My 2004 Post on Meteor Coverage Area Calculation

In August of 2004, I posted a message to this group about a calculation to
find the ground coverage radius to see a meteor at 60K feet above the
surface of the earth. Someone, from the British Isles I think, responded
with the formulas and maybe a Java script. I'm trying to find those formulas
again. I thought I had bookmarked his site.



Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet

Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"There is more to life than increasing its speed"
-- Mahatma Gandhi

Web Page: home.earthlink.net/~mtnviews

W. Watson wrote:
In August of 2004, I posted a message to this group about a calculation to
find the ground coverage radius to see a meteor at 60K feet above the
surface of the earth. Someone, from the British Isles I think, responded
with the formulas and maybe a Java script. I'm trying to find those formulas
again. I thought I had bookmarked his site.

If you're wondering about how far away on the ground one can see a
meteor at 60,000 feet (about 18 km), the formula is

d = 121*sqrt(h)

where h is the height in kilometers, and d is the distance from which an
object at that height is visible, also in kilometers. This formula
takes atmospheric refraction into account, using a rule of thumb
mentioned by Andrew Young.

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html

Brian Tung wrote:

W. Watson wrote:

In August of 2004, I posted a message to this group about a calculation to
find the ground coverage radius to see a meteor at 60K feet above the
surface of the earth. Someone, from the British Isles I think, responded
with the formulas and maybe a Java script. I'm trying to find those formulas
again. I thought I had bookmarked his site.


If you're wondering about how far away on the ground one can see a
meteor at 60,000 feet (about 18 km), the formula is

d = 121*sqrt(h)

where h is the height in kilometers, and d is the distance from which an
object at that height is visible, also in kilometers. This formula
takes atmospheric refraction into account, using a rule of thumb
mentioned by Andrew Young.

Actually, the formula I had in mind was similar to this idea, but I wanted
the radius when at 60K feet the meteor was 10 degrees above the horizon from
the observer's location. 15 degrees, etc.

Time for me to look in MatLab. I may have implemented it there.


Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"There is more to life than increasing its speed"
-- Mahatma Gandhi

Web Page: home.earthlink.net/~mtnviews

W. Watson wrote:
Actually, the formula I had in mind was similar to this idea, but I wanted
the radius when at 60K feet the meteor was 10 degrees above the horizon from
the observer's location. 15 degrees, etc.

Is the answer you get from assuming that the Earth is flat not precise
enough?

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html

Brian Tung wrote:

W. Watson wrote:

Actually, the formula I had in mind was similar to this idea, but I wanted
the radius when at 60K feet the meteor was 10 degrees above the horizon from
the observer's location. 15 degrees, etc.


Is the answer you get from assuming that the Earth is flat not precise
enough?

I believe the formula I used took into account a circular earth. The radius
I last calculated using it was something like 300 miles. I'm not sure a flat
earth wouldn't have given close to the same answer. Why I'm really
interested in finding the site is that the fellow had many more useful
calculations.

Well, this isn't really a pressing matter any longer. Eventually, I'll
likely find the method somewhere on my computer. If I really need to do the
calc for this case, I'll work out the details myself.


Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"There is more to life than increasing its speed"
-- Mahatma Gandhi

Web Page: home.earthlink.net/~mtnviews

W. Watson wrote:
I believe the formula I used took into account a circular earth. The radius
I last calculated using it was something like 300 miles. I'm not sure a flat
earth wouldn't have given close to the same answer. Why I'm really
interested in finding the site is that the fellow had many more useful
calculations.

Well, this isn't really a pressing matter any longer. Eventually, I'll
likely find the method somewhere on my computer. If I really need to do the
calc for this case, I'll work out the details myself.

In case you're still curious, I get

d = sqrt(2hR + [(R+h)tan(a)]^2) - (R+h)tan(a)

where h is the height of the meteor, R is the radius of the Earth (those
two measurements in consistent units), and a is the minimum angular
altitude of the meteor. Then d is in the same units as h and R. This
formula is an approximation, assuming that d is fairly small compared
to R. For a = 10 degrees and h = 60000 ft (= 18.3 km), it yields 97.3
km, as opposed to 101.8 km for a flat Earth.

To account for refraction, increase R by 15 percent. This yields 97.9
km, a slight increase.

--
Brian Tung
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html

Brian Tung wrote:

W. Watson wrote:

I believe the formula I used took into account a circular earth. The radius
I last calculated using it was something like 300 miles. I'm not sure a flat
earth wouldn't have given close to the same answer. Why I'm really
interested in finding the site is that the fellow had many more useful
calculations.

Well, this isn't really a pressing matter any longer. Eventually, I'll
likely find the method somewhere on my computer. If I really need to do the
calc for this case, I'll work out the details myself.


In case you're still curious, I get

d = sqrt(2hR + [(R+h)tan(a)]^2) - (R+h)tan(a)

where h is the height of the meteor, R is the radius of the Earth (those
two measurements in consistent units), and a is the minimum angular
altitude of the meteor. Then d is in the same units as h and R. This
formula is an approximation, assuming that d is fairly small compared
to R. For a = 10 degrees and h = 60000 ft (= 18.3 km), it yields 97.3
km, as opposed to 101.8 km for a flat Earth.

To account for refraction, increase R by 15 percent. This yields 97.9
km, a slight increase.

Good. Thanks very much.


Wayne T. Watson (Watson Adventures, Prop., Nevada City, CA)
(121.015 Deg. W, 39.262 Deg. N) GMT-8 hr std. time)
Obz Site: 39° 15' 7" N, 121° 2' 32" W, 2700 feet
--
"There is more to life than increasing its speed"
-- Mahatma Gandhi

Web Page: home.earthlink.net/~mtnviews