A proof for the Stationary Earth

[oxmixmudd]

Your analogy is false. If a boat is floating in a river (1) of velocity v, then the boat moves (2) also at v. If the boat is then propelled, the boat moves at v+vp. The vp will cause a pressure differential on the boat which moves against the river surface. Similarly, the Hornet moves against the atmosphere and will also feel a pressure. The only wat a plane does not feel pressure is when the plane always moves with the atmospheres W-E rotation. This would mean the plane would have to fly along with the atmosphere, like the boat floating in the water (1), rather than the boat moving through water (2).

If there is no force from the atmosphere acting on the S-N plane, then the plane must be always guided by the pilot to move W-E with the atmosphere. As this does not happen because flights take direct routes, then the atmosphere must be acting on the S-N plane.

JM

What is the wind equivalent of this force John. What does it equate to in miles per hour (or kilometers per hour if you prefer). not the total velocity change over the entire journey, but moment by moment - the actual force the pilot would encounter as he flies the plane?

You claim it is 'not' there. But how much of breeze is required for the pilot to notice it over the other winds and turbulance he encounters flying the plane? And is it big enough to notice? If it is too small to notice, then how do you know it is not happening?


Jim

[lilpixieofterror]

Eather flow.


JM

Yes, I forgot about magic Eather flow. The stuff that magically produces fake evidence for a rotating earth, but yet is unobserable and undetectable by anything. Of course, the much easier explanation that the earth really is rotating and that you're a nutcase trying desperately to defend nonsense is automatically discounted.

[lilpixieofterror]

The explanation for geostationary satellites

[ATTACH=CONFIG]15218[/ATTACH]

JohnMartin has none.

Sure he has, magical eather flow. You know, the stuff that magically produces all the fake evidence for a rotating earth, but is totally undetectable by any measurements.

[lilpixieofterror]

I watched a mythbusters episode where they pumped the air out of a railroad tanker car and the weight of the atmosphere imploded it. Air is freaking heavy.
A whole planetful of atmosphere moving at 1000kmh can definitely carry a jetliner along with it.

If JM doubts air pressure can cause bad things to happen. Here's a picture of an accident that happened when a maintenance guy made a mistake with pressure checking an aircraft:

Perhaps it was magic eather that blew up that aircraft or collapsed that tanker car.

[lilpixieofterror]

John, have you ever considered discussing your proof with an actual atmospherics scientist? Just pick any of them. They will quickly prove to you that every atmospherics scientist and meteorologist on Earth is part of the conspiracy against you! Watch out for those guys. They're sneaky.

Aeronautics engineers are just as bad. They'll tell you all sorts of things about flight that will violate JohnMartinian physics.

Everybody is part of the conspiracy (except John Martin).

[lilpixieofterror]

You are aware that fighter jets rarely travel at their max speed and on long distance flights, they are usually flying with a refueling aircraft, right?

[oxmixmudd]

Distance from Sydney to Singapore is 6311 km

Max speed of an F 18 Hornet fighter jet is 1,915 km/hr.

Approximate flight time from Sydney to Singapore is 6311/1,915 km/h = 3.30 hrs

The rotation W-E velocity difference between Sydney to Singapore is 277km/hr.

Average acceleration caused by the atmosphere on the Hornet is

277,000/(3.30x3600) = 14.38 m/s^2

No John. An average acceleration of 14.38m/s^2 operating over 3.3 hours would produce a velocity differential of ...

(1) a = 14.38 m/s^2
(2) v = 14.38*t m/s
(3) d = 7.19*t^2 m

So the final velocity shift associated with the acceleration you have described here would be 14.38 m/s^2 *(3.3*3600) s = 277,000 m/s = 277 km/s * 3600 s/h = 997,200 km/h

OOOOPS!!! Wow! 997,200 km/h is a whole heck of a lot faster than 277 km/h. I wonder where you went wrong? Well, let's see ...

To find actual acceleration required to produce a velocity of 277 km/h after 3.3 hours first lets get the velocity in meters per second (since we generally like to express acceleration in m/s^2)

277 km/h*1000m/km = 277,000 m/h * 1/3600 h/s = 23.32 m/s

Now from our basic 3 equations of motion listed above

v = a*t

so that means to find 'a', we set up using our value for v and t (in meters and seconds)

23.32 m/s = a m/s^2 * (3.3 * 3600) s

Now we solve for a:

a= 23.32/3.3*3600 = .0019 (all the units just cancel out and we have a simple constant)



Now what velocity breeze would that feel like? Well, velocity is d/t, and so to get an effective velocity, we need the distance that change in velocity will move the air in 1 second, and that is 1/2at^2 or (since t=1) .00095m.

So each second the change in atmospheric velocity would feel like a 'breeze' of .00095 m/s, or about .0034km/h

When is the last time you noticed a breeze of 3.4 thousandths of a km/h?


Jim

ETA: to be clear - that is the W-E breeze a supersonic F-18 Hornet would feel at full speed and due north. How much time does anyone suppose a pilot spends worrying about breezes of .0034 km/h while they are rocketing across the landscape at nearly 2 times the speed of sound?

[DesertBerean]

If JM doubts air pressure can cause bad things to happen. Here's a picture of an accident that happened when a maintenance guy made a mistake with pressure checking an aircraft:

Perhaps it was magic eather that blew up that aircraft or collapsed that tanker car.

Eeee-ouch. Scarier than Sparko's demo

[lilpixieofterror]

Eeee-ouch. Scarier than Sparko's demo

We had a briefing on flight accidents and that was among the incidents they discussed. From what they said, one of the escape doors blew across the flightline.

[lilpixieofterror]

What is the wind equivalent of this force John. What does it equate to in miles per hour (or kilometers per hour if you prefer). not the total velocity change over the entire journey, but moment by moment - the actual force the pilot would encounter as he flies the plane?

You claim it is 'not' there. But how much of breeze is required for the pilot to notice it over the other winds and turbulance he encounters flying the plane? And is it big enough to notice? If it is too small to notice, then how do you know it is not happening?


Jim

During alert take offs (which usually has 6 aircraft taking off one after another), the last two kind of struggle to get off the ground. Obviously though, launching 6 aircraft one after another doesn't seem to disturb the air too bad or for too long. It does take a few minutes until they are able to launch more aircraft.

Edited to add: As I recall though, it has more to do with oxygen levels than turbulence produced by the Aircraft, but I'm no flight engineer, so I don't know for sure.

[DesertBerean]

We had a briefing on flight accidents and that was among the incidents they discussed. From what they said, one of the escape doors blew across the flightline.

Flightline being the maintenance area?

[lilpixieofterror]

Flightline being the maintenance area?

Aircraft parking area. Maintenance is also done on the flightline too.

[DesertBerean]

Aircraft parking area. Maintenance is also done on the flightline too.

That's quite a distance to cover.

[lilpixieofterror]

That's quite a distance to cover.

Depends on the fightline. Some are rather large and some are kind of small. Also depends on the aircraft they are parking. Ours is rather large, but that's because it has parking for over 40 heavy aircraft.

[JohnMartin]

No John. An average acceleration of 14.38m/s^2 operating over 3.3 hours would produce a velocity differential of ...

(1) a = 14.38 m/s^2
(2) v = 14.38*t m/s
(3) d = 7.19*t^2 m

So the final velocity shift associated with the acceleration you have described here would be 14.38 m/s^2 *(3.3*3600) s = 277,000 m/s = 277 km/s * 3600 s/h = 997,200 km/h

OOOOPS!!! Wow! 997,200 km/h is a whole heck of a lot faster than 277 km/h. I wonder where you went wrong? Well, let's see ...

To find actual acceleration required to produce a velocity of 277 km/h after 3.3 hours first lets get the velocity in meters per second (since we generally like to express acceleration in m/s^2)

277 km/h*1000m/km = 277,000 m/h * 1/3600 h/s = 23.32 m/s

Now from our basic 3 equations of motion listed above

v = a*t

so that means to find 'a', we set up using our value for v and t (in meters and seconds)

23.32 m/s = a m/s^2 * (3.3 * 3600) s

Now we solve for a:

a= 23.32/3.3*3600 = .0019 (all the units just cancel out and we have a simple constant)



Now what velocity breeze would that feel like? Well, velocity is d/t, and so to get an effective velocity, we need the distance that change in velocity will move the air in 1 second, and that is 1/2at^2 or (since t=1) .00095m.

So each second the change in atmospheric velocity would feel like a 'breeze' of .00095 m/s, or about .0034km/h

When is the last time you noticed a breeze of 3.4 thousandths of a km/h?


Jim

It seems I've been corrected. I've muddled up the units.

F = 29,932 x 0.0019 = 56.87 N

Pressure on Hornet = 56.87/(38) = 1.50 Pa.

The pressure difference on the Hornet may be measured if the plane flies S-N, then turns back N-S.

JM