Eyeballing it doesn't work, especially when the distances are great.

[notbatman]

^^^ That's why they make the Nikon P1000, while I don't have the maximum resolution limit handy I'm confident it's measured in seconds.

BTW I do show how I get the the distance, it's based on 1 minute and I show my source for that including the page number.







This means that on a plain with an eye height of 0 a ~1 foot tall object will not be visible beyond ~1/2 nautical miles thus, with an eye height of ~1 foot the distance to the horizon vanishing point will be ~1/2 nautical miles. With a ~2 foot eye height the distance to the horizon is ~1 nautical miles. At ~6 feet the distance is 3 nautical miles.

I used (eye_height + eye_maximum_resolution_angle) / 2 to calculate this. Is there a better way?



edit:

I was trying to avoid a divide by zero error for a zero input and was fatigued and constantly disrupted. It looks like the output is off by 1/2 miles and 0.5 needs to be subtracted from the answer. Help me me fix this, I shouldn't have posted this without double checking but I did...



edit 2:

I'll change it to ((eye_height + eye_maximum_resolution_angle) / 2) - (eye_maximum_resolution_angle / 2) as a temporary fix.



edit 3:

You can see how I've obtained the distance to the horizon, even if the way I've calculated it is not optimal you can see where the values come from...

[BADecker]

^^^ What you are showing is that the eye has limits. The camera has greater ability than the eye. The simple telescope has greater ability than the camera. And an extremely powerful telescope has greater ability of sight than the simple telescope.

Now be patient with me for a moment, and don't just lay into me because of the distances I am using. What if the object is a parsec away? Even the best telescope would see it way less than a parsec. The eye, camera, and simple telescope would see it essentially the same distance away, way less than the best telescope.

But if you use trig triangulation, you can find a more accurate distance than any of them... if you have two measurements.

If the object is 20 parsecs away, all standard eyeballing through any instrument we have in existence would show a tremendously nearer object than the 20 parsecs. And the eyeballing would be wrong.

Triangulation has been done on distant objects using the double earth-to-sun distance, at opposite times of the year, as the measuring method. While the distances are not absolutely accurate, they are a "million" times more accurate than eyeballing with one of the instruments... which is way, way more accurate than eyeballing directly with the eye.

Why trust triangulation based on a distance of 186 million miles (the distance of the earth in it's orbit on the exact opposite side of the sun)? Because all kinds of other measurements have been done that show that 186 million miles is near accurate. These measurements include combining several parallaxes of several planets.



Your measurements are fine and good at earth-surface distances (much of the time). But they fail at inter-planetary, inter-stellar, and inter-galactic distances. And you admit it by stating the limits that eyeballing has. Your mistake lies in trying to shrink the universe so that it will match earth-surface eyeballing distances.

[notbatman]

No, no you can use for example a Nikon P1000 with an angular resolution limit in the seconds, at ~6 feet an eye with a 1 second resolution limit can see a ~1 foot object up to ~180 nautical miles away (if the atmosphere was removed). My calculations allow for any eye to be used, I just chose the human eye because there's 1 nautical mile per 1 minute, it's the standard.

The Sun and Moon are measuring ~3,100 nautical miles away, not millions. You claim they're millions of miles away and thus a 100 million dollar telescope is needed for accuracy but you're begging the question (petitio principii).

Stop assuming the Copernican model is correct, it's cut from the same cloth as evolution...

[BADecker]

No, no you can use for example a Nikon P1000 with an angular resolution limit in the seconds, at ~6 feet an eye with a 1 second resolution limit can see a ~1 foot object up to ~180 nautical miles away (if the atmosphere was removed). My calculations allow for any eye to be used, I just chose the human eye because there's 1 nautical mile per 1 minute, it's the standard.

The Sun and Moon are measuring ~3,100 nautical miles away, not millions. You claim they're millions of miles away and thus a 100 million dollar telescope is needed for accuracy but you're begging the question (petitio principii).

Stop assuming the Copernican model is correct, it's cut from the same cloth as evolution...

Your camera and 180 miles? You have a reasonably good telescope there in that camera.

The moon isn't millions of miles away. Easy to do calculations, that you can do yourself, show that the moon is about 250,000 miles away. These calculations use trig and parallaxes... with a partner if you want to get the measurement down in one night. There is no eyeballing about it... so, no eyeballing limitations.

The sun is a little different. But you can easily look up the parallax/trig methods on the Net if you want to do it yourself. You probably won't be as accurate small-scale as the professional astronomers. But you will see that it is millions of miles... with no eyeballing limitations about it.

There aren't any assumptions in trig, and you don't have to depend on un-clarity of sight, hoping that the telescopic lens on your camera will make up the difference.

[notbatman]

@BADecker,

   Measuring with a sextant isn't fucking "eyeballing it", you're throwing out direct measurement of the Sun and Moon, pretending it's not possible then giving a lame answer from NASA that's absolute bullshit. If the they're both 32 miles wide and both 3100 miles high then a sextant and the human eye work just fine for making an approximate measurement. You have to beg the question about the distance to the Sun and Moon being millions and hundreds of thousands miles away to table your parallax bullshit.

Go crawl in the oven and shut the door behind you.

[BADecker]

@BADecker,

   Measuring with a sextant isn't fucking "eyeballing it", you're throwing out direct measurement of the Sun and Moon, pretending it's not possible then giving a lame answer from NASA that's absolute bullshit.

Go crawl in the oven and shut the door behind you.

Measuring with a sextant is half eyeballing it. The sextant angle and line of sight are not eyeball things. The assumption that the distance is an eye-accurate/measurement-accurate thing is eyeballing it.

The "flatter" thing is that you would get upset because someone can show you the flaws in your thinking. The fact that you get upset shows that you are hanging onto religious/philosophical principles regarding FE. Religious motivations are strong motivations. So, you are forgiven - even pitied - for your bad language.

[notbatman]

^^^ I'm upset because you're being intellectually dishonest.


@odolvlobo,

   I'll do up a diagram for measuring the Sun and Moon (this may take me a while), the 3100 mile distance calc was not done by me so I'll do it and see what I get. BTW don't confuse angle U and angle X, only with angle X does 1 minute = 1 nautical mile. With the Sun and Moon however atmospheric refraction keeps them the same size all the way to the horizon so X may in fact equal U for these two objects. Their distance makes an accurate measurement in this manner difficult, as BAD keeps pointing out however, an error in measurement of few miles doesn't save the globe.

[odolvlobo]

@odolvlobo,
   I'll do up a diagram for measuring the Sun and Moon (this may take me a while), the 3100 mile distance calc was not done by me so I'll do it and see what I get. BTW don't confuse angle U and angle X, only with angle X does 1 minute = 1 nautical mile.

My math was wrong, so I deleted the post.

[BADecker]

^^^ I'm upset because you're being intellectually dishonest.

(Chuckle) As long as you can't show accuracy by eyeballing, why should it bother me at all? You are the one stuck with your religion and cult.

[notbatman]

My first attempt at calculating the distance to the Sun resulted in a height of 3068.4 nautical miles. I'm still going over the calculations...

[BADecker]

^^^ Write down every particle of your calc, and once you finish it, show it to us.

[notbatman]

This is what I've got so far:

[image]

edit:

Removed image/calculation due to horrific errors, I must have mixed up my notes.

[odolvlobo]

There is an inconsistency. If Q is 59.73 degrees and E is 1.99 nm and the angle marked 90 degrees is 90 degrees, then Z can't be 3068 nm.

Z = E x tan Q = 1.99 x 1.73 = 3.4 nm


I think you are making it overly complex. If the apparent size of the sun is 32 minutes and it is 32 nm wide, then the distance to the sun can be calculated with this:

distance = width / 2 / tan(angle / 2) = 32 / 2 / tan(0.533 / 2) = 3438 nm


Now that you know the distance to the sun, you can calculate its height:

height = distance * sin(Q) = 3438 * sin(59.73) = 2969 nm

There is a problem. If the sun is always 32 nm wide, then it must always be 3438 nm away, and if it is always 3438 nm then its height changes as Q changes. In other words, all three - width, height, and distance, cannot be constant.

[BADecker]

This is what I've got so far:

The snag in your equation is the eye's limit. If you use a telescope, the limit is reduced. The increased distance changes the angles and increases the distance and diameter of the sun. Since these things can be increased with increased vision, the whole formula collapses until you use maximum vision acuity. You only get this by getting within eye-limits of the sun.

[notbatman]

Keep in mind that I'm combining angles from both physical space and optical space and the diagram is less than clear in this regard. R & T also need to be moved over to the left side.

let me go over the possible inconsistencies, I was overcome with fatigue for no particular reason and passed out for 5 hours after posting that...


edit:

There is an inconsistency. If Q is 59.73 degrees and E is 1.99 nm and the angle marked 90 degrees is 90 degrees, then Z can't be 3068 nm.

Z = E x tan Q = 1.99 x 1.73 = 3.4 nm


I think you are making it overly complex. If the apparent size of the sun is 32 minutes and it is 32 nm wide, then the distance to the sun can be calculated with this:

distance = width / 2 / tan(angle / 2) = 32 / 2 / tan(0.533 / 2) = 3438 nm


Now that you know the distance to the sun, you can calculate its height:

height = distance * sin(Q) = 3438 * sin(59.73) = 2969 nm

There is a problem. If the sun is always 32 nm wide, then it must always be 3438 nm away, and if it is always 3438 nm then its height changes as Q changes. In other words, all three - width, height, and distance, cannot be constant.

First point, the distance to the sun can't be calculated from the sun's apparent size alone. As you can see I didn't use "I" or "U" at all in my calculation at all, only "Q" and "A".

Second point, like I mentioned there are two spaces combined into a single diagram/calculation. The left side is mostly optical space for calculating physical distance (X=angular size) and the right is mostly physical space for calculating optical angles (U=apparent size).

I hope my second point makes sense.

The square angle triangle formed by E-Z-O is apparent and Z can't be calculated on this side. The square angle triangle formed by D-Z-[unlabelled leg between sun and horizon] is physical and the distance is calculated on this side. The square angle triangle [unlabelled leg between horizon and the ground below the sun]-V-D is where the physical angle to the sun is calculated. So the left side is actually two triangles with P = T.

If that makes sense.

Finally the sun is always 32 minutes wide due to refraction and maintains the same diameter as if it was at 90° because of it. As you can see the calculations for the angular size (X) have not been completed yet. In the case of the sun and moon here X will equal U and refraction plays a role such that a refracted optical space is created in addition to the physical and optical spaces.

If that makes sense you get an award!

Overly complex, no, no not exactly...



@BADecker,

  I could be wrong but I think the aspect ratio is maintained when changing the resolution limit.




edit:
I updated the image to include the calculation for Z, I didn't realize it was missing. I also made a few other adjustments.

[odolvlobo]

Keep in mind that I'm combining angles from both physical space and optical space and the diagram is less than clear in this regard. R & T also need to be moved over to the left side.

First point, the distance to the sun can't be calculated from the sun's apparent size alone. As you can see I didn't use "I" or "U" at all in my calculation at all, only "Q" and "A".

You can if you know the effect of refraction. Do you know how to compute that?

...
Finally the sun is always 32 minutes wide due to refraction and maintains the same diameter as if it was at 90° because of it. As you can see the calculations for the angular size (X) have not been completed yet. In the case of the sun and moon here X will equal U and refraction plays a role such that a refracted optical space is created in addition to the physical and optical spaces.

I see. The sun appears to have the same size because of refraction. That raises a question for me. Why are the sun and moon affected by refraction in this way, but other objects are not?

[BADecker]

Finally the sun is always 32 minutes wide due to refraction and maintains the same diameter as if it was at 90° because of it.

Even if the sun is always 32 minutes wide, using a telescope directed at the sun shows far greater perspective than using the eyes alone. You have agreed with this in former posts.

This shows that neither the eyes or the telescope are necessarily accurate. A different method is necessary, one that determines either the distance or the size of the sun. It has to be far more accurate than the eyes or even the telescope. This is what modern astronomy has accomplished. It is also why your calc fails.