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minutes degrees etc

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minutes degrees etc

Old 30th Aug 2000, 23:13
  #1 (permalink)  
Ceppo
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Arrow minutes degrees etc

I've done a search but i can't seem to find what i want.

How would you calculate the distance (say)from London to New york?

1 deg = 60 mins

1 min = 1nm that way.

I've got a book which says
1 min = 60 mins
and
1 min = 1 nm

but on a map i have, 1 hour is 15 degrees. And it works out that NY is about 75 degs from london.

What's going on? it seems as though 60 mins are different to 1 hour time, is this true?

Help
 
Old 31st Aug 2000, 00:31
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stagger
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Great Circle Calculations...
http://www.chicago.com/airliners/gc-faq.html#$gc-calc
 
Old 31st Aug 2000, 00:48
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ORAC
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People will direct you to sites, but a minor explanation (others please amend as required.)

In simple terms. At the equator, One degree E-W = 60 Minutes = 60 NM. But the further towards the poles you go it gets less.

Secondly, if you fly between 2 points you fly a "great circle" route which is the shortest distance on a 3D route.

Look at the sites you are directed at and remember that the world is aglobe and everything is trying to drape their their projections over this in 2D.

Does anyone know a site where they show/display GC lines??
 
Old 31st Aug 2000, 01:36
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Luftwaffle
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Ceppo, here's why your calculation seems contradictory:

An hour is divided into sixty minutes.
A degree is divided into sixty minutes.
The two sorts of minutes have the same name, but are unrelated.

The "1 minute = 1 nm" rule applies to minutes of latitude. For every 60 nm north or south you go, you move through an arc of about one degree on the Earth's surface. This is approximate, and it only works in the north-south direction. (It works east-west at the equator, but you probably don't live there).

Your 1 hour = 15 degrees observation is accurate with respect to the rotation of the earth. The earth makes a complete 360 degree rotation in 24 hours, so in one hour it rotates through 15 degrees.

Neither rule can be used to calculate the distance from London to New York. If you still want to do that, look at the Great Circle site suggested.

[This message has been edited by Luftwaffle (edited 30 August 2000).]
 
Old 31st Aug 2000, 01:45
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autothrottle
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If I remember correctly even in the ATPL syllabus you are definately NOT required to calculate the distance between two points that neither lie on the same meridian or on the same parallel.
However these points might (or might not )prove useful:
1)1 degree of latitude along any meridian will always represent 60 nm.
2)1 degree of longitude will ONLY represent 60 nm at the equator.The distance will vary as the cosine of the latitude.
3)The DEPARTURE FORMULA (the formula relating change of longitude to distance is ;

DEPARTURE DISTANCE(NM)=DIFF IN LONG(MINS)*COS LAT.
If you use the average latitude with this you get an answer that is roughly correct.
Remember that this formula is really only accurate if both points are at the same lat and therefore by definition is a Rhumb line distance.

Hope this is of some help,

Cheers AUTOTHROTTLE
 
Old 31st Aug 2000, 02:43
  #6 (permalink)  
Ceppo
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Talking

I love you people

Thanks SOOOOOOOO much

It was all helpful and very much appreciated.

 
Old 31st Aug 2000, 15:05
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Teroc
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Autothrottle,
Isnt the Dept formula only good for 600nm or less ? or is that the Mid lat sailing ?

Time to dust off the books i think...
 
Old 31st Aug 2000, 19:17
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stagger
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ORAC wrote...

"Does anyone know a site where they show/display GC lines??"

Err...yes...at the site I directed Ceppo to in my last posting.

Here's a link to the specific page with GC flight path display...
http://www.chicago.com/airliners/gc.html
 
Old 31st Aug 2000, 19:52
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autothrottle
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Teroc,
Not sure but I have dusted off my books and it doesn't mention that the dep formula is only good for 600nm or less.I suppose as the distance increases then it becomes less accurate!
Anyway looking in those books brings back terrible memories!!!!!

cheers for now
AUTOTHROTTLE
 
Old 31st Aug 2000, 23:22
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Jetlagged
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Hi Ceppo,
here's an equation that I found for GCD between two points on a sphere.
GCD={ACOS([SIN latone*SIN lattwo]+[COS latone*COS lattwo*COS(longone-longtwo)])}*59.99985

Later
Bye
 
Old 1st Sep 2000, 20:07
  #11 (permalink)  
autothrottle
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Red face

OH DEAR!THATS ME OUT FOLKS,FARRR TOO TECHNICAL NOW!!!

CHEERS AUTOTHROTTLE
 
Old 2nd Sep 2000, 01:21
  #12 (permalink)  
Ceppo
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In that equation, what does A (before ACOS that is) represent?

Also I take it lat and long are in degrees yes? and one more thing? what is the 59.99985??

I was asking this because some of my friends for interviews have been asked to roughly estimate the distance from london to new york.....assuming (big assumptions)

london and NY are on the same angle of lat,

The earth is perfectly spherical.

The only way i can think is to say....LON-NY 75 deg. 1 deg = 60 mins 1 min = 1 nm so 60 * 75 which is roughly 4 and a half thousand miles (but that seems way too much)

any thoughts?
 
Old 2nd Sep 2000, 02:05
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Jetlagged
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Hi Ceppo, The above equation will calculate the distance between any two pooints on a perfect sphere, the earth is not a perfect sphere hence the 59.99985, use 60 if you want to. The ACOS means ARC COS, if you convert cos 30 deg to numerical value you get .87 to get it back to 30 deg push ACOS on the calculator. Most calculators will allow you to convert standard degrees into decimal degress far calculation purposes, use the HMMSS feature supplied for this.

Later
Bye
 
Old 2nd Sep 2000, 06:05
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Luftwaffle
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Ceppo, you're on the right track.

Your estimate of 75 degrees => (60 * 75) nm => 4500 nm would be valid at the equator, but New York and London aren't at the equator. New York is at about 40 degrees latitude, and London around 50, making 45 N a useful average.

Very useful, in fact, as 45 is halfway between the equator and the pole. So is the distance half that at the equator? No. As Autothrottle ahas already pointed out, the arc distance of a great circle varies with the cosine of the latitude. The cosine of 45 is about 0.7 -- you remember this from from calculating crosswind components, or or banked stall speeds, right? So based on our assuptions, the great circle distance from New York to London is approximately 0.7 * 4500 nm = 3150 nm.

It's always smart to double check your estimates using a completely different method. How about multiplying the speed of the aircraft you aspire to fly by its flight time from London to New York.

Plugging the actual latitudes and longitudes of New York and London into a great circle calculator that assumes a spherical Earth, gave me a distance of 5571 km, which is 3008 nm, so we're within 5 percent, sounds like an okay estimate.

Of course, no interviewer will use this question now, because they all read PPRuNe, didn't you know?
 
Old 2nd Sep 2000, 21:25
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Ceppo
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Excellent, thanks a lot, much appreciated.

If inteviewers stopped asking questions seen on PPRuNe or changed their aptitude tests, it would be a bit stupid because they'd end up having to change them for every single recruitment they hold. It'd be pointless anyway because many wannabes don't even know PPRuNe exists.

I remember this from the BA aptitudes. most of the others didn't have the first clue as to what to expect and if you hae a swift look in the wannabes section BA is printed all over the place. Also when the bloke asked "who's heard of PPRuNe" and a couple of us kind of half put our hands up, when he walked out for a second and the usual chatter began, several people in the room asked us .....what's PPRuNe?

I think using PPRuNe to gather information is acceptable. If you're gathering info for an interview, what's better than an internet site where a lot of it is readily available? In the end, the interviews, aptitudes etc still have to be passed by the person in question.

strayed a bit haven't I .... hmm well
Thanks again and see yer laters.

Ceppo
 

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