Curvature of the FE

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Sir, you're mistaken. Your attempt does not perfectly explain the phenomenon. For example, the apparent size of the sun in your analogy would decrease, but we know from experience it does not.

The apparent size of the sun actually increases as it nears the horizon. On the next sunset look at the sun along the horizon and notice how the sun appears much larger near the horizon than it does overhead at zenith. This is due to the effect Dr. Rowbotham describes in Chapter 10 of Earth Not a Globe.

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IT is well known that when a light of any kind shines through a dense medium it appears larger, or rather gives a greater "glare," at a given distance than when it is seen through a lighter medium. This is more remarkable when the medium holds aqueous particles or vapour in solution, as in a damp or foggy atmosphere. Anyone may be satisfied of this by standing within a few yards of an ordinary street lamp, and noticing the size of the flame; on going away to many times the distance, the light upon the atmosphere will appear considerably larger. This phenomenon may be noticed, to a greater or less degree, at all times; but when the air is moist and vapoury it is more intense. It is evident that at sunrise, and at sunset, the sun's light must shine through a greater length of atmospheric air than at mid-day; besides which, the air near the earth is both more dense, and holds more watery particles in solution, than the higher strata through which the sun shines at noonday; and hence the light must be dilated or magnified, as well as modified in colour.

This effect is present in the following image: http://i13.tinypic.com/4zvjjlx.jpg

Notice how the lights, particularly the ones in the upper right, are magnified much larger than their normal size. If portrayed in a vacuum, without this effect upon the atmosphere, the distant bulbs would not take up a single pixel of the image.

You're really wrong this time. I measured the apparent sizes just two days ago. The Sun appeared the same size at sunrise, midmorning, afternoon, evening, and sunset. I was careful to allow the common mistake just looking and perhaps being fooled by perspective. I used my upright thumb on my outstretched arm, like artists commonly do, to compare the Sun to my thumb. The Sun also appeared to be the size of my thumbnail.

I completely reject your analysis of the picture. I have seen many such effects that you attribute to such thick atmosphere in nighttime photographs. Lens flare, overexposure, filters, and capturing the area illuminated vice the light itself could each readily explain the effects. You state that "If portrayed in a vacuum, without his effect upon the atmosphere, the distant bulbs would not take up a single pixel of the image." but provide no proof for your claim. I challenge you to demonstrate what you claim.

Oh and you also manage to fail to respond to my other comments about your previous post. I have to wonder for what reason you avoid debating these issues.

Apparent curvature from a high altitude means nothing.

If it means nothing then why did you waste your time looking for it while flying in a commercial airplane, as you claimed to have done?
If no curvature present: Proof of FE
If some curvature present: Proof of FE
If complete globe visible: Proof of conspiracy

You wouldn't fault us if we insisting on arriving at the Truth by our own journey though, would you?

of course not. inquiry is one thing, blatant idiocy is another.

Not always.  But that would mean a rate of constant turn, Don't you think that a pilot would discover the fact that he's been pressing the left or right rudder with the same pressure every time the air was calm?

Here's a tidbit I tossed off a while back:

Assume:
800 MPH aircraft = 357 m/s
radius equator: 6445 mi = 10.37 E6 m

Calculating the acceleration felt from turning in the circle assumed above:
angular acceleration = v^2 / r  = 357^2/10.37E6 = 0.0123 m/s^2

That's not very much acceleration.  Now calculate the degree of roll needed to keep the airplane at altitude while making this circle:

Assume:
acceleration due to gravity = 9.8 m/s
acceleration due to turn = 0.0123 m/s

Degree roll = arctan(0.0123/9.8 ) = 0.072

That is a very small amount of roll, even at such high speeds.  You would NOT notice such a small turn.  A pilot would follow his instruments which probably have an accuracy on the order of 0.5 degrees or higher, correct me if I'm wrong. 

Quote from: CSSGHLNN on May 19, 2007, 08:24:58 PM

Not always.  But that would mean a rate of constant turn, Don't you think that a pilot would discover the fact that he's been pressing the left or right rudder with the same pressure every time the air was calm?

Here's a tidbit I tossed off a while back:

Assume:
800 MPH aircraft = 357 m/s
radius equator: 6445 mi = 10.37 E6 m

Calculating the acceleration felt from turning in the circle assumed above:
angular acceleration = v^2 / r  = 357^2/10.37E6 = 0.0123 m/s^2

That's not very much acceleration.  Now calculate the degree of roll needed to keep the airplane at altitude while making this circle:

Assume:
acceleration due to gravity = 9.8 m/s
acceleration due to turn = 0.0123 m/s

Degree roll = arctan(0.0123/9.8 ) = 0.072

That is a very small amount of roll, even at such high speeds.  You would NOT notice such a small turn.  A pilot would follow his instruments which probably have an accuracy on the order of 0.5 degrees or higher, correct me if I'm wrong. 

Ah, no.
You've confused angular acceleration with acceleration at an angle.

Here are the results of my analysis:

Distance at Equator: 6,378,000m      
Speed   510 MPH   (I though this might be a more acceptable speed.)       
      230 m/s

Latitude	Distance (m)	Time (s)	Acceleration (m/s^2)
79 N	190,000	830	0.39
45 N	800,000	3,500	0.093
0.0	1,600,000	7,000	0.046
45 S	2,400,000	10,000	0.033
79 N	3,000,000	13,000	0.025

I chose to stay at two significant digits.

I decided to send the plane on a due east (or west) heading and have it travel one-fourth of the
way around the FE.

The distance calculation assumes that latitudes are equidistant, and therefore proportional to each other.

The time calculation is based on v=d/t. The acceleration is based on vector addition of the velocity changes, one back + one north = square root of 2 at 45 degrees, divided by the time.

The differences in flight times would be difficult to explain, varying by more than 10 times in one case.

The accelerations are significant enough to cause rolling of round objects such as pens since many coefficients of rolling friction are less than the accelerations in the table above. I don't have engineering estimates, but I also believe that glossy magazines laid open on the
seat tray would slide, for example.

I hope that helps. It is a challenging problem.
Gulliver

I also misused some math to further my point.   

I am evil. 

Quote from: Tom Bishop on May 17, 2007, 10:16:34 AM

Apparent curvature from a high altitude means nothing.

If it means nothing then why did you waste your time looking for it while flying in a commercial airplane, as you claimed to have done?
If no curvature present: Proof of FE
If some curvature present: Proof of FE
If complete globe visible: Proof of conspiracy

That basically sums it up.