What would it take to fake the ISS?

Personal note:
Linking this to keep track: https://www.theflatearthsociety.org/forum/index.php?topic=78484.msg2118105#msg2118105
As far as I can tell, barring startling discoveries I think this proves the ISS, if fake, must be above 85km. Likely a fair way above, but that's the threshold.

In lieu of getting more data on dynamic viscosity, another angle to examine would be mass. Thanks to that height we have a lower bound for the size of the ISS, and thanks to the lower air density a limited number of means of propulsion. If we can determine the lightest material it could be made from, using an estimate of volume from the length and measurements of images captured from Earth, we can gauge what would be required to make such an object stay aloft when it cannot orbit.

There's also models that change things up, to mention, the most notable being dome models where the ISS can be said to be a projection. Could be fun working out the energy that'd require.

Quote from: NotSoSkeptical on November 24, 2018, 05:43:48 PM

Why don't you use the Reynolds Calculator here:  https://www.grc.nasa.gov/www/k-12/airplane/reynolds.html

The phrase 'euqal to the dynamic viscosity' for one, the whole page feels a bit low-grade. I did come across that, but when I tried to confirm values via direct calculation the numbers came out off; I think that basically does just what my formula does, but with less data points as they clearly didn't spend too much effort on the rest of the page. I'm instinctively wary of any page that starts off with obvious MS paint. As an easier approximation though, sure, probably should've mentioned it.

My biggest concern though is lack of knowledge; it doesn't mention the points to which it's accurate, which does limit its usefulness. I can easily tell something goes wrong because if I plug in a high altitude and low speed it gives a low number, indicating higher drag from what I've seen, so...

I don't think a lower number necessarily equates to higher drag.

While I'm not a fan of wikipedia, it provides some interesting information on it: https://en.wikipedia.org/wiki/Reynolds_number

Quote from: Jane on November 24, 2018, 05:54:03 PM

Quote from: NotSoSkeptical on November 24, 2018, 05:43:48 PM

Why don't you use the Reynolds Calculator here:  https://www.grc.nasa.gov/www/k-12/airplane/reynolds.html

The phrase 'euqal to the dynamic viscosity' for one, the whole page feels a bit low-grade. I did come across that, but when I tried to confirm values via direct calculation the numbers came out off; I think that basically does just what my formula does, but with less data points as they clearly didn't spend too much effort on the rest of the page. I'm instinctively wary of any page that starts off with obvious MS paint. As an easier approximation though, sure, probably should've mentioned it.

My biggest concern though is lack of knowledge; it doesn't mention the points to which it's accurate, which does limit its usefulness. I can easily tell something goes wrong because if I plug in a high altitude and low speed it gives a low number, indicating higher drag from what I've seen, so...

I don't think a lower number necessarily equates to higher drag.

While I'm not a fan of wikipedia, it provides some interesting information on it: https://en.wikipedia.org/wiki/Reynolds_number

Go by: https://en.wikipedia.org/wiki/Drag_coefficient#Background

My understanding is that for more aerodynamic shapes, the drag coefficient basically only depends on that number, and going by the link I found (and its claim that high kinematic viscosity=high drag and vice versa was corroborated by other resources) that's the direction it goes. There likely are situations where the correllation isn't so clear-cut, but we dismiss them because we want to use the most aerodynamic shape possible.

Quote from: NotSoSkeptical on November 24, 2018, 06:29:10 PM

Quote from: Jane on November 24, 2018, 05:54:03 PM

Quote from: NotSoSkeptical on November 24, 2018, 05:43:48 PM

Why don't you use the Reynolds Calculator here:  https://www.grc.nasa.gov/www/k-12/airplane/reynolds.html

The phrase 'euqal to the dynamic viscosity' for one, the whole page feels a bit low-grade. I did come across that, but when I tried to confirm values via direct calculation the numbers came out off; I think that basically does just what my formula does, but with less data points as they clearly didn't spend too much effort on the rest of the page. I'm instinctively wary of any page that starts off with obvious MS paint. As an easier approximation though, sure, probably should've mentioned it.

My biggest concern though is lack of knowledge; it doesn't mention the points to which it's accurate, which does limit its usefulness. I can easily tell something goes wrong because if I plug in a high altitude and low speed it gives a low number, indicating higher drag from what I've seen, so...

I don't think a lower number necessarily equates to higher drag.

While I'm not a fan of wikipedia, it provides some interesting information on it: https://en.wikipedia.org/wiki/Reynolds_number

Go by: https://en.wikipedia.org/wiki/Drag_coefficient#Background

My understanding is that for more aerodynamic shapes, the drag coefficient basically only depends on that number, and going by the link I found (and its claim that high kinematic viscosity=high drag and vice versa was corroborated by other resources) that's the direction it goes. There likely are situations where the correllation isn't so clear-cut, but we dismiss them because we want to use the most aerodynamic shape possible.

If I'm reading that correctly, a lower number means less drag.

For a streamlined body to achieve a low drag coefficient, the boundary layer around the body must remain attached to the surface of the body for as long as possible, causing the wake to be narrow. A high form drag results in a broad wake. The boundary layer will transition from laminar to turbulent if Reynolds number of the flow around the body is sufficiently great. Larger velocities, larger objects, and lower viscosities contribute to larger Reynolds numbers.[9]

A low drag coefficient is low drag.  Low drag allows for a narrow wake which allows for laminar flow which is principle in low Reynolds Numbers.


Additionally:

The Reynolds number will be low for small objects, low velocities, and high viscosity fluids.

All of those have low drag.

If I'm reading that correctly, a lower number means less drag.

Only if it's an increasing formula.

Larger velocities, larger objects, and lower viscosities contribute to larger Reynolds numbers.[9]

This is the important part. If you spend much time google aerodynamic flow you find it gets a little non-intuitive. "There might be a few cases where viscous drag increases with Reynolds number (like laminar airfoils which lose their laminar bucket when the Reynolds number goes up), but in general the observation is true."
If we focus on the best case scenario, which is what we need to be doing here, we can assume laminar flows regardless.

My understanding is that there's no one value where the flow transitions independently of any other concerns, that's why aerodynamics can be such a pain to work with, it seems to almost all be determined by a case by case basis. This is never going to be a 100% accurate description when we're trying to keep as general as possible, so we don't try for one, we just try for a bound. The most hypothetically ideal aerodynamic shape with a perpetually laminar flow that never shifts from that. All that we really need is a ballpark figure of what works to compare with.

A low drag coefficient is low drag.  Low drag allows for a narrow wake which allows for laminar flow which is principle in low Reynolds Numbers.

Additionally:

Quote

The Reynolds number will be low for small objects, low velocities, and high viscosity fluids.

All of those have low drag.

At hypersonic speeds, the Reynolds number starts to lose its significance and many othe factors come into play.

By 85,000 m altitude, you are above the limit where aerodynamic lift can support much of the weight of an aircraft. 
Before the 100 km was decided as the Kármán line there was much discussion on whether it should be at 80 km or 100 km.

Here are some relevant references:
         Kármán line.
         The edge of space: Revisiting the Karman Line.
and

... flew 2000 miles per hour (3,200 km/h) at 126,000 feet (38,500 m), or 24 miles up. At this altitude and speed aerodynamic lift still carries 98 per cent of the weight of the plane, and only two per cent is carried by centrifugal force, or Kepler Force, as the space scientists call it.

But at 300,000 feet (91,440 m) or 57 miles up this relationship is reversed because there is no longer any air to contribute lift. Only centrifugal force prevails. This is certainly a physical boundary ...

(Quote from Von Kármán's autobiography, from Wikipedia)

And on a flat earth there is no "centrifugal force, or Kepler Force".

Have fun!

My biggest concern though is lack of knowledge; it doesn't mention the points to which it's accurate, which does limit its usefulness. I can easily tell something goes wrong because if I plug in a high altitude and low speed it gives a low number, indicating higher drag from what I've seen, so...

My understanding is that the Reynolds number is the velocity times a length (which can be the square root of an area of importance for drag) divided by the viscosity.
Likewise the drag force is proportional to the velocity squared times an area, as well as the density of the fluid and the drag coefficient. Assuming we have a newtonian fluid, then that means it is proportional to the Reynolds number squared times the drag coefficient.

That makes me assume high Reynolds number should be high drag.

The question then is how does the drag coefficient change with velocity (or Reynolds number).

From what I can quickly find, it seems that for low Re, there is an inverse linear relationship, meaning you end up with roughly drag is proportional to the Reynolds number.
But then at high Re, you end up with it much higher than that expected trend, with it flattening off and having a transition, as it moves to turbulent flow.
This results in the drag being much closer to just proportional to the reynolds number squared.

So that makes me think that while the correlation isn't perfect, high Re=high drag.

The time this doesn't hold is when you switch between fluids.
Fluids with low viscosities have a higher Re, but a lower drag.
The leftover parts of the equation is density/kinematic viscosity^2 = 1/(density*dynamic viscosity).
So the question then becomes how does changing this effect it.


My other concern is how well this holds in the supersonic regime and above, and how well it holds in very low density environments.

Also, should this be here or in the science and tech section?

Things to do
If we can work out the distance along the ground the ISS appears to take, then on a flat Earth that easily tells us how much distance it covers ..

Calculate a minimum altitude, and from that an estimate as to mass to work out what forces (at a basic level) would be needed.

Relatively easy to do, assuming you have some rudimentary observing equipment.

Some web sites (like heavens-above.com) give accurate data on the predicted track of the ISS at a location. For this exercise, it is immaterial how the calculations are made, and what model is used to generate the data. The only thing that matters is that the resultant data reflects what actually happens.

Verify for yourself that the data is roughly correct. That is, generate an ISS pass prediction, and observe with telescope or binoculars on a mount with azimuth and elevation circles that 1) the ISS reaches the same maximum elevation angle as the prediction within about a degree and 2) the time at which it does so agrees with the prediction within about a second.

Do this as many times as it takes to convince yourself that the data is correct. Depending on how paranoid you're feeling, make predictions for different locations (at least 10 miles away) and repeat. Have trusted friends within 50 miles or so perform the same measurement, if you have any friends you trust.

Choose a prediction where the ISS will be going directly overhead. Station a trusted friend 50-60 miles along the predicted track. (I assume you can trust car odometers to be able to tell when someone is roughly 50-60 miles away so you don't have to use Google Maps.) Record the time at which the ISS goes directly over your location. Have your friend record the same, or be on the phone with him at the same time (that way you can verify you're both seeing the same bright object traveling the same direction). 50-60 miles divided by time gives ground speed. Don't worry, the difference between flat earth and round earth will be a small error in this calculation.

At another time, choose a prediction where the ISS goes directly over your location. Station a trusted friend 50-60 miles perpendicular to the predicted track. You verify that the path takes it directly overhead (90 degrees); your friend measures the maximum elevation angle. Using basic geometry, calculate the altitude. Again, flat versus round makes only a small difference.

Since I have been using predictors to spot and track the ISS for decades, and the results have always turned out as predicted, I trust them and can use their data to make a preliminary calculation of speed and altitude.

Ground speed: About 5 miles per second
Altitude: (Flat Earth) 264 miles (Round Earth) 245 miles

If you don't have friends to help you, choose any predicted pass to make these calculations, and spot verify with observations from one location or the other. Repeat as often as necessary with different path locations and observer locations to convince yourself that no one can be "faking" the predictions.

Even if you want to put unnecessarily huge error bars on these observations (a situation that can be remedied by statistics of multiple observations), you're still left with an object with a speed in excess of a mile per second and an altitude in excess of 100 miles.

That what you're looking for?

A lot.

First all the countries in the world that are involved with it would have to cover it up.  And all those people working on the ground would have to swear to secrecy that it didn't exist.  There are thousands of people around the world who would have to comply with this never-tell policy.  And from my experience, you cannot keep a secret that big for a long time.

Second...  you would have to fake out all the journalists who would is looking to make themselves famous. 

Once you get past those things, which is practically impossible, you would need to have a bunch of good video like hollywood style or even more so that would show the astronauts in the space station. 

Finally, you would need to have the congress and all the law makers around the world agree to cover it up.  There is lots of money going to the space program and the space station.  All the time these programs are being challenged by people to shut them down.  Why would they fight for it if it was not real? 

Just my 2 cents on it.

And from my experience, you cannot keep a secret that big for a long time.

I am intrigued by your back story.

Second...  you would have to fake out all the journalists who would is looking to make themselves famous. 

Once you get past those things, which is practically impossible, you would need to have a bunch of good video like hollywood style or even more so that would show the astronauts in the space station. 

More focused on the practicalities for now, the rest gets way too nebulous to both with. I put my mass calculations aside for one day and now I can't find them, so hey, that's fun. And forgot about this thread. I do that a lot.

Finally, you would need to have the congress and all the law makers around the world agree to cover it up.  There is lots of money going to the space program and the space station.  All the time these programs are being challenged by people to shut them down.  Why would they fight for it if it was not real?

Basic conspiracy answer, they still get the money, so long as faking is cheaper than the reality...