# Nav Problem...

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**nav problem**

hello founder,

here some reflections for your nav problem.

first, as in many such problems, the answer lies in the question & you must very carefully read the question over & over trying to understand what's in the backmind of the questioner.

- when stated : autopilot is coupled to INS(inertial navigation system), it simply means the aeroplane is flying a great circle(orthodromic) route = shortest distance.

- now to memorise/visualise the relative position of orthodrome route vs loxodrome route(loxodrome route= following a parallel or flying a constant latitude), remember the acronym:" P O L E "

P = [ north or (south )pole]

O= (orthodrome)

L= (loxodrome)

E= (equator)

so, in our example,the ortho route(concave to the equator, as said in the previous post by outofsynch), is north of the loxo route & going from 60N 30W to 60N 20W, we are flying a great circle route eastbound & moving gradually up on a more northern latitude. at midpoint(25W), also called "the vertex", the latitude is the most northern. past midpoint the latitude decreases again to reach 60N in wpt 2.

the question now is, what is our latitude at midpoint?or in other terms what is the greatest distance between the ortho & loxo route in nm knowing that 1 nm = 1 minute of latitude?

there is a barbaric formula for this : f = a x distance wpt1 to wpt2 /230

f= greatest vertical distance in nm, between ortho & loxo route(this is what we are looking for).

a= angle between ortho route & loxo route & is yet another formula called the "givry correction" : = 1/2 delta longitude x sinus mean latitude or 1/2 x 10° x sin60°= 4.33°

loxo distance( i make a small but acceptable error here) between wpt 1 to wpt 2 = delta longitude x cosinus 60° = 10° x 1/2 = 5° x 60' = 300nm.(at 60N, due to the convergence of the meridians, the loxo distance is halved: cos60°= 1/2)

so f= 4.33 x 300nm/230 = 5.65nm or 5.65' of latitude

finally i would answer at midpoint our latitude is 60 05.65N( or 60 05.7N in your example)

here some reflections for your nav problem.

first, as in many such problems, the answer lies in the question & you must very carefully read the question over & over trying to understand what's in the backmind of the questioner.

- when stated : autopilot is coupled to INS(inertial navigation system), it simply means the aeroplane is flying a great circle(orthodromic) route = shortest distance.

- now to memorise/visualise the relative position of orthodrome route vs loxodrome route(loxodrome route= following a parallel or flying a constant latitude), remember the acronym:" P O L E "

P = [ north or (south )pole]

O= (orthodrome)

L= (loxodrome)

E= (equator)

so, in our example,the ortho route(concave to the equator, as said in the previous post by outofsynch), is north of the loxo route & going from 60N 30W to 60N 20W, we are flying a great circle route eastbound & moving gradually up on a more northern latitude. at midpoint(25W), also called "the vertex", the latitude is the most northern. past midpoint the latitude decreases again to reach 60N in wpt 2.

the question now is, what is our latitude at midpoint?or in other terms what is the greatest distance between the ortho & loxo route in nm knowing that 1 nm = 1 minute of latitude?

there is a barbaric formula for this : f = a x distance wpt1 to wpt2 /230

f= greatest vertical distance in nm, between ortho & loxo route(this is what we are looking for).

a= angle between ortho route & loxo route & is yet another formula called the "givry correction" : = 1/2 delta longitude x sinus mean latitude or 1/2 x 10° x sin60°= 4.33°

loxo distance( i make a small but acceptable error here) between wpt 1 to wpt 2 = delta longitude x cosinus 60° = 10° x 1/2 = 5° x 60' = 300nm.(at 60N, due to the convergence of the meridians, the loxo distance is halved: cos60°= 1/2)

so f= 4.33 x 300nm/230 = 5.65nm or 5.65' of latitude

finally i would answer at midpoint our latitude is 60 05.65N( or 60 05.7N in your example)

*Last edited by blackmail; 14th Dec 2006 at 07:54.*

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I'm impressed by the knowledge of people here at PPRuNe, Thanx Blackmail and outofsynch, I tried other problems of the same type with the method you described and it works!! =)

Thanx...

Thanx...

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[QUOTE=blackmail;3018507]hello founder,

route(

Great job, blackmail, but I have to point something that, I believe, have slipt on the keyboard as your writing went too enthusiastic:

The loxodrome is the route that crosses all meridians at an equal angle; the parallell and the equator are particular loxodromes as they cross all meridians at an angle of 90°; out of those 90°, the curve on the sphere will be a loxodroïd, closing to the pole without ever reaching it.

Just for the sake of rookies who might flunk the question at a nav exam.

L

route(

**loxodrome route= following a parallel or flying a constant latitude**), remember the acronym:" P O L E "Great job, blackmail, but I have to point something that, I believe, have slipt on the keyboard as your writing went too enthusiastic:

The loxodrome is the route that crosses all meridians at an equal angle; the parallell and the equator are particular loxodromes as they cross all meridians at an angle of 90°; out of those 90°, the curve on the sphere will be a loxodroïd, closing to the pole without ever reaching it.

Just for the sake of rookies who might flunk the question at a nav exam.

L

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**nav problem**

hi lemper,

very good point. in our example, the loxodrome is indeed special & equal to the parallel going through 60N from wpt 1 to wpt 2. the equator is also a special parallel as it is the only parallel that equals a great circle, supposed the earth is considered a perfect globus, which, of course, in reality it is not.

two more observations. first, problems as that occupies us here, have always some special/odd features in the question itself & it depends on the sharpness of the respondents mind to see this through & hind already in the good direction to find the correct answer. a good drawing to visualise the problem, associated with some basic formulae is always a good start. then, look back & reflect.

secondly, to elaborate on the "P O L E" acronym, this acronym can also be used when projecting ortho/loxo routes on common used maps.

to remember how the concave feature of these routes are, following may help:

lambert charts(used in mid latitudes) : ortho: concave towards its contact parallel. loxo : concave towards its pole.

mercator(used on equator & lower lat's) : ortho : concave towards equator( a special case of contact parallel). loxo : straight line.

stereopolair(used in mid to high lat's): ortho & loxo : both concave towards the pole.

i stop here, hoping not to have made a mistake & if so, please correct me.

kind regards,

bm

very good point. in our example, the loxodrome is indeed special & equal to the parallel going through 60N from wpt 1 to wpt 2. the equator is also a special parallel as it is the only parallel that equals a great circle, supposed the earth is considered a perfect globus, which, of course, in reality it is not.

two more observations. first, problems as that occupies us here, have always some special/odd features in the question itself & it depends on the sharpness of the respondents mind to see this through & hind already in the good direction to find the correct answer. a good drawing to visualise the problem, associated with some basic formulae is always a good start. then, look back & reflect.

secondly, to elaborate on the "P O L E" acronym, this acronym can also be used when projecting ortho/loxo routes on common used maps.

to remember how the concave feature of these routes are, following may help:

lambert charts(used in mid latitudes) : ortho: concave towards its contact parallel. loxo : concave towards its pole.

mercator(used on equator & lower lat's) : ortho : concave towards equator( a special case of contact parallel). loxo : straight line.

stereopolair(used in mid to high lat's): ortho & loxo : both concave towards the pole.

i stop here, hoping not to have made a mistake & if so, please correct me.

kind regards,

bm

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Hey Blackmail,

you seem to have taken nav lectures a very long time ago, as it seems that modern flying schools just browse over these concepts.

No mistake of course, just that little freudian slip before.

While I am at it, I never found a better way to teach these "magic thingies" than pulling strings on a globe.

Then you have to put the winds and the politics to explain why we so seldom fly those (not so) great Circles.

Merry X-mas.

you seem to have taken nav lectures a very long time ago, as it seems that modern flying schools just browse over these concepts.

No mistake of course, just that little freudian slip before.

While I am at it, I never found a better way to teach these "magic thingies" than pulling strings on a globe.

Then you have to put the winds and the politics to explain why we so seldom fly those (not so) great Circles.

Merry X-mas.

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It seems that Spherical Trigonometry isn’t taught as it was before. If (for the sake of those interested in older pilot skills) anyone is interested in a pure Spherical Trig solution to the problem, here goes.

First, build a spherical triangle, the vertices being N60°W030°, N60°W020°, and the North Pole (N90° and all Longitudes). Find the Co-Latitude of both points, i.e. both are 30°, being 90°-60°, as both latitudes are the same.

Next Find the Diff Long, i.e. the difference in longitude, i.e. 30°W – 20°W = 10°

Now, we have a triangle with two sides being 30°, and the Included Angle being 10°.

If we call the sides A and B, and the included angle C, then to find the distance, D –

Distance = D X 60 = 299.7142835 nm.

For the example, both Latitudes were equal, thus the angle between the meridians and the Initial and Final Track are the same. Because both Latitudes were equal, the highest latitude will be at the mid meridian (not so if the Latitudes are not the same), thus we wish to find the Latitude at the mid point, half the total distance.

Distance to mid point = 4.995238059° / 2 = 2.497619029°, call it H

To find the Angle between the Initial Meridian and the Initial Track, (call it E) –

Now we have a new triangle, side A the Co-Latitude of the origin, side H the half distance to destination, and angle E the included angle between the two sides. The third side may now be found, the Co-Latitude of the mid position –

As INS Co-Ordinates are in DMT format,

Answer

Next step, let’s take on WGS84

Regards, and Happy New Year,

Old Smokey

First, build a spherical triangle, the vertices being N60°W030°, N60°W020°, and the North Pole (N90° and all Longitudes). Find the Co-Latitude of both points, i.e. both are 30°, being 90°-60°, as both latitudes are the same.

Next Find the Diff Long, i.e. the difference in longitude, i.e. 30°W – 20°W = 10°

Now, we have a triangle with two sides being 30°, and the Included Angle being 10°.

If we call the sides A and B, and the included angle C, then to find the distance, D –

**D = Arc Cos (Cos A Cos B + Sin A Sin B Cos C) = 4.995238059°**Distance = D X 60 = 299.7142835 nm.

For the example, both Latitudes were equal, thus the angle between the meridians and the Initial and Final Track are the same. Because both Latitudes were equal, the highest latitude will be at the mid meridian (not so if the Latitudes are not the same), thus we wish to find the Latitude at the mid point, half the total distance.

Distance to mid point = 4.995238059° / 2 = 2.497619029°, call it H

To find the Angle between the Initial Meridian and the Initial Track, (call it E) –

**E = Arc Cos (Cos B - Cos A Cos D ) / (Sin A Sin D) = 85.66712608°**Now we have a new triangle, side A the Co-Latitude of the origin, side H the half distance to destination, and angle E the included angle between the two sides. The third side may now be found, the Co-Latitude of the mid position –

**CoLat = Arc Cos (Cos A Cos H + Sin A Sin H Cos E) = 29.90550141°****Latitude = 90 – CoLat = 90 - 29.90550141 = 60.09449859°N**As INS Co-Ordinates are in DMT format,

**Lat = N60°05.7’**Answer

**D**is correct (or the nearest).Next step, let’s take on WGS84

Regards, and Happy New Year,

Old Smokey