polarpoint

Hi

I have a line running from 0,0,0 to 120,5,5. I want to put a point 4" up the
line.

I use polarpoint to move out the 4" along the line and then I use the rise
that the 'Z' triangle makes to move it up. However it just doesn't come out
right.



Does anyone know of another command or any suggestions?



Thanks

Bob vetrano

 

I'm not sure, exactly, what you're trying to accomplish. If you want to go
4 units, parallel to the line, polarpoint should be exactly what you need,
something like:

vStPt = oLine.StartPoint
vEnPt = oLine.EndPoint
dAng = ThisDrawing.Utility.AngleFromXAxis(vStPt, vEnPt)

ThisDrawing.Utility.PolarPoint vStPt, dAng, 4

If you have something similar to this and you're not getting the point you
expect, check your coordinate system.

 

always easier for someone to point out whats wrong if they can see what
you're trying...ie the code.
other wise all that can be said is it sounds like you know where you want
your new point to be, so just calc the coordinates and put it there.
how are you calculating the rise?
how are you using polar point?
sounds like you could just use trig to calc the point then use that for the
insertion point of the point object.

 

You're talking about a 3d line, and the angle along the line.
I don´t think polarpoint will work for you here...
....unless, not sure it this will work, but maybe if you move the UCS so that
the XY plane contains the line, and then use the polarpoint.

 

You can calculate the length of the line between the two points or get it
from ACAD if it's a Line object. Then, no trig or anything involved,
interpolate the X, Y, and Z values separately between the StartPoint and
EndPoint. For instance, in your case the line's length is 120.2", so 4" is
3.33% along the line. So your point's X would be 0.0 + 3.33% * (120 -
0.0). And so on for Y & Z.

James

 

hi
what i am trying to do is put a circle 4" along a 3D line which is angled
and rising. attached is the code. there are a lot of msgboxs to check points
and angles. Perhaps I'm over complicating the situation.

thanks bob Vetrano

Sub drawline()
Dim drawline As AcadLine
Dim startpt As Variant
Dim endpt As Variant
Dim SelectedPoint As Variant
Dim slopeLineLen As Double
Dim slopeRun As Double, slopeRise As Double, slopeAngle As Double
Dim DeltaSlopeX As Double
Dim slopePlanAngle As Double
Dim Circle1 As AcadCircle
Dim r As Double
Dim Ccenter As Variant
Dim LeftSetBack As Double
Dim DeltaLeftSetBackX As Double
Dim DeltaLeftSetBackY As Double
Dim PlanDeltaLeftSetBackX As Double
Dim PlanDeltaLeftSetBackY As Double
Dim PathStartpt As Variant
Dim pointy As Variant

Dim ucsobj As AcadObject
Dim AngleY As Double

With ThisDrawing.Utility
.GetEntity drawline, SelectedPoint, "select line"
End With
startpt = drawline.StartPoint ' start point of Drawline
endpt = drawline.EndPoint ' end point of Drawline
slopeLineLen = drawline.Length ' length of Drawline
slopePlanAngle = drawline.Angle ' Angle in plan of Drawline

AngleY = 90 * 3.14159 / 180 + slopePlanAngle
'MsgBox "... " & AngleY & ">>>"
pointy = ThisDrawing.Utility.PolarPoint(startpt, AngleY, 8)

'MsgBox "... " & startpt(0) & ">>>"
'MsgBox "... " & startpt(1) & ">>>"
'MsgBox "... " & startpt(2) & ">>>"
'MsgBox "... " & pointy(0) & ">>>"
'MsgBox "... " & pointy(1) & ">>>"
'MsgBox "... " & pointy(2) & ">>>"

'used this circle as a test to see if I had a point(pointY) that is
perpendicular to drawline.

r = 1
Ccenter = ThisDrawing.Utility.PolarPoint(startpt, AngleY, 4)
Set Circle1 = ThisDrawing.ModelSpace.AddCircle(Ccenter, r)

' get error when trying to set ucs

Set ucsobj = ThisDrawing.UserCoordinateSystems.Add(startpt, endpt, pointy,
"DrawlineUcs")


slopeRise = endpt(2) - startpt(2) ' 'Z' distance of Drawline
DeltaSlopeX = endpt(0) - startpt(0) ' delta 'X' distance of
Drawline
slopeAngle = Atn(slopeRise / DeltaSlopeX) ' Angle in the world of
Drawline

LeftSetBack = 4 ' setback from end of line

DeltaLeftSetBackX = Cos(slopeAngle) * LeftSetBack ' get delta X of the 'Z'
triangle

DeltaLeftSetBackY = Sin(slopeAngle) * LeftSetBack ' get delta y of the 'Z'
triangle

PlanDeltaLeftSetBackX = Cos(slopePlanAngle) * LeftSetBack ' get delta X of
the line in plan (top view)
PlanDeltaLeftSetBackY = Sin(slopePlanAngle) * LeftSetBack ' get delta y of
the line in plan (top view)

'want to place this circle 4" up the line that is angled and
rising


r = 1
Ccenter = ThisDrawing.Utility.PolarPoint(startpt, slopePlanAngle, 8)
Set Circle1 = ThisDrawing.ModelSpace.AddCircle(Ccenter, r)

'MsgBox "... " & slopeRise & " >>> "
'MsgBox "... " & DeltaSlopeX & " >>> "
'MsgBox "... " & slopeAngle & " >>> "
'MsgBox "... " & slopePlanAngle & " >>> "
'MsgBox "... " & DeltaLeftSetBackX & " >>> "
'MsgBox "... " & DeltaLeftSetBackY & " >>> "
'MsgBox "... " & PlanDeltaLeftSetBackX & " >>> "
'MsgBox "... " & PlanDeltaLeftSetBackY & " >>> "



End Sub

 

In that case, I'd suggest the same as James.

 

just a thought, but maybe if you construct a 4 unit radius circle at the end point, then get the intersectwith with acExtendthisentity and then get the new endpoint and then draw your circle there...

 

Bob,

What is your final goal, your overall task? Are you really trying to draw
circles, or do you want to create UCS's, or what?

In order to find the plane perpendicular to the line, I wouldn't use any
angle calculations. Instead, I would use cross products. They have the
handy property that when you cross 2 vectors (say, first vector from the
startpoint to the endpoint of the line, and second vector a random one), the
resulting vector will be perpendicular to both of the first (i.e. if you
cross X & Y axis, you get the Z-axis). If you THEN cross your new vector
with your original vector, you get a vector that is perpendicular to your
line. Scale this vector to a length of 4, add it to your startpoint, and
you have a point that's 4" away from your startpoint in a direction
perpendicular to your line.


HTH,
James

 

James
I knew I was making it more complicated then then I had to. That worked
great. I also want to thank everyone that replied and offered assistants. I
learned something from everyone

bob v

 

I'm glad you could make sense of my explanation. I was trying not to write
a book, but still give you a complete thought. But since you used this
method, I'll add a couple of things you want to check to keep your
cross-products from failing or giving you a useless answer...

-- make sure the first line and your random line each have length > 0
(this sounds obvious, but this can be tough to track down b/c you get
(0,0,0) as an answer)
-- make sure that first line and your random vector are not parallel (in
either the same or opposite directions)

HTH,

James