Visible by Height

I don't think it would be anywhere near 10k feet...more like about a couple of thousand feet...

http://www.ringbell.co.uk/info/hdist.htm

https://www.dailymail.co.uk/sciencetech/article-2159142/The-Burj-Khalifa-skyscraper-stands-tall-visitors-watch-sun-set-twice-minute.html

Here is a handy calculator,with the math to explain it.

Of course,this disregards any obstructions,and assumes one is on the ocean and/or knows their height above sea level.

http://www.ringbell.co.uk/info/hdist.htm

<...How high...?...>

It depends on the substances available, really.

Here's a rough calculation... 1 degree on the equator is 25000/360, about 70 miles. The radius of the earth is about 4000 miles. By similar triangles d/70 = 70/4000. d=4900/4000. About 1 1/4 miles, 6480 feet.

It takes time to get to <...6480 feet...> from mean sea level, though. So by the time one had arrived there, the sun would have dipped a little deeper, and one would need to go up a bit more. There's an iterative calculation coming on...

So It Looks like Tornado here understood my question.

I have an astronomy program, so I can at any time of year, learn how far below the horizon the Sun is "X" minutes after Sunset. So that number is easily obtained.

What I am trying to calculate is how fast is the rate that the Earths shadow is rising after Sunset. In June this Rise is MUCH slower than in say March. Due to the path of the Sun.

This is for one of our High altitude balloon flights. We want to launch the balloon as soon as possible after sundown, But we do not want the rising balloon to rise up too fast and get sunlight heating by going up high and fast enough that it sees the Sun again.

So the need to calculate the speed that the Earth's shadow is rising after Sunset.

Joe

I have given you a GA but your atmospheric refraction value, while valid over sea, can be wildly out over a desert where hot air can distort the apparent position of the sun to below rather than above it's true position. If the land temperature is high enough to create a mirage then the value you quoted is certainly incorrect and that may apply in to a lesser degree at all temperatures. Another variable is when the the sun sets behind a mountain range where the sun falls below the horizon while still high enough for atmospheric refraction not to come into play. You would need to factor in your height above sea level plus the height and distance of the mountain range.

Without knowing the terrain and temperature no reliable value can be calculated and even knowing this information the calculation would be so complex as to be impractical.

10,00ft above sea level 122.5 miles

20,000ft ASL* 173.3 miles

30,000ft ASL* 212.3 miles

Reference the link I provided previously.

*ASL=Above Sea Level

No, 20,000 feet = 3.79 miles

???? Where did you come up with that number?It it does not make sense.

The original link I provided is accurate,and shows the formula used. Check it out.

I am talking the distance to horizon,for a given heigth above sea level.

Think about it:You are in a plane at 20,000 feet over the ocean,and you can only see 3.79 miles?

At 48 feet height to the horizon on the ocean is a little over 8.5 miles.

Old sailors used this to estimate distance (from the guy in the crows nest) to other ships and to range their cannons.

_____________________________________________________________________

This is a rough guide to determine the distance of the horizon based on the observer's height above mean sea level. The screen will work in Metric or Imperial measurements. Enter the height above Sea Level either in Meters or Feet. Press the Calculate button and the distance of the horizon will be displayed in Kilometers when the observer's height is in meters or Miles when the observer height is entered in feet.

For your convenience,I have posted the link again below.

Distance to the Horizon Calculator

Observation Height in Meters in Feet

Distance to the horizon 8.5 Miles

The Mathematics behind this Calculation

This calculation should be taken as a guide only as it assumes the earth is a perfect ball 6378137 meters radius. It also assumes the horizon you are looking at is at sea level. A triangle is formed with the center of the earth (C) as one point, the horizon point (H) is a right angle and the observer (O) the third corner. Using Pythagoras's theorem we can calculate the distance from the observer to the horizon (OH) knowing CH is the earth's radius (r) and CO is the earth's radius (r) plus observer's height (v) above sea level.

Sitting in a hotel room 10m above sea level a boat on the horizon will be 11.3km away. The reverse is also true, whilst rowing across the Atlantic, the very top of a mountain range 400m high could be seen on your horizon at a distance of 71.4 km assuming the air was clear enough.

https://cr4.globalspec.com/edit/editcomment?c=1312085

Oh sorry. I used Google math for what 20,000 feet equaled, not how far you can see. Your post 6 didn't make sense either.

"How Sew?"said the tailor in Hong Kong.

"Smoke 'em if you got 'em."

What standard are you using,SG?

Sunset is defined by when the upper edge is on the horizon.

What is the definition of sunrise/sunset and moonrise/... | Old Farmer's Almanac

Formula for distance D to the horizon (in miles) at altitude h (in feet):

D = 1.2246 * sqrt(h)

(This formula assumes a parabolic approximation to a circle, very good at relatively small elevations relative to earth radius.)

If you were on the equator, 1 degree of longitude = 69 miles. So moving 69 miles toward the setting sun results in one degree increase in elevation of the sun.

Given your elevation in feet, calculate the distance to the horizon in miles and divide by 69 to get the sun's increase in elevation in degrees. This is the same as the degrees below the horizon at ground level.

10000 D=122.46 miles Deg=1.77 deg

20000 D=141.42 miles Deg=2.5 deg

30000 D=212.1 miles Deg = 3.074 deg

40000 D=282.92 miles Deg = 5 deg

So there would be a window of sunlight stretching around 3000 ft from just visible to fully visible, rising at around 3000 ft per minute....?

So there would be a window of sunlight stretching around 3000 ft from just visible to fully visible, rising at around 3000 ft per minute....?

In #8 we are connecting sun elevation (in degrees) with altitude at a fixed time just using geometry. If you put 'time' into the equation, it gets a lot more complicated.

The sun "moves' 15 degrees per hour or 4 minutes per degree, but its path is only perpendicular to the horizon at the equator during an equinox. The angle of its path with respect to the horizon is a function of both time of year and latitude. Near the arctic circle, the time it takes to drop 1 degree below the horizon is measured in days.

The sun elevation changes linearly with time, but the "sunset altitude" increases as the square of the elapsed time. The second plot shows delta time on the equator/equinox. At other times and places the elapsed time will be longer. (Sunsets are noticeably more rapid in the tropics.)

Scaled to Equator/Equinox

Excellent....so now we would need the amount of heating of a gradual emergence of the Sun above the horizon...of course in practicality you would need to take into account cooling of the atmosphere at altitude and that there would be an equalization of heating and cooling at a certain altitude perhaps...air temperature dropping about 3.6°F every thousand feet...and air density dropping as well...but solar radiation increasing...this along with the surface reflectivity of the balloon and square footage of exposure....all assuming a clear sky...

Good thoughts Eagle,

More or less, the amount of extra lift provided by incoming solar radiation is between 10 and 15 percent of the total lift already in the balloon.

Hence the need for it NOT to see the sun as it is rising. Getting into the sun would change the amount of lift desired, and make the balloon rise faster than desired.

We would need to know the rate of rise, which I suspect is not uniform...

..."The balloons typically have an ascent rate between 250 to 350 meters/minute(about 8.5-12 mph) and the average burst height is about 30 km(98k ft)."...

At this slow of a rate of rise you could probably launch shortly after sunrise...

https://www.weather.gov/media/upperair/Documents/Radiosonde%20Ascent%20Rates.pdf

close but We will want to launch after sunset, and NOT want to re enter into sunlight.

But still launch as close to sunset as possible

...."At this slow of a rate of rise you could probably launch shortly after sunrise.sunset."...

1 mph is 88 ft per minute....10 mph would be 880 ft per minute...."The sun "moves' 15 degrees per hour or 4 minutes per degree"...

It would be critical to know wind direction and speed...Latitude and launch elevation...I would say at 10 min after sunset in a still or parallel or opposite to sunset wind you would be safe...any wind towards the point of sunset would add time...