Are Hardcore Conspiracy Theorists Retarded?

[mcveteran81]

Some *perspective* examples from a cursory google search by image.

Btw I don't care if it's a circle or flat.

I'm not sure how much truth is in this next one, but I'll post it anyway. Would like to see others reasoning on it;

 

[rj144]

I was speaking of Newton's law of universal gravitation. Not sure where I got third law from when I wrote that earlier.

This article talks about it:

https://www.google.com/amp/s/amp.th...3/oct/13/newtons-universal-law-of-gravitation

"It encapsulates the idea that all the particles of matter in the universe attract each other through the force of gravity – Newton's law tells us how strong that attraction is. The equation says that the force (F) between two objects is proportional to the product of their masses (m1 and m2), divided by the square of the distance between them. The remaining term in the equation, G, is the gravitational constant, which has to be measured by experiment and, as of 2007, US scientists have measured it at 6.693 × 10−11cubic metres per kilogram second squared."

I was just wondering aloud about how a model could mimic the earth's gravitational pull up to a certain point before the earth's pull overcomes it.

More or less ramblings. But, basically, a hemisphere 200 miles in diameter would feel earth-like up to a certain distance and it would be interesting how It's characteristics mimic earth up until fail point .

If's fine. What do you mean by "feel earth-like" and "up until fail point"? Gravity-wise, curvature-wise...?

 

[splendica]

If's fine. What do you mean by "feel earth-like" and "up until fail point"? Gravity-wise, curvature-wise...?

Maybe both. Just ramblings.

 

[rj144]

Maybe both. Just ramblings.

Curvature wise, the actual equation for how much the earth curves is R - (R^2-x^2)^(1/2) where x is the distance you travel and R is the radius of the sphere. However, since the radius of the Earth is so large this is approximately equal to x^2/(2*R). So, if x = 1 mi and the radius of Earth is 3959 mi, you get (1 mi)^2/(3959 mi)*(5280 feet/mi)*(12 inches/feet) = 8.8 inches about.

On a 200 mile Earth, we need the full equation. So, for 1 mile, we would have (200 mi) - ((200 mi^2)- (1 mi)^2)^(1/2)*(5280 feet/mi) = 13.20 feet drop for every mile.

For gravity, is the mass of the Earth the same density or same mass?

 

[splendica]

Curvature wise, the actual equation for how much the earth curves is R - (R^2-x^2)^(1/2) where x is the distance you travel and R is the radius of the sphere. However, since the radius of the Earth is so large this is approximately equal to x^2/(2*R). So, if x = 1 mi and the radius of Earth is 3959 mi, you get (1 mi)^2/(3959 mi)*(5280 feet/mi)*(12 inches/feet) = 8.8 inches about.

On a 200 mile Earth, we need the full equation. So, for 1 mile, we would have (200 mi) - ((200 mi^2)- (1 mi)^2)^(1/2)*(5280 feet/mi) = 13.20 feet drop for every mile.

For gravity, is the mass of the Earth the same density or same mass?

I'd suggest density, not mass. But it would be interesting calculating for both.

 

[rj144]

I'd suggest density, not mass. But it would be interesting calculating for both.

So, density = p = m/V. Thus, m = p V = p 4/3 pi R^3 and the gravitational acceleration on earth is g = G m/R^2 = G (p 4/3 pi R^3)/R^2 = G p 4/3 pi R. Thus, if the density is the same, it scales as R, gravity is 19.79 times less on this planet compared to our Earth.

If mass is the same, from g = G m/R^2, the gravity would be (19.79)^2 = 391.64 times greater than our Earth.

 

[splendica]

So, density = p = m/V. Thus, m = p V = p 4/3 pi R^3 and the gravitational acceleration on earth is g = G m/R^2 = G (p 4/3 pi R^3)/R^2 = G p 4/3 pi R. Thus, if the density is the same, it scales as R, gravity is 19.79 times less on this planet compared to our Earth.

If mass is the same, from g = G m/R^2, the gravity would be (19.79)^2 = 391.64 times greater than our Earth.

Interesting, that's what I was roughly considering in my initial ramblings. The gravity based on density might be enough to overcome the earth's if there was a distance between the sphere and the earth. And in the second case there definitely would be enough, but maybe too much to be functional.

Given that 13 feet per mile is the curvature, I think perception would be similar to life on earth (ie. not perceiving things upside down).

Thanks for the efforts there, I was too lazy and occupied.

 

[rj144]

Interesting, that's what I was roughly considering in my initial ramblings. The gravity based on density might be enough to overcome the earth's if there was a distance between the sphere and the earth. And in the second case there definitely would be enough, but maybe too much to be functional.

Given that 13 feet per mile is the curvature, I think perception would be similar to life on earth (ie. not perceiving things upside down).

Thanks for the efforts there, I was too lazy and occupied.

Well to not be pulled off from the Earth, the center of the Earth would need to be (19.79)^(1/2)*200 mi = 889 miles or further away from the surface of the sphere.

 

[splendica]

Well to not be pulled off from the Earth, the center of the Earth would need to be (19.79)^(1/2)*200 mi = 889 miles or further away from the surface of the sphere.

Ah, well done. So the hypothetical sphere can't work within our atmosphere except by mass. And in that case there could be an adjustment so that It's closer to that of the earth's. Cool stuff.

 

[Deleted member 110072]

Again the depth of the ridiculousness of this theory is that its FAR beyond simply calling a few NASA photographs fake, the earth being spherical is tied to a vast amount of scientific observation going back thousands of years.

 

[facePuncher7]

Curvature wise, the actual equation for how much the earth curves is R - (R^2-x^2)^(1/2) where x is the distance you travel and R is the radius of the sphere. However, since the radius of the Earth is so large this is approximately equal to x^2/(2*R). So, if x = 1 mi and the radius of Earth is 3959 mi, you get (1 mi)^2/(3959 mi)*(5280 feet/mi)*(12 inches/feet) = 8.8 inches about.

On a 200 mile Earth, we need the full equation. So, for 1 mile, we would have (200 mi) - ((200 mi^2)- (1 mi)^2)^(1/2)*(5280 feet/mi) = 13.20 feet drop for every mile.

For gravity, is the mass of the Earth the same density or same mass?

Do you know the arc:cord ratio off the top of your head?

*Edit* appreciate your posts btw, good information.

 

[rj144]

Do you know the arc:cord ratio off the top of your head?

*Edit* appreciate your posts btw, good information.

No, I figured it out.

 

[facePuncher7]

No, I figured it out.

All g.

I haven't read a majority of this thread. Do you work with spatial data?

 

[rj144]

All g.

I haven't read a majority of this thread. Do you work with spatial data?

No, I don't.

 

[splendica]

Some *perspective* examples from a cursory google search by image.

Btw I don't care if it's a circle or flat.

I'm not sure how much truth is in this next one, but I'll post it anyway. Would like to see others reasoning on it;

I picture a marble rolling the distance of a gymnasium floor. Your face is against the floor. You see the marble roll til it's a speck across the room. But you never see the marble appear to disappear below the hardwood.

With water equalizing it's level, the sun should never look to sink past the level. It should only become smaller and smaller, and possibly moving to the right or left, but, eventually it should disappear due to the concentration of gasses and It's increasing tininess and direction of movement to the right or left.

 

[rj144]

I picture a marble rolling the distance of a gymnasium floor. Your face is against the floor. You see the marble roll til it's a speck across the room. But you never see the marble appear to disappear below the hardwood.

With water equalizing it's level, the sun should never look to sink past the level. It should only become smaller and smaller, and possibly moving to the right or left, but, eventually it should consistently disappear because of gasses and distance.

Exactly... it's completely ridiculous if they're not trolling.

 

[facePuncher7]

I picture a marble rolling the distance of a gymnasium floor. Your face is against the floor. You see the marble roll til it's a speck across the room. But you never see the marble appear to disappear below the hardwood.

With water equalizing it's level, the sun should never look to sink past the level. It should only become smaller and smaller, and possibly moving to the right or left, but, eventually it should disappear due to the concentration of gasses and It's increasing tininess and direction of movement to the right or left.

Unless you went to one of those private schools with a 3000mile long gym floor.

 

[Shuque]

In my opinion anyone who willfully assumes they know what is real or not without any hesitation or looking into opposing opinion is a fool. As such they are fooling themselves and proving to lack any kind of ability to discern or critically think on an independent and individual basis.

For example out of all the poster's that have given there opinion itt, how many of you are astronauts? Has anyone of you gone into space? Let's cut out the shenanigans, none of you are and never will be. With that being said I'm not here to say the world is flat nor am I saying the world is round. Because honestly I've never been to outer space personally myself to make that conclusive of an answer. So that means I have to depend on the word of others to believe what is plausibly true or not. (hence how a sucker is born every minute / day) & (That shit would never fly in court for testimony if I appealed to authority. Expertise has to be proven.)

I look at it like this. If I accept 100% of the time that what NASA is telling me is true, then I'm just allowing myself to be a sucker eventually. Further along;

I.E So does that mean that if NASA tells me pigs are blue and have wings to fly, that I should believe that's true simply because NASA said so? Of course this works both ways. I don't assume the world is flat, nor do I assume the world is round. Given a totality of everything I honestly have no opinion on the matter.

I see it as several things on the part of people as a whole.

Willful ignorance
Willful pride
Willful fear
Willful belief in experts
Willful belief in masses
Willful belief in popularity
Willful belief in appealing to authority
Willful desire to avoid ridicule
Fearful of what you may or may not find.

I'm not the smartest guy that's going to sit next to you on bar stool at the bar, but I far from being *retarded*

For instance;
My high School SAT score was 1350
My military ASVAB was a 92 out of 100
My MOS was 6386 (Top secret clearance)
I have Bachelor of Science in Electrical Engineering & Bachelor of Science in Public Health.
I was a Police Officer for 7 years and then transitioned into Firearms examiner (Of which I have been the past 5 years)
I make as a base salary 120k a year without considering all the OT I make for call outs, on call, and Court.

Oh and btw I do thoroughly enjoy my share of conspiracy theories. Cheers.

You should of payed more attention when they explained the scientific method when you got your BAs.

That way you would understand that when it comes to science, you are not simply taking an expert's word or simply believing them...

 

[Deleted member 110072]

A lot of these arguments basically depend on "common sense" when in reality is everyday observation on a small local level that does not immediately make a lot of natural laws obvious.

The classic example to me would be the claims that the speed of light being an upward limit is false because if you had a giant poll millions of miles long you could poke venus instantly with it. Basically depending on human observation that force is transmitted along a long poll instantly but this observation is both imprecise and on a vastly smaller scale billions of times smaller than the claim.

 

[JudasAnthony]

Forget it globetards
Your pseudoscience and occult religious theory's are crumbling before you.