Ground/Air Nautical Miles

I have never really seen the terms used with light aircraft flight
planning (G.A), only with heavy aircraft flight planning. Whats the
difference between GNM and ANM?

"M.Lopresti" wrote in message
...
I have never really seen the terms used with light aircraft flight
planning (G.A), only with heavy aircraft flight planning. Whats the
difference between GNM and ANM?


A trip of 100 nm over the ground, in an hour, if into a 10 knot direct
headwind, would be a trip of 110 nm relative to still air. For logical
range calculations, the ANM concept is necessary.

John Lowry
Flight Physics

On Wed, 3 Nov 2004 at 12:33:39 in message
.net, John T Lowry
wrote:
A trip of 100 nm over the ground, in an hour, if into a 10 knot direct
headwind, would be a trip of 110 nm relative to still air.

Are you sure about that?

Aircraft flying at 200k effective speed over the ground 190 knots
agreed?
Time taken = 100/190 = 0.5263157 hours
Effective distance at 200k TAS is 200*0.5263157 = 105 .26 nm

try a very low flight speed of 50k

Effective speed 50 -10 = 40k
Time take = 100/40 = 2.5 hours
effective distance = 50*2.5 = 125 nm

Try 1000k
effective speed over ground 990k
time taken = 100/990 = 0.10101010hours
effective distance = 1000*0.1010101 = 101.1nm

If you imagine that the headwind is the same as the aircraft speed than
it makes no progress at all and covers an infinite distance through the
air to cover that 100nm.
--
David CL Francis

Divide by zero error! A division by zero is undefined, so I think
John's statement is correct. Even if you take the limit as the
effective speed goes to zero, the result is infinity, which is also,
theoretically, an undefineable number since it leads to inherent
contradictions in the strict sense. If we were to take ANM as a strict
mathematical concept, then I fear we would be laughed at. But
practically speaking, you're never going to have an effective ground
speed of 0 in an airplane.

On Thu, 4 Nov 2004 at 18:23:24 in message
. com, Raul Ruiz
wrote:
Divide by zero error! A division by zero is undefined, so I think
John's statement is correct. Even if you take the limit as the
effective speed goes to zero, the result is infinity, which is also,
theoretically, an undefineable number since it leads to inherent
contradictions in the strict sense. If we were to take ANM as a strict
mathematical concept, then I fear we would be laughed at. But
practically speaking, you're never going to have an effective ground
speed of 0 in an airplane.

That does not answer the fact that the original statement by John
appears to me to be wrong.

What is wrong with my calculations? I am perfectly willing to hear
corrections or that I was wrong.

A divide by zero error is a *computing* error that computers cannot
easily handle. There is nothing mathematically wrong with a result of
infinity.

In this case it merely represents the obvious fact that if you head into
a wind of the same strength as your cruising speed you will not get
anywhere. There is no contradiction in a result of infinity per se. Some
practical problems break down at that point. This one doesn't. The
meaning of infinity in this case is perfectly clear.

Finally a division of a real number by zero is *not* undefined, the
answer is unequivocally infinity. Zero divided by zero is undefined.

I await John's response.
--
David CL Francis

On Fri, 05 Nov 2004 15:33:21 GMT, David CL Francis
wrote:

On Thu, 4 Nov 2004 at 18:23:24 in message
.com, Raul Ruiz
wrote:
Divide by zero error! A division by zero is undefined, so I think
John's statement is correct. Even if you take the limit as the
effective speed goes to zero, the result is infinity, which is also,
theoretically, an undefineable number since it leads to inherent
contradictions in the strict sense. If we were to take ANM as a strict
mathematical concept, then I fear we would be laughed at. But
practically speaking, you're never going to have an effective ground
speed of 0 in an airplane.

That does not answer the fact that the original statement by John
appears to me to be wrong.

What is wrong with my calculations? I am perfectly willing to hear
corrections or that I was wrong.

A divide by zero error is a *computing* error that computers cannot
easily handle. There is nothing mathematically wrong with a result of
infinity.

In this case it merely represents the obvious fact that if you head into
a wind of the same strength as your cruising speed you will not get
anywhere. There is no contradiction in a result of infinity per se. Some
practical problems break down at that point. This one doesn't. The
meaning of infinity in this case is perfectly clear.

Finally a division of a real number by zero is *not* undefined, the
answer is unequivocally infinity. Zero divided by zero is undefined.

I await John's response.

I cant believe that you are asking this.
you always use the airspeed corrected for wind to determin time
enroute so that you can then determin the actual fuel consumed for the
leg.
John's response was correct.
Stealth Pilot

Brush up on your math...
http://www.math.utah.edu/~alfeld/math/0by0.html

David CL Francis wrote in message ...
On Wed, 3 Nov 2004 at 12:33:39 in message
.net, John T Lowry
wrote:
A trip of 100 nm over the ground, in an hour, if into a 10 knot direct
headwind, would be a trip of 110 nm relative to still air.

Are you sure about that?

Aircraft flying at 200k effective speed over the ground 190 knots
agreed?
Time taken = 100/190 = 0.5263157 hours
Effective distance at 200k TAS is 200*0.5263157 = 105 .26 nm

What you say confirms what John T Lowry said. He said that if you
spend an hour taking that 100nm (measured on the ground) trip into
a 10nm headwind, you add 10nm to your total air distance travelled.
Your calculations show that if you spend approximately one-half
hour going into that headwind, you'll add approximately one half
that distance, or about 5 nm.

Actually, John's original statement could have been generalized to
say that a trip of X nm over the ground in an hour, if into a 10 knot
direct headwind, would be a trip of X+10 nm relative to still air.

It could be further generalized to say that a trip of X nm over the
ground in T hours, into a Y knot headwind, will be a trip of X + T*Y
nm relative to still air. Using your numbers, we get X=100, T=.526,
Y=10, X + T*Y = 105.26nm effective distance, just as you claim.

--Rich

On Fri, 5 Nov 2004 at 11:16:17 in message
, Todd Pattist
wrote:

David CL Francis wrote:
That does not answer the fact that the original statement by John
appears to me to be wrong.

It's not wrong.

Nice to meet you here again Todd. I agree I was wrong, in that the
problem he postulated was different from what I assumed. I think my
calculations were right though.

I apologise to those concerned

However John's actual statement now I read it more carefully seems to
imply that given the wind speed you must find the TAS at which you
must fly to get there in an hour!

Is this the calculation that is intended? Even more trivial than my
calculations!

John wrote:
"A trip of 100 nm over the ground, in an hour, if into a 10
knot direct headwind, would be a trip of 110 nm relative to
still air."

Thus we have an unknown cruise TAS cruise speed
Let that be V

We have a 100 nm distance and a head wind of 10k
We have a time of flight of exactly one hour

Therefore 100/(V-10)= 1

and V -10 = 100

it follows that V = 110k

So at a TAS of 110k you travel a ground distance of 100nm against a wind
of 10k and surprise, surprise you than fly 110 'air' nm

More or less self evident so I am unclear what that achieves?

You're wrong, here's why:

You wrote :

"Aircraft flying at 200k effective speed over the ground 190
knots Time taken = 100/190 = 0.5263157 hours
Effective distance at 200k TAS is 200*0.5263157 = 105 .26
nm"

That is still mathematically correct I hope?

John assumed an airspeed that would take 1 hour exactly to
fly 100nm in one hour into a 10 knot headwind. During that
hour, the air moved 10 nm so the aircraft covered 110nm of
air and 100 nm of ground.

So he did, but that means he is implying a cruise speed that will get
you there in one hour.

Perhaps I am naive but it seemed to me you needed to calculate the
effect of head winds on a given cruise speed not a cruise speed to give
constant time? So if your aircraft had to fly against a 50k wind it
simply has to increase its cruise speed to 150k and it covers a 100nm in
one hour just the same. Is that useful thing to know? In that case it
is of course true that the aircraft would have covered 150 'air' miles.

You assumed a speed that would take 1 hour to fly 100nm in
no wind, and then calculated the time. (actually you chose
200 knots, I'm not sure why). The assumptions were
different. For John's case, the aircraft was not flying 100
knots - it was flying at 110 knots and covered 110 air nm
and 100 ground nm.

I chose three different speeds I believe, 200k, 50k and 1000k just to
show the differences.

I apologise again to those concerned for my error in reading the
original statement..

More interesting would be to know the galls per hour consumed at various
speeds and then create a table of cruise speeds against head wind
components (Positive or negative) and then determine the best cruise
speed for each wind speed for minimum Gallons per mile perhaps?

--
David CL Francis

Not to complex the issue up, but you also have to do a table for various
altitudes - head/tailwinds - power settings in the data set. I've done this for
the 182; given Excel and a rainy Saturday, it wasn't too hard.

Jim



David CL Francis
shared these priceless pearls of wisdom:


-More interesting would be to know the galls per hour consumed at various
-speeds and then create a table of cruise speeds against head wind
-components (Positive or negative) and then determine the best cruise
-speed for each wind speed for minimum Gallons per mile perhaps?



Jim Weir (A&P/IA, CFI, & other good alphabet soup)
VP Eng RST Pres. Cyberchapter EAA Tech. Counselor
http://www.rst-engr.com