Parabolic arch: focus, directrix and the tangent at quarter span
A footbridge over a road has a parabolic intrados springing from two abutments 60 m apart on the same level, with the crown 12 m above the line joining the springing points. On site the arch was set out with two circular arcs meeting near the crown; the surveyor reports a visible flat spot there and asks for a check. The drawing office must (a) locate the focus and directrix of the parabola that fits the given span and rise with the vertex at the crown, (b) give the classical construction of the tangent at the quarter-span point, using focus and directrix only, so the handrail brackets can be set normal to the intrados, and (c) supply numbers to verify the construction. Everything is drawn at 1:100 on A1, and the checker will not accept a tangent obtained by reading coordinates off a table - the bisector construction must be shown and justified.
1. Locate the focus and the directrix of the parabola that fits the span and rise. Give the values in metres and their positions on the 1:100 sheet.
Read the worked answer
Put the origin at the crown, the axis vertical, x horizontal and y measured downward: y = x²/(4a).
The springing points are at x = ±30 m (half of 60 m) and y = 12 m below the crown:
12 = 30²/(4a) → 4a = 900/12 = 75 → a = 18.75 m.
So:
- Focus F on the axis, 18.75 m below the crown - that is 6.75 m below the springing line, since 18.75 − 12 = 6.75 m.
- Directrix a horizontal line 18.75 m above the crown, i.e. 18.75 + 12 = 30.75 m above springing level.
- Latus rectum 4a = 75 m, a useful span check.
At 1:100 these become 187.5 mm below the crown for the focus and 187.5 mm above the crown for the directrix. Both are far outside the drawn arch, which is normal: for a shallow arch the focus sits deep and the directrix sits high, so the drawing must be laid out on a sheet with enough margin. Sanity check: every point of the intrados must be exactly as far from F as from the directrix - verify at the springing, where the distance to F is √(30² + 6.75²) = 30.75 m, equal to the directrix distance.
2. Give the classical construction of the tangent at the quarter-span point, and of the normal there for the handrail bracket. Justify why the construction works.
Read the worked answer
Quarter span is x = 15 m from the axis; on the curve y = 15²/75 = 3 m below the crown (9 m above springing). Call this point P.
Construction (focus and directrix only):
- Through P draw the line parallel to the axis - vertical - and take the ray pointing away from the focus, i.e. upward here.
- Join P to F.
- Bisect the angle between that upward vertical ray and PF. The bisector is the tangent at P.
- Erect the perpendicular at P; that is the normal, and it is the direction the bracket must be set to.
Justification. The parabola is the locus of points equidistant from focus and directrix, and the tangent at a point makes equal angles with the focal radius and the line parallel to the axis. Equivalently: a ray arriving parallel to the axis reflects into the focus, so the tangent must bisect the angle between the incoming ray and PF. The bisector is therefore the unique line with that equal-angle property, which is exactly the tangent.
Do not take the tangent as parallel to the chord between the springing and the crown: that chord has slope 12/30 = 0.4 by coincidence only at a different point, and at the abutment the true tangent is at arctan(0.8) = 38.7°, not 21.8°.
3. Verify the construction numerically, and explain why the two-arc set-out produced a flat spot near the crown.
Read the worked answer
Verification. At P(15, 3), with F(0, 18.75) and y positive downward:
- PF = √(15² + (18.75 − 3)²) = √(225 + 248.06) = 21.75 m.
- The tangent at P meets the axis at T. The subtangent is bisected by the vertex, so with the ordinate foot M at 3 m below the crown and the vertex V at the crown, T lies 3 m above the crown.
- TF = 3 + 18.75 = 21.75 m.
TF = PF confirms the bisector construction. The slope, dy/dx = 2x/75 = 0.4, gives a tangent angle of 21.8° to the horizontal - matching the bisector direction to the drawn accuracy.
The flat spot. A circular arc through the crown and both springing points has its centre on the axis with R = (30² + 12²)/(2 × 12) = 1044/24 = 43.5 m. The parabola's radius of curvature at the crown is 2a = 37.5 m. The substitute arcs are 6 m too flat exactly where the arch is most visible, and two arcs of different centres also meet with a tangent break unless the joint is placed where the slopes match. The fix is to set out points from y = x²/75 at, say, 3 m intervals and fair a curve through them, or to use several arcs whose radii are computed to match curvature at each joint.