topic Re: in VBA Forum

Perpendicular

Anonymous — Fri, 04 Aug 2000 18:05:56 GMT

Greetings All,
This is my situation: I have a block that is X distance from a curved
pline.
What I would like to do is draw a construction line from the insertion
point of the block perpendicular to the pline. Getting the InsPnt isn't
a problem(first point of line), but how do I get that perpendicular
point(second point of line)? Any thoughts or ideas are thanked in
advance.
Eric Nagel

Re: Perpendicular

Anonymous — Fri, 04 Aug 2000 21:06:19 GMT

Hi Eric,
I thought I was going to have a formula or some code for you, but it
doesn't look I'll have time today. It can be done, it's just a matter of
doing the math and writing the code. Using the acadpolyline.coordinates
array of points, create 3 lines to form a triangle between the points. Then
create a fourth line from the bulge to the center of the longest line. The
two shorter legs of the overall triangle form the bases of two isometric
triangles. The fourth line gives you the angle of the isometric sides. The
peak of the triangles (which will have to be trigged out) will be located on
the center point of the arc. The length of the sides, the radius of the arc.
Then you can use the polarpoint method to create a line through the center
point at a distance of (the radius + the distance form the insertionpoint to
the center point). Your line will be perpindicular to the arc. I know it
sounds rough, but it's just a matter of adding a few more lines than
intended and deleting them afterward. Here's a jpeg of what I'm trying to
relay, hope it helps (darn mail convertor darkens it up). If you don't have
it done by Monday, I'll try to find some time (Mondays are always busier
than Fridays) and come up with a formula and maybe even a little code.
-Josh
[Image]

Eric Nagel wrote:

> Greetings All,
> This is my situation: I have a block that is X distance from a curved
> pline.
> What I would like to do is draw a construction line from the insertion
> point of the block perpendicular to the pline. Getting the InsPnt isn't
> a problem(first point of line), but how do I get that perpendicular
> point(second point of line)? Any thoughts or ideas are thanked in
> advance.
> Eric Nagel

Re: Perpendicular

Anonymous — Tue, 08 Aug 2000 20:31:52 GMT

Here it is Eric,
The line will always fall perpendicular to the inside (concave) portion
of the arc, even if the arc doesn't extend far enough.

Option Explicit
Private Function Radians(Degrees As Double) As Double
Radians = Degrees / 180 * 3.141592654
End Function
Private Function Degrees(Radians As Double) As Double
Degrees = Radians / 3.141592654 * 180
End Function
Sub Perp2Arc()
Dim Parc As AcadLWPolyline
Dim Blk As AcadBlockReference
Dim Ent As AcadEntity
Dim Ent2 As AcadEntity
Dim ArcPt, BlkPt
Dim Bulg As Double
Do 'Select a polyline arc
Set Ent = Nothing
'On Error Resume Next
ThisDrawing.Utility.GetEntity Ent, ArcPt, vbCr & "Select a PolyLine Arc
: "
If Err Then
If ThisDrawing.GetVariable("ERRNO") = 52 Then
Err.Clear
Exit Sub
End If
End If
If Not Ent Is Nothing Then
If TypeOf Ent Is AcadLWPolyline Then
Set Parc = Ent
Bulg = Parc.GetBulge(0) 'verify pline arc - won't work unless
the 1st segment is an arc
If Bulg = 0# Then
MsgBox "The PolyLine You Selected Is Not an Arc. Try Again."

Else
Exit Do
End If
Else
MsgBox "You Did Not Select a PolyLine Arc. Try Again."
End If
End If
Loop
Do 'Select a block reference
Set Ent2 = Nothing
ThisDrawing.Utility.GetEntity Ent2, BlkPt, vbCr & "Select a Block : "
If Err Then
If ThisDrawing.GetVariable("ERRNO") = 52 Then
Err.Clear
Exit Sub
End If
End If
If Not Ent2 Is Nothing Then
If TypeOf Ent2 Is AcadBlockReference Then
Set Blk = Ent2
Exit Do
Else
MsgBox "You Did Not Select a Block. Try Again."
End If
End If
Loop
'begin construction lines & math
Dim Lin1 As AcadLine
Dim Lin2 As AcadLine
Dim Ang As Double
Dim ClockWise As Boolean
Dim RadAng As Double
Dim IsoAng As Double
Dim Radius As Double
Dim CentrPt, PolarPt, PolyPts As Variant
Dim Sp(0 To 2) As Double
Dim Ep(0 To 2) As Double
PolyPts = Parc.Coordinates 'elements 0 & 1 are the startpoint, 2 & 3 are
the endpoint
Sp(0) = PolyPts(0): Sp(1) = PolyPts(1)
Ep(0) = PolyPts(2): Ep(1) = PolyPts(3)
Set Lin1 = ThisDrawing.ModelSpace.AddLine(Sp, Ep)
Ang = Atn(Bulg) * 4 'arctangent of bulge times 4 = angle of arc
IsoAng = Degrees(Ang) 'determine angle of the two base angles
If IsoAng < 0 Then
IsoAng = 180 + IsoAng
ClockWise = True
Else
IsoAng = 180 - IsoAng
End If
IsoAng = IsoAng / 2
IsoAng = Radians(IsoAng)
If ClockWise Then
RadAng = Lin1.Angle + IsoAng + Radians(180) 'angle from endpt of arc to
centerpoint
Else
RadAng = Lin1.Angle - IsoAng + Radians(180) 'angle from endpt of arc to
centerpoint
End If
Radius = (Lin1.Length / 2) / Cos(IsoAng) 'determine radius
CentrPt = ThisDrawing.Utility.PolarPoint(Ep, RadAng, Radius) 'determine
center point
Set Lin2 = ThisDrawing.ModelSpace.AddLine(Blk.InsertionPoint, CentrPt)
PolarPt = ThisDrawing.Utility.PolarPoint(CentrPt, Lin2.Angle, Radius)
'use lin2.angle + radians(180) to force the line to fall to the outside of
the arc
Lin1.Delete
Lin2.Delete
Set Lin2 = ThisDrawing.ModelSpace.AddLine(Blk.InsertionPoint, PolarPt)
End Sub

-Josh

Eric Nagel wrote:

> Greetings All,
> This is my situation: I have a block that is X distance from a curved
> pline.
> What I would like to do is draw a construction line from the insertion
> point of the block perpendicular to the pline. Getting the InsPnt isn't
> a problem(first point of line), but how do I get that perpendicular
> point(second point of line)? Any thoughts or ideas are thanked in
> advance.
> Eric Nagel

Re: Perpendicular

Anonymous — Tue, 29 May 2001 13:42:33 GMT

Hi All,

I read a previous message posted by Monica Eisfeld on 3/19/01 titled "Draw
Perpendicular Line?" I am trying to do something similar. I have
contacted Monica and she unfortunately never had a chance to work on this
tool. Josh from Minkwitz design responded by saying:

"The first thing you have to do is to determine where
the point is relative
to the line or vise versa. For instance if the point is
considered 0,0,0 in the
cartesian coordinate system, which quadrant would the
line fall in? After that,
perpendicularly is a breeze. You just add the
appropriate angle (in radians) to
the angle property of the line."

"A fairly simple way to pin down the location of the
line relative to the
point would be to create temporary lines form the
startpoint and endpoint of
the selected line to the selected point. You could
use the angle property of
those lines to pin down the quadrant."

Could someone point me in the right direction on how to code this? I
am simply trying to have the user select a point on a line, select a second line
and have it automatically draw a new line from the selected point, perpendicular
to the second line that was picked.

I have tried to do this by setting the object snaps prior to drawing the
new line, but for some reason, it try's to draw the new line from the
selected point to the perpendicular of the first point selected.

Any thoughts?

Rob

Re:

Anonymous — Tue, 29 May 2001 19:24:41 GMT

Rob, here is a way I came up with to
do it using good ol geometry and some algebra (but without any trig!). It fails
when the line is either vertical or horizontal though, but that should be easy
to work around. Have a look at the code, watch out for word wrap! I'm sure there
are many neat ways to do it with math - this is my "simple"
version...

Dim oSpace As
AcadObject
Dim oLine1 As AcadLine, oLine2 As AcadLine
Dim
Pt1 As Variant, Pt2 As Variant
Dim m1 As Double, m2 As Double, dX As
Double, dY As Double, b1 As Double

'GET THE CURRENT
SPACE
If ActiveDocument.ActiveSpace = acModelSpace
Then
Set oSpace = ActiveDocument.ModelSpace

Else
Set oSpace = ActiveDocument.PaperSpace
End
If

'PICK OUR POINT
Pt1 = ThisDrawing.Utility.GetPoint(,
"Pick first point :")
'PICK OUR LINE (Pt2 IS NEVER USED)

ThisDrawing.Utility.GetEntity oLine1, Pt2, "Perpendicular to: "

'CALCULATE SLOPE OF THE EXISTING LINE
m1 =
(oLine1.StartPoint(1) - oLine1.EndPoint(1)) / (oLine1.StartPoint(0) -
oLine1.EndPoint(0))
'USE GEOMETRY: PERP LINES SLOPES' ARE NEGATIVE
RECIPROCALS
m2 = -1 / m1
'SOLVE FOR Y-ICEPT OF LINE1, WHICH
WE'LL NEED LATER
b1 = oLine1.StartPoint(1) - (m1 *
oLine1.StartPoint(0))

'WRITING THE SLOPE FORMULA FOR THE PERP LINE LEAVES AN "X" AND A
"Y" UNKNOWN
'BY RE-ARRANGING WE CAN PLUG IN THE EQUATION OF THE LINE
FOR "Y" AND SOLVE
'IN TERMS OF "X" (SCARY, DO IT ON PAPER IF YOU WANT
TO SEE WHERE IT COMES FROM!)
dX = ((Pt1(0) * m2) - (Pt1(1) - b1)) /
(m2 - m1)
'NOW THAT WE HAVE "X" WE CAN SOLVE FOR "Y"
dY =
((Pt1(0) * m2) - (dX * m2) - Pt1(1)) * -1
Pt2(0) = dX: Pt2(1) =
dY
'FINALLY ADD THE LINE
Set oLine2 = oSpace.AddLine(Pt1,
Pt2)

Just to give you a "verbal" rundown
of the math sequence in case you want to work it out...

- Calculate slope of line
(m1)

- Slopes of perp lines are always
negative reciprocals so m2 = -1/m1

- y=mx+b :solve for the y-icept
(b)

- look at slope formula for perp.
line (3-y) / (4-x) = 2 :subst eq. of line for y (y = mx+b)

- solve for x

- solve for y

- all done!

Regards,

Jacob Dinardi

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"Rob Outman" <
href="mailto:routman@nakata.com">routman@nakata.com> wrote in message

href="news:B483578BDBC88ECC6B607EB14B182525@in.WebX.maYIadrTaRb">news:B483578BDBC88ECC6B607EB14B182525@in.WebX.maYIadrTaRb...

Hi All,

I read a previous message posted by Monica Eisfeld on 3/19/01 titled
"Draw Perpendicular Line?" I am trying to do something similar. I
have contacted Monica and she unfortunately never had a chance to work on this
tool. Josh from Minkwitz design responded by saying:

"The first thing you have to do is to determine where
the point is relative
to the line or vise versa. For instance if the point
is considered 0,0,0 in the
cartesian coordinate system, which quadrant
would the line fall in? After that,
perpendicularly is a breeze. You just
add the appropriate angle (in radians) to
the angle property of the
line."

"A fairly simple way to pin down the location of
the line relative to the
point would be to create temporary lines form the
startpoint and endpoint of
the selected line to the selected point. You
could use the angle property of
those lines to pin down the
quadrant."

Could someone point me in the right direction on how to code this?
I am simply trying to have the user select a point on a line, select a second
line and have it automatically draw a new line from the selected point,
perpendicular to the second line that was picked.

I have tried to do this by setting the object snaps prior to drawing the
new line, but for some reason, it try's to draw the new line from
the selected point to the perpendicular of the first point
selected.

Any thoughts?

Rob

Re:

Anonymous — Wed, 30 May 2001 05:15:36 GMT

This works in 2D or 3D and will calculate the
theoretical perpendicular point. I included a test routine after the
PointPerpendicular.

Hope this helps.

Function PointPerpendicular(PointToProject As
Variant, StartPoint As Variant, EndPoint As Variant) As Variant
' This
routine calculates the projection of a point (PointToProject)
' onto a
theoretical line between StartPoint and EndPoint using the
' dot product of
two vectors.

' The resultant vector of the following
algorithm
' represents the point's distance and direction from
StartPoint.

size=2>'
/ U . V \
' projection of U on V = | ------- |
V
'
\ ||V||^2 /
'
' where:
'    U = vector between
StartPoint and PointToProject
'    V = vector between
StartPoint and EndPoint
'    U.V = dot product of two
vectors
'    ||V|| = length of vector V

    ' define
vectors
    Dim u(0 To 2) As Double
    u(0)
= PointToProject(0) - StartPoint(0)
    u(1) =
PointToProject(1) - StartPoint(1)
    u(2) = PointToProject(2)
- StartPoint(2)
    Dim v(0 To 2) As
Double
    v(0) = EndPoint(0) -
StartPoint(0)
    v(1) = EndPoint(1) -
StartPoint(1)
    v(2) = EndPoint(2) -
StartPoint(2)

    ' ensure vectors are not
parallel
    If u(0) = v(0) And u(1) = v(1) And u(2) = v(2)
Then
        MsgBox "The three points
define a straight line", vbCritical, "Error: ThreePointUCS: Invalid
Input"
        Exit
Function
    End If

    ' calculate dot product of U and
V
    Dim uDOTv As Double
    uDOTv = u(0) *
v(0) + u(1) * v(1) + u(2) * v(2)

    ' calculate length of
V
    Dim vLength As Double
    vLength =
(v(0) ^ 2 + v(1) ^ 2 + v(2) ^ 2) ^ 0.5

    ' calculate projected
vector
    Dim projectUonV(0 To 2) As
Double
    projectUonV(0) = (uDOTv / vLength ^ 2) *
v(0)
    projectUonV(1) = (uDOTv / vLength ^ 2) *
v(1)
    projectUonV(2) = (uDOTv / vLength ^ 2) *
v(2)

    ' return
results
    Dim ProjectedPoint(0 To 2) As
Double
    ProjectedPoint(0) = StartPoint(0) +
projectUonV(0)
    ProjectedPoint(1) = StartPoint(1) +
projectUonV(1)
    ProjectedPoint(2) = StartPoint(2) +
projectUonV(2)
    PointPerpendicular = ProjectedPoint
End
Function

Sub testPointPerpendicular()

Dim ln1 As AcadLine
    Dim p1 As
Variant
    Dim pt As Variant
    p1 =
ThisDrawing.Utility.GetPoint(, "Select point")

ThisDrawing.Utility.GetEntity ln1, pt, "Select second
line"
    pt = PointPerpendicular(p1, ln1.StartPoint,
ln1.EndPoint)
    Dim l1 As AcadLine
    Set
l1 = ThisDrawing.ModelSpace.AddLine(p1, pt)

l1.Update
    Set l1 = Nothing
End Sub

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Rob Outman <
href="mailto:routman@nakata.com">routman@nakata.com> wrote in message

href="news:B483578BDBC88ECC6B607EB14B182525@in.WebX.maYIadrTaRb">news:B483578BDBC88ECC6B607EB14B182525@in.WebX.maYIadrTaRb...

Hi All,

I read a previous message posted by Monica Eisfeld on 3/19/01 titled
"Draw Perpendicular Line?" I am trying to do something similar. I
have contacted Monica and she unfortunately never had a chance to work on this
tool. Josh from Minkwitz design responded by saying:

"The first thing you have to do is to determine where
the point is relative
to the line or vise versa. For instance if the point
is considered 0,0,0 in the
cartesian coordinate system, which quadrant
would the line fall in? After that,
perpendicularly is a breeze. You just
add the appropriate angle (in radians) to
the angle property of the
line."

"A fairly simple way to pin down the location of
the line relative to the
point would be to create temporary lines form the
startpoint and endpoint of
the selected line to the selected point. You
could use the angle property of
those lines to pin down the
quadrant."

Could someone point me in the right direction on how to code this?
I am simply trying to have the user select a point on a line, select a second
line and have it automatically draw a new line from the selected point,
perpendicular to the second line that was picked.

I have tried to do this by setting the object snaps prior to drawing the
new line, but for some reason, it try's to draw the new line from
the selected point to the perpendicular of the first point
selected.

Any thoughts?

Rob

Re:

Anonymous — Wed, 30 May 2001 05:54:42 GMT

Thanx Guys,

This is exactly what I am looking for. They
both work great! It's amazing how two completely different code sets can
do the exact same thing!

Thanx again,

Rob

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"Rob Outman" <
href="mailto:routman@nakata.com">routman@nakata.com> wrote in message

href="news:B483578BDBC88ECC6B607EB14B182525@in.WebX.maYIadrTaRb">news:B483578BDBC88ECC6B607EB14B182525@in.WebX.maYIadrTaRb...

Hi All,

I read a previous message posted by Monica Eisfeld on 3/19/01 titled
"Draw Perpendicular Line?" I am trying to do something similar. I
have contacted Monica and she unfortunately never had a chance to work on this
tool. Josh from Minkwitz design responded by saying:

"The first thing you have to do is to determine where
the point is relative
to the line or vise versa. For instance if the point
is considered 0,0,0 in the
cartesian coordinate system, which quadrant
would the line fall in? After that,
perpendicularly is a breeze. You just
add the appropriate angle (in radians) to
the angle property of the
line."

"A fairly simple way to pin down the location of
the line relative to the
point would be to create temporary lines form the
startpoint and endpoint of
the selected line to the selected point. You
could use the angle property of
those lines to pin down the
quadrant."

Could someone point me in the right direction on how to code this?
I am simply trying to have the user select a point on a line, select a second
line and have it automatically draw a new line from the selected point,
perpendicular to the second line that was picked.

I have tried to do this by setting the object snaps prior to drawing the
new line, but for some reason, it try's to draw the new line from
the selected point to the perpendicular of the first point
selected.

Any thoughts?

Rob