I'd never seen this before

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 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.

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?

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.

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.

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.

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.

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.

Mxsmanic wrote in
news
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

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

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