I'd never seen this before

[Kyle Boatright]

On Christmas Eve, I got to go flying. I took a friend of ours on a 200 nm
cross country to drop her off with her family for the holidays. On the good
side, the visibility was as good as I've ever seen in the Southeast.
Probably 150 miles or more. I could see the Smokey Mtns waay north. Also,
on the way "out", there was a 25 knot tailwind, so that portion of the trip
went pretty quickly.

On the way home, I flew down low (3000 -3500' MSL) to duck most of the wind,
but still enjoyed the good visibility. Part of the way home, I saw a large
tower a long way in front of me. I was surprised, because the tower
appeared to be much higher than my cruising altitude. I didn't remember any
extremely tall towers NE of Atlanta, but studied the sectional several times
looking for what my eyes were seeing. Never did find a particularly tall
tower on the sectional.

Anyway, it probably took me 20 minutes to reach the tower and by the time I
reached it, it was well below my altitude, (which hadn't changed). After a
little thought, I realized that the curvature of the earth had resulted in
an illusion that the tower was extremely tall when viewed from a distance,
but was only 1000' AGl (or 1800' MSL) in reality.

I've heard of this phenomena causing confusion between aircraft approaching
each other at high altitude, but never realized it would occur with towers
and the like when visibility was good. It'll probably be a long time
before the visibility around here is good enough to see a tower from 50
miles away again.

[Bertie the Bunyip[_19_]]

"Kyle Boatright" > wrote in
:

> On Christmas Eve, I got to go flying. I took a friend of ours on a
> 200 nm cross country to drop her off with her family for the holidays.
> On the good side, the visibility was as good as I've ever seen in the
> Southeast. Probably 150 miles or more. I could see the Smokey Mtns
> waay north. Also, on the way "out", there was a 25 knot tailwind, so
> that portion of the trip went pretty quickly.
>
> On the way home, I flew down low (3000 -3500' MSL) to duck most of the
> wind, but still enjoyed the good visibility. Part of the way home, I
> saw a large tower a long way in front of me. I was surprised, because
> the tower appeared to be much higher than my cruising altitude. I
> didn't remember any extremely tall towers NE of Atlanta, but studied
> the sectional several times looking for what my eyes were seeing.
> Never did find a particularly tall tower on the sectional.
>
> Anyway, it probably took me 20 minutes to reach the tower and by the
> time I reached it, it was well below my altitude, (which hadn't
> changed). After a little thought, I realized that the curvature of
> the earth had resulted in an illusion that the tower was extremely
> tall when viewed from a distance, but was only 1000' AGl (or 1800'
> MSL) in reality.
>
> I've heard of this phenomena causing confusion between aircraft
> approaching each other at high altitude, but never realized it would
> occur with towers and the like when visibility was good. It'll
> probably be a long time before the visibility around here is good
> enough to see a tower from 50 miles away again.


Well, even at middling altitudes I find it hard to tell the altitude of
other airplanes. I also find it difficult to tell the height of a tower,
even relative to my own altitude. if there's a trick to it aside from
makrking the windscreen with a magic marker, it's eluded me!
The introduction of TCAS was a revelation. NEarby airplkanes that you
might have sron were a couple of thousand above turned out to be lower
than us and vice versa. It's not just me, everyone says the same thing.
I can't see how the curvature of the earth would do it, though. Seems to
me no matter which way you slice that the tower would look lower.




Bertie

[Tina]

That part of the country is pretty flat, so something a thousand feet
agl might look really high from a distance. From your altitude, did it
appear higher than the Smokies? That would do a number on your
perception.




On Jan 1, 8:27*am, Bertie the Bunyip > wrote:
> "Kyle Boatright" > wrote m:
>
>
>
>
>
> > On Christmas Eve, I got to go flying. *I took a friend of ours on a
> > 200 nm cross country to drop her off with her family for the holidays.
> > *On the good side, the visibility was as good as I've ever seen in the
> > Southeast. Probably 150 miles or more. I could see the Smokey Mtns
> > waay north. * Also, on the way "out", there was a 25 knot tailwind, so
> > that portion of the trip went pretty quickly.
>
> > On the way home, I flew down low (3000 -3500' MSL) to duck most of the
> > wind, but still enjoyed the good visibility. *Part of the way home, I
> > saw a large tower a long way in front of me. *I was surprised, because
> > the tower appeared to be much higher than my cruising altitude. *I
> > didn't remember any extremely tall towers NE of Atlanta, but studied
> > the sectional several times looking for what my eyes were seeing.
> > Never did find a particularly tall tower on the sectional.
>
> > Anyway, it probably took me 20 minutes to reach the tower and by the
> > time I reached it, it was well below my altitude, (which hadn't
> > changed). * After a little thought, I realized that the curvature of
> > the earth had resulted in an illusion that the tower was extremely
> > tall when viewed from a distance, but was only 1000' AGl (or 1800'
> > MSL) in reality.
>
> > I've heard of this phenomena causing confusion between aircraft
> > approaching each other at high altitude, but never realized it would
> > occur with towers and the like when visibility was good. * It'll
> > probably be a long time before the visibility around here is good
> > enough to see a tower from 50 miles away again.
>
> Well, even at *middling altitudes I find it hard to tell the altitude of
> other airplanes. I also find it difficult to tell the height of a tower,
> even relative to my own altitude. if there's a trick to it aside from
> makrking the windscreen with a magic marker, it's eluded me!
> The introduction of TCAS was a revelation. NEarby airplkanes that you
> might have sron were a couple of thousand above turned out to be lower
> than us and vice versa. It's not just me, everyone says the same thing.
> I can't see how the curvature of the earth would do it, though. Seems to
> me no matter which way you slice that the tower would look lower.
>
> Bertie- Hide quoted text -
>
> - Show quoted text -

[Mxsmanic]

Kyle Boatright writes:

> Anyway, it probably took me 20 minutes to reach the tower and by the time I
> reached it, it was well below my altitude, (which hadn't changed). After a
> little thought, I realized that the curvature of the earth had resulted in
> an illusion that the tower was extremely tall when viewed from a distance,
> but was only 1000' AGl (or 1800' MSL) in reality.

The curvature of the planet won't do this; it makes things seem lower, not
higher, just as a tower behind a hill might not appear as tall as it does once
you reach the crest of the hill.

However, some atmospheric effects can make things seem larger or taller than
they are from a distance.

At an altitude of 3000 feet AGL over smooth terrain, you'll be able to see the
top of a 1000' tower (but not the whole thing) from up to 92 nm away.

[Ron Wanttaja]

On Tue, 01 Jan 2008 18:19:29 +0100, Mxsmanic > wrote:

> Kyle Boatright writes:
>
> > Anyway, it probably took me 20 minutes to reach the tower and by the time I
> > reached it, it was well below my altitude, (which hadn't changed). After a
> > little thought, I realized that the curvature of the earth had resulted in
> > an illusion that the tower was extremely tall when viewed from a distance,
> > but was only 1000' AGl (or 1800' MSL) in reality.
>
> The curvature of the planet won't do this; it makes things seem lower, not
> higher, just as a tower behind a hill might not appear as tall as it does once
> you reach the crest of the hill.

No. Pilots are accustomed to judging the altitude of other aircraft based on
whether it's above or below the horizon. This is great for collision avoidance,
but doesn't work at longer ranges. ANYTHING located at the viewer's horizon
will appear above the altitude of the observer, and it *is* due to the curvature
of the Earth. See:

http://www.wanttaja.com/los.jpg

Normally, though, details at such distances are obscured by haze.

Ron Wanttaja

[Kyle Boatright]

"Tina" > wrote in message
...
That part of the country is pretty flat, so something a thousand feet
agl might look really high from a distance. From your altitude, did it
appear higher than the Smokies? That would do a number on your
perception.

No, it did not look higher than the Smokies, which were well off to my right
on the return trip. It did look higher (and probably was) than any of the
hills within about 50 miles of Atlanta.




On Jan 1, 8:27 am, Bertie the Bunyip > wrote:
> "Kyle Boatright" > wrote
> m:
>
>
>
>
>
> > On Christmas Eve, I got to go flying. I took a friend of ours on a
> > 200 nm cross country to drop her off with her family for the holidays.
> > On the good side, the visibility was as good as I've ever seen in the
> > Southeast. Probably 150 miles or more. I could see the Smokey Mtns
> > waay north. Also, on the way "out", there was a 25 knot tailwind, so
> > that portion of the trip went pretty quickly.
>
> > On the way home, I flew down low (3000 -3500' MSL) to duck most of the
> > wind, but still enjoyed the good visibility. Part of the way home, I
> > saw a large tower a long way in front of me. I was surprised, because
> > the tower appeared to be much higher than my cruising altitude. I
> > didn't remember any extremely tall towers NE of Atlanta, but studied
> > the sectional several times looking for what my eyes were seeing.
> > Never did find a particularly tall tower on the sectional.
>
> > Anyway, it probably took me 20 minutes to reach the tower and by the
> > time I reached it, it was well below my altitude, (which hadn't
> > changed). After a little thought, I realized that the curvature of
> > the earth had resulted in an illusion that the tower was extremely
> > tall when viewed from a distance, but was only 1000' AGl (or 1800'
> > MSL) in reality.
>
> > I've heard of this phenomena causing confusion between aircraft
> > approaching each other at high altitude, but never realized it would
> > occur with towers and the like when visibility was good. It'll
> > probably be a long time before the visibility around here is good
> > enough to see a tower from 50 miles away again.
>
> Well, even at middling altitudes I find it hard to tell the altitude of
> other airplanes. I also find it difficult to tell the height of a tower,
> even relative to my own altitude. if there's a trick to it aside from
> makrking the windscreen with a magic marker, it's eluded me!
> The introduction of TCAS was a revelation. NEarby airplkanes that you
> might have sron were a couple of thousand above turned out to be lower
> than us and vice versa. It's not just me, everyone says the same thing.
> I can't see how the curvature of the earth would do it, though. Seems to
> me no matter which way you slice that the tower would look lower.
>
> Bertie- Hide quoted text -
>
> - Show quoted text -

[Tina]

Gee, wrong again. I haven't run the numbers, there may be a line of
sight, but the only way a human pilot would see a tower top under
these conditions is at night if it had a bright flashing light on top
of it. Real eyeballs in the daytime would not be able to see it, even
if it was in the line of sight.


On Jan 1, 12:19*pm, Mxsmanic > wrote:
> Kyle Boatright writes:
> > Anyway, it probably took me 20 minutes to reach the tower and by the time I
> > reached it, it was well below my altitude, (which hadn't changed). * After a
> > little thought, I realized that the curvature of the earth had resulted in
> > an illusion that the tower was extremely tall when viewed from a distance,
> > but was only 1000' AGl (or 1800' MSL) in reality.
>
> The curvature of the planet won't do this; it makes things seem lower, not
> higher, just as a tower behind a hill might not appear as tall as it does once
> you reach the crest of the hill.
>
> However, some atmospheric effects can make things seem larger or taller than
> they are from a distance.
>
> At an altitude of 3000 feet AGL over smooth terrain, you'll be able to see the
> top of a 1000' tower (but not the whole thing) from up to 92 nm away.

[Bertie the Bunyip]

On 1 Jan, 17:19, Mxsmanic > wrote:
> Kyle Boatright writes:
> > Anyway, it probably took me 20 minutes to reach the tower and by the time I
> > reached it, it was well below my altitude, (which hadn't changed). * After a
> > little thought, I realized that the curvature of the earth had resulted in
> > an illusion that the tower was extremely tall when viewed from a distance,
> > but was only 1000' AGl (or 1800' MSL) in reality.
>
> The curvature of the planet won't do this; it makes things seem lower, not
> higher, just as a tower behind a hill might not appear as tall as it does once
> you reach the crest of the hill.
>
> However, some atmospheric effects can make things seem larger or taller than
> they are from a distance.
>
> At an altitude of 3000 feet AGL over smooth terrain, you'll be able to see the
> top of a 1000' tower (but not the whole thing) from up to 92 nm away.


Sez the gy who has never flown


Ever

Fjukkwit


Bertie

[Tina]

You can be fairly sure he used someone else's equations for line of
sight. I'd bet a significant sum he could not derive them himself. He
and Euclid would not have gotten along.




On Jan 1, 4:52*pm, Bertie the Bunyip >
wrote:
> On 1 Jan, 17:19, Mxsmanic > wrote:
>
>
>
>
>
> > Kyle Boatright writes:
> > > Anyway, it probably took me 20 minutes to reach the tower and by the time I
> > > reached it, it was well below my altitude, (which hadn't changed). * After a
> > > little thought, I realized that the curvature of the earth had resulted in
> > > an illusion that the tower was extremely tall when viewed from a distance,
> > > but was only 1000' AGl (or 1800' MSL) in reality.
>
> > The curvature of the planet won't do this; it makes things seem lower, not
> > higher, just as a tower behind a hill might not appear as tall as it does once
> > you reach the crest of the hill.
>
> > However, some atmospheric effects can make things seem larger or taller than
> > they are from a distance.
>
> > At an altitude of 3000 feet AGL over smooth terrain, you'll be able to see the
> > top of a 1000' tower (but not the whole thing) from up to 92 nm away.
>
> Sez the gy who has never flown
>
> Ever
>
> Fjukkwit
>
> Bertie- Hide quoted text -
>
> - Show quoted text -

[Mxsmanic]

Ron Wanttaja writes:

> No. Pilots are accustomed to judging the altitude of other aircraft based on
> whether it's above or below the horizon. This is great for collision avoidance,
> but doesn't work at longer ranges. ANYTHING located at the viewer's horizon
> will appear above the altitude of the observer, and it *is* due to the curvature
> of the Earth. See:
>
> http://www.wanttaja.com/los.jpg
>
> Normally, though, details at such distances are obscured by haze.

But the original post made no mention of the horizon.

On a flat surface of infinite extent, the horizon is always at eye level, no
matter what your position. On a flat surface of finite extent, it is always
slightly below eye level, depending on how far away the edge of the surface
is. On a spherical planet, the horizon is still lower; to find its distance
(assuming smooth terrain), add your altitude to the radius of the planet,
square it, subtract the radius of the planet squared, and take the square root
of the result. At an altitude of 3000 feet above smooth terrain (such as
water, or a dry lake), the horizon on Earth is 58 nm away. If you are six
feet tall and standing on the surface with smooth terrain (or if you are in a
rowboat on a calm ocean, for example), the horizon is just 2.6 nm distant.

Your diagram is interesting, but since it dramatically overstates the height
of towers and dramatically understates the size of the planet, it's a bit
misleading. Your towers would be several times higher than the orbit of the
International Space Station, and the aircraft would be in outer space.

On a flat surface, anything moving down in your field of view is something
you'll fly over, and anything moving up is something you'll hit. The
curvature of the Earth complicates this, but the curvature is gentle enough
that anything affected by it is too far away to be an immediate hazard,
anyway. At 3000' AGL, you could see Mount Everest from 239 nm away, but since
a small plane might take two hours reach it, you'd have plenty of time to
evaluate it as a hazard.

Thought experiments like this can be interesting. People often say that the
Concorde was wonderful because you could see the curvature of the Earth, but
the fact is that you can see the curvature from anywhere, even a hill
overlooking the beach. It just gets more obvious as you move further away
from the surface. From an airliner at 39,000 feet, the view extends for well
over 210 miles in every direction. I was once amused to discover on a flight
from Phoenix to Los Angeles that I could see both cities from my window at the
same time at the midpoint of the flight.

[Mxsmanic]

Tina writes:

> Gee, wrong again. I haven't run the numbers, there may be a line of
> sight, but the only way a human pilot would see a tower top under
> these conditions is at night if it had a bright flashing light on top
> of it. Real eyeballs in the daytime would not be able to see it, even
> if it was in the line of sight.

An object 1000 feet in size would be visible from about 1100 nm away, under
ideal conditions. A flashing light could be visible from any distance,
depending on its brightness. I _did_ run the numbers.

[Bertie the Bunyip]

On 1 Jan, 22:07, Tina > wrote:
> You can be fairly sure he used someone else's equations for line of
> sight. I'd bet a significant sum he could not derive them himself. He
> and Euclid would not have gotten along.


Him and counting using popsicle sticks wouldn't get along!

Bertie

[Bertie the Bunyip]

On 1 Jan, 22:15, Mxsmanic > wrote:
> Ron Wanttaja writes:
> > No. *Pilots are accustomed to judging the altitude of other aircraft based on
> > whether it's above or below the horizon. *This is great for collision avoidance,
> > but doesn't work at longer ranges. *ANYTHING located at the viewer's horizon
> > will appear above the altitude of the observer, and it *is* due to the curvature
> > of the Earth. *See:
>
> >http://www.wanttaja.com/los.jpg
>
> > Normally, though, details at such distances are obscured by haze.
>
> But the original post made no mention of the horizon.
>
> On a flat surface of infinite extent, the horizon is always at eye level, no
> matter what your position. *On a flat surface of finite extent, it is always
> slightly below eye level, depending on how far away the edge of the surface
> is. *On a spherical planet, the horizon is still lower; to find its distance
> (assuming smooth terrain), add your altitude to the radius of the planet,
> square it, subtract the radius of the planet squared, and take the square root
> of the result. *At an altitude of 3000 feet above smooth terrain (such as
> water, or a dry lake), the horizon on Earth is 58 nm away. *If you are six
> feet tall and standing on the surface with smooth terrain (or if you are in a
> rowboat on a calm ocean, for example), the horizon is just 2.6 nm distant.
>
> Your diagram is interesting, but since it dramatically overstates the height
> of towers and dramatically understates the size of the planet, it's a bit
> misleading. *Your towers would be several times higher than the orbit of the
> International Space Station, and the aircraft would be in outer space.
>
> On a flat surface, anything moving down in your field of view is something
> you'll fly over, and anything moving up is something you'll hit. *The
> curvature of the Earth complicates this, but the curvature is gentle enough
> that anything affected by it is too far away to be an immediate hazard,
> anyway. *At 3000' AGL, you could see Mount Everest from 239 nm away, but since
> a small plane might take two hours reach it, you'd have plenty of time to
> evaluate it as a hazard.
>
> Thought experiments like this can be interesting. *People often say that the
> Concorde was wonderful because you could see the curvature of the Earth, but
> the fact is that you can see the curvature from anywhere, even a hill
> overlooking the beach. *It just gets more obvious as you move further away
> from the surface. *From an airliner at 39,000 feet, the view extends for well
> over 210 miles in every direction. *I was once amused to discover on a flight
> from Phoenix to Los Angeles that I could see both cities from my window at the
> same time at the midpoint of the flight.


the only curavture you've ever seen is that of your butt as you
inserted your head all those years ago.

Bertie

[[email protected]]

Mxsmanic > wrote:
> Tina writes:

> > Gee, wrong again. I haven't run the numbers, there may be a line of
> > sight, but the only way a human pilot would see a tower top under
> > these conditions is at night if it had a bright flashing light on top
> > of it. Real eyeballs in the daytime would not be able to see it, even
> > if it was in the line of sight.

> An object 1000 feet in size would be visible from about 1100 nm away, under
> ideal conditions. A flashing light could be visible from any distance,
> depending on its brightness. I _did_ run the numbers.

Yeah, maybe an object 1000 feet in diameter but not a tower 1000 feet
tall and 3 to 6 feet wide dipwad.


--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Tina > wrote:
> You can be fairly sure he used someone else's equations for line of
> sight. I'd bet a significant sum he could not derive them himself. He
> and Euclid would not have gotten along.

True and the equations are easy to find on the internet, but they are
all rough approximations.

There is the geometric horizon which assumes the Earth is a perfectly
round billiard ball and the optical horizon which attempts to account
for the fact that the atmosphere bends light and increases the
distance around 10% depending on state of the atmosphere between the
two points.

Given all the ambiguities in the problem, numbers like 92 instead of
"approximatly 90" just show someone can punch numbers into a calculator
without any understanding of the true nature of the problem.

What a surprise.

--
Jim Pennino

Remove .spam.sux to reply.

[Tina]

MX's calculations remember something that was said in an undergraduate
physics course I took: "Assume a spherical cow. . . "

On Jan 1, 6:05*pm, wrote:
> Tina > wrote:
> > You can be fairly sure he used someone else's equations for line of
> > sight. I'd bet a significant sum he could not derive them himself. He
> > and Euclid would not have gotten along.
>
> True and the equations are easy to find on the internet, but they are
> all rough approximations.
>
> There is the geometric horizon which assumes the Earth is a perfectly
> round billiard ball and the optical horizon which attempts to account
> for the fact that the atmosphere bends light and increases the
> distance around 10% depending on state of the atmosphere between the
> two points.
>
> Given all the ambiguities in the problem, numbers like 92 instead of
> "approximatly 90" just show someone can punch numbers into a calculator
> without any understanding of the true nature of the problem.
>
> What a surprise.
>
> --
> Jim Pennino
>
> Remove .spam.sux to reply.

[Ron Wanttaja]

On Tue, 01 Jan 2008 23:15:48 +0100, Mxsmanic > wrote:

> Ron Wanttaja writes:
>
> > No. Pilots are accustomed to judging the altitude of other aircraft based on
> > whether it's above or below the horizon. This is great for collision avoidance,
> > but doesn't work at longer ranges. ANYTHING located at the viewer's horizon
> > will appear above the altitude of the observer, and it *is* due to the curvature
> > of the Earth. See:
> >
> > http://www.wanttaja.com/los.jpg
> >
> > Normally, though, details at such distances are obscured by haze.
>
> But the original post made no mention of the horizon.

But the original post was made by a pilot, who could naturally be assumed to
estimate relative altitudes the way pilots do.

> Your diagram is interesting, but since it dramatically overstates the height
> of towers and dramatically understates the size of the planet, it's a bit
> misleading.

Certainly, because a diagram to scale couldn't illustrate anything. While the
distances involved are exaggerated, the visual effects are the same.

The original poster didn't express fear that he was going to hit this far-off
object; he merely reported a curious observation where an object that was
definitely below his aircraft appeared, when first seen, to extend above his
flight path. You stated that this was *not* due to the curvature of the Earth,
I say it is.

> Your towers would be several times higher than the orbit of the
> International Space Station, and the aircraft would be in outer space.

The International Space Station *isn't* in outer space? Better tell NASA....

Ron Wanttaja

[WingFlaps]

On Jan 2, 11:19 am, Mxsmanic > wrote:
> Tina writes:
> > Gee, wrong again. I haven't run the numbers, there may be a line of
> > sight, but the only way a human pilot would see a tower top under
> > these conditions is at night if it had a bright flashing light on top
> > of it. Real eyeballs in the daytime would not be able to see it, even
> > if it was in the line of sight.
>
> An object 1000 feet in size would be visible from about 1100 nm away, under
> ideal conditions. A flashing light could be visible from any distance,
> depending on its brightness. I _did_ run the numbers.

Nonsense. If the observer were at 1000' the top of a 1000' tower would
be visible ~75 miles away. A flashing light would not be visible from
"any distance". Think about it -what if it were on the other side of
the planet!

Cheers MC

[Mxsmanic]

Ron Wanttaja writes:

> The International Space Station *isn't* in outer space? Better tell NASA....

They already know, since they have to boost it periodically in order to
compensate for drag from the atmosphere.

[Mxsmanic]

writes:

> Yeah, maybe an object 1000 feet in diameter but not a tower 1000 feet
> tall and 3 to 6 feet wide dipwad.

There are few thousand-foot towers that are only a metre wide.

[Tina]

nice catch re my typo -- but it would have to be a an interesting
tower to be seen from 75 miles with unaided vision. You'd have trouble
resolving an ordinary building at that range

On Jan 1, 8:37 pm, WingFlaps > wrote:
> On Jan 2, 11:19 am, Mxsmanic > wrote:
>
> > Tina writes:
> > > Gee, wrong again. I haven't run the numbers, there may be a line of
> > > sight, but the only way a human pilot would see a tower top under
> > > these conditions is at night if it had a bright flashing light on top
> > > of it. Real eyeballs in the daytime would not be able to see it, even
> > > if it was in the line of sight.
>
> > An object 1000 feet in size would be visible from about 1100 nm away, under
> > ideal conditions. A flashing light could be visible from any distance,
> > depending on its brightness. I _did_ run the numbers.
>
> Nonsense. If the observer were at 1000' the top of a 1000' tower would
> be visible ~75 miles away. A flashing light would not be visible from
> "any distance". Think about it -what if it were on the other side of
> the planet!
>
> Cheers MC

[Mxsmanic]

Tina writes:

> You can be fairly sure he used someone else's equations for line of
> sight. I'd bet a significant sum he could not derive them himself. He
> and Euclid would not have gotten along.

It's just simple trig. In fact, it's just solving for different sides of a
right triangle, as should be obvious from the description I gave.

[Tina]

prove it.
show your work

On Jan 1, 8:53 pm, Mxsmanic > wrote:
> Tina writes:
> > You can be fairly sure he used someone else's equations for line of
> > sight. I'd bet a significant sum he could not derive them himself. He
> > and Euclid would not have gotten along.
>
> It's just simple trig. In fact, it's just solving for different sides of a
> right triangle, as should be obvious from the description I gave.

[[email protected]]

Mxsmanic > wrote:
> Ron Wanttaja writes:

> > The International Space Station *isn't* in outer space? Better tell NASA....

> They already know, since they have to boost it periodically in order to
> compensate for drag from the atmosphere.

Correct answer to the wrong question.

The International Space Station orbits at about 400 km; the official
definition of outer space is 100 km.


--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Mxsmanic > wrote:
> writes:

> > Yeah, maybe an object 1000 feet in diameter but not a tower 1000 feet
> > tall and 3 to 6 feet wide dipwad.

> There are few thousand-foot towers that are only a metre wide.

Not at the bottom, but there are few towers that are more than a meter
or so wide at the top, which is the part one would see at a distance
dipwad.

Also towers are not solid, they are a skeleton frame work, so what
you have to see at a distance is tubing on the order of a couple
of inches in diameter.

--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Mxsmanic > wrote:
> Tina writes:

> > You can be fairly sure he used someone else's equations for line of
> > sight. I'd bet a significant sum he could not derive them himself. He
> > and Euclid would not have gotten along.

> It's just simple trig. In fact, it's just solving for different sides of a
> right triangle, as should be obvious from the description I gave.

Wrong; the optical line of sight is different than the geometric line of
sight because the atmosphere bends light.


--
Jim Pennino

Remove .spam.sux to reply.

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Tina writes:
>
>> You can be fairly sure he used someone else's equations for line of
>> sight. I'd bet a significant sum he could not derive them himself. He
>> and Euclid would not have gotten along.
>
> It's just simple trig. In fact, it's just solving for different sides
> of a right triangle, as should be obvious from the description I gave.
>

Nope


Bertie

[Ron Wanttaja]

On Wed, 02 Jan 2008 02:52:09 +0100, Mxsmanic > wrote:

> Ron Wanttaja writes:
>
> > The International Space Station *isn't* in outer space? Better tell NASA....
>
> They already know, since they have to boost it periodically in order to
> compensate for drag from the atmosphere.

Hay-el, if you use that as a criteria, the Shuttle doesn't go into outer space,
either. You get measurable atmospheric drag out to 1000 km or more. The
internationally-agreed boundary for space starts is at 100 km.

Ron Wanttaja

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> writes:
>
>> Yeah, maybe an object 1000 feet in diameter but not a tower 1000 feet
>> tall and 3 to 6 feet wide dipwad.
>
> There are few thousand-foot towers that are only a metre wide.
>

Wrong again.



Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Tina writes:
>
>> Gee, wrong again. I haven't run the numbers, there may be a line of
>> sight, but the only way a human pilot would see a tower top under
>> these conditions is at night if it had a bright flashing light on top
>> of it. Real eyeballs in the daytime would not be able to see it, even
>> if it was in the line of sight.
>
> An object 1000 feet in size would be visible from about 1100 nm away,
> under ideal conditions. A flashing light could be visible from any
> distance, depending on its brightness. I _did_ run the numbers.


Nope, you're an idiot


Bertie
>

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Ron Wanttaja writes:
>
>> The International Space Station *isn't* in outer space? Better tell
>> NASA....
>
> They already know, since they have to boost it periodically in order
> to compensate for drag from the atmosphere.

Nope


Bertie
>

[Mxsmanic]

writes:

> The International Space Station orbits at about 400 km; the official
> definition of outer space is 100 km.

Which "official" definition did you have in mind?

[Mxsmanic]

Ron Wanttaja writes:

> Hay-el, if you use that as a criteria, the Shuttle doesn't go into outer space,
> either. You get measurable atmospheric drag out to 1000 km or more.

Yes.

> The internationally-agreed boundary for space starts is at 100 km.

100 km above the surface of the sun is still a pretty wild place. Which
international agreement did you have in mind?

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> writes:
>
>> The International Space Station orbits at about 400 km; the official
>> definition of outer space is 100 km.
>
> Which "official" definition did you have in mind?
>

I think he said the definition that says that outer space is 100 km.

Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Ron Wanttaja writes:
>
>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into
>> outer space, either. You get measurable atmospheric drag out to 1000
>> km or more.
>
> Yes.
>
>> The internationally-agreed boundary for space starts is at 100 km.
>
> 100 km above the surface of the sun is still a pretty wild place.
> Which international agreement did you have in mind?

Apparently the one that says outer space is 100 km up.


Bertie
>

[John Mazor[_2_]]

"Mxsmanic" > wrote in message
...
> Ron Wanttaja writes:
>
>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into outer space,
>> either. You get measurable atmospheric drag out to 1000 km or more.
>
> Yes.
>
>> The internationally-agreed boundary for space starts is at 100 km.
>
> 100 km above the surface of the sun is still a pretty wild place. Which
> international agreement did you have in mind?

If your familiarity with Wikipedia extended beyond authoring articles on flypaper and
being rejected for editor status, you'd know the definition as set by the Fédération
Aéronautique Internationale:

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

[Andy Hawkins]

Hi,

In article >,
Bertie the > wrote:
> Well, even at middling altitudes I find it hard to tell the altitude of
> other airplanes.

I read somewhere that if the other plane appears above the horizon, it's
above you. Similarly, if it appears below the horizon, it's below you.

Obviously this will change depending on whether you're climbing or
descending, but as a general rule it sounds like it could make sense.

Haven't really had the chance to test it myself since I read it though.

Andy

[Bertie the Bunyip[_19_]]

Andy Hawkins > wrote in
:

> Hi,
>
> In article >,
> Bertie the > wrote:
>> Well, even at middling altitudes I find it hard to tell the altitude
>> of other airplanes.
>
> I read somewhere that if the other plane appears above the horizon,
> it's above you. Similarly, if it appears below the horizon, it's below
> you.
>
> Obviously this will change depending on whether you're climbing or
> descending, but as a general rule it sounds like it could make sense.
>
> Haven't really had the chance to test it myself since I read it
> though.
>

Well, we often can't see the horizon, suppose that;s the difficulty. It's
easier low down, though. Lots easier.

Bertie

[Bertie the Bunyip[_19_]]

"John Mazor" > wrote in
news:rWPej.1335$v_4.524@trnddc03:

>
> "Mxsmanic" > wrote in message
> ...
>> Ron Wanttaja writes:
>>
>>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into
>>> outer space, either. You get measurable atmospheric drag out to
>>> 1000 km or more.
>>
>> Yes.
>>
>>> The internationally-agreed boundary for space starts is at 100 km.
>>
>> 100 km above the surface of the sun is still a pretty wild place.
>> Which international agreement did you have in mind?
>
> If your familiarity with Wikipedia extended beyond authoring articles
> on flypaper and being rejected for editor status, you'd know the
> definition as set by the Fédération Aéronautique Internationale:
>
> http://en.wikipedia.org/wiki/Karman_line
>
>
>


Bwawhawhahwhahwh!

You're kidding about the flypaper, right?

Berie
>

[John Mazor[_2_]]

"Bertie the Bunyip" > wrote in message
.. .
> "John Mazor" > wrote in
> news:rWPej.1335$v_4.524@trnddc03:
>
>>
>> "Mxsmanic" > wrote in message
>> ...
>>> Ron Wanttaja writes:
>>>
>>>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into
>>>> outer space, either. You get measurable atmospheric drag out to
>>>> 1000 km or more.
>>>
>>> Yes.
>>>
>>>> The internationally-agreed boundary for space starts is at 100 km.
>>>
>>> 100 km above the surface of the sun is still a pretty wild place.
>>> Which international agreement did you have in mind?
>>
>> If your familiarity with Wikipedia extended beyond authoring articles
>> on flypaper and being rejected for editor status, you'd know the
>> definition as set by the Fédération Aéronautique Internationale:
>>
>> http://en.wikipedia.org/wiki/Karman_line
>
> Bwawhawhahwhahwh!
>
> You're kidding about the flypaper, right?

Nope.

And it was a half-ass effort at that. He didn't even bother using his photographic
"talents" to contribute an original pic of flypaper, he just linked to an existing generic
wiki shot.

[Mxsmanic]

John Mazor writes:

> If your familiarity with Wikipedia extended beyond authoring articles on flypaper and
> being rejected for editor status, you'd know the definition as set by the Fédération
> Aéronautique Internationale:
>
> http://en.wikipedia.org/wiki/Karman_line

What makes their definition special?

[george]

On Jan 3, 6:41 am, Andy Hawkins > wrote:
> Hi,
>
> In article >,
> Bertie the > wrote:
>
> > Well, even at middling altitudes I find it hard to tell the altitude of
> > other airplanes.
>
> I read somewhere that if the other plane appears above the horizon, it's
> above you. Similarly, if it appears below the horizon, it's below you.
>
> Obviously this will change depending on whether you're climbing or
> descending, but as a general rule it sounds like it could make sense.
>
> Haven't really had the chance to test it myself since I read it though.
>

I stopped at putting the finger on the wind shield over the other
aircraft.
If it moves from under my finger I keep on course.
If it doesn't appear I change course, altitude or airspeed until it
does appear.
The horizon trick seems valid though I've never tried (or even thought
of it until now)
Thanks

[Rich Ahrens[_2_]]

Bertie the Bunyip wrote:
> "John Mazor" > wrote in
> news:rWPej.1335$v_4.524@trnddc03:
>
>> "Mxsmanic" > wrote in message
>> ...
>>> Ron Wanttaja writes:
>>>
>>>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into
>>>> outer space, either. You get measurable atmospheric drag out to
>>>> 1000 km or more.
>>> Yes.
>>>
>>>> The internationally-agreed boundary for space starts is at 100 km.
>>> 100 km above the surface of the sun is still a pretty wild place.
>>> Which international agreement did you have in mind?
>> If your familiarity with Wikipedia extended beyond authoring articles
>> on flypaper and being rejected for editor status, you'd know the
>> definition as set by the Fédération Aéronautique Internationale:
>>
>> http://en.wikipedia.org/wiki/Karman_line
>
> Bwawhawhahwhahwh!
>
> You're kidding about the flypaper, right?

No, he isn't. Too bad Anthony doesn't take his own advice with which he
edited the UNICOM entry. He wrote, "Flight simulation trivia doesn't
belong in a real-world discussion of aviation."

[Tina]

As in many things, arbitrary choices are made and agreed upon. You can
choose to accept those definitions or not, but if you don't you have
trouble communicating with the community of experts who do use the
words in the way other professionals understand them to be used. The
phrase near infrared comes to mind, and now we can add outer space as
evidence of your lack of comprehensive abilities. Other than literate
postings and playing sim games, is there anything for which you can
claim expert status.

Oh, I forgot the most obvious ones -- you inspire Bertie.


An enduring trait among true professionals is their willingness to
admit when they are wrong. Insecure people seem to have trouble with
that.

I am still waiting to see you derive, using right triangles, line of
sight distances over a horizon defined by ones altitude over a sphere.

Quick -- google to the rescue!



On Jan 2, 2:27 pm, Mxsmanic > wrote:
> John Mazor writes:
> > If your familiarity with Wikipedia extended beyond authoring articles on flypaper and
> > being rejected for editor status, you'd know the definition as set by the Fédération
> > Aéronautique Internationale:
>
> >http://en.wikipedia.org/wiki/Karman_line
>
> What makes their definition special?

[Gig601XLBuilder]

Mxsmanic wrote:
> John Mazor writes:
>
>> If your familiarity with Wikipedia extended beyond authoring articles on flypaper and
>> being rejected for editor status, you'd know the definition as set by the Fédération
>> Aéronautique Internationale:
>>
>> http://en.wikipedia.org/wiki/Karman_line
>
> What makes their definition special?


Because, much like your ignorance, it is internationally recognized.

[Bertie the Bunyip[_19_]]

"John Mazor" > wrote in
news:WxRej.87645$NL5.49053@trnddc05:

>
> "Bertie the Bunyip" > wrote in message
> .. .
>> "John Mazor" > wrote in
>> news:rWPej.1335$v_4.524@trnddc03:
>>
>>>
>>> "Mxsmanic" > wrote in message
>>> ...
>>>> Ron Wanttaja writes:
>>>>
>>>>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into
>>>>> outer space, either. You get measurable atmospheric drag out to
>>>>> 1000 km or more.
>>>>
>>>> Yes.
>>>>
>>>>> The internationally-agreed boundary for space starts is at 100 km.
>>>>
>>>> 100 km above the surface of the sun is still a pretty wild place.
>>>> Which international agreement did you have in mind?
>>>
>>> If your familiarity with Wikipedia extended beyond authoring
>>> articles on flypaper and being rejected for editor status, you'd
>>> know the definition as set by the Fédération Aéronautique
>>> Internationale:
>>>
>>> http://en.wikipedia.org/wiki/Karman_line
>>
>> Bwawhawhahwhahwh!
>>
>> You're kidding about the flypaper, right?
>
> Nope.
>
> And it was a half-ass effort at that. He didn't even bother using his
> photographic "talents" to contribute an original pic of flypaper, he
> just linked to an existing generic wiki shot.
>

Good grief.


Well, thank god he;s out there doig it for the sake of civilisation, I
say.

Bertie

[Bertie the Bunyip[_19_]]

Rich Ahrens > wrote in news:477be7de$0$1115$804603d3
@auth.newsreader.iphouse.com:

> Bertie the Bunyip wrote:
>> "John Mazor" > wrote in
>> news:rWPej.1335$v_4.524@trnddc03:
>>
>>> "Mxsmanic" > wrote in message
>>> ...
>>>> Ron Wanttaja writes:
>>>>
>>>>> Hay-el, if you use that as a criteria, the Shuttle doesn't go into
>>>>> outer space, either. You get measurable atmospheric drag out to
>>>>> 1000 km or more.
>>>> Yes.
>>>>
>>>>> The internationally-agreed boundary for space starts is at 100 km.
>>>> 100 km above the surface of the sun is still a pretty wild place.
>>>> Which international agreement did you have in mind?
>>> If your familiarity with Wikipedia extended beyond authoring
articles
>>> on flypaper and being rejected for editor status, you'd know the
>>> definition as set by the Fédération Aéronautique Internationale:
>>>
>>> http://en.wikipedia.org/wiki/Karman_line
>>
>> Bwawhawhahwhahwh!
>>
>> You're kidding about the flypaper, right?
>
> No, he isn't. Too bad Anthony doesn't take his own advice with which
he
> edited the UNICOM entry. He wrote, "Flight simulation trivia doesn't
> belong in a real-world discussion of aviation."
>

And it doesn't get more trivial than him.


Bertie

[Ross]

Marc J. Zeitlin wrote:
> Ron Wanttaja wrote:
>
>> Certainly, because a diagram to scale couldn't illustrate anything.
>> While the distances involved are exaggerated, the visual effects
>> are the same.
>
>
> Ron, thanks for the diagram and the explanation. I frequently fly with
> my wife out here (socal) in the mountains, and she will regularly look
> out at a ridge 50-100 miles away from us when we're cruising at
> 8500-9500 ft. and say "those mountains look higher than us - are we
> going to hit them?", or the functional equivalent. I glance at the map,
> say "well, they're 7500 ft high, so we're 1K-2K ft above them, but
> acknowledge that they DO look higher than our altitude, judging from the
> horizon position.
>
> Of course, when we get there, we find that we're above the ridge, just
> like the map says (it's never wrong [so far]) and we shrug our shoulders
> and go "huh - how do you like that".
>
> Now I have the explanation for her (and me) as to why it looks like it
> does - thanks!
>

I had the same issue flying to Tucson from the east one time. The
mountains looked much higher than I was. But looking at the charts and
the MEAs, I was fine at my altitude. As we motored along, they went
right below us.

--

Regards, Ross
C-172F 180HP
KSWI

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> John Mazor writes:
>
>> If your familiarity with Wikipedia extended beyond authoring articles
>> on flypaper and being rejected for editor status, you'd know the
>> definition as set by the Fédération Aéronautique Internationale:
>>
>> http://en.wikipedia.org/wiki/Karman_line
>
> What makes their definition special?
>

What makes you "special"?


Bertie

[Bertie the Bunyip[_19_]]

Gig601XLBuilder > wrote in
:

> Mxsmanic wrote:
>> John Mazor writes:
>>
>>> If your familiarity with Wikipedia extended beyond authoring
>>> articles on flypaper and being rejected for editor status, you'd
>>> know the definition as set by the Fédération Aéronautique
>>> Internationale:
>>>
>>> http://en.wikipedia.org/wiki/Karman_line
>>
>> What makes their definition special?
>
>
> Because, much like your ignorance, it is internationally recognized.
>

Boom! headshot!

Bertie

[Rich Ahrens[_2_]]

Tina wrote:
> As in many things, arbitrary choices are made and agreed upon. You can
> choose to accept those definitions or not, but if you don't you have
> trouble communicating with the community of experts who do use the
> words in the way other professionals understand them to be used. The
> phrase near infrared comes to mind, and now we can add outer space as
> evidence of your lack of comprehensive abilities. Other than literate
> postings and playing sim games, is there anything for which you can
> claim expert status.

Well, he has claimed expertise in breastfeeding...

[John Mazor[_2_]]

"Mxsmanic" > wrote in message
...
> John Mazor writes:
>
>> If your familiarity with Wikipedia extended beyond authoring articles on flypaper and
>> being rejected for editor status, you'd know the definition as set by the Fédération
>> Aéronautique Internationale:
>>
>> http://en.wikipedia.org/wiki/Karman_line
>
> What makes their definition special?

YGBSM. (No, make that YASM.)

It applies to the topic under discussion and it's accepted internationally - which is far
more than we can say for your aviation credentials.

Do you get blackouts when you jink this hard to avoid admitting that you've been shot down
in flames (so to speak) again?

[John Mazor[_2_]]

"Bertie the Bunyip" > wrote in message
.. .
> Gig601XLBuilder > wrote in
> :
>
>> Mxsmanic wrote:
>>> John Mazor writes:
>>>
>>>> If your familiarity with Wikipedia extended beyond authoring
>>>> articles on flypaper and being rejected for editor status, you'd
>>>> know the definition as set by the Fédération Aéronautique
>>>> Internationale:
>>>>
>>>> http://en.wikipedia.org/wiki/Karman_line
>>>
>>> What makes their definition special?
>>
>>
>> Because, much like your ignorance, it is internationally recognized.
>
> Boom! headshot!

He's too hard-headed to even have noticed that.

[JGalban via AviationKB.com]

Bertie the Bunyip wrote:
>>> If your familiarity with Wikipedia extended beyond authoring articles
>>> on flypaper and being rejected for editor status, you'd know the
>[quoted text clipped - 3 lines]
>>
>> What makes their definition special?
>
>What makes you "special"?
>

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

--
Message posted via http://www.aviationkb.com

[Morgans[_2_]]

"Ron Wanttaja" <> wrote

> Hay-el, if you use that as a criteria, the Shuttle doesn't go into outer
> space,
> either. You get measurable atmospheric drag out to 1000 km or more. The
> internationally-agreed boundary for space starts is at 100 km.

Speaking of boosts back to orbit, I have always been surprised that it is
done with such a sudden jolt of the (whatever type) thrusters they use.

Don't jolts like that screw up some of the weightless experiments? Why are
there not lower powered thrusters that would boost the space vehicles in a
smoother, slower manner?
--
Jim in NC

[[email protected]]

Morgans > wrote:

> "Ron Wanttaja" <> wrote

> > Hay-el, if you use that as a criteria, the Shuttle doesn't go into outer
> > space,
> > either. You get measurable atmospheric drag out to 1000 km or more. The
> > internationally-agreed boundary for space starts is at 100 km.

> Speaking of boosts back to orbit, I have always been surprised that it is
> done with such a sudden jolt of the (whatever type) thrusters they use.

> Don't jolts like that screw up some of the weightless experiments? Why are
> there not lower powered thrusters that would boost the space vehicles in a
> smoother, slower manner?

What weightless experiments?

Because of the way the station is oriented, there is no place on it
with a true weightless envirionment; micro gravity yes, zero gravity no.


--
Jim Pennino

Remove .spam.sux to reply.

[Ron Wanttaja]

On Wed, 2 Jan 2008 17:28:30 -0500, "Morgans" > wrote:

>
> "Ron Wanttaja" <> wrote
>
> > Hay-el, if you use that as a criteria, the Shuttle doesn't go into outer
> > space,
> > either. You get measurable atmospheric drag out to 1000 km or more. The
> > internationally-agreed boundary for space starts is at 100 km.
>
> Speaking of boosts back to orbit, I have always been surprised that it is
> done with such a sudden jolt of the (whatever type) thrusters they use.
>
> Don't jolts like that screw up some of the weightless experiments? Why are
> there not lower powered thrusters that would boost the space vehicles in a
> smoother, slower manner?

I really don't know much about Space Station operation; I was on the team with a
(winning) proposal back in the '80s, and haven't worked on it since. But
looking quickly at some data, it has about 200,000 pounds of mass (feel free to
convert to slugs :-). Unless they've got some big honking engines, I suspect
the ride is pretty soft. Back of the envelope calculation, I get about 1/20th
of a G if they're using the Shuttle OMS engines with the Shuttle docked.

Most of the experiments have to be tolerant of some microgravity variation,
anyway.

Ron Wanttaja

[Mxsmanic]

Tina writes:

> As in many things, arbitrary choices are made and agreed upon. You can
> choose to accept those definitions or not, but if you don't you have
> trouble communicating with the community of experts who do use the
> words in the way other professionals understand them to be used.

Which "experts" and "professionals" do you have in mind?

> An enduring trait among true professionals is their willingness to
> admit when they are wrong. Insecure people seem to have trouble with
> that.

Another one of their enduring traits is a reluctance to attack others.

> I am still waiting to see you derive, using right triangles, line of
> sight distances over a horizon defined by ones altitude over a sphere.

I didn't realize you needed an explanation.

If A is your eyepoint, and B is where your line of sight touches the horizon,
and C is the center of the Earth, ABC form a right triangle. You know the
distances CA and CB, so all you have to do is solve for AB. Even for someone
who hates math as much as I do, it's pretty straightforward.

[WingFlaps]

On Jan 3, 9:53 pm, Mxsmanic > wrote:
> Tina writes:
> > As in many things, arbitrary choices are made and agreed upon. You can
> > choose to accept those definitions or not, but if you don't you have
> > trouble communicating with the community of experts who do use the
> > words in the way other professionals understand them to be used.
>
> Which "experts" and "professionals" do you have in mind?
>
> > An enduring trait among true professionals is their willingness to
> > admit when they are wrong. Insecure people seem to have trouble with
> > that.
>
> Another one of their enduring traits is a reluctance to attack others.
>
> > I am still waiting to see you derive, using right triangles, line of
> > sight distances over a horizon defined by ones altitude over a sphere.
>
> I didn't realize you needed an explanation.
>
> If A is your eyepoint, and B is where your line of sight touches the horizon,
> and C is the center of the Earth, ABC form a right triangle. You know the
> distances CA and CB, so all you have to do is solve for AB. Even for someone
> who hates math as much as I do, it's pretty straightforward.

And yet you got it wrong. Care to explain why?

Cheers

[Mxsmanic]

WingFlaps writes:

> And yet you got it wrong. Care to explain why?

You're the one who says it's wrong, so it's up to you to explain why.

[Bertie the Bunyip[_19_]]

"John Mazor" > wrote in news:zJSej.16112$DG4.3791
@trnddc04:

>
> "Bertie the Bunyip" > wrote in message
> .. .
>> Gig601XLBuilder > wrote in
>> :
>>
>>> Mxsmanic wrote:
>>>> John Mazor writes:
>>>>
>>>>> If your familiarity with Wikipedia extended beyond authoring
>>>>> articles on flypaper and being rejected for editor status, you'd
>>>>> know the definition as set by the Fédération Aéronautique
>>>>> Internationale:
>>>>>
>>>>> http://en.wikipedia.org/wiki/Karman_line
>>>>
>>>> What makes their definition special?
>>>
>>>
>>> Because, much like your ignorance, it is internationally recognized.
>>
>> Boom! headshot!
>
> He's too hard-headed to even have noticed that.
>
>
>

True.

Bertie

[Bertie the Bunyip[_19_]]

"JGalban via AviationKB.com" <u32749@uwe> wrote in news:7d9fafc9f4e18@uwe:

> Bertie the Bunyip wrote:
>>>> If your familiarity with Wikipedia extended beyond authoring articles
>>>> on flypaper and being rejected for editor status, you'd know the
>>[quoted text clipped - 3 lines]
>>>
>>> What makes their definition special?
>>
>>What makes you "special"?
>>
>
> http://en.wikipedia.org/wiki/Short_bus
>

You saying he edited this?


Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Tina writes:
>
>> As in many things, arbitrary choices are made and agreed upon. You
>> can choose to accept those definitions or not, but if you don't you
>> have trouble communicating with the community of experts who do use
>> the words in the way other professionals understand them to be used.
>
> Which "experts" and "professionals" do you have in mind?
>
>> An enduring trait among true professionals is their willingness to
>> admit when they are wrong. Insecure people seem to have trouble with
>> that.
>
> Another one of their enduring traits is a reluctance to attack others.
>
>> I am still waiting to see you derive, using right triangles, line of
>> sight distances over a horizon defined by ones altitude over a
>> sphere.
>
> I didn't realize you needed an explanation.
>
> If A is your eyepoint, and B is where your line of sight touches the
> horizon, and C is the center of the Earth, ABC form a right triangle.
> You know the distances CA and CB, so all you have to do is solve for
> AB. Even for someone who hates math as much as I do, it's pretty
> straightforward.
>

Nope

Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> WingFlaps writes:
>
>> And yet you got it wrong. Care to explain why?
>
> You're the one who says it's wrong, so it's up to you to explain why.
>

Nope


Bertie

[[email protected]]

Mxsmanic > wrote:
> WingFlaps writes:

> > And yet you got it wrong. Care to explain why?

> You're the one who says it's wrong, so it's up to you to explain why.

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.

--
Jim Pennino

Remove .spam.sux to reply.

[Tina]

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. That is what one is trying to
determine, even in this very simple minded example.

It's simple minded, and yet, it seems, too complex for you to
understand.

So no, you can't on your own solve this. Quick, google to the answer!




glles, On Jan 3, 3:53 am, Mxsmanic > wrote:
> Tina writes:
> > As in many things, arbitrary choices are made and agreed upon. You can
> > choose to accept those definitions or not, but if you don't you have
> > trouble communicating with the community of experts who do use the
> > words in the way other professionals understand them to be used.
>
> Which "experts" and "professionals" do you have in mind?
>
> > An enduring trait among true professionals is their willingness to
> > admit when they are wrong. Insecure people seem to have trouble with
> > that.
>
> Another one of their enduring traits is a reluctance to attack others.
>
> > I am still waiting to see you derive, using right triangles, line of
> > sight distances over a horizon defined by ones altitude over a sphere.
>
> I didn't realize you needed an explanation.
>
> If A is your eyepoint, and B is where your line of sight touches the horizon,
> and C is the center of the Earth, ABC form a right triangle. You know the
> distances CA and CB, so all you have to do is solve for AB. Even for someone
> who hates math as much as I do, it's pretty straightforward.

[Mxsmanic]

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.

[[email protected]]

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.

--
Jim Pennino

Remove .spam.sux to reply.

[WingFlaps]

On Jan 3, 10:49 pm, Mxsmanic > wrote:
> WingFlaps writes:
> > And yet you got it wrong. Care to explain why?
>
> You're the one who says it's wrong, so it's up to you to explain why.

So you still think a 1000' tower can be seen over 1000 miles away? Was
this another MS flight sim experience?

Bwhahhahahhahaha.

Now, don't be petulant- just try to engage some common sense -does it
even sound plausible? Could the empire state building really be seen
1/3 of the way across the Atlantic? Can people in London see the
Eiffel tower or people in Paris see the PO tower in London?

Cheers

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> 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


Nope



Bertie

[Bertie the Bunyip[_19_]]

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

[flynrider via AviationKB.com]

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.aspx/aviation/200801/1

[Mxsmanic]

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.

[Mxsmanic]

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?

[Tina]

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?

[John Mazor[_2_]]

"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.

[John Mazor[_2_]]

"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?

[[email protected]]

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.

[[email protected]]

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.

[Tina]

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.

[Tina]

Jim, you said

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


I don't think so. If R = 1 and D = 1, then the root of the sum of the
squares isn't 2, but 1.414. .


but for sure distance to the horizon is ((R + d) ^2 - R ^2) ^.5

maybe you were thinking (R + d)^2 as R^2 + 2Rd + d^2, so that the root
is really of (2Rd +d^2)

This would be fun, remembering some of the approximations when R>>d
(except in Mx's universe)

[[email protected]]

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

Probably infinitesimally less than 1.

I thought his 30 arc seconds and explaination of why it is so seemed a bit
bogus, so a quick Google reveals:

The theoretical resolving power of the eye is determined by the aperature
diameter (the pupil of the eye) and the wavelength of the light.

MX stated it was the size of the sensors.

The theoretical resolution of the eye with green light is about 20 arc
seconds.

But wait, there's more.

The actual resolving power of the human eye with 20/20 vision is typically
considered to be about one arc minute.

See http://www.tvtechnology.com/features/Tech-Corner/Hoffner_features.shtml
and http://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=1405458

--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Tina > wrote:
> Jim, you said

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


> I don't think so. If R = 1 and D = 1, then the root of the sum of the
> squares isn't 2, but 1.414. .

The equation is just the hypotenuse of a right triagle equals the sum
of the squares of the other two sides.

I have no idea why you think the root of the sum of the squares should
be 2.

If R = 1, and d = 1, h = 0.414.

There are no approximations in this.

--
Jim Pennino

Remove .spam.sux to reply.

[Tina]

remember too that resolving power doesn't mean seeing a single object,
but the ability to see that there are two lines with a space between
them -- to resolve them. The eye can see a single object that is much
less than a second of arc. Look at a star in a not very busy part of
the night sky. Resolving power would be more related to distinguishing
individual stars in a complex area, like the milky way.

Mx knows all of this, of course. No, wait. He nos this.


On Jan 3, 8:35*pm, wrote:
> Tina > wrote:
> > what are the chances mx wouldn't know what 30 seconds of arc is if a
> > clock fell on him?
>
> Probably infinitesimally less than 1.
>
> I thought his 30 arc seconds and explaination of why it is so seemed a bit
> bogus, so a quick Google reveals:
>
> The theoretical resolving power of the eye is determined by the aperature
> diameter (the pupil of the eye) and the wavelength of the light.
>
> MX stated it was the size of the sensors.
>
> The theoretical resolution of the eye with green light is about 20 arc
> seconds.
>
> But wait, there's more.
>
> The actual resolving power of the human eye with 20/20 vision is typically
> considered to be about one arc minute.
>
> Seehttp://www.tvtechnology.com/features/Tech-Corner/Hoffner_features.shtml
> andhttp://www.pubmedcentral.nih.gov/articlerender.fcgi?artid=1405458
>
> --
> Jim Pennino
>
> Remove .spam.sux to reply.

[Tina]

The equation in error was written as

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

I demonstrated this is an error by setting R and d to 1. The square
root of the sum of those squares is 1.414, yet the equal sign would
indicate it shoud be 2.

That is straightforward.

Never the less, the final answer that was posted was correct. As it
happens, that identity was not used later when the correct answer was
derived.

But both that post, and mine, provided the correct final equation, for
the geometric line of sight, but as to the visual one, it's subject to
both the assumptions that were llisted, and some that were not.

The good news is, ain't no pilot gonna worry about this stuff. No
airplane pilot, anyhow,but those who pilot boats on blue water think
about those things pretty often, We use them, for example, to
estimate distances off shore.

[Rip]

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.
>
>
>>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.

Anthony, you poor boy. Your understanding of the resolution of the human
eye is limited by your poor research techniques. As usual, you conducted
"research" until you arrived at the answer you sought, with no attempt
to ascertain whether there might be other factors involved. Allow me to
enlighten you. The resolving power of the human optical system is not
limited by the photoreceptors, but by the focusing system. If this were
not true, useful humans (defined as those other than yourself) would not
have bothered to invent the microscope or telescope. Granted, you may be
suffering from some cyborg-unique condition in which you have a single
binary photoreceptor in each (I take the liberty of presuming you have
two) eye. However, that is not the case for homo sapiens who actually
fly real aircraft.

Additionally, your grasp of simple trigonometry, let alone spherical
trig, is appallingly primitive. Do try to refrain from instructing your
betters until you have something useful to add. I stress the word
useful. That does not include meager attempts to grasp a subject by
"googling" it, or any other foul thing you might do in front of your
"simulator".

[Bertie the Bunyip[_19_]]

"flynrider via AviationKB.com" <u32749@uwe> wrote in
news:7dac9b47ee29d@uwe:

> 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.

Ah, OK. I just got a list of the crap he's edited in Wikipedia and it is as
long as it is disturbing.
>
>
> 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.
>


That should keep Larry from getting all netkkkopy on your ass.


Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> 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.
>

Wow, he gets better all the time. Breathtaking.




Bertie

[[email protected]]

Tina > wrote:
> The equation in error was written as

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

What error? That is the Pythagorean theorem.

Get a piece of paper, a straight edge, something to draw a big circle
and draw it out.

The line that represents the tower is of total length R + h, i.e. the
radius of the circle plus the tower height and is the hypotenuse of a
right triangle.

The other two sides of the triangle are:

The distance to the horizon, d, which is the line tangent to the circle
and whose length is from the tangent point to the top of the "tower".

The radius line, R, that intercepts the point where the distance line
intercepts the circle whose length is the radius of the circle.

> I demonstrated this is an error by setting R and d to 1. The square
> root of the sum of those squares is 1.414, yet the equal sign would
> indicate it shoud be 2.

> That is straightforward.

How is that?

Substituting 1 for R and d in the equation gives:
sqrt (1^2 + 1^2) = 1 + h
Evaluating 1^2 gives:
sqrt (1 + 1) = 1 + h
Adding 1 + 1 gives:
sqrt (2) = 1 + h
Taking the square root of 2 gives:
1.41 = 1 + h
Subtracting 1 from both sides gives:
..41 = h
QED

> Never the less, the final answer that was posted was correct. As it
> happens, that identity was not used later when the correct answer was
> derived.

What identity? The whole thing is the Pythagorean theorem.

> But both that post, and mine, provided the correct final equation, for
> the geometric line of sight, but as to the visual one, it's subject to
> both the assumptions that were llisted, and some that were not.

Yes.

> The good news is, ain't no pilot gonna worry about this stuff. No
> airplane pilot, anyhow,but those who pilot boats on blue water think
> about those things pretty often, We use them, for example, to
> estimate distances off shore.

Yep, and there are any number of rules of thumb to get a ballpark estimate.

--
Jim Pennino

Remove .spam.sux to reply.

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> 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?
>

The bit with the alleged words in it.


Bertie

[Tina]

Jim, you may wish to google Pythagorean. or you can take my word for
this: it is NOT what you had written and the equation .

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

is not true in the general case.

You are probably thinking of

A^2 + B^2 = C^2, where A and B are the two sides forming the right
angle. That, solving for the long side, is

C = sqrt(A^2 + B^2), a far cry from the identity mentioned above.

It's easy to forget these things, but in this case I have not.







On Jan 4, 12:45*am, wrote:
> Tina > wrote:
> > The equation in error was written as
> > > > > sqrt (R^2 + d^2) = R + h
>
> What error? That is the Pythagorean theorem.
>
> Get a piece of paper, a straight edge, something to draw a big circle
> and draw it out.
>
> The line that represents the tower is of total length R + h, i.e. the
> radius of the circle plus the tower height and is the hypotenuse of a
> right triangle.
>
> The other two sides of the triangle are:
>
> The distance to the horizon, d, *which is the line tangent to the circle
> and whose length is from the tangent point to the top of the "tower".
>
> The radius line, R, that intercepts the point where the distance line
> intercepts the circle whose length is the radius of the circle.
>
> > I demonstrated this is an error by setting R and d to 1. The square
> > root of the sum of those squares is 1.414, yet the equal sign would
> > indicate it shoud be 2.
> > That is straightforward.
>
> How is that?
>
> Substituting 1 for R and d in the equation gives:
> sqrt (1^2 + 1^2) = 1 + h
> Evaluating 1^2 gives:
> sqrt (1 + 1) = 1 + h
> Adding 1 + 1 gives:
> sqrt (2) = 1 + h
> Taking the square root of 2 gives:
> 1.41 = 1 + h
> Subtracting 1 from both sides gives:
> .41 = h
> QED
>
> > Never the less, the final answer that was posted was correct. As it
> > happens, that identity was not used later when the correct answer was
> > derived.
>
> What identity? The whole thing is the Pythagorean theorem.
>
> > But both that post, and mine, provided the correct final equation, for
> > the geometric line of sight, but as to the visual one, it's subject to
> > both the assumptions that were llisted, and some that were not.
>
> Yes.
>
> > The good news is, ain't no pilot gonna worry about this stuff. No
> > airplane pilot, anyhow,but those who pilot boats on blue water think
> > about those things pretty often, We use them, for example, *to
> > estimate *distances off shore.
>
> Yep, and there are any number of rules of thumb to get a ballpark estimate..
>
> --
> Jim Pennino
>
> Remove .spam.sux to reply.

[Mxsmanic]

John Mazor writes:

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

The key is the angular size of the object; whether this is vertical or
horizontal matters little.

> IOW you don't know.

I do know. If it's at least 30 seconds across, it's visible.

> Show the math for the limits of resolution.

The math is based on the size of cone cells in the retina. It works out to 30
seconds of arc.

> And remember, whichever tower you are using,
> you must use the narrower dimension, which would be width, not height.

Why?

[Mxsmanic]

writes:

> I thought his 30 arc seconds and explaination of why it is so seemed a bit
> bogus, so a quick Google reveals:

Google covers every site on the Web (almost), and isn't an authority in
itself.

> The theoretical resolving power of the eye is determined by the aperature
> diameter (the pupil of the eye) and the wavelength of the light.
>
> MX stated it was the size of the sensors.

It is.

> The theoretical resolution of the eye with green light is about 20 arc
> seconds.

Green light is easy to see. The density of green-light receptors in the eye
is about twice that of red or blue.

> The actual resolving power of the human eye with 20/20 vision is typically
> considered to be about one arc minute.

The actual resolving power under ideal conditions is as stated. Since
conditions are rarely ideal, in real-world conditions the resolving power is
often lower. Indeed, one minute is still rather optimistic.

[Mxsmanic]

Rip writes:

> Anthony, you poor boy. Your understanding of the resolution of the human
> eye is limited by your poor research techniques. As usual, you conducted
> "research" until you arrived at the answer you sought, with no attempt
> to ascertain whether there might be other factors involved.

No. I studied eye physiology long ago as part of another course of study that
required this knowledge.

> The resolving power of the human optical system is not
> limited by the photoreceptors, but by the focusing system.

No. The focusing system actually provides better resolution than the
photoreceptors, so the limiting factor is the size of the photoreceptors.

> If this were not true, useful humans (defined as those other than
> yourself) would not have bothered to invent the microscope or telescope.

I don't see the relevance of this.

> Additionally, your grasp of simple trigonometry, let alone spherical
> trig, is appallingly primitive. Do try to refrain from instructing your
> betters until you have something useful to add. I stress the word
> useful. That does not include meager attempts to grasp a subject by
> "googling" it, or any other foul thing you might do in front of your
> "simulator".

The trig required to determine the line-of-sight distance is indeed very
simple. If you see an error in the explanation I gave, feel free to correct
it. While the above seems to indicate that you have a problem with my
explanation, I find it curious that you don't explain what that problem is.

[Mxsmanic]

Tina writes:

> 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.

Yes. And since the light of sight is always tangent to the planet's surface
where it touches the horizon, this is in fact always a right triangle.

> 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.

I did. That's how you solve for the unknown side of the triangle, which is
the distance to the horizon.

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

I did, but I think I used nautical miles for the result.

> What part don't YOU understand?

I understood all of it from the beginning. But do keep talking.

[Mxsmanic]

writes:

> 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.

What "equations"? It's just the Pythagorean theorem: a^2 + b^2 = c^2.

[Tina]

Opps!

I looked at one of your earlier posts, maybe on 1/1, and you in fact
did have it right.




On Jan 4, 8:53*am, Mxsmanic > wrote:
> Tina writes:
> > 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.
>
> Yes. *And since the light of sight is always tangent to the planet's surface
> where it touches the horizon, this is in fact always a right triangle.
>
> > 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.
>
> I did. *That's how you solve for the unknown side of the triangle, which is
> the distance to the horizon.
>
> > Be sure to use consistant units: you do know there are 5280 feet in a
> > mile, don't you?
>
> I did, but I think I used nautical miles for the result.
>
> > What part don't YOU understand?
>
> I understood all of it from the beginning. *But do keep talking.

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Tina writes:
>
>> 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.
>
> Yes. And since the light of sight is always tangent to the planet's
> surface where it touches the horizon, this is in fact always a right
> triangle.
>
>> 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.
>
> I did. That's how you solve for the unknown side of the triangle,
> which is the distance to the horizon.
>
>> Be sure to use consistant units: you do know there are 5280 feet in a
>> mile, don't you?
>
> I did, but I think I used nautical miles for the result.
>
>> What part don't YOU understand?
>
> I understood all of it from the beginning. But do keep talking.
>
You don't understan anything.

Bertie

[Bertie the Bunyip[_19_]]

Tina > wrote in news:3e24cbaf-ddf0-4237-8262-
:

> Opps!
>
> I looked at one of your earlier posts, maybe on 1/1, and you in fact
> did have it right.
>


Rule number 1?


Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> Rip writes:
>
>> Anthony, you poor boy. Your understanding of the resolution of the
>> human eye is limited by your poor research techniques. As usual, you
>> conducted "research" until you arrived at the answer you sought, with
>> no attempt to ascertain whether there might be other factors
>> involved.
>
> No. I studied eye physiology long ago as part of another course of
> study that required this knowledge.
>



****ing around with Wiki is not "studying"



Bertie

>

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> writes:
>
>> I thought his 30 arc seconds and explaination of why it is so seemed
>> a bit bogus, so a quick Google reveals:
>
> Anthony covers every site on the Web (almost), and isn't an authority
> in itself.



Better


Bertie

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> John Mazor writes:
>
>> So now the tower is 1000' across as well as 1000' high?
>
> The key is the angular size of the object; whether this is vertical or
> horizontal matters little.
>
>> IOW you don't know.
>
> I do know. If it's at least 30 seconds across, it's visible.
>


Yeh. that's geting to be a popular design for towers.




Bertie

[[email protected]]

Mxsmanic > wrote:
> John Mazor writes:

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

> The key is the angular size of the object; whether this is vertical or
> horizontal matters little.

Wrong, it is both the horizontal and vertical.

A simple example shows you to be wrong; from how far away could one
see a 1000' human hair?

> > IOW you don't know.

> I do know. If it's at least 30 seconds across, it's visible.

Wrong.

> > Show the math for the limits of resolution.

> The math is based on the size of cone cells in the retina. It works out to 30
> seconds of arc.

Wrong; the math is based on the lens aperature (pupil for eyes) and the
light frequency and gives the theoretical maximum resolution. See any
Google article on resolution.

And the real resolution of the average 20/20 human eye based on research
is 60 arc seconds.

> > And remember, whichever tower you are using,
> > you must use the narrower dimension, which would be width, not height.

> Why?

Take a 4' piece of sewing thread, hang it on a wall and back away until
you can't see it.

Make the string 8' long and repeat.

Can you now see it from twice as far away?

No.

If you had read the entire Google artical you used to come up with
your answers, you would know why.


--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Mxsmanic > wrote:
> writes:

> > I thought his 30 arc seconds and explaination of why it is so seemed a bit
> > bogus, so a quick Google reveals:

> Google covers every site on the Web (almost), and isn't an authority in
> itself.

Correct but totally irrelevant.

What matters is the authority of the site that Google leads you to.

Another red herring post.

> > The theoretical resolving power of the eye is determined by the aperature
> > diameter (the pupil of the eye) and the wavelength of the light.
> >
> > MX stated it was the size of the sensors.

> It is.

It isn't and the laws of optics say you are full of ****.

> > The theoretical resolution of the eye with green light is about 20 arc
> > seconds.

> Green light is easy to see. The density of green-light receptors in the eye
> is about twice that of red or blue.

True but irrelevant to the issue.

> > The actual resolving power of the human eye with 20/20 vision is typically
> > considered to be about one arc minute.

> The actual resolving power under ideal conditions is as stated. Since
> conditions are rarely ideal, in real-world conditions the resolving power is
> often lower. Indeed, one minute is still rather optimistic.

Wrong.

If you had bothered to follow the link you cut and read the research paper
you would know that.

You have been shown to be wrong about everything you have said about
the human eye and been provided with references that show why you are
wrong, but your ego won't allow you to accept that truth.


--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Mxsmanic > wrote:
> Rip writes:

> > Anthony, you poor boy. Your understanding of the resolution of the human
> > eye is limited by your poor research techniques. As usual, you conducted
> > "research" until you arrived at the answer you sought, with no attempt
> > to ascertain whether there might be other factors involved.

> No. I studied eye physiology long ago as part of another course of study that
> required this knowledge.

And obviously either payed no attention or failed horribly since it is
trivial to show that everything you have said about the human eye is
wrong.


--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Tina > wrote:
> Jim, you may wish to google Pythagorean. or you can take my word for
> this: it is NOT what you had written and the equation .

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

> is not true in the general case.

> You are probably thinking of

> A^2 + B^2 = C^2, where A and B are the two sides forming the right
> angle. That, solving for the long side, is

> C = sqrt(A^2 + B^2), a far cry from the identity mentioned above.

The two equation are identical.

> It's easy to forget these things, but in this case I have not.

You have the equation correct, but have forgotten basic algebra.

The following two equations are identical:

y = x^2
sqrt (y) = x

For the original equation:

Let R = A
Let d = B
Let (R + h) = C

Substituting into my original equation we get:

sqrt (A^2 + B^2) = C

Square both sides of the equation and we get:

(A^2 + B^2) = C^2

The parentheses are now irrelevant, so when removed we have:

A^2 + B^2 = C^2

Substitute back in the original values we get:

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

QED


See http://en.wikipedia.org/wiki/Pythagorean_theorem




--
Jim Pennino

Remove .spam.sux to reply.

[[email protected]]

Mxsmanic > wrote:
> writes:

> > 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.

> What "equations"? It's just the Pythagorean theorem: a^2 + b^2 = c^2.

Strange you could't just say that until it was already posted.


--
Jim Pennino

Remove .spam.sux to reply.

[Tina]

I reread your definitions of the symbols, and do stand corrected:
you're right.


On Jan 4, 12:15*pm, wrote:
> Tina > wrote:
> > Jim, you may wish to google Pythagorean. or you can take my word for
> > this: it is NOT what you had written and the equation .
> > sqrt (R^2 + d^2) = R + h
> > is not true in the general case.
> > You are probably thinking of
> > A^2 + B^2 = C^2, where A and B are the two sides forming the right
> > angle. That, solving for the long side, is
> > C = sqrt(A^2 + B^2), a far cry from the identity mentioned above.
>
> The two equation are identical.
>
> > It's easy to forget these things, but in this case I have not.
>
> You have the equation correct, but have forgotten basic algebra.
>
> The following two equations are identical:
>
> y = x^2
> sqrt (y) = x
>
> For the original equation:
>
> Let R = A
> Let d = B
> Let (R + h) = C
>
> Substituting into my original equation we get:
>
> sqrt (A^2 + B^2) = C
>
> Square both sides of the equation and we get:
>
> (A^2 + B^2) = C^2
>
> The parentheses are now irrelevant, so when removed we have:
>
> A^2 + B^2 = C^2
>
> Substitute back in the original values we get:
>
> R^2 + d^2 = (R + h)^2
>
> QED
>
> Seehttp://en.wikipedia.org/wiki/Pythagorean_theorem
>
> --
> Jim Pennino
>
> Remove .spam.sux to reply.

[Mxsmanic]

writes:

> Strange you could't just say that until it was already posted.

Nothing prevented me from saying it earlier. I just thought everyone already
knew. I didn't see a reference to it until I had mentioned it.

[[email protected]]

Mxsmanic > wrote:
> writes:

> > Strange you could't just say that until it was already posted.

> Nothing prevented me from saying it earlier.

Then why didn't you?

> I just thought everyone already knew.

Obviously not since there were numerous posts about it.

> I didn't see a reference to it until I had mentioned it.

Liar.

--
Jim Pennino

Remove .spam.sux to reply.

[John Mazor[_2_]]

"Mxsmanic" > wrote in message
...
> John Mazor writes:
>
>> So now the tower is 1000' across as well as 1000' high?
>
> The key is the angular size of the object; whether this is vertical or
> horizontal matters little.

How many towers are 1000' x 1000' ?

>> IOW you don't know.
>
> I do know. If it's at least 30 seconds across, it's visible.

And in the specific case cited, the answer would be...?

>> Show the math for the limits of resolution.
>
> The math is based on the size of cone cells in the retina. It works out to 30
> seconds of arc.

IOW you can google on limits of human eye resolution but can't do the math.

>> And remember, whichever tower you are using,
>> you must use the narrower dimension, which would be width, not height.
>
> Why?

I was hoping you'd do your usual Anthony evasion on that one.

To keep it in terms that even you might understand, we'll do a thought experiment that
doesn't require any calculations whatsoever.

Let's take your 1000' x 1000' tower and position ourselves at the distance where it is
just barely visible at the limits of resolution for the human eye. We'll ignore curvature
of the earth's suface because you shifted away from that when you introduced the concept
of limits of resolution. We'll also ignore the fact that we ought to be using a round
object because the difference between a diagonal and a diameter is negligible in this
exercise. Since you are conversant in computer speak, let's think of 1000' x 1000' as the
minimum-sized pixel that can be seen at that distance. You can see it only because the
pixel is just above the eyeball's limit of resolution both horizontally and vertically.

Now make it a more realistic "tower" - 1000' high but, say, 200' wide. Can you still see
it? Really? No, but let's not stop there. Make it 1' wide. Can you still see it?
Really? Make it the width of a pencil, and then the thickness of a human hair. Gone,
gone, gone.

Once you decrease the pixel size below the limits of resolution in any direction, you
could no longer can see it. So "whichever tower you are using, you must use the narrower
dimension." QED.

Just for jollies, let's hold a contest on what MadMax's response will be:

1. You're wrong because there's no such thing as a 1000' tower the width of a human hair.
2. Technically, a pixel is defined as...
3. I'm talking about a 1000' x 1000' tower so I'm still right.
4. You still can see the tower in the dimension where it's 1000' because 1000' still is
resolvable at that distance.
5. Introduce an irrelevant factoid to try to change the subject.
6. Complete silence.

[John Mazor[_2_]]

> wrote in message ...
> Mxsmanic > wrote:
>> John Mazor writes:
>
>> > So now the tower is 1000' across as well as 1000' high?
>
>> The key is the angular size of the object; whether this is vertical or
>> horizontal matters little.
>
> Wrong, it is both the horizontal and vertical.
>
> A simple example shows you to be wrong; from how far away could one
> see a 1000' human hair?
>
>> > IOW you don't know.
>
>> I do know. If it's at least 30 seconds across, it's visible.
>
> Wrong.
>
>> > Show the math for the limits of resolution.
>
>> The math is based on the size of cone cells in the retina. It works out to 30
>> seconds of arc.
>
> Wrong; the math is based on the lens aperature (pupil for eyes) and the
> light frequency and gives the theoretical maximum resolution. See any
> Google article on resolution.
>
> And the real resolution of the average 20/20 human eye based on research
> is 60 arc seconds.
>
>> > And remember, whichever tower you are using,
>> > you must use the narrower dimension, which would be width, not height.
>
>> Why?
>
> Take a 4' piece of sewing thread, hang it on a wall and back away until
> you can't see it.
>
> Make the string 8' long and repeat.
>
> Can you now see it from twice as far away?
>
> No.
>
> If you had read the entire Google artical you used to come up with
> your answers, you would know why.

Ahh, you beat me to it!

And I had laid that trap every so carefully.

[Bertie the Bunyip[_19_]]

Mxsmanic > wrote in
:

> writes:
>
>> Strange you could't just say that until it was already posted.
>
> Nothing prevented me from saying it earlier. I just thought everyone
> already knew. I didn't see a reference to it until I had mentioned
> it.
>

Yeh, right.


Bertie

[Bertie the Bunyip[_19_]]

"John Mazor" > wrote in news:nkvfj.30$qV.27@trnddc03:

>
> "Mxsmanic" > wrote in message
> ...
>> John Mazor writes:
>>
>>> So now the tower is 1000' across as well as 1000' high?
>>
>> The key is the angular size of the object; whether this is vertical
>> or horizontal matters little.
>
> How many towers are 1000' x 1000' ?


I've been trying to work in a gag about the comparitve size of his but for
ages on this thread, but I'm just gettign nowhere.

Bertie

[Gig601XLBuilder]

John Mazor wrote:

> 6. Complete silence.
>
>
>

If you get that I demand that everyone add your post as a signature line
to all messages.

[John Mazor[_2_]]

"Gig601XLBuilder" > wrote in message
...
> John Mazor wrote:
>
>> 6. Complete silence.
>
> If you get that I demand that everyone add your post as a signature line to all
> messages.

I humbly accept your encomium. Does anyone know what the Vegas betting line is against
#6?

FWIW, years ago when I started posting here my sig was:

-- John Mazor
"The search for wisdom is asymptotic."

"Except for Internet newsgroups, where it is divergent..."
-- R J Carpenter

I don't know if the Richard Carpenter who posts here is the same chap, but he kindly gave
me permission to add his witty retort to my initial observation. I may revive it since
nothing has changed in the ensuing years except the names of some of the divergers.

[Al G[_1_]]

"Tina" > wrote in message
...
> The equation in error was written as
>
>> > > sqrt (R^2 + d^2) = R + h
>
> I demonstrated this is an error by setting R and d to 1. The square
> root of the sum of those squares is 1.414, yet the equal sign would
> indicate it shoud be 2.
>
> That is straightforward.
>
> Never the less, the final answer that was posted was correct. As it
> happens, that identity was not used later when the correct answer was
> derived.
>
> But both that post, and mine, provided the correct final equation, for
> the geometric line of sight, but as to the visual one, it's subject to
> both the assumptions that were llisted, and some that were not.
>
> The good news is, ain't no pilot gonna worry about this stuff. No
> airplane pilot, anyhow,but those who pilot boats on blue water think
> about those things pretty often, We use them, for example, to
> estimate distances off shore.

Back in radar school, the horizon was said to be approx 1.25 X
sqrt(altitude)

10,000' cruise means 125nm to the radar horizon(line of sight)

To see a tower 1,000nm out, you'd have to be around 640,000'

Good luck.

Al G

[[email protected]]

Al G > wrote:


> Back in radar school, the horizon was said to be approx 1.25 X
> sqrt(altitude)

> 10,000' cruise means 125nm to the radar horizon(line of sight)

> To see a tower 1,000nm out, you'd have to be around 640,000'

> Good luck.

Ah yes, the radar horizon.

Since this topic has gone on so long, might as well add a bit of general
information.

There is the geometric horizon, which assumes the Earth is a big
billiard ball.

Then there is the optical horizon, which because the atmosphere bends
light, is around 10% greater than the optical horizon with the same
billiard ball assumption.

Now since both light and radio signals are electromagnetic radiation,
there is the radio horizon.

Because the atmospheric bending is roughly proportional to wavelength,
the lower the frequency, the more the bending.

So, the radar, or microwave, horizon is a bit farther out than the
optical horizon.

And the VHF horizon, as in aircraft COM, is a bit farther out yet.

If you go down low enough in frequency, there is no horizon, as the
bending is sufficient to cause the signal to follow the curve of the
Earth all they way around if you have enough power to overcome the
losses along the way.

Generally, such things are only of interest to those that do radar
or microwave tropospheric scatter communications, but I feel there
may be some readers interested in some random tidbits of knowledge.


--
Jim Pennino

Remove .spam.sux to reply.

[John Mazor[_2_]]

> wrote in message ...
> Al G > wrote:
>
>
>> Back in radar school, the horizon was said to be approx 1.25 X
>> sqrt(altitude)
>
>> 10,000' cruise means 125nm to the radar horizon(line of sight)
>
>> To see a tower 1,000nm out, you'd have to be around 640,000'
>
>> Good luck.
>
> Ah yes, the radar horizon.
>
> Since this topic has gone on so long, might as well add a bit of general
> information.
>
> There is the geometric horizon, which assumes the Earth is a big
> billiard ball.
>
> Then there is the optical horizon, which because the atmosphere bends
> light, is around 10% greater than the optical horizon with the same
> billiard ball assumption.
>
> Now since both light and radio signals are electromagnetic radiation,
> there is the radio horizon.
>
> Because the atmospheric bending is roughly proportional to wavelength,
> the lower the frequency, the more the bending.
>
> So, the radar, or microwave, horizon is a bit farther out than the
> optical horizon.
>
> And the VHF horizon, as in aircraft COM, is a bit farther out yet.
>
> If you go down low enough in frequency, there is no horizon, as the
> bending is sufficient to cause the signal to follow the curve of the
> Earth all they way around if you have enough power to overcome the
> losses along the way.
>
> Generally, such things are only of interest to those that do radar
> or microwave tropospheric scatter communications, but I feel there
> may be some readers interested in some random tidbits of knowledge.
> --
> Jim Pennino

At least some of us love the random tidbits of information that can surface here,
especially when the conversation wanders a bit. Good post, thanks.

-- John Mazor
"The search for wisdom is asymptotic."

"Except for Internet newsgroups, where it is divergent..."
-- R J Carpenter

[Morgans[_2_]]

"John Mazor" > wrote
>
> At least some of us love the random tidbits of information that can
> surface here, especially when the conversation wanders a bit. Good post,
> thanks.
Yes, the tidbits are nice, but the cost is so high, don't you think?

Would it be possible to agree that the twit has been shown as foolish and
ignorant as needs be for the current time, and let him have the inevitable
last word on this subject?

It would be nice to move on without him in any of the discussions on the
subject, and try to ignore him and his irrelevant comments for a while,
perhaps?

Just a thought to consider, I thought.
--
Jim in NC

[John Mazor[_2_]]

"Morgans" > wrote in message ...
>
> "John Mazor" > wrote
>>
>> At least some of us love the random tidbits of information that can surface here,
>> especially when the conversation wanders a bit. Good post, thanks.
> Yes, the tidbits are nice, but the cost is so high, don't you think?
>
> Would it be possible to agree that the twit has been shown as foolish and ignorant as
> needs be for the current time, and let him have the inevitable last word on this
> subject?
>
> It would be nice to move on without him in any of the discussions on the subject, and
> try to ignore him and his irrelevant comments for a while, perhaps?
>
> Just a thought to consider, I thought.
> --
> Jim in NC

Well, in addition to the random factoids, some of us get entertainment value out of
whacking the loons. That's not something you'd necessarily be proud to own up to - maybe
on a par with admitting that you eat ice cream right out of the box - but it's harmless
fun. Besides, people like Anthony wouldn't stop mouthing off even if he got zero
responses so why not exercise your debating skills in the cause of keeping the record
straight?

Just because a newsgroup has "aviation" in the name doesn't give the serious,
aviation-minded participants proprietary ownership over the channel. Stick with moderated
groups if you want that. Trying to police an open newsgroup is like it used to be trying
to keep the motormouths and goofballs off Channel 19 in the heyday of CB radio. The
Internet, like CB, is a permanent open mike for anyone and everyone. UseNet has one
advantage over CB in that you always can killfile whomever you don't like, although in 24
years on the Internet I've never killfiled anyone. I learned to live with it by becoming
adept at skimming the posts and skipping anything I don't want.

In a perfect world we wouldn't even have to have this conversation, but in a perfect world
I'd be as rich as Bill Gates, as handsome as Pierce Brosnan (okay, at my age I'll settle
for Sean Connery) and between the babes and the high life I wouldn't even have time to be
posting here. I think most of us agree with you in principle, but remember that constant,
useless railing against human nature also degrades the signal-to-noise ratio.