Fun arithmatix with EinsteinX

I have a utm gridded map so I estimate the kilometer distance, multiply by 6 and move the decimal place over one place (to left) to get miles. That's about the only caculatin' I do on the trail.

Thank you for reminding me how much I detest the spherical coordinate system :).

Adam

If one is talking about actual hiking distance between two points, one also has to take into account topography. Something that looks like "one mile" can be many times that — for example — if the trail includes 96 switchbacks.

EinsteinX,
You and your stupid Kilometers.
Why?
Now it's going to take me twice as long to figure this out.
Thank's a lot for rubbing it in that us Americans have to multiply everything by .6214 just to make any sense of it.

And on top of that, the only way I can make sense of anything you can so easily make sense of is to use Nautical Miles for the distance.
So in reality this is going to take me 3x longer to figure out.

O.K.,

So, using Pi, 42,000 km gives the earth’s diameter 13,369 km.
Since there are 360* in a circle, 10* is 1/36th of distance.

So 1/36th of 42,000 divided by Pi of the 10* or 600nm or 1,111.91km (http://www.metric-conversions.org/length/uk-nautical-miles-to-kilometers.htm ) added over the strait line distance is 353.932km.

Now you just find out what the strait line distance traveled from 10* LAT 0* LON and 0* LAT 10* LON is and add 353.932 on to that.

So in 600nm = 1,111.91km and the strait line distance going from 10* LAT 0* LON and 0* LAT 10* LON is 850nm or 1575.2km, (http://ostermiller.org/calc/triangle.html).

Then just add the 353.932 to 1575.2 and you have a total distance of 1,929.132km

O.K.,
The 353.932 is not correct as it is the addition for 600km not the 850km that you travel across the plane.

The correct addition would be 501.4 km giving the corrected answer of 2076.6 kilometers…

Aaron, firstly I don't care if you calculate the distance in miles, parsecs, light seconds or the length of you thumb. You're eventual answer will only differ a factor of the answer in km and will be easily calculated. Anyway, it seemed like I made a small error in the circumference of the earth. My aging mind remembered 42000km, when in fact, the circumference is only 40050km on average [1].

I also wonder why you " just find out what the strait line distance traveled from 10* LAT 0* LON and 0* LAT 10* LON is and add 353.932 on to that"??? Your answer of 1575km is pretty close, you're answer of 1929km is 354km to much. [2]

[1] http://wiki.answers.com/Q/What_is_the_polar_circumference_of_the_earth_or_is_it_the_same_as_the_equatorial_circumference

[2]http://williams.best.vwh.net/avform.htm#Dist

To keep the fun going…

With the same parameters as above what is the angle traveled between both points as seen from the centre of the earth?

Have fun, Eins

Hey EinsteinX
O.K. I was wrong.
I was using the distance of 850 km over a strait line and then adding the distance of what the circumfrence would be over the strait line.

Even though you weren't correct I'd be really interested in how you calculated that because I just don't see how it works. What I understand is you make a triangle in a normal two dimensional plane with one side going from the centre of the earth to 0*,10* and the other side going from the centre of the earth to 10*,0*, right? Than you calclate the shortes straight line distance between these two points (which would be a rather substantial tunnel). And from that you seem to be able to calculate the arc distance along the surface of the earth, could you tell me how you do that, because I don't see how that is done.

Thanx, Eins

I think we must make simplifying assumptions, lets forget the the poles are flattened a little and assume any great circle is 40050kilometers taking einsteins figure. Imagine 3 lines from the centre of the earth to 0,0,point 1, 10,0 point 2 and 0,10 point 3. The surface arc distance between point 1 and 2 and point 1 and 3 is 40050/36= 1112.5 kilometers. If you constructed a triangle of straight lines between these surface points it would be something like a 45 degree isocelese triangle so the ratio of the angles between the lines at the centre of the earth will be 10, 10, 10 x 1.414 approximately. Therefore I would estimate the distance from 0,10 to 10,0 along a great circle route to be 1112.5 x 1.414= 1573 kilometers.
On that basis the angle between the points 2 and 3 from the centre of the earth should be about 14 degree