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I found that the contrail corresponded remarkably well with flight AWE808, which flew from Hawaii to Phoenix; it showed the change in course to the northeast at the same location, further confirming my conclusion.”
Why not e-mail or write to Dr Minnis and ask him your questions about the contrail?
Where did you get your "given data" from?
Originally posted by SarK0Y
reply to post by Arbitrageur
after sunlight reflecting, here: www.abovetopsecret.com...
after color fluctuations of incandescent point, here: www.abovetopsecret.com...
...do the engines all rotate the same direction? Or do the engines rotate in opposite directions on opposite wings?
Where did you get your "given data" from?
And I don't think you can measure light intensity in watts,
and I don't see where you took the shape of an airplane into account.("area of hemispherical surface is...") what about the wings? They aren't hemispherical, and in fact have some flattish areas.
I see. Physics was one of my majors so I don't really have to learn it. Now that I know what you're trying to do, I can explain the correct way to do the math and the physics. Here's the math you gave:
Originally posted by SarK0Y
reply to post by Arbitrageur
and I don't see where you took the shape of an airplane into account.("area of hemispherical surface is...") what about the wings? They aren't hemispherical, and in fact have some flattish areas.
learn how electromagnetic waves travel through space(environments) & you'll get what i was talking about "area of hemispherical surface is..."
You assumed the light source was 60km away but this is not the distance to the sun.
Originally posted by SarK0Y
reply to post by tommyjo
well, let's do math of the event i took rough data to simplify situation, but that's towards upper limit of possible numbers.
Reflection and glare doesn't always appear. Why do you think that every aircraft or any other object filmed in the sky will automatically generate glare or reflection?
------------------------------------------------
given data.
distance to object 60, 000 m's
reflecting area Ra= 100 m2
light intensity I=1, 000 W per m2
----------------------------------------------------
so, area of hemispherical surface is S=(2*3.14)*60000^2=2.260800000*10^10; light intensity at cam from reflecting area is i=Ra*I/S=0.000004423213022 W per m2.
Thanks for the answer. I actually tried to research it before asking and I did find the propeller torque stuff, but not the answer I was looking for which you just provided. I thought they were all clockwise facing forward as the pilot sits, but you're saying the Rolls Royce N1 are counterclockwise?
Originally posted by weedwhacker
NOW....to answer, again.....engines rotate the same direction (although Rolls Royce N1 Fans ARE opposite than GE or P&W!!!), on modern jets, they all rotate the same direction.
ALL of the above, about "P-factor: and such? Not the same with big jets. THRUST is what we feel, no torque.....
In my opinion there is absolutely no doubt that what was captured on video off the coast of California was a missile launch, was clearly observed by NORAD, assessed by a four-star General in minutes, and passed to the President immediately.
You assumed the light source was 60km away but this is not the distance to the sun.
1000/(93000200/93000000)^2
aw doesn't apply to a non-isotropic light source, which would apply to an airplane wing reflecting sunlight
You wanted to apply the inverse square law, so the sun is the isotropic light source. You compare the 1000 watt per square meter intensity at 93 million miles versus 93 million miles plus 200 miles when the sunlight reflects off a flat surface, like a mirror, or a plane wing.
Originally posted by SarK0Y
wuhhhhhhhhhhh! why do we need distance to the Sun?? really, we need to know sunlight Power which was reached to the Earth en.wikipedia.org...
That's the math of what I just said in English, applying the inverse square law.hmmmmm.... what is that??
1000/(93000200/93000000)^2
look at the page of the textbook I posted, it's the bullet that says the inverse square law only applies to isotropic light sources (which means it doesn't apply to non-isotropic light sources). The sun is an isotropic source, because the light rays go out in all directions. Sunlight reflecting from a flat mirror or a flat plane wing is not isotropic, the light rays are nearly parallel so they don't spread out hardly at all except for imperfections in the reflecting surface.wuhhhhh! where do you get it??????
aw doesn't apply to a non-isotropic light source, which would apply to an airplane wing reflecting sunlight
Because the sun is so far away, the rays of sunlight are nearly parallel to one another.