Angle/Slope of Tangent to a curve vb.net

[russell84]

Is anyone able to help

In vb.net how do i find the angle of the tangent line to a curve at a given point relative to the X axis?

Also would this have to be quadrant sensitive?

[CarlB]

Don't know vb.net, but the function/expression you need is the "arc tangent", so look for something like that. In lisp it's "atan". And yes it's quadrant sensitive, positive tangent would give you a 0-90 degree result, negative is for a 90-180 degrees.

**edit** found this, might help:

This example uses the Atan method of the Math class to calculate the value of pi.

Public Function GetPi() As Double
    ' Calculate the value of pi.
    Return 4.0 * Math.Atan(1.0)
End Function

[SEANT]

The general methodology for 3d curves would be:

1. Select the curve
2. Pick a point(Point3d) on the curve with (DatabaseServices.Curve.GetClosestPointTo)
3. Get first derivative(Geometry.Vector3d) with(DatabaseServices.Curve.GetFirstDerivative (Geometry.Point3d))
4. Get angle with System.Math.Atan(Geometry.Vector3d.Y/Geometry.Vector3d.X)

[russell84]

ahh thank you very much

[russell84]

so how do i calculate what quadrant it is in so that i can calculate the 'Normal to the tangent'.?

[Lee Mac]

Tan is positive for quadrants 1 and 3 and negative for quadrants 2 and 4, hence a positive result indicates a vector at 0-pi/2 to the x-axis, plus or minus pi; and negative pi/2 - pi, plus or minus pi.

I suppose you could add coding to correct the angle by pi radians so that you will only get an answer between 0 and pi, then the normal to the tangent will just be the arctan + pi/2.

Is this a purely 2D problem, or are you interested in the Binormal to the tangent also?

[russell84]

No its purely 2d.

Thanks for that - i changed the code around so it will work now.

This is what i have.

Public Function CalcAngle(ByVal ObjCurve As AcadNetDbServices.Curve, ByVal Dist As Double)
 
Dim Radians As Double
Dim Degrees As Double
Dim Point1 As AcadNetGeometry.Point3d = ObjCurve.GetPointAtDist(Dist)
Dim firstderiv As AcadNetGeometry.Vector3d = ObjCurve.GetFirstDerivative(Point1)
Radians = Math.Atan2(Cdbl(firstderiv.Y) , Cdbl(firstderiv.X))
Degrees = Radians * (180 / Math.PI)

'If in positive quadrants then = 180 - abs.Degrees
'If in Negative quadrants then = Abs.Degrees - 180
If Degrees < 0 Then
Degrees = (Math.Abs(Degrees)) + 180
Else
Degrees = 180 - (Math.Abs(Degrees))
End If
CalcAngle = Degrees

End Function

[CarlB]

Not sure why you would set Degrees = 180 - angle for any positive result. Say it was initially calculated at 45 degrees, wouldn't that be a good answer already? But then I don't know what you're using it for...

[Lee Mac]

I don't know much about vb.net (in fact I know absolutely nothing about it), but when you find the first derivative, does that not return a vector? - Then you would know exactly the direction of the tangent from the vector.

[russell84]

derivative gives the slope of the tangent.

using atan2 gets the arc tangent to it.

also i had to use Degrees = 180 - Angle in the positive quadrant because i was getting answers like 175 & 115 when it should be 5 & 65 etc

Thanks for your help guys.

This works