Lesson 4 · 40 min

Certification and Performance Requirements

Some requirements are not negotiable with anyone. To carry passengers, an aircraft must be certified against airworthiness rules, and several of those rules are, in effect, performance requirements: how long a runway it needs, how slowly it approaches, and how well it climbs with an engine failed. This lesson turns three of them into numbers a designer can use.

Learning objectives

Identify the certification basis of an aircraft (Parts 23 and 25 and their equivalents) and how it becomes a source of requirements.
Explain the balanced field length and the takeoff field length of a transport aircraft.
Use the 60% rule and \(V_{\text{REF}} \ge 1.23\,V_{SR}\) to turn a landing requirement into a field length and a maximum wing loading.
Compute the thrust-to-weight ratio needed to meet a one-engine-inoperative climb gradient, and explain why twins need more thrust than four-engine aircraft.

The certification basis

Each country's aviation authority publishes airworthiness standards, and an aircraft type is certified against a specific set of them, its certification basis. The main families are harmonized closely:

Normal category

US 14 CFR Part 23, EASA CS-23. Aeroplanes with up to 19 passenger seats and a maximum takeoff weight up to 19 000 lb (8618 kg): light aircraft, business turboprops, commuters.

Transport category

US 14 CFR Part 25, EASA CS-25. Airliners and large business jets. The field-length and climb rules in this lesson are Part 25's.

Elsewhere

Transport Canada's Airworthiness Manual chapters 523 and 525 correspond to Parts 23 and 25. Military aircraft are qualified against their own standards and specifications (for example, the US MIL-HDBK-516 criteria and MIL-STD-1797 for flying qualities).

The rules set minimums for safety. They do not say how long a runway the aircraft should need: that is the customer's requirement. They say how that runway length must be measured, with what margins and with which engine failed. That is why the regulations appear in the conceptual designer's toolkit: they decide what a "2100 m field length" actually demands.

Takeoff: the balanced field

A transport aircraft must be able to lose its most critical engine at the worst moment of the takeoff run and still either stop on the runway or continue and climb away. The pilot's decision speed is \(V_1\): an engine failure before \(V_1\) means stop, after \(V_1\) means go.

The accelerate-stop distance (accelerate with all engines to \(V_1\), then brake to a stop) grows with \(V_1\).
The accelerate-go distance (lose an engine at \(V_1\), continue on the remaining engines to a height of \(35\ \text{ft}\)) shrinks as \(V_1\) rises, since less of the run is made on reduced thrust.

Choosing \(V_1\) where the two are equal gives the shortest runway that covers both: the balanced field length. Part 25's takeoff field length is, in essence, the greater of the balanced field length and 115% of the all-engines distance to \(35\ \text{ft}\).

Figure 4.1 The balanced field, schematically, for an illustrative twin. Move the decision speed \(V_1\): the runway needed is the larger of the accelerate-stop and accelerate-go distances, and is shortest where they cross. Module 5 estimates these distances from \(\WS\), \(\TW\) and \(\CLmax\) with empirical methods.

Landing: the 60% rule and the approach speed

Two Part 25 rules shape the landing requirement:

The landing distance is measured from \(50\ \text{ft}\) above the threshold to a full stop, approaching at a reference speed \(V_{\text{REF}}\) of at least \(1.23\) times the reference stall speed in the landing configuration, \(V_{SR}\). (Older rules used \(1.3\,V_S\), with a stall speed defined differently.)
Airline operating rules then require the demonstrated landing distance to fit within 60% of the runway, so the landing field length is the distance divided by 0.6. On a wet runway it is 15% longer again.

Landing field length and the approach-speed limit on wing loading

\[ s_{\text{field}} = \frac{s_{\text{land}}}{0.6}, \qquad V_{SR} = \frac{V_{\text{REF}}}{1.23}, \qquad \left(\frac{W}{S}\right)_{\text{land}} \le \frac{\tfrac12\rho\,V_{SR}^2\,\colL{C_{L_{\max},\text{land}}}}{g} \]

The last is the stall-speed equation of Module 1, Lesson 4, solved for \(\WS\) (in kg/m²). It applies at the landing weight; divide by \(W_{\text{land}}/\Wo\) to get the takeoff wing loading.

Airports and air traffic control group aircraft into approach categories by their threshold speed, roughly \(V_{\text{REF}}\) at maximum landing weight: A below 91 kt, B 91 to 120 kt, C 121 to 140 kt, D 141 to 165 kt, E 166 to 210 kt. Most airliners are C or D. A customer may require, say, category C, which is an upper limit on \(V_{\text{REF}}\).

Example 4.1 — From the approach category to the wing

A new narrow-body must be approach category C (\(V_{\text{REF}} \le 140\ \text{kt}\)) at sea level, and its flaps are expected to give \(C_{L_{\max}} = 2.6\) in the landing configuration. (a) What is the largest landing wing loading? (b) If the maximum landing weight is 85% of \(\Wo\), what takeoff wing loading does that allow? (c) Tests show a landing distance of \(1350\ \text{m}\). What field length is required, dry and wet?

Show solution

\[ V_{\text{REF}} = 140(0.5144) = 72.02\ \text{m/s}, \qquad V_{SR} = \frac{72.02}{1.23} = 58.55\ \text{m/s} \] \[ \text{(a)}\ \left(\frac{W}{S}\right)_{\text{land}} \le \frac{\tfrac12(1.225)(58.55)^2(2.6)}{9.81} = 556.6\ \text{kg/m}^2 \] \[ \text{(b)}\ \frac{W}{S} = \frac{556.6}{0.85} = 654.8\ \text{kg/m}^2 \] \[ \text{(c)}\ s_{\text{field}} = \frac{1350}{0.6} = 2250\ \text{m dry}, \qquad 1.15(2250) = 2588\ \text{m wet} \]

For comparison, an A320 at its maximum takeoff weight has a wing loading of about \(640\ \text{kg/m}^2\). One customer requirement and one regulation have already fixed the wing area to within a few percent, before the wing is even drawn.

Figure 4.2 The largest landing wing loading allowed by an approach speed, for several maximum lift coefficients, with the approach-category boundaries. Better flaps (a higher \(C_{L_{\max}}\)) allow a smaller wing for the same approach speed, which is why so much effort goes into high-lift systems.

Climb with an engine failed

After an engine fails at \(V_1\), the aircraft must climb away over obstacles. Part 25 (section 121) sets minimum climb gradients, the height gained per distance flown, for each part of the departure and for a go-around, with the critical engine inoperative. The requirement is stricter for aircraft with more engines, because they lose a smaller share of their thrust when one fails.

Minimum climb gradients, 14 CFR / CS 25.121 (one engine inoperative unless noted)
Segment	Configuration	2 engines	3 engines	4 engines

In a steady climb at a small angle \(\gamma\), the excess of thrust over drag lifts the weight: \(T - D = W\sin\gamma \approx W G\), where \(G\) is the gradient. Dividing by \(W\), with \(D/W = 1/(L/D)\), gives the thrust-to-weight ratio needed on the remaining engines. The full-thrust \(\TW\) of the aircraft is \(N/(N-1)\) times that:

Thrust-to-weight ratio for a one-engine-inoperative climb gradient

\[ \frac{T}{W} = \frac{N}{N-1}\left(\frac{1}{\colL{L/D}} + G\right) \]

\(L/D\) in the climb configuration (flaps at the takeoff setting, gear up for the second segment; typically 8 to 12 for a transport), \(G\) as a decimal (2.4% is 0.024). For an all-engines requirement, drop the factor \(N/(N-1)\).

Example 4.2 — Why twins have more thrust

A transport has \(L/D = 10\) in the second-segment configuration. Find the thrust-to-weight ratio needed to meet the second-segment gradient as a twin, a three-engine and a four-engine aircraft.

Show solution

\[ N = 2:\ \ \frac{T}{W} = \frac{2}{1}(0.100 + 0.024) = 0.248 \] \[ N = 3:\ \ \frac{T}{W} = \frac{3}{2}(0.100 + 0.027) = 0.191 \] \[ N = 4:\ \ \frac{T}{W} = \frac{4}{3}(0.100 + 0.030) = 0.173 \]

A twin loses half its thrust when an engine fails, so it must carry 43% more installed thrust than a four-engine aircraft to climb the same way. In return, it has fewer engines to buy and maintain, and the extra thrust gives it a shorter takeoff and a faster climb when both engines work. These values are at the climb speed and temperature; corrected to the static sea-level thrust the engine is rated at, they are larger still.

Figure 4.3 Thrust-to-weight ratio needed for each climb requirement against the climb-configuration \(L/D\), for two, three and four engines. Choose a segment; the marked point is the chosen \(L/D\) and number of engines.

Ceilings and other performance requirements

The service ceiling is the altitude at which the best rate of climb has fallen to \(100\ \text{ft/min}\) (\(0.5\ \text{m/s}\)); the cruise ceiling uses \(300\ \text{ft/min}\) (\(1.5\ \text{m/s}\)). An airliner's requirement is usually an initial cruise altitude it must reach at MTOW, with that residual climb rate.
Military requirements add maneuver performance: a sustained turn load factor at a given speed and altitude, or a specific excess power \(P_s = V(T - D)/W\), the rate at which the aircraft can gain energy height.
General aviation adds its own: Part 23 aircraft have long been held to low stall speeds (61 kt for single-engine aeroplanes), which limits their wing loading just as the approach category limits an airliner's.

Each of these, like the landing and climb rules above, becomes a line on the \(\TW\)–\(\WS\) chart of Module 5. The design point must satisfy all of them at once.

Check your understanding

Key takeaways

The certification basis (Part 23 or 25, or equivalents) turns safety rules into performance requirements.
Takeoff field length: the greater of the balanced field length (engine failure at \(V_1\)) and 115% of the all-engines distance to 35 ft.
Landing: \(V_{\text{REF}} \ge 1.23\,V_{SR}\); field length \(= s_{\text{land}}/0.6\) (×1.15 wet); an approach-speed limit is a wing-loading limit, \((W/S) \le \tfrac12\rho V_{SR}^2\CLmax/g\).
One engine out: \(\TW = \frac{N}{N-1}\left(1/(L/D) + G\right)\), with \(G\) = 2.4%, 2.7%, 3.0% in the second segment for 2, 3, 4 engines.
Next, Lesson 5 describes the mission the aircraft must fly.