UNSOLVABLE!?!?!?! latitude longitude range bearing and degrees

[Skylar289]

Hey guys and gals, here is a good one,

Given 3 latitudes and longitudes, can you calculate the bearing from the first lat/long to the second lat/long. 2nd to 3rd, then 3rd to first?

The ULTIMATE goal is to create an irregular polygon given multiple lat longs, and determine the area.

I don't necessarily care about the terrain elevation data anomalies, but a rough guestimation.

I've searched the threads and done the research, and the trig and math is there, but I cant seem to figure it out.

This is in Degrees Decimal

Lat1 -29.952092°
Long 1 -54.974860°

Lat2 -29.949798°
Long 2 -55.012654°

Lat3 -30.003882°
Long 3 -55.087920°

Lat4 -30.039345°
Long 4 -55.105729°

Lat5 -30.117079°
Long 5 -55.078748°

Lat6 -30.117432°
Long 6 -55.033185°


Obviously the lats and longs need to be interchangeable. I attempted to get the distance via the haversine formula, then take the distance and bearing from point 1 and 3 to determine the angle of 2, and re-construct that way....but I obviously can't hence the request for help.


Thanks!

Sky

 

[marious]

Excellent formula to calculate an area by coordinates by Mr. pgc01

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AreaPolingon

*	Q	R	S	T	U	V	W
1	*	*	*	Area (Km2)	135.9181234	*	*
2	*	*	*	OFFSET	-3.6395622	*	*
3	*	*	*	*	*	*	*
4	*	LAT	LON	LAT KM	LON KM	XAYB+	XBYA-
5	A	0.002294	-0.03779	0.2550928	0	0.928426	0
6	B	0	0	0	3.6395622	0	21.88884
7	C	0.054084	0.075266	6.0141408	10.887678	75.79432	108.4154
8	D	0.089547	0.093075	9.9576264	12.6026847	99.62022	234.4307
9	E	0.167281	0.066094	18.6016472	10.0044144	104.4798	186.4913
10	F	0.167634	0.020531	18.6409008	5.6166975	0	1.432779
11	G	0.002294	-0.03779	0.2550928	0	0	0
12	*	*	*	*	*	280.8228	552.659
13	area = 1/2(XAYB+XBYC+XCYD+XDYE+XEYF+XFYA-XBYA-XCYB-XDYC-XEYD-XFYE-XGYF-XAYG)	*	*	*	*	*	*

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Spreadsheet Formulas

Cell Formula U1 =ABS((V12-W12)/2) R5 =O5-$O$2 S5 =P5-$P$2 T5 =R5*111.2 U5 =S5*96.3+3.63956219999982 V5 =T5*U6 W5 =T6*U5 R6 =O6-$O$2 S6 =P6-$P$2 T6 =R6*111.2 U6 =S6*96.3+3.63956219999982 V6 =T6*U7 W6 =T7*U6 R7 =O7-$O$2 S7 =P7-$P$2 T7 =R7*111.2 U7 =S7*96.3+3.63956219999982 V7 =T7*U8 W7 =T8*U7 R8 =O8-$O$2 S8 =P8-$P$2 T8 =R8*111.2 U8 =S8*96.3+3.63956219999982 V8 =T8*U9 W8 =T9*U8 R9 =O9-$O$2 S9 =P9-$P$2 T9 =R9*111.2 U9 =S9*96.3+3.63956219999982 V9 =T9*U10 W9 =T10*U9 R10 =O10-$O$2 S10 =P10-$P$2 T10 =R10*111.2 U10 =S10*96.3+3.63956219999982 V10 =T10*U11 W10 =T11*U10 R11 =O11-$O$2 S11 =P11-$P$2 T11 =R11*111.2 U11 =S11*96.3+3.63956219999982 V11 =T11*U12 W11 =T12*U11 V12 =SUM(V5:V11) W12 =SUM(W5:W11) <TBODY> </TBODY>

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Excel tables to the web >> Excel Jeanie HTML 4
　
　
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[cyrilbrd]

Hi cyrilbrd

The formula I posted id to calculate the area of a plane polygon. You'd have to do some calculations to get some coordinate values that would make sense it terms of for ex. km.
A latitude degree is approximately the same wherever (with a maximum of 1% difference), but a longitude degree varies a lot.

What I thought could be done, since the OP just wants a rough estimate, is to consider that the polygon is approximately plane in a small vicinity of a point, which is the case in the example (less than 1º in both latitude and longitude. In this case we transform the spherical polygon into a plane polygon, using for a latitude degree and a longitude degree the corresponding values in the vicinity of the medium point (circa 30º 55º). That's about 96.3 km for 1º latitute and 111.2 km for 1º longitude.

This will give a rough estimate of the area of the polygon (I think).

So, a quick way (not much thought into this)

- transform the coordinates subtracting the minimum of the latitude and longitude.
- convert to km
- use the formula to get the area

Something like this:

TABLE

As I said this does not have much thought into it. Does this make sense to you both? Please comment.

Hi, thanks for your feedback pgc01.
Using your solution yields an area that is only 24 meters higher than by using Decimal Degrees and Heron's formula. I do not know for the OP but this is quite good considering the limitations.
This is a very nice approach. As the OP stated, elevation is not considered.

To be able to use such solution for other coordinates, the users would just have to adjust such as:

Φ	ΔLAT	ΔLONG
0°	110.574 km	111.320 km
15°	110.649 km	107.551 km
30°	110.852 km	96.486 km
45°	111.132 km	78.847 km
60°	111.412 km	55.800 km
75°	111.618 km	28.902 km
90°	111.694 km	0.000 km

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As always you come up with the best solution.
Thanks again for your assistance.

 

[pgc01]

Hi Cyrillbird

Thank you for your comments.

This is a simple solution if you don't mind that it's not too exact.

The problems are

- it must be a small polygon, so that we can consider it as if it were in a plane and not lose much precision. In this case we had less than 1 degree width/height and so it was OK

- as I posted and you commented, we need to have values for 1 degree of lat/long in the vicinity of the point. This means a table, like you posted, of some formula to get it. If I have the time during the weekend I'll see if we can get those values with a formula.

If the OP doesn't mind using vba, then a solution like yours will be preferable, it's clean and works everywhere.

 

[pgc01]

Cyrillbird

cyrilbrd, sorry about your name in the last post. Must have been one of those automatic substitutions and I didn't notice.

 

[cyrilbrd]

cyrilbrd, sorry about your name in the last post. Must have been one of those automatic substitutions and I didn't notice.

No worries!

If the OP doesn't mind using <acronym title="visual basic for applications">vba</acronym>, then a solution like yours will be preferable, it's clean and works everywhere.

Re VBA, Please kindly enlighten us with your opinion whenever you will have the time, I would be interested to see your opinion on the matter.


Cheers.