NavList:
A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding
Re: Refraction and Artificial Horizons
From: Greg R_
Date: 2006 May 23, 18:36 -0500
Frank wrote:
> You don't need to apply a dip correction when you use an artificial
> horizon, but in fact you do need to apply refraction.
You're right, I guess I'm more rusty on this stuff than I'd realized. :-)
--
GregR
----- Original Message ----
From: Frank Reed <FrankReedCT@aol.com>
To: NavList <NavList@fer3.com>
Sent: Tuesday, May 23, 2006 12:50:27 PM
Subject: [NavList 304] Re: Refraction and Artificial Horizons
You don't need to apply a dip correction when you use an artificial
horizon, but in fact you do need to apply refraction.
For a specific case: suppose I measure the angle between a star and its
image reflected in a pan of mercury and find that it's 90 degrees
exactly. What's the true altitude? First, of course, I divide by two.
At 45 degrees altitude, the refraction is just about 1.0 minutes of
arc. Refraction always lifts the stars, so if I could remove the
atmosphere, the star is lower: 44d 59' in altitude. That's the altitude
I would use in celestial navigation calculations. If you measure the
same altitude from the sea horizon, you would find approximately 45d 4'
from an altitude of 16 feet, or 45d 5' from an altitude of 25 feet, and
so on (dip is nearly 1' * sqrt(ht in feet)).
-FER
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From: Greg R_
Date: 2006 May 23, 18:36 -0500
Frank wrote:
> You don't need to apply a dip correction when you use an artificial
> horizon, but in fact you do need to apply refraction.
You're right, I guess I'm more rusty on this stuff than I'd realized. :-)
--
GregR
----- Original Message ----
From: Frank Reed <FrankReedCT@aol.com>
To: NavList <NavList@fer3.com>
Sent: Tuesday, May 23, 2006 12:50:27 PM
Subject: [NavList 304] Re: Refraction and Artificial Horizons
You don't need to apply a dip correction when you use an artificial
horizon, but in fact you do need to apply refraction.
For a specific case: suppose I measure the angle between a star and its
image reflected in a pan of mercury and find that it's 90 degrees
exactly. What's the true altitude? First, of course, I divide by two.
At 45 degrees altitude, the refraction is just about 1.0 minutes of
arc. Refraction always lifts the stars, so if I could remove the
atmosphere, the star is lower: 44d 59' in altitude. That's the altitude
I would use in celestial navigation calculations. If you measure the
same altitude from the sea horizon, you would find approximately 45d 4'
from an altitude of 16 feet, or 45d 5' from an altitude of 25 feet, and
so on (dip is nearly 1' * sqrt(ht in feet)).
-FER
--~--~---------~--~----~------------~-------~--~----~
To post to this group, send email to NavList@fer3.com
To , send email to NavList-@fer3.com
-~----------~----~----~----~------~----~------~--~---