How to calculate angle of a line

[Killer42]

Hi all.

Sorry to be brief, but have to leave.

If I have the coordinates of two points (on the computer screen), how do I calculate the angle of the line which joins them?

I know, my terms are vague and in some cases completely wrong (for instance it would be an "interval", not a line) but hopefully you get the idea.

[Shashi Sadasivan]

Trignometry :)
you know the cordinates of the points,
Get the height and the width of the that line and use tan

angle = tan inverse ( X / Y)

[Killer42]

Trignometry :)
you know the cordinates of the points,
Get the height and the width of the that line and use tan

angle = tan inverse ( X / Y)

Thanks. Haven't done a lot of trig for a few years.

I had already worked out the horizontal and vertical distance (X and Y), and I routinely use these to calculate the straight-line distance (good old Pythagorus), but didn't know how to work out the angle.

I'm still not sure. I need to be able to go the full circle. That is, angles all the way around to 360 degrees (or just under, since that would be the same as zero of course). Will this work? Not sure what you mean by "inverse", in this context. (Be gentle, it has been a long time...)

[Shashi Sadasivan]

I'm still not sure. I need to be able to go the full circle. That is, angles all the way around to 360 degrees (or just under, since that would be the same as zero of course). Will this work?

Depends on your frame of reference (my frame of reference is the x axis)

Not sure what you mean by "inverse", in this context. (Be gentle, it has been a long time...)

Well... lets assume that we now see @ as theta (the general angle variable)
tan @ = X/Y;
so @ = tan inverse (X/Y)
i have to type in inverse, because i cant type in -1 in superscript :D

Hope the math comes along well for you :D

cheers

[bartonc]

Hi all.

Sorry to be brief, but have to leave.

If I have the coordinates of two points (on the computer screen), how do I calculate the angle of the line which joins them?

I know, my terms are vague and in some cases completely wrong (for instance it would be an "interval", not a line) but hopefully you get the idea.

Here's how I'd do it:

>>> from math import atan, degrees
>>> bottom, left = 0, 0
>>> top, right = 10.0, 20.0
>>> degs = degrees(atan((top - bottom)/(right - left)))
>>> print degs
26.5650511771
>>> 

[Shashi Sadasivan]

Double radians = Math.Atan((x2-x1)/(y2-y1));
Double degrees = radians *  180 / Math.PI;

Thats c# :D

[Killer42]

Thanks for that, people.

Since I need to cover the full circle (in other words, the angle could be up to 360 degrees), I've had to do some real kludges. Hope someone can point out a better way.

• Calculate (absolute) DistanceX & DistanceY
• Calculate Ratio: DistanceX / DistanceY
• Calculate Degrees = Atn(Ratio) * RadsToDegs
  This provides a value between 0 and 90, which is not good enough. So...
• Determine which "quadrant" we're in (lower-right = 0, upper-right = 1, upper left = 2, lower left = 3) by checking signs of horizontal & vertical distances.
• Adjust the angle based on the quadrant.
  • Quadrant 0: 360 - Degrees
  • Quadrant 1: 90 - Degrees
  • Quadrant 2: 90 + Degrees
  • Quadrant 3: 270 - Degrees
• And that's it. Simple, huh. :)

This does produce what looks like the right answer, or close enough. But as you can see, it's rather messy so far.

Note, using the absolute values of the distances dates back to before you people provided the Atan solution, so I'll be revisitng that to see what I can do better.

[Shashi Sadasivan]

Well...I would say calculate the degrees,
the values of X=(x2-x1) and Y=(y2-y1) will indicate which quadrant the angle is being measured to.


Assume that the quadrants are being split as in picture.
X(sign)__Y(sign) ___ Quadrant
__+_______+________ 0
__+_______- ________1
__-_______-_________2
__-_______+_________3

(Excuse the undescores, as that gets trimmed off in the posts)

The angle is measured always to the X axis, so then apply the required arithmetic to get the desired angle from your frame of ref, and angle of ref (ie, whether clockwise or anti-clockwise).

[Killer42]

Thanks Shashi. That's pretty much what I'm doing. I've just complicated it a little by using the absolute values of X and Y to begin with. I plan to change that.

[Shashi Sadasivan]

I would suggest to keep the values of X and Y as they are.
the angles yould come as positive or negative depending which quadrant there are in (0 and 2 will be positive and 1,3 negative)

Soall you then need to find which qadrant they are in
and then
(90 * (quadrant number +1)) + angle = resulting required angle

the resulting angle starts from the X axis at quadrant 0, and works it was anti clock wise from there.

cheers :) (woo hoo,, i never knew i could start liking trignometry....no...wait, i started avoiding it when my games designer friend came to ask me doubts :P)

[Killer42]

Ok, thanks for that.

One thing, though. This is something which keeps cropping up here lately. The word is "questions", not "doubts".

[Shashi Sadasivan]

Oh, no it was doubts,
Because he was good at geometry and trignometry, and he would come and pile everything up on me, and explaining the humungus geometry, seemed that he would almost recreate the entire earth in detail.
And expected me to follow him, and call out when the error is occouring !!

so doubts...not questions !!! (questions would have been easier than doubts)

[bartonc]

Thanks for that, people.

Since I need to cover the full circle (in other words, the angle could be up to 360 degrees), I've had to do some real kludges. Hope someone can point out a better way.

• Calculate (absolute) DistanceX & DistanceY
• Calculate Ratio: DistanceX / DistanceY
• Calculate Degrees = Atn(Ratio) * RadsToDegs
  This provides a value between 0 and 90, which is not good enough. So...
• Determine which "quadrant" we're in (lower-right = 0, upper-right = 1, upper left = 2, lower left = 3) by checking signs of horizontal & vertical distances.
• Adjust the angle based on the quadrant.
  • Quadrant 0: 360 - Degrees
  • Quadrant 1: 90 - Degrees
  • Quadrant 2: 90 + Degrees
  • Quadrant 3: 270 - Degrees
• And that's it. Simple, huh. :)

This does produce what looks like the right answer, or close enough. But as you can see, it's rather messy so far.

Note, using the absolute values of the distances dates back to before you people provided the Atan solution, so I'll be revisitng that to see what I can do better.

I developed these for my Garmin GPS interface:

 
def QuadHeading(east, north):
    """Given the vector, convert to degrees in a quadrant.
       N is zero, S is 180, E is positive, W is negitive."""
    return degrees(atan2(east, north))
 
def QuadToCirc(heading):
    """Convert quadrant to degrees of a circle."""
    return (heading, abs(heading + 360))[heading < 0] # index the list on bool

[Killer42]

For future reference, here's the function I ended up putting together in VB6. You'll note that I've defined named constants for everything rather than hard-coding any values. This is based on the fact that constants have traditionally been faster to reference than variables. I don't actually know whether this is still the case in VB6.

Option Explicit
DefLng A-Z
 
Private DistanceX As Single, DistanceY As Single
Private Const Zero As Long = 0
Private Const NotQuiteZero As Single = 0.00001
'Private Const One As Long = 1
'Private Const Two As Long = 2
'Private Const Three As Long = 3
Private Const Ninety As Long = 90
Private Const TwoSeventy As Long = 270
Private Const Pi As Single = 3.14159265358979
Private Const RadsToDegs As Single = 180 / Pi
 
Public Function DirectionOfLine(ByVal From_X As Single, ByVal From_Y As Single, ByVal To_X As Single, ByVal To_Y As Single) As Single
  ' Given two points, return the angle of the line from P1 to P2.
  DistanceX = To_X - From_X
  DistanceY = To_Y - From_Y
  If DistanceY = Zero Then DistanceY = NotQuiteZero ' Prevent div-by-zero error.
  If DistanceY < Zero Then
    DirectionOfLine = Ninety + Atn(DistanceX / DistanceY) * RadsToDegs
  Else
    DirectionOfLine = TwoSeventy + Atn(DistanceX / DistanceY) * RadsToDegs
  End If
End Function

[Killer42]

Actually, here's a shorter (and probably very slightly faster) version...

Option Explicit
DefLng A-Z
 
Private DistanceX As Single, DistanceY As Single
Private Const NotQuiteZero As Single = 0.00001
Private Const Ninety As Long = 90
Private Const TwoSeventy As Long = 270
Private Const RadsToDegs As Single = 57.29578
 
 
Public Function DirectionOfLine2(ByVal From_X As Single, ByVal From_Y As Single, ByVal To_X As Single, ByVal To_Y As Single) As Single
  ' Given two points, return the angle of the line from P1 to P2.
  If From_Y = To_Y Then To_Y = From_Y - NotQuiteZero ' Prevent div-by-zero error.
  If To_Y < From_Y Then
    DirectionOfLine2 = Ninety + Atn((To_X - From_X) / (To_Y - From_Y)) * RadsToDegs
  Else
    DirectionOfLine2 = TwoSeventy + Atn((To_X - From_X) / (To_Y - From_Y)) * RadsToDegs
  End If
End Function

[hopalongcassidy]

Hi all.

Sorry to be brief, but have to leave.

If I have the coordinates of two points (on the computer screen), how do I calculate the angle of the line which joins them?

I know, my terms are vague and in some cases completely wrong (for instance it would be an "interval", not a line) but hopefully you get the idea.

Try arctan((y2-y1)/(x2-x1))

By the way, I read all of the replies and every one of them is wrong. The tangent of an angle is the change in Y divided by the change in X. The change in X divided by the change in Y is the cotangent!!!

[Killer42]

Try arctan((y2-y1)/(x2-x1))

Um... isn't that what I've got?

By the way, I read all of the replies and every one of them is wrong. The tangent of an angle is the change in Y divided by the change in X. The change in X divided by the change in Y is the cotangent!!!

Thanks for the correction. I have to admit though, all I'm really concerned about is whether the code works. Which it does.

[cowpie]

Since this is the first link that pops up on Google when I search how to do this, I thought I'd include a much easier method.

Double radians = Math.Atan2((y2-y1), (x2-x1));
Double degrees = radians * 180 / Math.PI;
 
if(degrees < 0) {
   degrees += 360; // degrees are now 0-360
}

[Verrazano]

recently today my friend and I worked this out originally I thought of using the four quadrants of a graph to get the angle, but the way I had done it ended with me having to build a large table to get values back.
so after about an hour we came up with this:
cos^-1(adjacent/hypotenuse) this translates to being: let one point be your origin and the other point be the x and y values you get the hypotenuse by obviously using the Pythagorean theorem. So lets say the coordinates are (0,0) and (4,4) using the formula above it would look like this.
cos^-1(4/sqrt(4^2+4^2)) = 45 degrees which is correct since the slope of the line is 1/1.
hope this helps,
Verrazano

[Killer42]

Cowpie, what language is that? VB.Net?

I am/was working in VB6. By the way, this question is from 4 years ago. :-)