Posted 16 April 2011 - 12:57 AM
Posted 16 April 2011 - 01:00 AM
I posted this in General Chat as I am not asking for code, I just need the mathematical formula to figure out the problem.
I am two points, A and B. Given both the x and y coordinates and assuming the distance from x1 to x2 is 1P, what formula could I use to figure out the distance, in a straight line, from both points with completely different x coord's and y coord's. I figure I would need to figure the slope, but I have no idea where to go from there.
This is why I wish I paid more attention in Algebra.
Posted 16 April 2011 - 01:03 AM
X1:=0, Y1:=0, X2:=3, Y2:=3 distance := sqrt((y2-y1)**2+(X2-X1)**2) MsgBox % Distance " should equal the square-root of 18:`n" sqrt(18)
Posted 16 April 2011 - 01:03 AM
Posted 16 April 2011 - 01:12 AM
Posted 16 April 2011 - 01:27 AM
Posted 16 April 2011 - 03:04 AM
hahhah. Yeah, I was trying to take the easy way out.@codybear:
1.)Save image as
2.)upload to AutoHotkey.net
3.)???
4.)Profit
Posted 16 April 2011 - 11:48 AM
a4u
Posted 16 April 2011 - 05:23 PM
Pythagorean Theorem & Quadratic Formula - two things my Algebra II teacher said to Never Forget :wink:This is why I wish I paid more attention in Algebra.
PEG
Posted 05 January 2012 - 03:30 PM
distcords(X1,Y1,X2,Y2)
{
Return sqrt((Y2-Y1)**2+(X2-X1)**2)
;~ Round(sqrt((Y2-Y1)**2+(X2-X1)**2))
}Usage: distcords(0,0,100,100) would return ~141.421... (the rounded version 141.421 = 141)
Posted 05 January 2012 - 04:33 PM
Posted 05 January 2012 - 11:31 PM
Posted 06 January 2012 - 01:40 PM
Posted 06 January 2012 - 04:23 PM
I think she means that for real navigation calculations, complexer math would be requiered, as the world is not flat and even not a sphere.Mabye it's too early in the morning but I can't translate anything you just said... :lol:
Pythagoras requires a (2D) right-angled triangle.
Posted 06 January 2012 - 04:52 PM
