Computer Science Canada

How to get the direction (in degrees) between two points?
Author:  s_climax [ Tue May 04, 2004 5:32 pm ]
Post subject:  How to get the direction (in degrees) between two points?
ssia
Author:  Mazer [ Tue May 04, 2004 6:09 pm ]
Post subject:  Re: How to get the direction (in degrees) between two points
s_climax wrote:
ssia

What the hell?

Anyways...
you can use arctand to give you an angle in degrees by supplying it with a slope. Of course that's not enough (most likely); that will just give you something between -90 and 90. You'll need to figure out the actual angle by checking which quadrant the second point is in with relation to the first point.
Author:  s_climax [ Tue May 04, 2004 6:18 pm ]
Post subject: 
ssia stand for "subject says it all"

Isn't there just some code I can use with sin/cos?
Author:  Mazer [ Tue May 04, 2004 6:30 pm ]
Post subject: 
Why would you want that?
You wanted a way to get an angle between two points. Given that you have points A (x1, y1) and B (x2, y2) you can get the angle from A to B using arctand (a.k.a. "tan inverse").

code:

angle := arctand ((y2 - y1) / (x2 - x1))

Angle, in this case should be a float. This is also assuming that B is above and to the right of A on a cartesian plane. If it's to the left you'll get a negative angle. And of course make sure to check if the x values are equal so you don't get a division by zero error.
Author:  Cervantes [ Tue May 04, 2004 7:29 pm ]
Post subject: 
code:

function find_angle (x1, y1, x2, y2 : real) : real
    if x2 = x1 then
        if y2 >= y1 then
            result 90
        elsif y2 < y1 then
            result 270
        end if
    else
        if x2 > x1 and y2 >= y1 then %QUAD 1
            result (arctand ((y2 - y1) / (x2 - x1)))
        elsif x2 > x1 and y2 < y1 then %QUAD 2
            result 360 + (arctand ((y2 - y1) / (x2 - x1)))
        elsif x2 < x1 and y2 < y1 then %QUAD 3
            result 180 + (arctand ((y2 - y1) / (x2 - x1)))
        elsif x2 < x1 and y2 >= y1 then %QUAD 4
            result 180 + (arctand ((y2 - y1) / (x2 - x1)))
        end if
    end if
end find_angle

Mouse.ButtonChoose ("multibutton")
var x1, y1, x2, y2 : real
var mx, my, btn : int
var firsthit := false

loop
    cls
    mousewhere (mx, my, btn)

    if btn = 1 and firsthit = false then
        firsthit := true
        x1 := mx
        y1 := my
    end if
    if btn = 100 then
        firsthit := false
    end if
    if firsthit = true then
        x2 := mx
        y2 := my
        drawline (round (x1), round (y1), round (x2), round (y2), black)
        locate (1, 1)
        put find_angle (x1, y1, x2, y2)
    end if
    View.Update
    delay (10)
end loop

that's how. Neutral
Author:  s_climax [ Tue May 04, 2004 8:24 pm ]
Post subject: 
If the line is horizontal and going to the right, shouldn't the angle be 360 or 0 not 270?
Author:  Cervantes [ Thu May 06, 2004 4:48 pm ]
Post subject: 
no it shouldn't be.
and in this program and in normal cardinal directions, a horizontal line going to the right isn't 270, its 90.
Author:  Tony [ Thu May 06, 2004 4:53 pm ]
Post subject: 
umm... no...

horizontal line that is directly to the right is 0 Laughing
up is 90
left is 180
down is 270

do you know why? because cos(0) = 1
Author:  s_climax [ Thu May 06, 2004 5:16 pm ]
Post subject: 
Thanks for clarifying that tony. I was right.
Author:  Cervantes [ Thu May 06, 2004 5:19 pm ]
Post subject: 
Cervantes wrote:
cardinal directions


Confused

anywho, a line going to the right still wasn't 270.
Author:  s_climax [ Fri May 14, 2004 8:34 pm ]
Post subject: 
Will this work instead?

code:

xDistance := personX - chaserX
yDistance := personY - chaserY

angle := tan (yDistance / xDistance)
Author:  Tony [ Fri May 14, 2004 9:45 pm ]
Post subject: 
no, lol... first of all if xDistance = 0 you get runtime error (div 0) and even if you don't, it's arctan, not tan... that returns the ratio based on the angle (in radians)
Author:  guruguru [ Fri May 14, 2004 11:14 pm ]
Post subject: 
My example was simply that, and example. Clearly there are errors that can occur (dividing by zero), but they can be tested for and then dealt with.

Sorry, I thought that Turing did it in degrees... my bad. Anyways, to convert to degrees, simple multiply the answer by 180/Pi.

As well, I realize that it works for only one quadrant. Again, this was an EXAMPLE, so you can easily modify it!!!

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