Tube cross-section

Question

By default, Tube has round cross-section (and we define its diameter). Is it possible to draw Tube along some known path in 3D with, e.g. square cross-section?

Thank you.

EDIT

Tried to follow answer from here. I have set of points (just an example, other arcs may have all the coordinates non-zero):

pts = {{0.`, 0.`, 0.`}, {7.497665500157259`*^-14, 0.00036555534709634656`, 0.041887873232098444`}, {2.9988378071419946`*^-13, 0.0014621100333863195`,  0.08376298663421493`}, {6.746528662189823`*^-13, 0.0032893300277940163`,  0.12561258426325045`}, {1.1991697498398906`*^-12, 0.005846658724918654`,  0.16742391794868797`}, {1.8732746539195436`*^-12, 0.009133317114586991`,  0.20918425117592362`}, {2.696762233490476`*^-12, 0.013148304019154402`,  0.25088086296604745`}, {3.669381638827055`*^-12, 0.017890396398482316`,  0.2925010517508915`}, {4.790836591859004`*^-12, 0.02335814972249914`,  0.33403213924216535`}, {6.0607854764232425`*^-12, 0.029549898411231134`,  0.37546147429349974`}, {7.478841442326563`*^-12, 0.03646375634216925`,  0.41677643675422266`}, {9.044572523187444`*^-12, 0.04409761742481735`,  0.45796444131369324`}, {1.0757501768021084`*^-11, 0.052449156242246726`,  0.499012941335023`}, {1.26171073865276`*^-11, 0.06151582875946169`,  0.5399094326770161`}, {1.4622822908039118`*^-11, 0.07129487309836018`,  0.5806414575031648`}, {1.6774037354077335`*^-11, 0.08178331037905351`, 0.6211966080765395`}, {1.9070095424469`*^-11, 0.09297794562728895`,  0.6615625305394178`}, {2.1510297696962602`*^-11, 0.10487536874769866`,  0.7017269286765003`}, {2.4093900840285483`*^-11, 0.11747195556257856`,  0.7416775676605682`}, {2.6820117840576448`*^-11, 0.13076386891588063`,  0.7814022777794407`}, {2.9688118241124915`*^-11, 0.14474705984208247`, 0.8208889581430966`}}

According to that answer first we need to have path in parametric form, to use FrenetSerretSystem. I tried to use Interpolation @ pts, but this gives me error and no result.

Then I attempted to find FindGeometricTransform between successive points in pts and apply these transformation functions to 4 points of initial rectangle (analogue of nlist in that answer). And (sub)-question here: in this my simple example initial point is (0,0,0) and inititial direction is along z-axis, so we choose some 4 points in x-y plain. And what to do in case of arbitrary initial point and direction of arc?

tf = #[[2]] & /@ (FindGeometricTransform[{#[[2]]}, {#[[1]]}] & /@ Partition[pts, 2, 1])

Then I thought I can use BSplineSurface on these computed points, but I get empty Graphics3D object.

bend = Join[ComposeList[tf, {-0.05, -0.05, 0}], ComposeList[tf, {-0.05, 0.05, 0}], ComposeList[tf, {0.05, -0.05, 0}], ComposeList[tf, {0.05, 0.05, 0}]]

Graphics3D[BSplineSurface[bend]]

So, where am I wrong? And can somebody explain in simple steps how to use that answer in my case.

EDIT2

Suppose, we have set of points with arbitrary initial one, as this:

pts1={{4.38673, -4.49861, 5.9078}, {4.36436, -4.52904, 5.93017}, {4.34115, -4.5582, 5.95338}, {4.31716, -4.58604, 5.97737}, {4.29241, -4.61254, 6.00212}, {4.26695, -4.63763, 6.02759}, {4.2408, -4.6613, 6.05373}, {4.21402, -4.6835, 6.08051}, {4.18665, -4.70419, 6.10789}, {4.15872, -4.72334, 6.13582}, {4.13028, -4.74093, 6.16426}, {4.10138, -4.75693, 6.19317}, {4.07205, -4.77131, 6.22249}, {4.04235, -4.78405, 6.2522}, {4.01232, -4.79513, 6.28224}, {3.982, -4.80454, 6.31256}, {3.95144, -4.81225, 6.34311}, {3.9207, -4.81827, 6.37386}, {3.88981, -4.82257, 6.40475}, {3.85882, -4.82515, 6.43574}, {3.82779, -4.82601, 6.46678}}

How to modify functions in answer of SquareOne?

EDIT3

Many thanks to SquareOne and J.M. for helpful comments. Now I can draw pictures like this:

Answer 1

So it does not appear to be doable with Tube but you can write your own if you need to. Here's some stuff to get you started:

squareTube[{p1 : {_, _, _}, p2 : {_, _, _}}, r_: .1] :=
  Module[{
    originVector = p2 - p1,
    zAxis = {0, 0, 1},
    zCross, zAngle, zVector,
    xAxis = {1, 0, 0},
    xCross, xAngle, xVector,
    graphic},
   zAngle = VectorAngle[originVector, zAxis];
   If[! (zAngle == 0 || zAngle == \[Pi]),
    zCross = originVector~Cross~zAxis;
    zVector = RotationMatrix[zAngle, zCross].originVector,
    zCross = {0, 0, 0};
    zVector = originVector
    ];
   xAngle = VectorAngle[zVector, xAxis];
   If[! (xAngle == 0 || xAngle == \[Pi]),
    xCross = zVector~Cross~xAxis;
    xVector = RotationMatrix[xAngle, xCross].zVector,
    xCross = {0, 0, 0};
    xVector = zVector
    ];
   With[{starts = 
      Join @@ Table[{-r, y, z}, {y, {r, -r}}, {z, {r, -r}}]},
    With[{ends = Table[{2 r, 0, 0} + xVector + v, {v, starts}]},
     graphic = {Prism@Join[Most@starts, Most@ends], 
       Prism@Join[Rest@starts, Rest@ends]}
     ]
    ];
   If[xCross != {0, 0, 0},
    graphic = Rotate[graphic, -xAngle, xCross]
    ];
   If[zCross != {0, 0, 0},
    graphic = Rotate[graphic, -zAngle, zCross]
    ];
   {EdgeForm[None],Translate[graphic, p1]}
   ];
squareTube[Line[p : {__}], r_: .1] :=

  squareTube[#, r] & /@ MapThread[List, {Most@p, Rest@p}];

This simply pulls and rotates each line segment so that it lies along the x-axis starting at the origin, creates a rectangular tube from the origin to the end of the segment (extended by the radius of the tube), and then rotates and translates this back into place.

Here's an example:

With[{p = {
    {0, 0, 0},
    {0, 0, 1},
    {0, 1, 1},
    {0, 1, 0},
    {0, 0, 0}
    }
  },
 squareTube[Line@p, .1]
  // Graphics3D
 ]

Note that figuring out how to do the joins will be pretty nasty:

With[{p = Partition[RandomReal[10, 3*5], 3]},
 squareTube[Line@p, .1]
  // Graphics3D
 ]

Answer 2

Show[Graphics3D[
   Line[{{0, 0, 0}, {3, 0, 0}, {3, 0, 3}, {0, 0, 3}, {0, 0, 0}}]] /. 
  Line -> (Tube[#, 0.2] &), PlotRange -> All, Labeled -> False, 
 Axes -> False]