Best way to unproove the globe model

Back again but this time not a question but an obvious observation.
I'm no scientist but by common sense and using physical proof i came with this though.
I haven't seen videos or topics about it, but it completly destroy the globe model, i'll try to explain without any illustration.

From what we've been told, earth spins at 1000 miles/h and rotate the sun at some higher speed but not important for the explanation.
So for any spinning objects the centrifuge force arround the edge is relative to the speed, so basically anything at this speed would be ejected, however the argument of having a gravity force that is stronger make us stay on the ground so that is why everything stays in place. Now the gravity force is equal anywhere on the globe. This would mean if i stand on the equator the centrifuge force would be at is max and if i stand on the north pole the force would be pretty much zero, this means there is no opposite force there.
so why do we feel the attraction to the earth the same in different locations?
Technically the gravity must be very strong for people at the equator to resist the centrifuge force, this mean at the north pole with no opposite force you would be smashed to the ground.

This is probably the most basic common sense, and scientifically make completly sense.

Told that to a non believer friend and he just froozed

The centrifugal force isn't that substantial, not enough to make a big impact. The measured change is you weigh about 0.53% less on the equator than on the poles, a barely noticeable change

If you had two opposing forces on you,you should feel both,

Havoc is right.  Centrifugal force only causes a reduction of .03 m/s² at the equator. 

If you had two opposing forces on you,you should feel both,

Again, it's such a small difference that it isn't really noticeable. It's the difference between weighing 150 lbs on the poles and 149.2 at the equator. It's a change but it's negligible. 

From what we've been told, earth spins at 1000 miles/h

The surface speed at the equator, sure.  It's still only one revolution every 24 hours.  Spin anything at the rate of one complete revolution per day and then ponder how much "centrifuge force" there would be.

The centrifugal force isn't that substantial, not enough to make a big impact. The measured change is you weigh about 0.53% less on the equator than on the poles, a barely noticeable change

My new diet plan . . . move to the Equator.   

What you just stated is in fact an argument that favors RE:
Gravity at the equator has indeed been measured to be lower than on the poles, which supports a spinning ball.

Back again but this time not a question but an obvious observation.
I'm no scientist but by common sense and using physical proof i came with this though.
I haven't seen videos or topics about it, but it completly destroy the globe model, i'll try to explain without any illustration.

From what we've been told, earth spins at 1000 miles/h and rotate the sun at some higher speed but not important for the explanation.
So for any spinning objects the centrifuge force arround the edge is relative to the speed, so basically anything at this speed would be ejected, however the argument of having a gravity force that is stronger make us stay on the ground so that is why everything stays in place. Now the gravity force is equal anywhere on the globe. This would mean if i stand on the equator the centrifuge force would be at is max and if i stand on the north pole the force would be pretty much zero, this means there is no opposite force there.
so why do we feel the attraction to the earth the same in different locations?
Technically the gravity must be very strong for people at the equator to resist the centrifuge force, this mean at the north pole with no opposite force you would be smashed to the ground.

This is probably the most basic common sense, and scientifically make completly sense.

Told that to a non believer friend and he just froozed

No you are not just not a scientist but also a lazy bastard You can calculate the centripetal forces  on earth with high school math easily if you even care

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

See? That's the advantage of using math. You are actually able to back up your claims. The centrifugal force has been calculated. It is minimal. Feel free to disprove it.

Quote from: sceptimatic on June 29, 2016, 02:08:46 AM

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

See? That's the advantage of using math. You are actually able to back up your claims. The centrifugal force has been calculated. It is minimal. Feel free to disprove it.

How has it been calculated to prove a spinning globe?
A force may have been calculated at certain spot on Earth.
A force of pressure can be calculated anywhere on a high and low pressure differential on a stationary Earth.
Your maths is bollocks for a globe. Give it up.

Ok ok didn't know that, i'll find some wiki page to do the calculation.
Now let try to be a bit more tricky, let say im at 45° from the axe of rotation.
The gravity force would still point to the center, however the centrifuge force direction would still be toward 90°, so this mean this force doesnt pull you up but at 45°, should this made light object tend to move to equators?

Theoretically, yes.
However, two things are to mention:
First, as it has been calculated, this Force is really small even at the equator, the further away you get from that line, the smaller it becomes.
Second, this effect was already in Action as our Earth formed, so it is not perfectly spherical but slightly oblate. This means that the ground is Level regarding to gravity and centrifugal Force combined. I might be able to express that in a better way when I am not on my phone 

Theoretically, yes.
However, two things are to mention:
First, as it has been calculated, this Force is really small even at the equator, the further away you get from that line, the smaller it becomes.
Second, this effect was already in Action as our Earth formed, so it is not perfectly spherical but slightly oblate. This means that the ground is Level regarding to gravity and centrifugal Force combined. I might be able to express that in a better way when I am not on my phone 

Sorry to be picky, i really try to make sense to all of this thing and I want to sleep at night without questioning the flat earth or globe
But more i dig more i found out that you have to rely on some other guys who created those formulas that no one truly understand or tested themselve and are so approximative.
And alway end up with "it is barely noticeable" so let ignore it (for convienance), it is a bit like if i was building some high tech things and i say it is 1mm bigger than it should but who cares ...
It is frustrating to feel like it is a non ending debate xD
But in my belief, to prove a theory you have to try to brake it, and if you can't this would be the truth.

Well, if no one could rely on previous works we would not make any progress.
But in this case:
The formula for centipetal acceleration can be derived from Newtons laws and can be verified by experiment easily.
The Radius of the earth can be derived from Maps, the rotational period is your own experience.

Your approach is good, try to disprove re theory, you will find that it is consistent with observations.

Quote from: Kami on June 29, 2016, 02:11:06 AM

Quote from: sceptimatic on June 29, 2016, 02:08:46 AM

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

See? That's the advantage of using math. You are actually able to back up your claims. The centrifugal force has been calculated. It is minimal. Feel free to disprove it.

How has it been calculated to prove a spinning globe?
A force may have been calculated at certain spot on Earth.
A force of pressure can be calculated anywhere on a high and low pressure differential on a stationary Earth.
Your maths is bollocks for a globe. Give it up.

LOL , listen :

The centripetal acceleration at the equator is given by 4 times pi squared times the radius of the Earth divided by the period of rotation squared (4*pi2*r/T2). The period of rotation is 24 hours (or 86400 seconds) and the radius of the Earth is about 6400 km. This means that the centripetal acceletation at the equator is about 0.03 m/s2 (metres per seconds squared). Compare this to the acceleration due to gravity which is about 10 m/s2 and you can see how tiny an effect this is - you would weigh about 0.3% less at the equator than at the poles!

There is an additional effect due to the oblateness of the Earth. The Earth is not exactly spherical but rather is a little bit like a "squashed" sphere, with the radius at the equator slightly larger than the radius at the poles (this shape can be explained by the effect of centripetal acceleration on the material that makes up the Earth, exactly as described above). This has the effect of slightly increasing your weight at the poles (since you are close to the centre of the Earth and the gravitational force depends on distance) and slightly decreasing it at the equator.

Taking into account both of the above effects, the gravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator.

Quote from: sceptimatic on June 29, 2016, 02:17:02 AM

Quote from: Kami on June 29, 2016, 02:11:06 AM

Quote from: sceptimatic on June 29, 2016, 02:08:46 AM

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

See? That's the advantage of using math. You are actually able to back up your claims. The centrifugal force has been calculated. It is minimal. Feel free to disprove it.

How has it been calculated to prove a spinning globe?
A force may have been calculated at certain spot on Earth.
A force of pressure can be calculated anywhere on a high and low pressure differential on a stationary Earth.
Your maths is bollocks for a globe. Give it up.

LOL , listen :

The centripetal acceleration at the equator is given by 4 times pi squared times the radius of the Earth divided by the period of rotation squared (4*pi2*r/T2). The period of rotation is 24 hours (or 86400 seconds) and the radius of the Earth is about 6400 km. This means that the centripetal acceletation at the equator is about 0.03 m/s2 (metres per seconds squared). Compare this to the acceleration due to gravity which is about 10 m/s2 and you can see how tiny an effect this is - you would weigh about 0.3% less at the equator than at the poles!

There is an additional effect due to the oblateness of the Earth. The Earth is not exactly spherical but rather is a little bit like a "squashed" sphere, with the radius at the equator slightly larger than the radius at the poles (this shape can be explained by the effect of centripetal acceleration on the material that makes up the Earth, exactly as described above). This has the effect of slightly increasing your weight at the poles (since you are close to the centre of the Earth and the gravitational force depends on distance) and slightly decreasing it at the equator.

Taking into account both of the above effects, the gravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator.

Not with centripetal force or gravity you don't. You do with denpressure at those places.

Quote from: aisantaros on June 29, 2016, 05:35:31 AM

Quote from: sceptimatic on June 29, 2016, 02:17:02 AM

Quote from: Kami on June 29, 2016, 02:11:06 AM

Quote from: sceptimatic on June 29, 2016, 02:08:46 AM

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

See? That's the advantage of using math. You are actually able to back up your claims. The centrifugal force has been calculated. It is minimal. Feel free to disprove it.

How has it been calculated to prove a spinning globe?
A force may have been calculated at certain spot on Earth.
A force of pressure can be calculated anywhere on a high and low pressure differential on a stationary Earth.
Your maths is bollocks for a globe. Give it up.

LOL , listen :

The centripetal acceleration at the equator is given by 4 times pi squared times the radius of the Earth divided by the period of rotation squared (4*pi2*r/T2). The period of rotation is 24 hours (or 86400 seconds) and the radius of the Earth is about 6400 km. This means that the centripetal acceletation at the equator is about 0.03 m/s2 (metres per seconds squared). Compare this to the acceleration due to gravity which is about 10 m/s2 and you can see how tiny an effect this is - you would weigh about 0.3% less at the equator than at the poles!

There is an additional effect due to the oblateness of the Earth. The Earth is not exactly spherical but rather is a little bit like a "squashed" sphere, with the radius at the equator slightly larger than the radius at the poles (this shape can be explained by the effect of centripetal acceleration on the material that makes up the Earth, exactly as described above). This has the effect of slightly increasing your weight at the poles (since you are close to the centre of the Earth and the gravitational force depends on distance) and slightly decreasing it at the equator.

Taking into account both of the above effects, the gravitational acceleration is 9.78 m/s2 at the equator and 9.83 m/s2 at the poles, so you weigh about 0.5% more at the poles than at the equator.

Not with centripetal force or gravity you don't. You do with denpressure at those places.

Yawn..... ok I play along , then explain your denpressure, or debunk my math or something, but no more bs please

scepti, as I already told you, I measured the weight of a small block of iron in near-vacuum, it did not change!
Pressure has nothing to do with weight.

I think OP needs to retake grammar and spelling, along with basic earth science.

There you go, isn't that just beautiful?

http://gnomeexperiment.com/
http://www.dailymail.co.uk/sciencetech/article-2117738/Long-weigh-gnome-The-bizarre-experiment-garden-ornament-travels-world-measure-gravity.html

There you go, isn't that just beautiful?

http://gnomeexperiment.com/
http://www.dailymail.co.uk/sciencetech/article-2117738/Long-weigh-gnome-The-bizarre-experiment-garden-ornament-travels-world-measure-gravity.html

Nice to know, thanks for sharing.

Why not do the same experiment on writing down sun positions every hours for 24hours on the same day in different locations. This would close any debate on how the sun moves arround us.

Quote from: Definitely Not Official on June 29, 2016, 12:46:11 PM

There you go, isn't that just beautiful?

http://gnomeexperiment.com/
http://www.dailymail.co.uk/sciencetech/article-2117738/Long-weigh-gnome-The-bizarre-experiment-garden-ornament-travels-world-measure-gravity.html

Nice to know, thanks for sharing.

Why not do the same experiment on writing down sun positions every hours for 24hours on the same day in different locations. This would close any debate on how the sun moves arround us.

That's among the things stronomers have been doing for thousands of years.

Hey, Scepi might have stumbled onto something here;

From over 1000 mph to zero at the poles

If you move from a larger radius to a smaller radius on a rotating body, for example towards the pole away from the equator then you will experience conservation of angular momentum effects.
On a small scale, say a kids roundabout, this will accelerate you and the roundabout as you clamber towards the hub.
On a larger scale, say the globe, this will deflect moving objects such as air masses (the wind) or projectiles as they move towards the poles and deflect them in the opposite direction if they move towards the equator.
Coriolis Effect!
Well done scepti.

Quote from: Kami on June 29, 2016, 02:11:06 AM

Quote from: sceptimatic on June 29, 2016, 02:08:46 AM

Kevin: you know you are onto something. It's blatantly obvious we are not on a globe for this and many other reasons.
From over 1000 mph to zero at the poles, we are told and yet they always jump in and tell us that the centripetal force is minimal. Hahahahaha.
It's minimal because we are not spinning. You know this and so does every other rational person who does not bow to en masse peer pressure.

See? That's the advantage of using math. You are actually able to back up your claims. The centrifugal force has been calculated. It is minimal. Feel free to disprove it.

How has it been calculated to prove a spinning globe?
A force may have been calculated at certain spot on Earth.
A force of pressure can be calculated anywhere on a high and low pressure differential on a stationary Earth.
Your maths is bollocks for a globe. Give it up.

What on earth does this gibberish mean "A force of pressure can be calculated anywhere on a high and low pressure differential on a stationary Earth."

As a purely hypothetical case (of course) could you please calculate for us the centripetal acceleration at the surface of the globe:
1) on the equator,
2) at 45° North or South latitude and
3) at either pole.
You can assume that the earth rotates 360° in 23.934 hours and the radius of earth is 3,959 miles.

If you can't hack I'm sure someone will volunteer.

I am confused why this is still an issue about centripetal acceleration...

Ac is Centripetal acceleration, R is radius in meters which is 6.371x10^6 or 6,371,000m, T is 1 day or 23.934 hours or 86162 seconds

Ac=V^2/R, V can equal 2piR/T, so Ac=(2piR/T)^2/R, 2piR/T=464.35, that squared is 215626.9, divide that by R again and the final result is a whopping....0.0338 m/s/s. Which compared to the 9.8 m/s/s of gravity is negligible.

If you had two opposing forces on you,you should feel both,

If what you mean is you feeling as if you're pulled apart by two forces, in this case you're incorrect, because every particle in your body is affected by the same forces, so the distance between them doesn't change, and you can't feel anything. Just think about it.

I am confused why this is still an issue about centripetal acceleration...

Ac is Centripetal acceleration, R is radius in meters which is 6.371x10^6 or 6,371,000m, T is 1 day or 23.934 hours or 86162 seconds

Ac=V^2/R, V can equal 2piR/T, so Ac=(2piR/T)^2/R, 2piR/T=464.35, that squared is 215626.9, divide that by R again and the final result is a whopping....0.0338 m/s/s. Which compared to the 9.8 m/s/s of gravity is negligible.

All those calculations looks cool i have to agree.
But i really can't put head arround the fact that gravity is stronger than spinning at a such speed.
Let say i manage to build a 100m radius bar, even at 20km/h i would really feel a strong pull if i was at the extemity right?
So 0.0338 m/s/s must really represent a strong force if i was on this bar of 6371km radius spinning at a 1000miles/h, even if at the center the rpm is ridiculoudly slow.
So i cant even imagine what must be 9.8 m/s/s and still we are standing with no effort.
Where does those calculation come from by the way? Who ever build such a machine to test those calculations?
Sorry i don't mean to try to be smart ass, because i am definitly not smart on those matters, but it is difficult to understand those numbers against reality.

Kevin, speed does not cause force, Acceleration causes force.

Every change of direction can also be regarded as acceleration.

A spinning ball permanently changes direction. That's why it causes force. We call it Centrifugal Force.

Earth does one rotation in one day. That means the change of direction=acceleration is SMALL, that's why the centrifugal force caused by it is small. The speed on its surface is not relevant, the thing relevant is the angular velocity. And an angular velocity of 360 per day is rather slow.