Pentahedron and Non-Rectangular Sections Study by Joe Bartok

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When I compare the Golden Rhombus and the current Offset
Rotated Rafters this Compound Angle Formula works all around:

μ = arctan (sin **Angle at Rafter Peak** ÷ tan **Blade Bevel
Angle**)

Golden Rhombus (my study half-split the roof angle but the
formula works just as well for your equal width Hips solution):

μ = arctan (sin (90° – 27.73230°) ÷ tan (90° – 47.05622°) = 43.56300°

β = 90° – R1 = 90° – 27.73230° = 62.26770°

α = C5 = 16.04506°

70° Offset 12/12 Side Rafter ... Claw Angle Version:

μ = arctan (sin 28.71825° ÷ tan 5.36467°) = 78.94194° (Angle on Upper Shoulder
of Offset Rafter)

β = 28.71825° (Upper Claw Angle)

α = 13.99545° (Rotated Rafter Backing Angle)

70° Offset 12/12 Side Rafter ... Plumb Line Version:

μ = arctan (sin (90° – 43.21918°) ÷ tan 16.86990°) = 67.40640° (Angle on Upper
Shoulder of Offset Rafter)

β = 90° – R1 = 90° – 43.21918° = 46.78082°

α = 13.99545° (Rotated Rafter Backing Angle)

Looking good, but a long way to go yet.
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Development of a Pentahedron describing a Non-Rectangular
Section, given μ, β and α. After the first few steps the rest of the drawing
falls into place.

Setting a different reference length = 1 has produced a few
new relationships:

ρ = (90° – BEV) – arctan (cos YDIH ÷ tan β)

Blade Bevel for YDIH = arcsin (tan β ÷ tan (BEV + ρ))

MIT = arctan (sin μ tan β ÷ (cos α cos μ tan β + sin μ sin
α))

Test firing the formulas on real Hip roof studies:

Golden Rhombus ... Half-split Roof Angle

μ = arctan (sin (90° – 27.73230°) ÷ tan (90° – 47.05622°) =
43.56300°

β = 90° – R1 = 90° – 27.73230° = 62.26770°

α = C5 = 16.04506°

ρ = Offset Rafter Slope Angle – arctan (tan Main Hip Slope
Angle sin Blade Bevel

ρ = 30.79574° – arctan (tan 27.73230° sin (90° – 47.05622°)
= 11.08972°

MIT = arctan (sin 43.56300° tan 62.26770° ÷ (cos 16.04506°
cos 43.56300° tan 62.26770° + sin 43.56300° sin 16.04506°)) = 40.86615°

70° Offset 12/12 Side Rafter ... Claw Angle Version:

μ = arctan (sin 28.71825° ÷ tan 5.36467°) = 78.94194° (Angle
on Upper Shoulder of Offset Rafter)

β = 28.71825° (Upper Claw Angle)

α = 13.99545° (Offset Rafter Backing Angle)

ρ = (90° – Trace Angle on Main Hip) – arctan (tan (90° –
Upper Claw Angle) sin Blade Bevel)

ρ = (90° – 55.50766°) – arctan (tan (90° – 28.71825°) sin
5.36467°) = 24.80855°

MIT = arctan (sin 78.94194 tan 28.71825 ÷ (cos 13.99545 cos
78.94194 tan 28.71825 + sin 78.94194 sin 13.99545)) = 57.74662°

70° Offset 12/12 Side Rafter ... Plumb Line Version:

μ = arctan (sin (90° – 43.21918°) ÷ tan 16.86990°) =
67.40640° (Angle on Upper Shoulder of Offset Rafter)

β = 90° – R1 = 90° – 43.21918° = 46.78082°

α = 13.99545° (Offset Rafter Backing Angle)

ρ = Main Hip Slope Angle – arctan (tan Offset Rafter Slope
Angle sin Blade Bevel)

ρ = 30.96374° – arctan (tan 43.21918 sin 16.86990°) =
15.71018°

MIT = arctan (sin 67.40640° tan 46.78082 ÷ (cos 13.99545 cos
67.40640° tan 46.78082 + sin 67.40640° sin 13.99545)) = 57.74661°

There doesn’t appear to be a pattern for the formulas for ρ
in terms of the roof angles, but:

... 30.79574° and 30.96374° are really 90° – Trace Angles
for their respective rafters

... Main Hip Slope Angle and Offset Rafter Slope Angle are
90° – Upper Claw Angle are the complements of the angles at their respective rafter
peaks.

ρ = 90° – Trace Angle – arctan( sin Blade Bevel ÷ tan Angle
at Rafter Peak)???

Penathedron and Non-Rectangular Section Test

Intersecting Hip Rafters – Warlock Cut, Upper Shoulder Hip B

Angle on Hip B Upper Left Shoulder = 37.65287°

Blade Bevel @ 37.65287° = 55.39851°

μ = arctan (sin 37.65287° ÷ tan 55.39851°) = 22.85244° =
Angle on Hip B Upper Right Shoulder ... expected that, the rafter section is
rectangular

β = 37.65287° = Angle on Hip B Upper Left Shoulder

α = 0° ... at first glance it might seem like this should be
45°, but there is no backing or rotation angle here

MIT = 37.65287°

BEV = 43.14862° (Trace Angle on Hip A)

ρ = 0°

Projected Right Angle = 136.85138°

Supplementary Angle = 43.14862° (= Trace Angle on Hip A)

Blade Bevel for XDIH = 55.39850°

Blade Bevel for YDIH = 26.71762° (Blade Bevel @ 22.85244°)

Blade Bevel for ZDIH = 55.39850° (= Blade Bevel for XDIH)

The code returned angles for a tetrahedron (which we already
know works here). This isn't telling me anything new or lending insight as to
an easier means of drawing the intersection.