Lesson 6 · 40 min

Taper Ratio and the Planform

With area, aspect ratio and sweep chosen, one number completes the shape of a straight-tapered wing: the taper ratio. It decides how the lift is spread along the span, and so where the wing stalls first and how much it weighs. With it the planform can be drawn, and its mean aerodynamic chord located: the reference length for every moment and center-of-gravity position that follows.

Learning objectives

Explain why an elliptic span loading is ideal, and how taper approaches it.
Estimate the span loading of a tapered wing with Schrenk's approximation, and find where it stalls first.
Compute the root and tip chords, the mean aerodynamic chord and its spanwise position.
Convert sweep between chord lines.

Taper ratio and span loading

The taper ratio is the tip chord over the root chord, \(\lambda = c_t/c_r\). Theory gives the ideal: for a given span and lift, the induced drag is least when the lift per unit span varies elliptically, from a maximum at the root to zero at the tips. An untwisted wing with an elliptic planform has exactly that loading (the Spitfire's wing is the famous example), but its curved edges are expensive to build.

A straight taper comes close. A rectangular wing (\(\lambda = 1\)) carries too much lift near the tips and has about 7% more induced drag than an elliptic wing of the same aspect ratio; a taper ratio of about 0.4 to 0.5 brings an unswept wing within about 1% of the elliptic ideal. Taper also saves weight: it moves lift inboard, where the wing is deepest, and shortens the outboard chords that would otherwise add area without much lift.

Sweep pushes the load outboard, so swept wings need more taper to stay near elliptic: taper ratios of 0.2 to 0.35 are usual for airliners. Too much taper overloads the tips and they stall first.

Schrenk's approximation (unswept, untwisted wing)

\[ \frac{c\,c_l}{C_L\,\bar c_g} = \frac{1}{2\bar c_g}\left[c(\eta) + \frac{4S}{\pi b}\sqrt{1 - \eta^2}\right], \qquad \eta = \frac{2y}{b} \]

The span loading is the average of the chord distribution and an elliptic distribution of the same area. Dividing by the local chord gives the local lift coefficient as a fraction of the wing's, \(c_l/C_L\): where it peaks, the wing stalls first.

Figure 6.1 Span loading from Schrenk's approximation (blue) against the elliptic ideal (gray dashed), and the local lift coefficient as a fraction of the wing's (orange). The dot marks the peak of \(c_l/C_L\), where stall begins. A rectangular wing stalls at the root, which is safe; a taper ratio near 0.45 is the most even; strong taper moves the peak toward the tip.

The planform's geometry

Given \(S\), \(A\), \(\lambda\) and the sweep of one chord line, the trapezoidal reference wing is fixed:

Trapezoidal wing

\[ b = \sqrt{AS}, \qquad c_r = \frac{2S}{b(1 + \lambda)}, \qquad c_t = \lambda c_r \] \[ \bar c = \frac{2}{3}c_r\,\frac{1 + \lambda + \lambda^2}{1 + \lambda}, \qquad \bar y = \frac{b}{6}\,\frac{1 + 2\lambda}{1 + \lambda} \] \[ \tan\Lambda_n = \tan\Lambda_m - \frac{4}{A}\left[(n - m)\frac{1 - \lambda}{1 + \lambda}\right] \]

\(\bar c\) is the mean aerodynamic chord (MAC) and \(\bar y\) its distance from the centerline; its leading edge is \(\bar y\tan\Lambda_{LE}\) behind the root leading edge. \(\Lambda_n\) is the sweep of the line at fraction \(n\) of the chord (0 the leading edge, 0.25 the quarter chord, 1 the trailing edge).

The mean aerodynamic chord is the chord of a rectangular wing that would have the same lift and pitching moment. Its quarter point is, to a good approximation, the wing's aerodynamic center at subsonic speed, and center-of-gravity limits are quoted in percent of it ("the CG range is 15 to 35% MAC"). Module 6 places the wing on the fuselage by its MAC, and Module 11 balances the aircraft about it.

Example 6.1 — The business jet's planform

Draw the business jet's wing: \(S = 28\ \text{m}^2\), \(A = 8\), \(\lambda = 0.35\), \(25^\circ\) of sweep at the quarter chord. Find the span, the root and tip chords, the MAC and its location, and the leading-edge sweep.

Show solution

\[ b = \sqrt{8(28)} = 14.97\ \text{m} \] \[ c_r = \frac{2(28)}{14.97(1.35)} = 2.772\ \text{m} \] \[ c_t = 0.35(2.772) = 0.970\ \text{m} \] \[ \bar c = \frac{2}{3}(2.772)\,\frac{1 + 0.35 + 0.1225}{1.35} = 2.015\ \text{m}, \qquad \bar y = \frac{14.97}{6}\,\frac{1.70}{1.35} = 3.141\ \text{m} \] \[ \tan\Lambda_{LE} = \tan 25^\circ - \frac{4}{8}\left[(0 - 0.25)\frac{0.65}{1.35}\right] = 0.4663 + 0.0602 = 0.5265 \] \[ \Lambda_{LE} = 27.8^\circ, \qquad \bar x_{LE} = 3.141(0.5265) = 1.654\ \text{m} \]

The MAC is 2.015 m long, 3.14 m from the centerline, and starts 1.65 m behind the root leading edge. Its quarter point, the wing's aerodynamic center, is \(1.654 + 0.25(2.015) = 2.16\ \text{m}\) behind the root leading edge. Module 3's sketch used an illustrative taper and less sweep; from here on this is the business jet's wing.

Figure 6.2 A trapezoidal wing drawn to scale from its area, aspect ratio, taper ratio and quarter-chord sweep, nose up, on a 1 m grid. The dashed line is the quarter chord; the violet line is the mean aerodynamic chord, with the aerodynamic center (dot) at its quarter point. It starts as the business jet's wing.

Check your understanding

Key takeaways

An elliptic span loading minimizes induced drag; a taper ratio near 0.45 comes close for an unswept wing, 0.2 to 0.35 for a swept one.
Too much taper overloads the tips, which then stall first; a rectangular wing stalls at the root.
\(c_r = 2S/[b(1 + \lambda)]\), \(\bar c = \tfrac23 c_r(1 + \lambda + \lambda^2)/(1 + \lambda)\), \(\bar y = (b/6)(1 + 2\lambda)/(1 + \lambda)\).
The wing's aerodynamic center is at the quarter point of the MAC; CG positions are quoted in % MAC.
Next, Lesson 7 adds twist, incidence, dihedral and the wing's vertical position.