I'd never seen this before

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 Rd
(except in Mx's universe)

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...features.shtml
and http://www.pubmedcentral.nih.gov/art...?artid=1405458

--
Jim Pennino

Remove .spam.sux to reply.

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.

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.

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.

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".

"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

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

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 a

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 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