I'd never seen this before

wrote in :

Mxsmanic wrote:
writes:

Because it makes the assumption that the Earth is round and smooth,
which it is not, and ignores the fact that the atmosphere bends
light.

The notion of stall speeds is wrong in that sense, too, but pilots
use the notion every day.

As always when shown to be incorrect, you respond with something
irrelevant to the original topic in a vain attempt to prove you have
self-worth.


And shows himself to be incorrect a second time.


Bertie

Bertie the Bunyip wrote:
What makes you "special"?

http://en.wikipedia.org/wiki/Short_bus

You saying he edited this?

Sorry, BB. Maybe I should have elaborated. I meant to convey that MX is
"special" because he rides the special bus.


P.S. Just to add some aviation content to this thread ; if you are ever in
Western Montana and drop in at the Seeley Lake strip, the courtesy car is a
"special" sized bus.

John Galban=====N4BQ (PA28-180)

--
Message posted via AviationKB.com
http://www.aviationkb.com/Uwe/Forums...ation/200801/1

WingFlaps writes:

So you still think a 1000' tower can be seen over 1000 miles away?

An object 1000' across can be seen at that distance under ideal conditions.
The resolving power of the eye is optimally 30 seconds of arc, a limit imposed
by the actual size of the receptor cells in the retina.

Was this another MS flight sim experience?

No, it is something I've studied in the past.

Now, don't be petulant- just try to engage some common sense -does it
even sound plausible?

Yes.

Could the empire state building really be seen 1/3 of the way across
the Atlantic?

As long as it is at least 30 seconds across, yes.

Can people in London see the Eiffel tower or people in Paris see
the PO tower in London?

If there are no obstructions in between, absolutely.

Tina writes:

The question not answered, assuming both a spherical cow and a round
smooth earth with no atmosphere, is how you, using right triangles,
know the distance to the horizon.

I've given an explanation. What part don't you understand?

You are in error again.

The calculation would be based on a triangle, one length is earth
radius, the other, earth radius plus tower height. The third length is
the geometric distance from tower top to earth horizon.


Oh, be sure to square (that means multiply it by itself) the two known
lengths, subtract one (earth radius squared) from the other (earth
radius plus tower heigth squared), and extract the sqare root.

Be sure to use consistant units: you do know there are 5280 feet in a
mile, don't you?

What part don't YOU understand?



.
On Jan 3, 5:57*pm, Mxsmanic wrote:
Tina writes:
The question not answered, assuming both a spherical cow and a round
smooth earth with no atmosphere, is how you, using right triangles,
know the distance to the horizon.

I've given an explanation. *What part don't you understand?

"Mxsmanic" wrote in message
...
WingFlaps writes:

So you still think a 1000' tower can be seen over 1000 miles away?

An object 1000' across can be seen at that distance under ideal conditions.

So now the tower is 1000' across as well as 1000' high?

The resolving power of the eye is optimally 30 seconds of arc, a limit imposed
by the actual size of the receptor cells in the retina.


Could the empire state building really be seen 1/3 of the way across the Atlantic?

As long as it is at least 30 seconds across, yes.

IOW you don't know.

Can people in London see the Eiffel tower or people in Paris see
the PO tower in London?

If there are no obstructions in between, absolutely.

Show the math for the limits of resolution. And remember, whichever tower you are using,
you must use the narrower dimension, which would be width, not height.

"Tina" wrote in message
...
You are in error again.

- The calculation would be based on a triangle, one length is earth
- radius, the other, earth radius plus tower height. The third length is
- the geometric distance from tower top to earth horizon.
-
- Oh, be sure to square (that means multiply it by itself) the two known
- lengths, subtract one (earth radius squared) from the other (earth
- radius plus tower heigth squared), and extract the sqare root.
-
- Be sure to use consistant units: you do know there are 5280 feet in a
- mile, don't you?
-
- What part don't YOU understand?

He's switched now to an argument about the limits of resolution of the eye. He's still
full of crap, but what else is new?.


On Jan 3, 5:57 pm, Mxsmanic wrote:
Tina writes:
The question not answered, assuming both a spherical cow and a round
smooth earth with no atmosphere, is how you, using right triangles,
know the distance to the horizon.

I've given an explanation. What part don't you understand?

Mxsmanic wrote:
WingFlaps writes:

So you still think a 1000' tower can be seen over 1000 miles away?

An object 1000' across can be seen at that distance under ideal conditions.
The resolving power of the eye is optimally 30 seconds of arc, a limit imposed
by the actual size of the receptor cells in the retina.

Yet another correct answer to the wrong question in yet another vain
attempt to prove your self worth through misdirection.

Towers aren't 1000' across at the top, they are a few feet across at the
tops.


--
Jim Pennino

Remove .spam.sux to reply.

Tina wrote:
You are in error again.

The calculation would be based on a triangle, one length is earth
radius, the other, earth radius plus tower height. The third length is
the geometric distance from tower top to earth horizon.

Once again he is correct as far as he goes, but I doubt he really derived
the equations, otherwise he would be posting them just to prove how
clever he is.

So, before he has a chance to Google the answer:

Let the Earth be a billiard ball of radius R.

Let the height of an object be h.

Let the distance to the horizon be d.

Draw a circle.

Draw a radial line from the center to the circle.

At the intersection of the radial line and the circle, draw a line tangent
to the circle.

Draw another radial line and extend it until it intersects the tangent line.

The two radial lines and the tangent line form a right triangle, so:

sqrt (R^2 + d^2) = R + h

R^2 + d^2 = (R + h)^2

R^2 + d^2 = R^2 + 2Rh + h^2

d = sqrt (2RH + h^2)



--
Jim Pennino

Remove .spam.sux to reply.

what are the chances mx wouldn't know what 30 seconds of arc is if a
clock fell on him?


On Jan 3, 7:05*pm, wrote:
Mxsmanic wrote:
WingFlaps writes:
So you still think a 1000' tower can be seen over 1000 miles away?
An object 1000' across can be seen at that distance under ideal conditions.
The resolving power of the eye is optimally 30 seconds of arc, a limit imposed
by the actual size of the receptor cells in the retina.

Yet another correct answer to the wrong question in yet another vain
attempt to prove your self worth through misdirection.

Towers aren't 1000' across at the top, they are a few feet across at the
tops.

--
Jim Pennino

Remove .spam.sux to reply.