Converting 2 angular measurements to distance on the earth.
Thread Starter
Joined: May 2003
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From: UK
Converting 2 angular measurements to distance on the earth.
If 2 aircraft at roughly the same latitude measure the sun in the sky at 2 different angle, how can you use that to estimate distance between the 2 aircraft ?
To my puny brain I can just take the number of arc minutes difference and call that nautical miles between them. Does that work ?
eg, Aircraft A measures the sun to be 10º above the horizon and aircraft B measures it as 5º. So, 5º difference between them equals 5ºx60" = 300nm separating the two. If so, why does that work ?
Does that work at all latitudes or would I have to make a correction (struggling to remember my atpl theory stuff 20 years ago). Also, what is the effect if the aircraft are at very different altitudes ?
Cheers
To my puny brain I can just take the number of arc minutes difference and call that nautical miles between them. Does that work ?
eg, Aircraft A measures the sun to be 10º above the horizon and aircraft B measures it as 5º. So, 5º difference between them equals 5ºx60" = 300nm separating the two. If so, why does that work ?
Does that work at all latitudes or would I have to make a correction (struggling to remember my atpl theory stuff 20 years ago). Also, what is the effect if the aircraft are at very different altitudes ?
Cheers
Joined: Jan 2008
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From: Reading, UK
If the aircraft are at the same latitude, then presumably it's the longitude difference you are considering.
In that case, bear in mind that one minute of longitude only equals a nautical mile at the equator.
Or am I misunderstanding what you're trying to do ?
In that case, bear in mind that one minute of longitude only equals a nautical mile at the equator.
Or am I misunderstanding what you're trying to do ?
Joined: Feb 2009
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From: Virginia
One thing that seems wrapped up in the question is whether you can use the angle to the visible horizon regardless of altitude. For the equation to work, you need the angle between the sun and the horizontal plane containing the aircraft. The higher the aircraft, the greater the angle between that plane and the visible horizon.
From a quick google search, you either need to correct for the "angle of dip" to the visible horizon, or use a sextant with a bubble attachment (basically an artificial horizon). I didn't try to run the numbers and find how big the altitude-based error would be if you didn't correct for dip.
From a quick google search, you either need to correct for the "angle of dip" to the visible horizon, or use a sextant with a bubble attachment (basically an artificial horizon). I didn't try to run the numbers and find how big the altitude-based error would be if you didn't correct for dip.
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From: Aus
How one airline Captain, previously navigator, found a light aircraft lost in the Pacific.
Mayday in December
Mayday in December
Joined: Feb 2002
Posts: 250
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From: at my computer
How one airline Captain, previously navigator, found a light aircraft lost in the Pacific.
From Wikipedia::
In June 2007 Captain Gordon Vette was awarded the ONZM (Officer of the New Zealand Order of Merit), recognising his services in assisting Justice Mahon during the Erebus Inquiry. Vette's book, Impact Erebus, provides a commentary of the flight, its crash and the subsequent investigations.
I can strongly recommend his book "Impact Erebus" and also the thread in this forum from many years ago.
Respect.
Joined: Jun 2009
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From: florida
Salute!
A good question up front, and as one said above, altitude may play a role. For most of us, I don't think it would be too much of a correction , but you could crank altitude into the equation.
Secondly, the basic idea would be fairly close if you recall that the "mile" per "minute" of longitude will be the cosine of the lattitude. So where I live, it's about 0.87 N,M, per minute.
When planning some timing and geographical aspects of a few attacks back in the day, we assumed a "flat earth" within an arbitrary distance of the tgt, and corrected the longitude "x" values on our plots using the cosine of lattitude. Seem to recall we could get relationship of tgt and IP within 20 or 30 feet of real world compared to a very accurate inertial frame that used a geoid model of the Earth, altitude, phase of the moon, Milankovitch orbital theory, etc.
Thank you megan and Terry. Many lessons to be learned and taken to heart from that sad accident. I had no idea what really happened until I found the posts here on pPrune.
Gums sends...
A good question up front, and as one said above, altitude may play a role. For most of us, I don't think it would be too much of a correction , but you could crank altitude into the equation.
Secondly, the basic idea would be fairly close if you recall that the "mile" per "minute" of longitude will be the cosine of the lattitude. So where I live, it's about 0.87 N,M, per minute.
When planning some timing and geographical aspects of a few attacks back in the day, we assumed a "flat earth" within an arbitrary distance of the tgt, and corrected the longitude "x" values on our plots using the cosine of lattitude. Seem to recall we could get relationship of tgt and IP within 20 or 30 feet of real world compared to a very accurate inertial frame that used a geoid model of the Earth, altitude, phase of the moon, Milankovitch orbital theory, etc.
Thank you megan and Terry. Many lessons to be learned and taken to heart from that sad accident. I had no idea what really happened until I found the posts here on pPrune.
Gums sends...
Joined: Nov 2005
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From: Zulu Time Zone
If 2 aircraft at roughly the same latitude measure the sun in the sky at 2 different angle, how can you use that to estimate distance between the 2 aircraft ?
To my puny brain I can just take the number of arc minutes difference and call that nautical miles between them. Does that work ?
To my puny brain I can just take the number of arc minutes difference and call that nautical miles between them. Does that work ?
eg, Aircraft A measures the sun to be 10º above the horizon and aircraft B measures it as 5º. So, 5º difference between them equals 5ºx60" = 300nm separating the two. If so, why does that work ?
Joined: Jun 2009
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From: florida
Salute!
Good point oggers.
OTOH, the question seemed to only require a measurement of the sun elevation from horizon or true vertical, but not corrected for the "azimuth" you described.
To get the best solution, I would use the time from pt 1 to pt 2 that the sun or star or .... reached the same "elevation" . So in that case the delta could be related to the one minute of longitude for each "clock" minute, corrected for the cosine.
Keep going oggers, somehow I feel we need to have at least an intro to celestial nav principles for some folks visiting here.
Gums sends...
Good point oggers.
OTOH, the question seemed to only require a measurement of the sun elevation from horizon or true vertical, but not corrected for the "azimuth" you described.
To get the best solution, I would use the time from pt 1 to pt 2 that the sun or star or .... reached the same "elevation" . So in that case the delta could be related to the one minute of longitude for each "clock" minute, corrected for the cosine.
Keep going oggers, somehow I feel we need to have at least an intro to celestial nav principles for some folks visiting here.
Gums sends...
Joined: Sep 2008
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From: Scotland
You will also be able to get an assumed position by reducing the sight. Starting from your sight, adjusted for dip, sextant error and so on you then compare it to the altitude you calculate the sun would be at if you are where you thought you were. The difference between the observed and calculated altitude is your intercept.
you end up with a single line of position (it’s actually a small circle you could be anywhere on but for practical purposes in this case can be a straight line) and you can also calculate the azimuth of the sun. The azimuth is perpendicular to your line of position and the difference between calculated and true sextant angles gives you how far down that azimuth from where you think you are your line of position crosses it. Where this azimuth crosses the line of position is your intercept terminal point and gives you a reasonable idea of where you might be. You could then just use either a plane or Mercator calculation to get the distance between your and your oppo’s position.
No experience doing celestial in aircraft though, all mine was at sea at rather more gentlemanly pace with a calculator and plenty of time to work the tables. The FAA navigator’s handbook is freely available online, and I guess most of us have plenty of time on our hands right now!
you end up with a single line of position (it’s actually a small circle you could be anywhere on but for practical purposes in this case can be a straight line) and you can also calculate the azimuth of the sun. The azimuth is perpendicular to your line of position and the difference between calculated and true sextant angles gives you how far down that azimuth from where you think you are your line of position crosses it. Where this azimuth crosses the line of position is your intercept terminal point and gives you a reasonable idea of where you might be. You could then just use either a plane or Mercator calculation to get the distance between your and your oppo’s position.
No experience doing celestial in aircraft though, all mine was at sea at rather more gentlemanly pace with a calculator and plenty of time to work the tables. The FAA navigator’s handbook is freely available online, and I guess most of us have plenty of time on our hands right now!
Thread Starter
Joined: May 2003
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From: UK
Thanks for replies folks.
It was in fact the Air New Zealand rescue of the Cessna over the Pacific that prompted my question.
Using only rough sun sightings and headings the AirNz Captain was able to estimate the Cessna to be around 210nm to his south west. Figuring out the approx direction is easy enough but how he came up with a distance was not clear to me. He estimated the difference in sun angle above the horizon to be around 4°. 4x60=240, corrected for latitude = around 210nm. Can I make a direct comparison like that between angles and distance ?
The rest of the story of him using vhf radio range to figure out position is well documented.
It was in fact the Air New Zealand rescue of the Cessna over the Pacific that prompted my question.
Using only rough sun sightings and headings the AirNz Captain was able to estimate the Cessna to be around 210nm to his south west. Figuring out the approx direction is easy enough but how he came up with a distance was not clear to me. He estimated the difference in sun angle above the horizon to be around 4°. 4x60=240, corrected for latitude = around 210nm. Can I make a direct comparison like that between angles and distance ?
The rest of the story of him using vhf radio range to figure out position is well documented.
