No, they have no cred because they're saying bullshit 24/24.

mainstream propaganda > reptilian propaganda

how can it be a globe LOL XDDDDDDDDDDDD

it wouldnt rain in australia if it was a globe. it would fall into space and we lose water. lots of water. earth is a flat not a globe. there is tons of proof.

why would you not believe in a proven fact? wtf. das like saying 2+2 aint 4 lol

no one can prove its a globe. just go pick a ball and try to make it rain everywhere at once with a glass of water. it wont happen. it will fall down from the bottom countries.

This is exactly right, The biggest clues are horizons are always flat, water seeks it's level, Pilots do not dip their airplane nose down to stay along the curve, The given measurements for earth's curve doesn't exist. Nasa states no actual picture of earth exist only CGI and also states it has to be CGI. Tons of moon landing lies and ISS mistakes the suggest it's a studio.

The list goes on and on, I don't feel like i'm constantly spinning why would anyone?

Lol. "I don't feel like I'm spinning"

You shouldn't since its a rate of 1/1440 rpm

This post was edited by Thor123422 on Aug 20 2018 09:26pm

Where do you get this information from? At the Equator the earth is spinning at 1,000+ MPH Allegedly

Uhh, if it spins once per day then the rate is 1/(mins in a day) rpm.

So calculate the expected cemtripetal force. The radius claimed is 6400000 meters, and speed is 450m/s (converted from 1000mph)

You can look up the centripetal acceleration equation and it's v^2/r
Which comes out to 0.031m/s^2

So.... such a slow acceleration that the friction of your feet on the ground is more than enough to stop it, by several orders of magnitude. Additionally since the measured (measured, not theoretical) downward acceleration of something dropped Is 9.8m/s^2 that means you wouldn't notice the spin anyway.

This post was edited by Thor123422 on Aug 20 2018 09:51pm

Really you can easily calculate the centripetal acceleration on any sphere at the equator pretty easily.

A rotation rate of X rotations per day turns into X/86400 rotations per second (24 hours/day * 60 minutes/hour * 60 seconds/minute = 86400 seconds/day)

Then you are traveling at the equator at a rate of C (the circumference) * X / 86400 meters per second.

The circumference is related to the radius of the sphere by C = 2*pi*r, so lets plug that in to get 2*pi*r*X/86400 meters/second.

So then plug that equation into the centripetal acceleration equation (v^2/r) and you get the following

X = # of rotations per day
r = radius of the sphere in meters

X^2 * 4 *pi^2 * r^2 * X^2 / 7464960000 meters/second^2

You can use this general formula to figure out the acceleration away from the surface of the sphere you would feel at the equator.

This post was edited by Thor123422 on Aug 20 2018 10:19pm