Transonic Cruise Margin on a Swept Business-Jet Wing
A 30-seat business jet is being cleared to raise cruise from M 0.80 to M 0.82 at FL410. The wing carries 25° quarter-chord sweep at the root and 20° outboard of the flap hinge; the wing box, section thickness and powerplant stay unchanged. Section tests in a pressurised tunnel at chord Reynolds number 3×10⁶ give a two-dimensional critical Mach number of 0.72 at the cruise lift coefficient and a drag-rise Mach number of 0.76. Flight test of the prototype at 25×10⁶ shows 1 g buffet onset at M 0.79, a 1.4 g buffet boundary at M 0.765, a shock sitting near x/c ≈ 0.55 on the outboard panel at M 0.80, and aileron effectiveness falling above M 0.78. The chief engineer asks whether M 0.82 is achievable without structural change, and if not, what the smallest modification is. Take-off, thrust and flutter margins are acceptable at all speeds discussed.
1. Estimate the Mach number at which the root and the outboard panel first reach their critical condition, and reconcile these figures with the three flight-test observations (1 g onset 0.79, 1.4 g boundary 0.765, outboard shock at 0.80). Which station controls the buffet boundary?
Read the worked answer
Apply the sweep rule to the component of Mach normal to the isobars: M_crit,wing ≈ M_crit,section / cos Λ.
- Root, Λ = 25°: 0.72/0.9063 = 0.794
- Outboard, Λ = 20°: 0.72/0.9397 = 0.766
So the outboard panel goes critical first, about 0.03 M below the root. That is exactly consistent with the flight data: the shock is seen outboard at x/c ≈ 0.55 while the root is still subcritical, and aileron effectiveness (an outboard, rear-loaded surface) degrades from 0.78 as the outboard shock strengthens and the separated region grows.
The 1 g onset at 0.79 falls between the two predictions and is close to the root value, but buffet onset is a force-break phenomenon that needs a finite shock strength, not merely the first appearance of supersonic flow. The 1.4 g boundary at 0.765 is the more informative number: at 1.4 g the outboard section lift coefficient rises, its own critical Mach number falls to roughly 0.70, giving 0.70/0.9397 = 0.745 on the wing - about 0.02 below the measured 0.765.
That 0.02 gap is the Reynolds deficit of the tunnel (3×10⁶ against 25×10⁶): the thicker tunnel boundary layer gives a larger displacement supervelocity and earlier shock-induced separation, so the tunnel critical and drag-rise Mach numbers are pessimistic. A second, smaller effect is that on a tapered wing the isobars are swept less than the geometric quarter-chord line, so the true normal Mach is a little higher than the geometric estimate.
Conclusion: the outboard panel, not the root, sets the buffet boundary, and the flight data are the more credible set for clearing the envelope.
2. Determine whether M 0.82 cruise is possible with the present planform. Compute the sweep required, check the drag-rise criterion as well as the critical-Mach criterion, and state the smallest change you would recommend.
Read the worked answer
Critical-Mach criterion. Require M cos Λ ≤ 0.72 at M = 0.82:
cos Λ ≤ 0.72/0.82 = 0.878 → Λ ≥ 28.6°
The outboard panel has only 20°, so it is about 9° short. At M 0.82 the outboard normal Mach with the present sweep is 0.82 × 0.9397 = 0.771, roughly 0.05 above the section critical value - a strong shock, separation and aileron snatch, not a margin issue but a handling-qualities issue.
Drag-rise criterion. M_dr,wing = 0.76/cos Λ. Root: 0.839. Outboard: 0.809. Cruise at 0.82 is already above the outboard drag-rise Mach with the present wing, so even if buffet were fixed, wave drag would climb steeply. At 30° outboard sweep, M_dr = 0.76/0.866 = 0.877, leaving 0.057 of margin at M 0.82.
Recommendation. The minimum credible change is to sweep the outboard panel from 20° to 30° (a cranked, inboard-shifted planform), which satisfies both criteria with about 0.01 M of critical-Mach margin and 0.05 M of drag-rise margin. Two caveats must be priced in: sweeping the tip moves lift inboard, raising root bending moment and wing weight, and it reduces torsional stiffness so flutter clearance must be repeated. A cheaper alternative is to thin the outboard section from t/c ≈ 0.10 to 0.085, which buys perhaps 0.02–0.03 in critical Mach - not enough on its own for 0.82. If neither change is funded, hold cruise at 0.80 and demonstrate the buffet margin there.
3. The tunnel was run at 3×10⁶ and flight is at 25×10⁶. Explain how this affects the critical and drag-rise Mach numbers you would quote, and state how the two data sets should be used in the clearance argument.
Read the worked answer
At the lower Reynolds number the boundary layer is thicker and its displacement effect larger, so the local supervelocity over the forward upper surface is higher for the same freestream Mach. Shock-induced separation also begins at a weaker shock because the separated shear layer is more vulnerable. Both effects push the measured critical Mach number and drag-rise Mach number down; the observed 0.02 discrepancy between the tunnel-predicted outboard value (0.745 at 1.4 g) and the flight value (0.765) is the right order of magnitude for a factor-of-eight Reynolds change on a conventional section.
A second Reynolds-dependent effect is transition: if the tunnel model runs with free transition, the trip location moves with Reynolds number and unit Reynolds number, so the same model can give different answers in different tunnels. Fixing the transition position with a trip makes the data repeatable but does not remove the displacement-thickness error.
Use of the two sets:
- The tunnel data are valid for ranking configuration changes (20° against 25° against 30° sweep, section A against section B), because the Reynolds error is largely common to all configurations.
- The tunnel data must not be used as an absolute fleet limit for Mmo, because it is conservative by an unknown 0.01–0.03 M.
- The absolute limit is set from flight test, at several altitudes (unit Reynolds number and hence section Reynolds number vary strongly with altitude at constant Mach).
- Where flight test cannot cover a point (e.g. hot-day, high-altitude corners), apply a Reynolds extrapolation based on the measured slope of onset Mach against log Re, and carry an explicit allowance in the margin rather than assuming the tunnel value.
In short: tunnel for shape, flight for numbers, with a stated Reynolds allowance on any interpolation.