### Author Topic: Math - 3D Solid Challenge  (Read 12419 times)

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#### David Hall

• Automatic Duh Generator
• King Gator
• Posts: 4043 ##### Re: Math - 3D Solid Challenge
« Reply #15 on: January 29, 2008, 04:53:37 PM »
Thats what I thought, but google didn't even know
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#### David Hall

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• Posts: 4043 ##### Re: Math - 3D Solid Challenge
« Reply #16 on: January 29, 2008, 04:53:58 PM »
what about the fact that the centerlines are off of the center of the upper part?
Everyone has a photographic memory, Some just don't have film.
They say money can't buy happiness, but it can buy Bacon and that's a close second

#### Cathy

• Bull Frog
• Posts: 251 ##### Re: Math - 3D Solid Challenge
« Reply #17 on: January 29, 2008, 05:04:44 PM »
Would putting some workpoints in your blocks help with some of this?  Like a point where the CLs cross, points where the pipes connect?

IRT -- odd I always use WRT (with respect to).  Is that just a difference is Aussie English and American English?  Or is just a difference in disciplines?

#### David Hall

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• Posts: 4043 ##### Re: Math - 3D Solid Challenge
« Reply #18 on: January 29, 2008, 05:08:15 PM »
I was thinking of calcing the workpoint in code, and then moving down and to the left/right from there to start the line.  If you look at the first pic, you can see some cross's inside the pipe, that is the start/stop point for the vertical anlge of pipe
Everyone has a photographic memory, Some just don't have film.
They say money can't buy happiness, but it can buy Bacon and that's a close second

#### MickD ##### Re: Math - 3D Solid Challenge
« Reply #19 on: January 29, 2008, 05:15:56 PM »
>>what about the fact that the centerlines are off of the center of the upper part?

The angle is still the same so I'd translate the pipe 11/16 to the left and the bottom T the amount of the offset from insert point to the left, repeat on right side.

I'd have a go at this but I'm busy with something else at the moment but here's the steps I would take using the math I posted earlier -

Calc the pipe lengths and pnt1 on the left side
Insert the top pipe at origin
Insert the left bottom T at pnt1 and translate it the offset amount to the left
Extrude the pipe, translate it to the origin and the angle then translate it left the top T offset amount
(and along the line a bit if needed to clear cross pipe as it's shorter than the actual line length)
Copy and mirror over top T c.l. for right side.

If I get time later I'll code something up.
Forth is like the Tao: it is a Way, and is realized when followed.
Its fragility is its strength; its simplicity is its direction - Michael Ham

"The E in Javascript stands for 'easy'."  Florin Pop tweet

#### David Hall

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• Posts: 4043 ##### Re: Math - 3D Solid Challenge
« Reply #20 on: January 29, 2008, 05:22:00 PM »
thanks
Everyone has a photographic memory, Some just don't have film.
They say money can't buy happiness, but it can buy Bacon and that's a close second

#### Bryco

• Water Moccasin
• Posts: 1864 ##### Re: Math - 3D Solid Challenge
« Reply #21 on: January 30, 2008, 10:55:23 AM »
I had some success with below code.
To the apex bock I added 2 regions on defpoints that are positiioned where you have the lines.
These regions are copied from the block itself then tranlated to the blockrefs position.
Then extruded.

Code: [Select]
`Sub Cmd()    Dim Ent As AcadEntity    Dim B As AcadBlock    Dim Br As AcadBlockReference    Dim P, m, Ret    Dim Ms As AcadBlock    Dim Obj(0) As Object    Dim C1 As AcadCircle, C2 As AcadCircle    Dim R1 As AcadRegion    Dim sBlock As String    Dim Dist As Double, Ht As Double    Dim Util As AcadUtility        Set Util = ThisDrawing.Utility    Set Ms = ThisDrawing.ModelSpace    sBlock = "5020760"    Util.GetEntity Ent, P, "Pick the 5020"    If Not TypeOf Ent Is AcadBlockReference Then Exit Sub    If Not Ent.Name = sBlock Then Exit Sub    Set Br = Ent    Set B = ThisDrawing.Blocks(sBlock)    For Each Ent In B        If TypeOf Ent Is AcadRegion Then        Set Obj(0) = Ent            Ret = ThisDrawing.CopyObjects(Obj, Ms)            Set R1 = Ret(0)            R1.Layer = "0"            R1.Color = 2            Exit For        End If    Next Ent    Dist = Util.GetDistance(, "Centerline distance")    If Dist < 17.0855 Then        MsgBox ("Distance must be larger than 17")        Exit Sub    End If    Ht = (Dist - 17.0855) / Cos(15 / 180 * PI) + 4    m = BlockRefMatrix(Br)    m = InverseMatrix(m)    R1.TransformBy (m)    Dim Tube As Acad3DSolid    Set Tube = Ms.AddExtrudedSolid(R1, Ht, 0)   End SubFunction BlockRefMatrix(oBref As AcadBlockReference) As Variant    Dim V, m(3, 3) As Double, M1, M2    Dim j As Integer    Dim X, Y, Z, ins    Dim Rot As Double        Rot = oBref.Rotation    Z = oBref.Normal    V = GetOcsFromNormal(Z)    X = V(0): Y = V(1)    PrintP X    PrintP Y    ins = oBref.InsertionPoint        M1 = RotZ(-Rot)    For j = 0 To 2        m(0, j) = X(j)        m(1, j) = Y(j)        m(2, j) = Z(j)    Next j    m(0, 3) = -ins(0): m(1, 3) = -ins(1): m(2, 3) = -ins(2)    m(3, 3) = 1    'PrintM M    If Rot = 0 Then        BlockRefMatrix = m    Else        M2 = M4xM4(m, M1)        PrintM M2        BlockRefMatrix = M2    End If            End Function`

#### MickD ##### Re: Math - 3D Solid Challenge
« Reply #22 on: January 30, 2008, 05:44:32 PM »
I had some success with below code.
To the apex bock I added 2 regions on defpoints that are positiioned where you have the lines.
These regions are copied from the block itself then tranlated to the blockrefs position.
Then extruded.

Great idea Bryco! saves quite a bit of matrix mojo. Perhaps you can put the other bottom blocks in as well as nested in the correct alignment then just translate it along the line vector the given amount after calc'ing with the new height?
Forth is like the Tao: it is a Way, and is realized when followed.
Its fragility is its strength; its simplicity is its direction - Michael Ham

"The E in Javascript stands for 'easy'."  Florin Pop tweet

#### Bryco

• Water Moccasin
• Posts: 1864 ##### Re: Math - 3D Solid Challenge
« Reply #23 on: January 30, 2008, 08:07:38 PM »
I haven't done much coding for the creation and editing of 3d objects so I tried a circle first, only to find a cylinder doesn't follow a path.
So yes the region saves a bit of matricising (Pres. Bush speak).
After an hour I thought I would have the hard part whooped but alas the correct matrix is eluding me for the funky blocks. (Dang it). Works for the blocks rotated in the same plane.
Mick, I think straight vector math will establish the insertion point of the other blocks and when that is translated using the same magical matrix David can say "Bob's me uncle"

#### MickD ##### Re: Math - 3D Solid Challenge
« Reply #24 on: January 30, 2008, 08:12:53 PM »
Do you mean to rotate the bottom T or to mirror it?

To create a mirror matrix you should be able to just negate some the vectors for axes and translation and transformBy.
If you are mirroring over the y axis, negate the x vector for example.

if it's the rotation of the blocks perhaps setting them up in the block drg in the correct location/orientation for simple insert (create a block for left and right say).
Forth is like the Tao: it is a Way, and is realized when followed.
Its fragility is its strength; its simplicity is its direction - Michael Ham

"The E in Javascript stands for 'easy'."  Florin Pop tweet

#### MickD ##### Re: Math - 3D Solid Challenge
« Reply #25 on: January 30, 2008, 08:36:21 PM »
It's even easier, I just noticed that ent's have a Mirror3d method. that may help.
Forth is like the Tao: it is a Way, and is realized when followed.
Its fragility is its strength; its simplicity is its direction - Michael Ham

"The E in Javascript stands for 'easy'."  Florin Pop tweet

#### MickD ##### Re: Math - 3D Solid Challenge
« Reply #26 on: January 30, 2008, 09:52:48 PM »
This is what I'm thinking -
Forth is like the Tao: it is a Way, and is realized when followed.
Its fragility is its strength; its simplicity is its direction - Michael Ham

"The E in Javascript stands for 'easy'."  Florin Pop tweet

#### Bryco

• Water Moccasin
• Posts: 1864 ##### Re: Math - 3D Solid Challenge
« Reply #27 on: January 31, 2008, 12:27:43 AM »
This works, finally. David the last part is mirror3d as Mick suggests.
Transpose rather than inverse threw for a while as usual (More brains reqd.)
There is a weird variety of math used here, but they all relate to normals somewhere.
Code: [Select]
`Option ExplicitConst PI As Double = 3.14159265358979Sub Dav()    Dim Ent As AcadEntity    Dim B As AcadBlock    Dim Br As AcadBlockReference    Dim P, M, M1, Ret    Dim Ms As AcadBlock    Dim Obj(0) As Object    Dim C1 As AcadCircle, C2 As AcadCircle    Dim R1 As AcadRegion    Dim sBlock As String    Dim Dist As Double, Ht As Double    Dim Util As AcadUtility    Dim Zero(2) As Double            Set Util = ThisDrawing.Utility    Set Ms = ThisDrawing.ModelSpace    sBlock = "5020760"    Util.GetEntity Ent, P, "Pick the 5020"    If Not TypeOf Ent Is AcadBlockReference Then Exit Sub    If Not Ent.Name = sBlock Then Exit Sub    Set Br = Ent    Set B = ThisDrawing.Blocks(sBlock)    For Each Ent In B        If TypeOf Ent Is AcadRegion Then        Set Obj(0) = Ent            Ret = ThisDrawing.CopyObjects(Obj, Ms)            Set R1 = Ret(0)            R1.Layer = "0"            R1.color = 2            Exit For        End If    Next Ent    Dist = Util.GetDistance(, "Centerline distance")    If Dist < 17.0855 Then        MsgBox ("Distance must be larger than 17")        Exit Sub    End If    Ht = (Dist - 17.0855) / Cos(15 / 180 * PI) + 4    M = BlockRefMatrix(Br)    M1 = Transpose(M)    R1.TransformBy (M1)    Dim Tube As Acad3DSolid    Set Tube = Ms.AddExtrudedSolid(R1, Ht, 0)    R1.Delete        Dim width As Double    Dim inspt(2) As Double    Dim Br2 As AcadBlockReference        width = (Tan(15 / 180 * PI) * (Dist - 17.0855)) + 4.0682    inspt(1) = -width: inspt(2) = -Dist        P = TransformPt2(M1, inspt)    Set Br2 = Ms.InsertBlock(Zero, "5141714", 1, 1, 1, 0)    Br2.Normal = Br.Normal    Br2.Rotation = Br.Rotation    Br2.Move Zero, P    End SubFunction BlockRefMatrix(oBref As AcadBlockReference) As Variant    Dim V, M(3, 3) As Double, M1, M2    Dim j As Integer    Dim X, Y, Z, ins    Dim Rot As Double        Rot = oBref.Rotation    Z = oBref.Normal    V = GetOcsFromNormal(Z)    X = V(0): Y = V(1)    ins = oBref.InsertionPoint        M1 = RotZ(-Rot)    For j = 0 To 2        M(0, j) = X(j)        M(1, j) = Y(j)        M(2, j) = Z(j)        M(3, j) = ins(j)    Next j    M(3, 3) = 1    If Rot = 0 Then        BlockRefMatrix = M    Else        M2 = M4xM4(M, M1)        BlockRefMatrix = M2    End If        End FunctionFunction Transpose(Matrix As Variant) As Variant    Dim iCnt As Integer, jCnt As Integer    Dim transMat(0 To 3, 0 To 3) As Double    Dim I As Integer, j As Integer    iCnt = UBound(Matrix, 1)    jCnt = UBound(Matrix, 2)    For I = 0 To iCnt        For j = 0 To jCnt            transMat(I, j) = Matrix(j, I)        Next j    Next I    Transpose = transMat    End FunctionFunction TransformPt2(M As Variant, P1 As Variant) As Variant    Dim I As Integer    Dim X As Double, Y As Double, Z As Double, d As Double    Dim P(3) As Double    Dim P2(2) As Double    For I = 0 To 2        P(I) = P1(I)    Next    P(3) = 1        For I = 0 To 3        X = X + P(I) * M(0, I)        Y = Y + P(I) * M(1, I)        Z = Z + P(I) * M(2, I)        d = d + P(I) * M(3, I)    Next    P2(0) = X: P2(1) = Y: P2(2) = Z    TransformPt2 = P2End FunctionFunction GetOcsFromNormal(N As Variant) As Variant    'Arbitrary Axis Algorithm in dxf help    'N is the normal vector.    'Wy is the world Y axis, which is always (0,1,0).    'Wz is the world Z axis, which is always (0,0,1).    Dim Wy(2) As Double    Dim Wz(2) As Double    Dim Nx As Double, Ny As Double    Dim Ax, Ay, Ocs(1) As Variant        N = NormaliseVector(N)    Wy(0) = 0: Wy(1) = 1: Wy(2) = 0    Wz(0) = 0: Wz(1) = 0: Wz(2) = 1    Nx = N(0): Ny = N(1)    If (Abs(Nx) < 1 / 64) And (Abs(Ny) < 1 / 64) Then         'Ax = Wy X N (where X is the cross-product operator).         Ax = Crossproduct(Wy, N)    Else         Ax = Crossproduct(Wz, N)    End If    Ocs(0) = Ax      Ay = Crossproduct(N, Ax)    Ocs(1) = Ay    GetOcsFromNormal = OcsEnd FunctionFunction RotZ(Ang As Double) As Variant   'Rotate by an angle around the z axis    Dim M    Dim CosAng As Double, sinAng As Double        CosAng = Cos(Ang): sinAng = Sin(Ang)    M = IDMatrix    M(0, 0) = CosAng    M(0, 1) = -sinAng    M(1, 0) = sinAng    M(1, 1) = CosAng          RotZ = MEnd FunctionFunction M4xM4(M1, M2) As Variant    'Matrix x matrix    Dim M(3, 3) As Double    Dim I As Integer, j As Integer    Dim k As Integer    Dim Sum As Double        For I = 0 To 3        For j = 0 To 3            For k = 0 To 3                Sum = Sum + M1(k, j) * M2(I, k)            Next k            M(I, j) = Sum            Sum = 0        Next j    Next I    M4xM4 = M    End FunctionFunction NormaliseVector(V As Variant) As Variant    Dim Unit As Double    Dim Vn(2) As Double    Unit = Sqr(V(0) * V(0) + V(1) * V(1) + V(2) * V(2))    Vn(0) = V(0) / Unit: Vn(1) = V(1) / Unit: Vn(2) = V(2) / Unit    NormaliseVector = VnEnd FunctionFunction Crossproduct(A, B) As Variant    Dim Ax As Double, Ay As Double, Az As Double    Dim Bx As Double, By As Double, Bz As Double    Dim Unit As Double    Dim c(2) As Double    'get CrossProduct    Ax = A(0): Ay = A(1): Az = A(2)    Bx = B(0): By = B(1): Bz = B(2)        c(0) = Ay * Bz - Az * By    c(1) = Az * Bx - Ax * Bz    c(2) = Ax * By - Ay * Bx        'Convert to unit normal    Unit = Sqr(c(0) * c(0) + c(1) * c(1) + c(2) * c(2))    c(0) = c(0) / Unit: c(1) = c(1) / Unit: c(2) = c(2) / Unit    Crossproduct = cEnd FunctionFunction IDMatrix()    Dim M(3, 3) As Double    Dim I As Integer, j As Integer    M(0, 0) = 1    M(1, 1) = 1    M(2, 2) = 1    M(3, 3) = 1    IDMatrix = M    End Function`

#### Bryco

• Water Moccasin
• Posts: 1864 ##### Re: Math - 3D Solid Challenge
« Reply #28 on: January 31, 2008, 12:39:12 AM »
Test dwg.

#### SEANT

• Bull Frog
• Posts: 344 ##### Re: Math - 3D Solid Challenge
« Reply #29 on: January 31, 2008, 07:02:10 AM »
Excellent work once again gentlemen.  A word of caution; if you persist in this development of one click design creation, managers may do the work themselves and figure they need CAD techs no longer. Bryco, I see youve expanded the scope of the project to include otherworldly alignments  certainly a welcome inclusion for those of us with chaotic structures.  The one recommendation I would make for dealing with these non-aligned setups, like the block in CmdrduhTest.dwg @ (99.9462, 265.5402, 0.4912), is to switch order of operation from:

Br2.Normal = Br.Normal
Br2.Rotation = Br.Rotation

To:

Br2.Rotation = Br.Rotation
Br2.Normal = Br.Normal

Sean Tessier