Lesson 6 · 40 min
Closing the Sizing Loop
All the ingredients are on the table: the crew and payload, the empty-weight trend, and a fuel fraction built from \(L/D\), \(C\) and the mission. This lesson puts them together and sizes the business jet, checks the answer, and then deals with a harder case: missions that drop part of their weight on the way.
Learning objectives
- Solve the sizing equation by iteration to a converged \(\Wo\), and split it into empty, fuel and fixed weights.
- Check a sizing result for consistency and plausibility.
- Size an aircraft whose mission drops payload, following its weight segment by segment.
Sizing the business jet
Lessons 1 to 5 gave everything the sizing equation needs:
The business jet
\[ \Wo = \frac{980}{1 - 0.2867 - 0.9727\,\Wo^{-0.06}} \]Crew and payload \(980\ \text{kg}\) (Lesson 1); fuel fraction 0.2867 (Lesson 5); jet-transport trend in kg (Lesson 2).
Example 6.1 — Iterate to convergence
Starting from \(10\,000\ \text{kg}\), iterate until \(\Wo\) changes by less than 1 kg. Then find the empty and fuel weights.
Show solution
| Iteration | Guess (kg) | \(\We/\Wo\) | New \(\Wo\) (kg) |
|---|---|---|---|
| 1 | 10 000 | 0.5598 | 6382 |
| 2 | 6382 | 0.5750 | 7088 |
| 3 | 7088 | 0.5714 | 6908 |
| 4 | 6908 | 0.5723 | 6951 |
| 5 | 6951 | 0.5721 | 6941 |
| 6 | 6941 | 0.5722 | 6943 |
| 7 | 6943 | 0.5721 | 6943 |
Check: \(980 + 1990 + 3972 = 6942\ \text{kg}\), equal to \(\Wo\) within rounding.
The guesses overshoot and undershoot, each time by less: the error shrinks by a factor of about 4 per iteration here. Convergence is fast because the empty-weight fraction changes slowly with weight. A spreadsheet does it in one column; the Sizing Calculator does it as you type.
Is the answer sensible?
A converged number is not necessarily a right one. Before going on:
- Add it up. Crew, payload, fuel and empty weight must sum to \(\Wo\).
- Compare with real aircraft. Light business jets with about eight seats weigh about 6 to 8 t at takeoff, so 6.9 t is plausible. A result far outside the range of existing aircraft of the class means an input is wrong or the class does not fit.
- Check the fractions. An empty-weight fraction outside about 0.3 to 0.7, or a denominator below about 0.1, is a warning: the design is close to the range wall and every error will be amplified.
- Record the assumptions. Class, technology factor, \(\LDmax\), \(C\), the mission. Every later module will revisit one of them.
Missions that drop payload
When a mission drops weight (bombs, cargo, sonobuoys, a fire-retardant load), the weight after the drop is not a fixed fraction of \(\Wo\): it is the weight before the drop minus a fixed number of kilograms. The fuel burned after the drop then depends on \(\Wo\), and the simple fuel fraction no longer exists. The remedy is to follow the weight through the mission:
- Guess \(\Wo\).
- Step through the segments: \(W_i = W_{i-1}(W_i/W_{i-1})\) for each flight segment, \(W_i = W_{i-1} - W_{\text{drop}}\) at the drop.
- The fuel burned is \(\Wo - W_{\text{drop}} - W_x\); add 6%: \(\Wf = 1.06(\Wo - W_{\text{drop}} - W_x)\).
- New \(\Wo = (W_{\text{crew}} + W_{\text{payload}} + \Wf)/(1 - \We/\Wo)\), with the payload including the dropped weight. Repeat.
Example 6.2 — Air-dropping cargo
A twin-turboprop transport (the twin-turboprop trend, \(\We/\Wo = 0.9228\,\Wo^{-0.05}\) in kg) with a crew of 3 (\(270\ \text{kg}\)) carries \(6000\ \text{kg}\) of cargo \(1000\ \text{km}\), air-drops it, and returns. Mission: takeoff 0.970, climb 0.985, cruise \(1000\ \text{km}\) (fraction 0.92867), drop, cruise \(1000\ \text{km}\) back (0.92867), 30-minute loiter (0.98474), landing 0.995. (The cruise and loiter fractions use \(\LDmax = 14\), \(c_P = 0.5\) and \(0.6\ \text{lb/(hp h)}\), \(\eta_p = 0.8\).) Size it, and compare with the same aircraft bringing the cargo back.
Show solution
At the converged \(\Wo = 23\,897\ \text{kg}\), following the weight:
\[ W_3 = 23\,897(0.970)(0.985)(0.92867) = 21\,204\ \text{kg} \] \[ W_4 = 21\,204 - 6000 = 15\,204\ \text{kg} \] \[ W_x = 15\,204(0.92867)(0.98474)(0.995) = 13\,835\ \text{kg} \] \[ \Wf = 1.06(23\,897 - 6000 - 13\,835) = 4306\ \text{kg} \] \[ \frac{\We}{\Wo} = 0.9228(23\,897)^{-0.05} = 0.5574 \] \[ \Wo = \frac{270 + 6000 + 4306}{1 - 0.5574} = 23\,897\ \text{kg} \ \checkmark \]From a guess of \(20\,000\ \text{kg}\) the iteration takes about ten steps (Figure 6.1). If the cargo stayed aboard, the ordinary fuel fraction would apply, \(1.06(1 - 0.970 \cdot 0.985 \cdot 0.92867^2 \cdot 0.98474 \cdot 0.995) = 0.2042\), and \(\Wo = 26\,040\ \text{kg}\): flying home empty saves 2.1 t of takeoff weight.
Check your understanding
Key takeaways
- Iterate \(\Wo = (W_{\text{crew}} + W_{\text{payload}})/(1 - \Wf/\Wo - \We/\Wo)\) from a guess; it converges in a handful of steps.
- The business jet: \(\Wo = 6943\ \text{kg}\), \(\We = 3972\ \text{kg}\), \(\Wf = 1990\ \text{kg}\).
- Check the sum, compare with real aircraft, watch the denominator, and record the assumptions.
- With a drop, follow the weights segment by segment: \(\Wf = 1.06(\Wo - W_{\text{drop}} - W_x)\).
- Next, Lesson 7 asks how the answer changes when the inputs do.