The fillet that does not fit the edges
A workshop in Terrassa is preparing a 1:1 drawing of a support plate whose two straight edges meet at a 60° vertex. The specification calls for a blending fillet of R20 tangent to both edges, with both tangent points lying on the edges themselves; each edge is drawn 30 mm long from the vertex. A junior draughtsman places the compass point on the internal bisector 40 mm from the vertex, sweeps an R20 arc, and finds that it touches neither edge within its drawn length - the contact points fall about 4.6 mm beyond the ends of the edges. He then corrects it by eye, sliding the centre until the arc looks tangent. The plate no longer closes against the adjoining side plate and leaves a 3 mm step at the joint. The section head wants to know whether the fault lies in the construction or in the specification, and what the largest fillet is that 30 mm edges can accept.
1. Show by calculation why the R20 fillet centred 40 mm from the vertex does not work on edges only 30 mm long, and state the largest radius that keeps both tangent points exactly at the ends of the edges.
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The half-angle at the vertex is 30°, and the centre of any circle tangent to both sides must lie on the bisector at a distance d = R / sin 30° = 20 / 0.5 = 40 mm from the vertex. The centre the draughtsman used is therefore correct: the fault is not in his bisector step.
The tangent point on each side lies at t = R / tan 30° = 20 / 0.5774 = 34.64 mm from the vertex. Because each edge is only 30 mm long, both tangent points fall 4.64 mm beyond the ends of the edges - the arc is tangent to the extensions of the edges, not to the edges themselves.
For tangency exactly at the ends, t = 30 mm, so R = 30 · tan 30° = 17.32 mm, with the centre at 34.64 mm on the bisector. The error is one of specification: the requested R20 is geometrically incompatible with a 30 mm edge length, and no amount of careful drawing can make it tangent.
2. Describe the construction and the checks that prove tangency, and explain what the by-eye correction destroyed.
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Construction: draw the internal bisector of the 60° angle. Locate O either at 40 mm along the bisector or as the intersection of two lines drawn parallel to the edges at 20 mm inside them. Drop perpendiculars from O onto each edge; the feet T1 and T2 are the tangent points. Draw the arc T1T2 from O with radius 20; its angular span is 180° − 60° = 120°.
Checks that prove tangency: (a) OT1 = OT2 = R = 20 mm; (b) OT1 is perpendicular to edge 1 and OT2 perpendicular to edge 2, because the radius at a tangent point is normal to the outline; (c) both radii make equal angles with the bisector, so T1 and T2 are symmetric; (d) the arc stays inside the angle and spans exactly 120°.
Sliding the centre by eye destroys check (b). The radius is no longer normal to the edge at the supposed tangent point, so the outline has a tangent discontinuity - geometrically a corner, not a blend. That corner is what produces the 3 mm step where the plate meets its neighbour, and no radius value can compensate for it.
3. Recommend a fix and state how the fillet should be dimensioned on the issued drawing.
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Options, in order of preference:
- Lengthen both edges to at least 35 mm and keep R20; this preserves the specified radius and restores real tangency. It is the right answer if the outline and the joint allow the extra material.
- Keep the 30 mm edges and reduce the fillet to R17, giving tangent points at 29.4 mm. R17.32 is the exact limit but cannot be drawn or verified reliably at 1:1, so it must be rounded.
- Change the vertex angle, which is normally not negotiable because the angle is set by the adjoining part.
Dimensioning: a single note "R20" (or R17) with a leader whose arrow touches the arc, using the radius symbol. The centre and the tangent points are construction entities and must not be dimensioned; adding them over-dimensions the profile and contradicts the radius. If the radius is not functional but merely a casting or edge-break convenience, mark it as a reference dimension in brackets.