find if an x and y co-ordinate between another co -ordinate.

[dilla1988]

hi guys,

So i have about 1000s of lat and long values.

for this set of data I have drawn a boundary wall so i wanna see if these co-ordinates are within that boundary. if yes it's simple "YES" if not "NO"

so the boundaries are:
A -37.921484 ,144.983256
B-37.925141, 145.209269
C-38.035877, 144.980122
D -38.039549, 145.206486

boundary wall looks like:
A B
C D

just few of the co-ords are

-37.6649	,145.3983
-37.6584,	145.3972
-37.9827,	145.1018
-37.9846,	145.0923
-37.9849,	145.0999
-37.9842,	145.0986
-37.7262,	144.8926

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So what's the best way to approach this? I'm getting a bit confused with the negative values.

 

[Kaper]

Hi,

First of all, your fence looks rather like:

DB

CA

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Having noticed that, appropraiate approach could be:
find equations of two close to horizontal pieces of fence DB and CA as y1=a1*x+b1 and y2=a2*x+b2
and equations for vertical pieces DC and BA as x1= c1*y+d1 and x2= c2*y+d2

Then for each point check if y is between y1=a1*x+b1 and y2=a2*x+b2 for given x
and whether
x is between x1= c1*y+d1 and x2= c2*y+d2 for given y

If both conditions are met - (x,y) is within the fence

See sample formulas in http://klubexcela.pl/images/files/xls/Testfence.xls
of course formulas from columns D:F could be combined into one, so in D8 one could write a bit longer direct formula:

=IF(AND(B8>=C8*$H$5+$I$5,B8<=C8*$H$6+$I$6,C8>=B8*$H$2+$I$2,C8<=B8*$H$3+$I$3),"YES","NO")

Where B8 and C8 are coordinates of tested point, and in H2:I5 are located parameters a...d of fence lines

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[mikerickson]

Consider a point X

If X is inside the quadrilateral, the Angle BAX has to be smaller than BAC
Similarly ACX < ACD, CDX < CDB and DBX < DBA

 

[Kaper]

Hi Mike,
Sure, but it will probably (correct me if I'm wrong) need more computing, at least if we try (as I'd do) to find angles based on the law of cosines.

Angle = arccos ( (BX^2-BA^2-AX^2) / 2*BA*AX )

and Pythagoras' theorem to calculate the length of the AX BA and BX

Length = sqrt( (X2-X1)^2 + (Y2-Y1)^2 )

 

[Rick Rothstein]

I think you will find the code I posted in my mini-blog here to be of use...

Test Whether A Point Is In A Polygon Or Not

 

[mikerickson]

Hi Mike,
Sure, but it will probably (correct me if I'm wrong) need more computing, at least if we try (as I'd do) to find angles based on the law of cosines.

Angle = arccos ( (BX^2-BA^2-AX^2) / 2*BA*AX )

and Pythagoras' theorem to calculate the length of the AX BA and BX

Length = sqrt( (X2-X1)^2 + (Y2-Y1)^2 )

I was thinking more of using the inner product to find angles.
Cos(Angle ABC) = (A-B).(C-B)/Len(A-B)/Len(B-C)

 

[dilla1988]

Hi Rick,

Greetings! Just this morning I came across your blog and the UDF. worked perfect!

thanks

I think you will find the code I posted in my mini-blog here to be of use...

Test Whether A Point Is In A Polygon Or Not

 

[dilla1988]

mikerickson and Kaper, Thanks for your input appreciate it!