distance between two points on the earth
 

I remember from my atpl training that to find the distance between two different longitudes on the same latitude we used the formulae for departure

ie 15w to 25w on the 60 deg lat = 600 x .5 = 300nm

but what if the 15w was on the 30 deg lat and the 25w was on the 60 deg lat.

how do you work out the distance??

do you take the mean lat ie 45

so 600 x cos 45(.707) = 424nm

Many thanks in advance

yes. I think!

This is roughly correct for a rhumb line distance.

Well the distance between 30N and 60N is 30*60=1800 nm so i think 424nm would be quite a shortcut... What you have calculated is the east-west average departure for that route, but you have forgotten to take into account the north-south component.

The rhumb line distance can be calulated by trigonometry by making a triangle on the Mercator chart between 30N15W, 60N25W and 60N15W. Since we are interested in measuring the hypotenus, let's make the chart scale equal to 1 unit = 1nm at 45N.

That gives us the short side (E/W) of the right angled triangle 424,26 units (10 degrees * 60min/degrees * cos 45). The long side (N/S) scale is not skewed so this is 1800 units.

The hypotenus is then 1849,32 units, which translates into 1849 nm rhumb line distance.

The great circle distance is always shorter than or equal to the rhumb line distance, with it being the same on a north-south route or along the equator, and the biggest difference found on east-west routes near the poles. That is to say, the great circle distance in this case is not more than 1849nm but not less than 1800nm (since this would be from 30N15W to 60N15W). Since this route is predominantly north-south in the mid-latitudes, it would be reasonable to believe that the GCD would be significantly closer to 1849nm than 1800nm.

The exact computation is best suited for fancy computers. However, using the Great Circle Mapper, it comes to a distance of 1845 for the great circle.

hi bfisk

I completely forgot about the north south component. thanks for reminding me

I worked out the RL distance using Pythagorus and I got 1872 nm

I think you went wrong for the 600 part of the calculation

It should have been 516nm 600 x cos30 = 516

so 516sq + 1800sq = Xsq

so x = 1872 for the rhumb line

you worked out the GC to be 1845 which seems correct for this distance

any idea how you actually work out the GC distance ??

Hmm, I'm getting myself badly tangled into this one. I see that RLD of 2500 or so cannot be correct. It's been a while, let me try to think this one out properly first :)

im getting confused too

i used http://williams.best.vwh.net/msail_mod.html to work out the rhumb line distance and it gave me 1846nm

then i used Javascript Great Circle Calculator to work out the GC distance and it gave me 1845nm

surely it has to be more than 1nm between the two

Well if the fancy computers say it's 1 nm between them it probably is :ok:

In the meantime, I had my head way too deep into this one, I think my explanation should be more or less correct now. My initial mistake was to make the calculation based on a E/W departure of 300nm; however, since the east/west movement is done throughout the flight, that distance will be greater. If it was done all at 30N it would be 519nm, as you wrote, and in my above calculation, I've assumed it to occur at the midpoint, ie 45N, giving a departure of 424nm. This also makes it easy to measure the hypotenus as you can design your chart to be scale accurate at that latitude.

From curiosity, the cos(45) is not the arithmetic average of cos(30) and cos(60). If we use the arithmetic average rather than the geographic average, we get a departure of 409,8nm and a consequent RLD of 1846,06 nm. This may be the way your computer calculated it.

But beyond this... I'm getting a headache. Long live the FMS :D

Edit: I see from that calculator you posted, that the results also do vary with which earth model you use (spherical or WGS84). Also, it probably does not account for altitude, I would guess. So I think we are down to splitting hair now.

Using 'normal' trigs ie COS, SIN etc you will only get an approximate answer as these apply to FLAT surfaces. To get an accurate answer you have to get into 'spherical' trig which is basically what is hidden inside IRS/INS & GPS but is outside the ATPL syllabus. Also bear in mind Earth is not a true sphere.

The most accurate computation I've come across for a great circle arc is that contained in Thaddeus Vincenty's 1975 paper in Survey Review - Directorate of Overseas Surveys of the Ministry of Overseas Development (Surrey, UK). Copy available here, http://www.ngs.noaa.gov/PUBS_LIB/inverse.pdf

I expect Mr. E. Williams' website (linked via post above) has employed Vincenty's formulæ [wikipedia entry here] as the paper is referenced on his GC Calc page. Mr. Chris Veness has also implemented a javascript calculation using Vincenty, at Geodesic distance between two Latitude/Longitude points using Vincenty ellipsoid formula in JavaScript (useful as he has provided the complete code).

Accuracy for the specified ellipsoid is claimed to be within half a millimetre.