 #### WellRoundedIndividual

• • 605
• Proverbs 13:20 is extremely relevant today. ##### Re: the ISS light in the sky is fake, right?
« Reply #20 on: January 17, 2019, 01:00:54 PM »
Go look up the diurnal parallax. Its basic geometry. Wow. Do you not see two right angle triangles in that diagram?

The angle that forms between the hypotenuse and the opposite sides of the triangle is known - the angle at which you are viewing from the telescope. You can then move your apparatus to achieve the angle again from a distance moved one way or another. You then measure that distance. Divide by two. You know have a known angle and a distance with which you can use  the basic SOHCAHTOA.
« Last Edit: January 17, 2019, 01:08:37 PM by WellRoundedIndividual »
BobLawBlah. #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #21 on: January 17, 2019, 01:13:41 PM »
Why would you post an isosceles triangle in a right triangle experiment?
Wht are you talking about the type of triangle when that isn't relevant to the point I'm explaining.
You could do the experiment when the ISS is at any point whether that point is between you or directly above one of you. It doesn't change my point.
This is like when you deliberately tried to put red herrings into the FE Map thread by talking about screen resolutions. I'm still waiting for the results of your tests about that...

Tom's point was the experiment pre-supposes a flat earth.
As I have explained, the earth is big. So if you're taking observations 100km apart then while yes, the curvature of the earth does introduce some error if you haven't accounted for it but the error is actually quite small. It doesn't invalidate the experiment completely if you don't but it does make it more accurate if you do.
If you are making your claim without evidence then we can discard it without evidence.

#### totallackey

• • 982 ##### Re: the ISS light in the sky is fake, right?
« Reply #22 on: January 17, 2019, 01:17:39 PM »
Go look up the diurnal parallax. Its basic geometry. Wow. Do you not see two right angle triangles in that diagram?
If the center line serves as a barrier, then yes.
The angle that forms between the hypotenuse and the opposite sides of the triangle is known - the angle at which you are viewing from the telescope. You can then move your apparatus to achieve the angle again from a distance moved one way or another. You then measure that distance. Divide by two. You know have a known angle and a distance with which you can use  the basic SOHCAHTOA.
Surveyors do not need to move their view in determining altitude of an object.

Determine the distance between you and the object having an unknown altitude. Find an object of known height, placing it in such a way as the top of the object obscures the view of the object being measured.

Do the ratios and you have the altitude of the unknown object.
« Last Edit: January 17, 2019, 01:25:17 PM by totallackey » #### WellRoundedIndividual

• • 605
• Proverbs 13:20 is extremely relevant today. ##### Re: the ISS light in the sky is fake, right?
« Reply #23 on: January 17, 2019, 01:19:57 PM »
Correct, but you are shifting context. We aren't surveyors in this example. We are referring to diurnal parallax. A method to calculate distance to far off object using geometry and trigonometry.

Side note: Are you drunk? What is "center is line..."
BobLawBlah.

#### totallackey

• • 982 ##### Re: the ISS light in the sky is fake, right?
« Reply #24 on: January 17, 2019, 01:24:02 PM »
Why would you post an isosceles triangle in a right triangle experiment?
Wht are you talking about the type of triangle when that isn't relevant to the point I'm explaining.
You could do the experiment when the ISS is at any point whether that point is between you or directly above one of you. It doesn't change my point.
This is like when you deliberately tried to put red herrings into the FE Map thread by talking about screen resolutions. I'm still waiting for the results of your tests about that...

Tom's point was the experiment pre-supposes a flat earth.
As I have explained, the earth is big. So if you're taking observations 100km apart then while yes, the curvature of the earth does introduce some error if you haven't accounted for it but the error is actually quite small. It doesn't invalidate the experiment completely if you don't but it does make it more accurate if you do.
You don't need your isosceles setup is the point.

All you need is the distance as measured on the ground to a point directly under the base of the object. And an object of known height. Position the object of known height so that it obscures all but the very top of the object having unknown height. Do the ratios and you have the altitude of the unknown object.

Still struggling with the screen resolution and monitor settings uh...still unable to comprehend how different resolutions render pixelation.

We will leave that to that thread though... #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #25 on: January 17, 2019, 01:33:34 PM »
All you need is the distance as measured on the ground to a point directly under the base of the object. And an object of known height. Position the object of known height so that it obscures all but the very top of the object having unknown height. Do the ratios and you have the altitude of the unknown object.

Correct. You can do the experiment in different ways. But all of those ways assume a flat earth, otherwise the base isn't a straight line and it isn't really a triangle.
My point was, and remains, that over distances of around 100km this doesn't actually introduce that much error into the method. But accounting for it will get you a more accurate result.
If you are making your claim without evidence then we can discard it without evidence.

#### totallackey

• • 982 ##### Re: the ISS light in the sky is fake, right?
« Reply #26 on: January 17, 2019, 04:40:09 PM »
All you need is the distance as measured on the ground to a point directly under the base of the object. And an object of known height. Position the object of known height so that it obscures all but the very top of the object having unknown height. Do the ratios and you have the altitude of the unknown object.

Correct. You can do the experiment in different ways. But all of those ways assume a flat earth, otherwise the base isn't a straight line and it isn't really a triangle.
My point was, and remains, that over distances of around 100km this doesn't actually introduce that much error into the method. But accounting for it will get you a more accurate result.
You do not assume anything when it is looking exactly the way it is to your own eyes. #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #27 on: January 17, 2019, 06:27:35 PM »
The base of the triangle is the earth. To do the calculations you are using the maths of a triangle. That maths only works if the base is flat. If the earth isn’t flat then you have to use different maths.
But, and I don’t know how else to explain this, over fairly short distances like 100km (which is less than 1% of the earth’s circumference) the error is fairly small. Ergo it doesn’t invalidate the experiment outlined in the OP.
If you are making your claim without evidence then we can discard it without evidence. #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #28 on: January 17, 2019, 06:34:03 PM »
I have explained it. I haven’t done the maths for you. Are you not able to maths? If not I will have a go for you later.
If you are making your claim without evidence then we can discard it without evidence. #### Tom Bishop

• Zetetic Council Member
•  • 6422
• Flat Earth Believer ##### Re: the ISS light in the sky is fake, right?
« Reply #29 on: January 17, 2019, 06:35:22 PM »
I have explained it. I haven’t done the maths for you. Are you not able to maths? If not I will have a go for you later.

If you can't understand it and the error levels, the difference between a 100 mile distant object and a 200 mile distant object, then you are in no position to tell us whether it is an accurate distance or not.
« Last Edit: January 17, 2019, 06:37:08 PM by Tom Bishop »

#### Jimmy McGill ##### Re: the ISS light in the sky is fake, right?
« Reply #30 on: January 17, 2019, 06:57:59 PM »
I have explained it. I haven’t done the maths for you. Are you not able to maths? If not I will have a go for you later.

If you can't understand it and the error levels, the difference between a 100 mile distant object and a 200 mile distant object, then you are in no position to tell us whether it is an accurate distance or not.

Tom I think it’s clear that everyone here, except perhaps you at the beginning of the conversation, understands the level of error in assuming a completely flat base for this calculation.
100km of curvature is negliblible in this experiment.
It’s an accurate distance. It’s the equivalent of going to the back yard and eyeballing a piece of lumber instead of getting it down to the exact angle you need down to the millions of an inch. #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #31 on: January 17, 2019, 07:56:52 PM »
I made a bunch of assumptions but if you’re 100km apart and the ISS was right overhead one person and the other person got an angle of 76 degrees then if you assume the earth is flat you’d get a height of 400km.

If the earth is actually a sphere and you were 100km apart and 1 degree round the circle (actually you’d be less on the earth, but for simplicity) then the radius of that globe earth would be 5729km.

The chord length (given by 2r sin A/2 - the angle here is 1 degree) is 92.13

So that’s the distance between the two people in a straight line through the curve of the earth.

The angle of the other two sides of the triangle formed by the centre of the earth and the two points on the earth where the angle at the centre is 1 degree is 89.3 degrees.

So that means the reading of 76 is 0.7 degrees off from the straight line chord through the circle.

So we now have a triangle where the base is 92.13 (the chord length) and the angles are actually 90.7 degrees and 76.7 degrees (adding the 0.7 degree correction).

That gives an actual height of 411.01km.

I may have got some of this wrong. Apologies for the lack of diagrams. Will try and add those if none of you know what I’m going on about!
« Last Edit: January 17, 2019, 08:08:17 PM by AllAroundTheWorld »
If you are making your claim without evidence then we can discard it without evidence. #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #32 on: January 18, 2019, 02:20:21 PM »
Right. Made a right balls up of the above, I've had another go...

So, A and B are 100km apart. The ISS is directly above A, B measures an angle of 76 degrees from the ground to the ISS. They assume that the ground between them is flat which forms a triangle. A-B is the base, 100km. The side of the triangle is going straight up from A to the ISS so it's a right angled triangle. The internal angle of the triangle A-B-ISS is 76 degrees which makes the other internal angle 14 degrees. Putting the known values into a triangle calculator you get a height of 401km Actually though, A and B are on a sphere and although they are 100km apart they are 1 degree apart on the circumference of the sphere. So the total circumference is 36000km
(360 degrees in a circle, 100km is 1 degree)

C = 2 pi r
so r = C / 2 pi
36000 / 2 pi = 5729.58km Chord length is given by 2r sin (a/2) - where r is the radus of the circle, a is the angle at the centre of the circle (angle ACB above). That is 1 degree, so:

2 x 5729.58 x sin(0.5) = 99.998

So this is the straight line length between A and B through the curve of the circle.
The triangle ABC is isosoles so if the angle ACB is 1 degree then CAB and CBA must both be 89.5 degrees. That means there is 0.5 degrees between the chord and the tangent to the circle at both ends which we must adjust for. So:

A observes the ISS directly above him but adding the adjustment makes the angle
ISS-A-B 90.5 degrees

B measures an angle of 76 degrees, adding the adjustment that makes the angle
A-B-ISS 76.5 degrees.

The straight line AB is 99.998 so now we cancalculate the other sides of the triangle. So...I'll be honest, this isn't what I was expectint but now the calculated height is 432km. I'll admit this is a much bigger error than I was expecting.

Have I done something wrong or is this just how it is?
If you are making your claim without evidence then we can discard it without evidence.

#### ChrisTP

• • 359 ##### Re: the ISS light in the sky is fake, right?
« Reply #33 on: January 18, 2019, 03:11:34 PM »
100km of distance in curvature is extremely negligible in this case, I mean here's a triangle using 3 points, it was a right angle until I moved the point on the bottom right down precisely to account for curvature and as you can see in this case it's only 2 pixels down in my image, you might end up with a difference of ~.0001 degrees in that angle if you don't account for curvature, I personally don't see the need unless you want to be absolutely perfect.. Tom is wrong most of the time. Hardly big news, don't you think? #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #34 on: January 18, 2019, 03:20:09 PM »
That was my gut instinct too but when I did the maths it did end up making more of a difference to the height calculation than I had anticipated.
Not orders of magnitude difference, it's not like a 93,000,000 mile away sun is suddenly 3,000 miles but it's about an 8% error
Unless I've got my maths or reasoning wrong which is entirely possible.
If you are making your claim without evidence then we can discard it without evidence.

#### ChrisTP

• • 359 ##### Re: the ISS light in the sky is fake, right?
« Reply #35 on: January 18, 2019, 03:26:58 PM »
Well accounting for the curvature and assuming that section of earth is a straight line I get these measurements in a triangle for the angles and distances EDIT I should mention that this is with the assumption that the ISS is 408 km up and the two other points are 100km a part, but I only moved the point directly down for the curvature rather than in tangent to the surface of the earth, so I guess it was a pointless diagram to show as the numbers would still be off by a tiny, tiny fraction. It's hard to determine to distance with such a small degree tbh, it would be better to try with a decent 400km across the earth rather than 100km
« Last Edit: January 18, 2019, 03:31:50 PM by ChrisTP »
Tom is wrong most of the time. Hardly big news, don't you think? #### AllAroundTheWorld

• • 2257 ##### Re: the ISS light in the sky is fake, right?
« Reply #36 on: January 18, 2019, 03:31:43 PM »
I went into a bit more detail, I assumed that the person got a measurement of 76 degrees from 100km and then did the maths as though that 100km was on a flat earth from the person who saw the ISS directly overhead. I then did the maths for what the height actually was given the same measurement if you accounted for a curve of 1 degree.
I was expecting the discrepancy to be quite small but it was actually much larger than I'd imagined.
I've explained my reasoning above, I may well have made a mistake somewhere (I did the first time!)
If you are making your claim without evidence then we can discard it without evidence.

#### ChrisTP

• • 359 ##### Re: the ISS light in the sky is fake, right?
« Reply #37 on: January 18, 2019, 03:36:37 PM »
I went into a bit more detail, I assumed that the person got a measurement of 76 degrees from 100km and then did the maths as though that 100km was on a flat earth from the person who saw the ISS directly overhead. I then did the maths for what the height actually was given the same measurement if you accounted for a curve of 1 degree.
I was expecting the discrepancy to be quite small but it was actually much larger than I'd imagined.
I've explained my reasoning above, I may well have made a mistake somewhere (I did the first time!)
Well without moving the point down for curvature the angle was still 76.228 degrees making a right angle triangle. Less than a degree out for that. accounting for the tangent would be half the amount anyway so 0.212
degrees difference. Sorry I can't be more precise this was just a quick lunch break exercise.
« Last Edit: January 18, 2019, 03:45:43 PM by ChrisTP »
Tom is wrong most of the time. Hardly big news, don't you think?

#### Nosmo

• • 2 ##### Re: the ISS light in the sky is fake, right?
« Reply #38 on: January 18, 2019, 10:11:22 PM »

So...I'll be honest, this isn't what I was expectint but now the calculated height is 432km. I'll admit this is a much bigger error than I was expecting.

Have I done something wrong or is this just how it is?

I think that you may be seeing a bigger discrepancy than you were expecting as there are two factors contributing to the discrepancy.
In the first instance you are taking the measurements and doing the math assuming that the Earth is flat.
In the second instance you are taking the measurements and doing the math assuming that the Earth is round.

If you want to see just what the difference is in using the arc length or the chord length for the base you need to calculate a third distance.
This would use the triangle that you used in the second example but use a side C length of 100 rather than 99.998.
(Angle C wont change between using arc or chord).

The rest of the discrepancy is then down to the different models.

#### sanshin66

• • 1 ##### Re: the ISS light in the sky is fake, right?
« Reply #39 on: February 06, 2019, 02:43:20 PM »
Hi, i'm uneducated about math and geometry, and I am wondering how you are able to calculate the angle in degrees, and wouldn't everything you look at in the sky come up with the same distance using this method?
First parallax is measured in relation to the background, the angle is measured in degrees, then divided by 2?
And the angle is solved with tan = o/a ?

How do you get distance from parallax, in relation to background into degrees and how can you solve TOA without OA