Hello, I've been wracking my feeble brain trying to find, or derive, an equation that lets one compute under what conditions can an observer on the ground, who has an obstruction of the horizon, see an object high above the Earth, for talks sake, a rocket.
I tried to draw a diagram to show you what the problem is:

http://s8.photobucket.com/albums/a30/sequesteredfishy/?action=view&current=DSC00082.jpg
How high above sea level does the rocket need to be before the over sized man can see it?

Obviously the height of the rocket and the angle between the persons eyes and the tip of the mountain/obstruction are important values but how to incorporate them is beyond my current ability.
I understand that one can normally find if it is possible for one to see a satellite/rocket high above sea level by adding the results of the following equations.

(1) the distance to the horizon for the observer on board the rocket
(2) the distance to the horizon for the ground observer.

s= rp(cos^-1)(rp/(rp + h)

Where
s= curved surface of the Earth below the observer.
rp= radius of planet
h= height of observer about sea level.
Note, your Calculator must be on ''radians''.

The below link has some info on what I'm trying to compute under the heading'' Optical adjustments and objects above the horizon'' but I have been unable to factor in an obstruction of the horizon for the ground observer.
http://en.wikipedia.org/wiki/Horizon


Secondly I've been attempting to find a method of converting a flares candela output
to the astronomical apparent Magnitude scale.
I've done some searching for the conversion and I found* that 1 candle power =0.981 candela. I think I need to translate candela into lumen and then lumen into lumen/m^2 (lux) and lastly lux into apparent magnitude but I'm not sure on this?

Let's say I wanted to find out how bright a M112A1 flare would seem to be if it was 10 km above my head at night and ignited. For example, how would an amateur astronomer who saw it classify it on the apparent magnitude scale. Using the apparent magnitude scale would they say it was brighter than Vega, Magnitude of 0, or brighter than Sirius, mag of -1.47.

The M112A1 weighing 227g emits 120 million candle power
120,000,000 candle power= 117,720,000 candela

¬''A light source that uniformly radiates one candela in all directions radiates a total of 4π lumens. If the source were partially covered by an ideal absorbing hemisphere, that system would radiate half as much luminous flux—only 2π lumens,''

I will assume that the flare is covered by a hemisphere and I will multiply 2 pi by 117,720,000, which equals 739,655,950 lumen, now I'm completely unsure on this but would I now have to factor in the area of the ground illuminated by the flare and divide the number of lumen by this area to get the lux? if this is the case I need to revise the equation below, yes?

apparent brightness= L/4(pi)(r^2)
r = distance from the source to observer
L =luminosity measured in Lumen
Brightness is measured in lux.


**As for converting Lux into apparent magnitude, I'm even more clueless.

I'd appreciate any helpful pointers or comments.


I found this usual converter but I'm not exactly sure what units I should be using:
http://www.unitconversion.org/unit_c...luminance.html

** http://www.bautforum.com/archive/index.php/t-32223.html (this source has an ambiguous formula for converting lux to apparent magnitude so I'm not sure on how to do this)
There's also this site with a useful apparent mag--> lux table.
http://stjarnhimlen.se/comp/radfaq.html but no formula.

http://zebu.uoregon.edu/~soper/Light/luminosity.html (on this site they use radiometry units and not photometry units so the light is total EM not apparent(preceived light).
¬ http://en.wikipedia.org/wiki/Luminosity
* http://www.onlineconversion.com/faq_10.htm

For your first problem, it sounds like you'd want to find the equation of the line from the path of the person's eyes to the top of the obstacle in the direction of the rocket, find the equation of the line for the rocket's trajectory, and find where the two lines intersect.

You'd want to treat the planet as a circle. Um, your variables that you have are the curved distance to the place directly under the rocket, the height of the rocket, the height of the person, the radius of the planet, and the angle at which the person's line of sight hits the top of the object.

Assuming that the person is on the top of the circle... let's first find what angle the rocket is being launched off at.

To get the distance in radians, we'd take s/rp. We can then convert that into the slope of the line of the rocket's trajectory (hopefully it's just going straight up).

slope intercept form:
y = mx + b
b = 0 (the equation of the line is going through the origin -- the center of the planet)
need to find m.

pi/2 - s/rp gives us the number of radians if you start at the right side of the circle.
The tangent will give you the slope of a line, so... we need that.

m = tan(pi/2 - s/rp)

So the line for the rocket's trajectory is
y = tan(pi/2 - s/rp) * x

Next, we need to find the equation of the person's line of sight to the top of the object. If we already have the angle, that will simplify things greatly, since we can just use the tangent function again to get the slope.

y = mx + b
b, the y-intercept, is the radius of the planet plus the person's distance off the surface.
b = rp + h
For m we just make a right triangle, and solve for the slope (ratio of opposite/adjacent legs). Um, what to call the angle. I guess just "a" will do, although they usually use theta or some other greek letter.
m = tan(a)
So your line would be
y = tan(a) * x + rp + h

Now you find where the lines of these two equations intersect.

tan(pi/2 - s/rp) * x = tan(a) * x + rp + h
tan(a) * x - tan(pi/2 - s/rp) * x + rp + h = 0
x * [tan(a) - tan(pi/2 - s/rp)] + rp + h = 0

x isn't really an important value to us. We want the height of the rocket, not its coordinate in terms of the center of the planet.

To get the desired value, we'd want to find the distance of the rocket from the center of the planet, then subtract the radius from that.

We can either use another trig function or the pythagorean theorem. I'm gonna use a trig function.

hypotenuse/adjacent is secant.
sec(pi/2 - s/rp) * x
That's the distance of the rocket from the center of the planet. I'll call the height of the rocket "hr".

hr = sec(pi/2 - s/rp) * x - rp

There we go. Now let's remove x from the equation.

hr + rp = sec(pi/2 - s/rp) * x
x = (hr + rp) / sec(pi/2 - s/rp)
or
x = (hr + rp) * cos(pi/2 - s/rp)

And so by substituting...

[(hr + rp) / sec(pi/2 - s/rp)] * [tan(a) - tan(pi/2 - s/rp)] + rp + h = 0

Whew... here is the final equation, not simplified in the least. I have a feeling I probably messed up somewhere... Solve for the variables you want. I'll do the height of the rocket for you.

[(hr + rp) / sec(pi/2 - s/rp)] * [tan(a) - tan(pi/2 - s/rp)] = - (rp + h)
(hr + rp) / sec(pi/2 - s/rp) = - (rp + h) / [tan(a) - tan(pi/2 - s/rp)]
hr + rp = - (rp + h) * sec(pi/2 - s/rp) / [tan(a) - tan(pi/2 - s/rp)]
hr = -(rp + h) * sec(pi/2 - s/rp) / [tan(a) - tan(pi/2 - s/rp)] - rp

... yeah, I probably screwed up somewhere. There are just too many complicated steps. One way to check it would be to plug in zero as the value for "a" and compare it to your other equation.