Point on arc with a given distance

Is there a way in VB to put a point on an arc that will be a given distance away from its startpoint? I know I could use vlax-curve-getPointAtDist. Unfortunately, I can't. So, I am looking for a VB/math way to do it.

 

I never deal with arcs so someone might have a better solution, but I thought I'd give this a crack.

What about creating a line that starts at the startpoint and proceeds in any direction for the desired distance. Then use the IntersectWith to determine if the endpoint of the line lies on the arc. Unless you're really lucky it probably won't to start with, so then rotate the line (say 90 degrees) and repeat the IntersectWith test.

If you're endpoint doesn't lie on the arc, but you do start getting some intersects along the line, then you can start doing like a binary search where you rotate the line back and forth whilst halving the angle of rotation until eventually your endpoint lands on the arc.

If on the other hand you rotate all the way through 360 degrees without intersecting the arc then you've probably jumped over it, so try again with rotation steps of half the size (ie 45 degrees). If you get all the way around again then halve it again and so on. Eventually when you start getting some hits you can proceed with the binary search.

Hope this helps.

Regards

Wayne Ivory
IT Analyst Programmer
Wespine Industries Pty Ltd

 

Hi

I'm sure there are more elegant solutions, but this code can give you an idea:

'-----
Function ArcPointAtDist(ArcObj As AcadArc, DstVal As Double) As Variant

Dim ArcRad As Double
Dim TmpAng As Double

With ArcObj
ArcRad = .Radius
TmpAng = .StartAngle + (DstVal / ArcRad)
ArcPointAtDist = ThisDrawing.Utility.PolarPoint(.Center, TmpAng, ArcRad)
End With

End Function
'-----

Cheers

 

...if the endpoint of the line lies on the arc.

Wayne, you've just created a circle ;-) An IntersectWith of a circle
(centered on the arc startpoint, and with radius equal to the desired
distance) and the arc will give a point that's a given straight-line
distance from the startpoint, and on the arc. If you don't get an
intersection, then either the included angle or the radius of the arc is too
small to have a point that far away.

James

 

Jürg,
Thanks! That's exactly what I needed.

 

Yes James I have gone full circle, but my solution was making discrete jumps (ie 90 degrees) so I may have stepped over a perfectly valid arc that was sitting at the 45-degree mark.

Anyhow, it looks like Jürg has come up with the "proper" solution, but it was an interesting diversion.

Regards

Wayne

 

Glad to help you...

Cheers

 

this code do that !

Public Function pointOnArc(pCentro As Variant, radio As Single, angBaseGrados As Single, distancia As Single, sentido As String, deltaAngGrados As Single, newAngleGrados As Single) As Variant
'calcula un punto sobre un arco en base a un centro de arco, radio de arco, el angulo
'inicial de grados y la distancia que se desea recorrer de arco
'sentido= cw es en el sentido de las manecillas del reloj
' = ccw contrario a las manecillas del reloj
Dim angRad As Single
Dim longCircle As Single
Dim p() As Double
Dim Pi As Double

Pi = 4 * Atn(1)
ReDim p(2)
longCircle = 2 * Pi * radio
deltaAngGrados = (distancia * 360) / longCircle
sentido = Trim(LCase(sentido))
If sentido = "cw" Then
newAngleGrados = angBaseGrados - deltaAngGrados
Else
newAngleGrados = angBaseGrados + deltaAngGrados
End If
angRad = gradosARadianes(CDbl(newAngleGrados)) 'CDbl convierte el single a double

p(0) = radio * Cos(angRad) + pCentro(0)
p(1) = radio * Sin(angRad) + pCentro(1)
p(2) = 0
pointOnArc = p
End Function