The true gain of increased use of CFRP or other lighter materials vs Al has not been easily visible for the 787 or 350 generation of frames. I had expected a decrease in OEW to MTOW ratio but this has not appeared, the 788 expected final ratio of 50,2% is clearly worse then the 15 years older 77W at 47,7%.

Astuteman brought me on the track that the reduced fuel consumption makes the OEW/MTOW ratio stay the same (or even slightly worse ) but the whole aircraft is "lighter".

This is of course correct, the ratio that gets better is payload/range vs MTOW and fuel burned. Thereby we have a very simple first measure of efficiency for a frame:

1. SIMILAR RANGE

Within the same range domain one can simply divide the payload with MTOW, this gives the first order efficiency factor at that range. As OEMs typically state the MTOW, the range at full typical cabin with no cargo (=fullPnoC) and the 210lb per pax+bag is set we can calculate this factor easily and correctly:

For the typical 8000nm class:

___________fullPnoC %_________Range nm @fullPnoC

788..........................9,4................~8000

789........................10,6................~8000

358..........................9,9................8490

359........................11,2................8100

351........................11,2................7990

77W........................9,9................7930

For the typical 6500nm class:

___________fullPnoC %_________Range nm @fullPnoC

781........................12,8................~6500

333........................11,8................5850

763E......................11,1................5975

The above says nothing about the earning vs. cost side which of course is the real measures (look at LAXDESIs analysis for that), but as a first check it is useful and we can use the first typical data the framers state .

2. COMPENSATED FOR RANGE DIFFERENCES

If one want one can start to compensate for the range differences, in that case one must use or assume an OEW. fullPnoC fuel used per nm is given as 90% (10% reserves) of MTOW-(OEW+fullPnoC) divided by range. If we don't have OEW we use 50% of MTOW as discussed in other threads. This gives:

___________fuel kg/nm_________Range comp vs 8k nm_____Comp fullCnoP %

788.......................10,4................~0.....................................9,4

789.......................11,3................~0.....................................10,6

358.......................11,4................-5,4t..................................12,0

359.......................11,6................-1,2t..................................11,6

351.......................13,0................~0.....................................11,1

77W......................16,9................1,2t...................................11.0

and for the midrange bunch:

___________fuel kg/nm_________Range comp vs 6k nm_____Comp fullCnoP %

781.......................12,9................-6,5t.................................15,4

333.......................13,1................2t.....................................11,0

763E.....................11,5................0......................................11,0

CAVEATS:

- Clearly one can see the deficiency of this simple ratio check, e.g. the 763E comes out equal to 333 but the ratio does not look at any costs or revenue i.e. those LD3 positions at anything other then fullPnoC range is not considered.

- The fuel average over distance is OK to do as the climb and descent fuel with 60% higher consumption is only for 4% of the distance, a wash.

- "fullPnoC ratio" is perhaps a bit cryptic, if we find it usefull one can also call it "Payload ratio".

Thoughts? Faults?

[Edited 2011-05-22 01:07:21]

Astuteman brought me on the track that the reduced fuel consumption makes the OEW/MTOW ratio stay the same (or even slightly worse ) but the whole aircraft is "lighter".

This is of course correct, the ratio that gets better is payload/range vs MTOW and fuel burned. Thereby we have a very simple first measure of efficiency for a frame:

1. SIMILAR RANGE

Within the same range domain one can simply divide the payload with MTOW, this gives the first order efficiency factor at that range. As OEMs typically state the MTOW, the range at full typical cabin with no cargo (=fullPnoC) and the 210lb per pax+bag is set we can calculate this factor easily and correctly:

For the typical 8000nm class:

___________fullPnoC %_________Range nm @fullPnoC

788..........................9,4................~8000

789........................10,6................~8000

358..........................9,9................8490

359........................11,2................8100

351........................11,2................7990

77W........................9,9................7930

For the typical 6500nm class:

___________fullPnoC %_________Range nm @fullPnoC

781........................12,8................~6500

333........................11,8................5850

763E......................11,1................5975

The above says nothing about the earning vs. cost side which of course is the real measures (look at LAXDESIs analysis for that), but as a first check it is useful and we can use the first typical data the framers state .

2. COMPENSATED FOR RANGE DIFFERENCES

If one want one can start to compensate for the range differences, in that case one must use or assume an OEW. fullPnoC fuel used per nm is given as 90% (10% reserves) of MTOW-(OEW+fullPnoC) divided by range. If we don't have OEW we use 50% of MTOW as discussed in other threads. This gives:

___________fuel kg/nm_________Range comp vs 8k nm_____Comp fullCnoP %

788.......................10,4................~0.....................................9,4

789.......................11,3................~0.....................................10,6

358.......................11,4................-5,4t..................................12,0

359.......................11,6................-1,2t..................................11,6

351.......................13,0................~0.....................................11,1

77W......................16,9................1,2t...................................11.0

and for the midrange bunch:

___________fuel kg/nm_________Range comp vs 6k nm_____Comp fullCnoP %

781.......................12,9................-6,5t.................................15,4

333.......................13,1................2t.....................................11,0

763E.....................11,5................0......................................11,0

CAVEATS:

- Clearly one can see the deficiency of this simple ratio check, e.g. the 763E comes out equal to 333 but the ratio does not look at any costs or revenue i.e. those LD3 positions at anything other then fullPnoC range is not considered.

- The fuel average over distance is OK to do as the climb and descent fuel with 60% higher consumption is only for 4% of the distance, a wash.

- "fullPnoC ratio" is perhaps a bit cryptic, if we find it usefull one can also call it "Payload ratio".

Thoughts? Faults?

[Edited 2011-05-22 01:07:21]

Non French in France

It seems the efficiency factor doesn't create that much debate or exitement. The above also gives another first estimate, the fuel burn. Here the fuel burns once again with the addition of fuel burn per seat:

___________fuel kg/nm_________fuel/seat and nm

788.......................10,4................0,046

789.......................11,3................0,040

358.......................11,4................0,041

359.......................11,6................0,037

351.......................13,0................0,036

77W......................16,9................0,046

and for the midrange bunch:

___________fuel kg/nm_________fuel/seat and nm

781.......................12,9................0,038

333.......................13,1................0,045

763E.....................11,5................0,053

The above is based on manufacturer figures in all corners except for OEW for the non EIS frames. The 50% of MTOW assumption for these covers that well, it is within a couple of tons and any diff in that order would not change these figures.

[Edited 2011-05-22 23:04:06]

___________fuel kg/nm_________fuel/seat and nm

788.......................10,4................0,046

789.......................11,3................0,040

358.......................11,4................0,041

359.......................11,6................0,037

351.......................13,0................0,036

77W......................16,9................0,046

and for the midrange bunch:

___________fuel kg/nm_________fuel/seat and nm

781.......................12,9................0,038

333.......................13,1................0,045

763E.....................11,5................0,053

The above is based on manufacturer figures in all corners except for OEW for the non EIS frames. The 50% of MTOW assumption for these covers that well, it is within a couple of tons and any diff in that order would not change these figures.

[Edited 2011-05-22 23:04:06]

Non French in France

- SchorschNG
**Posts:**259**Joined:**

Did you take ACAP values or what did you use as input?

"Fuel per seat" requires that you assume reserves. How did you solve that problem?

The A330-300 achieves 2.4kg/100km & Seat, which seems a bit optimistic but still close to plausible. What seat count did you assume?

"Fuel per seat" requires that you assume reserves. How did you solve that problem?

The A330-300 achieves 2.4kg/100km & Seat, which seems a bit optimistic but still close to plausible. What seat count did you assume?

From a structural standpoint, passengers are the worst possible payload. [Michael Chun-Yung Niu]

Quoting SchorschNG (Reply 2):The A330-300 achieves 2.4kg/100km & Seat, which seems a bit optimistic but still close to plausible. |

SK provided a figure of 3.5 liters per 100km for their A333 fleet.

In March 2010, Boeing claimed the following fuel burn for the 787 in their manufacturer OEM config:

787-8: 2.6 liters per 100 passenger kilometers

787-9: 2.5 liters per 100 passenger kilometers

Quoting SchorschNG (Reply 2):Did you take ACAP values or what did you use as input?
"Fuel per seat" requires that you assume reserves. How did you solve that problem? |

For the 787 and 350 I used manufacturers claims as there is no performance figures in the airport guide (ACAP), for 77W I used ACAP as Wikipedia and ACAP concurr, for 333 and 767 Wikipedia as the ACAP is issued for the 233t and 207t variants.

For reserves I used 200nm alternate, 30 min holding and 5% contingency fuel, this all adds up to 10% reserves on top of trip fuel ( according to Piano-X) which I have used for all fuel analysis. This is fair as e.g. Boeing for the 767 ACAP uses ATA domestic reserves which has no holding and I assume others use the international rules with 30 min holding.

Quoting Stitch (Reply 3):SK provided a figure of 3.5 liters per 100km for their A333 fleet. |

Equeates to 0,052 kg/nm and pax. This figure must be with SKs lower count of pax in their configurations.

Quoting Stitch (Reply 3):787-8: 2.6 liters per 100 passenger kilometers
787-9: 2.5 liters per 100 passenger kilometers |

Equates to 0,039 vs my 0,042 (corrected from first table, had old seat count in my spreadsheet) for 788 and 0,037 vs 0,040 for 789.

As stated my calculation includes climb and descent sections, that might explain some of the difference, further one needs to know when these claims were made (at what OEW). My calculations are with the latest figures, as they assume an OEW they have some margin for error but I think it is rather small. The figures for SK also show that one can not mix OEMs optimal broschyre values with their assumptions and practical airline figures, there is simply to much difference in terms of OEW, cost index procedures, cabin layouts with fewer seating optimised for revenue generation etc.

The proposed simple ratios gives a good base to reality check a lot of claims and assumptions, they use the broshyre values consistently and makes only a few non critical assumptions.

[Edited 2011-05-23 12:00:20]

Non French in France

You may have to do a chart to compare OEW/MTOW and normalize it with the size of the aircraft.

Otherwise you would not be able to identify whether the increase in efficiency is due to the composite or size.

bikerthai

Otherwise you would not be able to identify whether the increase in efficiency is due to the composite or size.

bikerthai

Intelligent seeks knowledge. Enlightened seeks wisdom.

Quoting bikerthai (Reply 5):Otherwise you would not be able to identify whether the increase in efficiency is due to the composite or size. |

Thanks Bikerthai, your are right. In the end I just want to see which frame is more efficient or not (made with whatever materials) and one would have to compare them within their size bracked as you point out, i.e:

788 vs 358

789 vs 358

77W vs 351

781 vs 333

and so on.

Non French in France

Quoting bikerthai (Reply 7):Sounds like there's enough there for a Master's Thesis |

For the complete comparison sure but for the simple ratio not really. The beauty of the ratio is that it is 100% based on the framers typical statement when he launches a frame or a variant.

Normally they state for the "bragging case:

- number of pax in a mixed cabin, with this we take it * 210lb and we have the carried payload.

- MTOW, there we have the mass to divide the carried payload with.

- the range, this we use to check the aircraft category. If it differs more then marginally from our chosen class standard (like 8k nm for the 788/350/77W class) we can normalise by counting the extra fuel burned/missing as payload adjustment.

[Edited 2011-05-24 12:23:00]

Non French in France

I just saw that LAXDESI allocates 9,1% fuel for Diversion, Hold and Contingency, then I will apply the same (he has probably done more poking then I on this).

He also states the "broschyre" range for the 788 as 7400nm with B stating 7650nm. Could someone point to wether this is with fullPnoC and the above reserves? For what tail number, after 20?

What would then be the best range to use for the 789?

He also states the "broschyre" range for the 788 as 7400nm with B stating 7650nm. Could someone point to wether this is with fullPnoC and the above reserves? For what tail number, after 20?

What would then be the best range to use for the 789?

Non French in France

For some reason I completely missed this thread. What if we compare the early 788 to A332HGW to get some sense of how the higher composite 788(242 seats) frame compares to A332HGW(242 seats).

___________fullPnoC %_________Range nm @fullPnoC

788............................10.11..............~7653nm(Piano X)

A332HGW....................9.70..............~7250nm

As for OEW/MTOW, the numbers are as follows:

788......................(252,500/502,500) = 0.50 (Piano X OEW)

332HGW..............(257,588/524,700) = 0.49

As for MSP/MTOW, the numbers are as follows:

788......................(102,500/502,500) = 0.20 (based on Piano X OEW)

332HGW..............(112,788/524,700) = 0.22

The last two metrics, for most part, do not show B788 to be any better than A332HGW. However, the first metric (max. passenger payload/MTOW) does indicate that the B788 can carry the same passenger payload at a much lesser MTOW and carry it much farther.

So how about the following metric to capture the lower fuel burn due to lighter weight, better aero. and sfc:

.......................Max. passenger payload/MTOW) X (Design range/1000)

788..................(50,820/502,500) X (7,653/1,000) = 0.77

332HGW..........(50,820/524,700) X (7,250/1,000) = 0.70

The above metric would suggest that 788 is better than 332HGW by nearly 10%.

Applying the above metric to A350-10 and 773ER:

.......................Max. passenger payload/MTOW) X (Design range/1000)

A350-10............(73,500/657,000) X (8000/1,000) = 0.89

B773ER............(76,650/775,000) X (7,930/1,000) = 0.78

The above metric would suggest that A350-10 would be about 14% better than 773ER, and it is consistent with expectation that A350-10, at specs, will be significantly better than B773ER.

___________fullPnoC %_________Range nm @fullPnoC

788............................10.11..............~7653nm(Piano X)

A332HGW....................9.70..............~7250nm

As for OEW/MTOW, the numbers are as follows:

788......................(252,500/502,500) = 0.50 (Piano X OEW)

332HGW..............(257,588/524,700) = 0.49

As for MSP/MTOW, the numbers are as follows:

788......................(102,500/502,500) = 0.20 (based on Piano X OEW)

332HGW..............(112,788/524,700) = 0.22

The last two metrics, for most part, do not show B788 to be any better than A332HGW. However, the first metric (max. passenger payload/MTOW) does indicate that the B788 can carry the same passenger payload at a much lesser MTOW and carry it much farther.

So how about the following metric to capture the lower fuel burn due to lighter weight, better aero. and sfc:

.......................Max. passenger payload/MTOW) X (Design range/1000)

788..................(50,820/502,500) X (7,653/1,000) = 0.77

332HGW..........(50,820/524,700) X (7,250/1,000) = 0.70

The above metric would suggest that 788 is better than 332HGW by nearly 10%.

Applying the above metric to A350-10 and 773ER:

.......................Max. passenger payload/MTOW) X (Design range/1000)

A350-10............(73,500/657,000) X (8000/1,000) = 0.89

B773ER............(76,650/775,000) X (7,930/1,000) = 0.78

The above metric would suggest that A350-10 would be about 14% better than 773ER, and it is consistent with expectation that A350-10, at specs, will be significantly better than B773ER.

Quoting LAXDESI (Reply 10):So how about the following metric to capture the lower fuel burn due to lighter weight, better aero. and sfc:
.......................(Max. passenger payload/MTOW) X (Design range/1000) 788..................(50,820/502,500) X (7,653/1,000) = 0.77 332HGW..........(50,820/524,700) X (7,250/1,000) = 0.70 The above metric would suggest that 788 is better than 332HGW by nearly 10%. |

The above metric is useful for a first order comparison when only the data on no. of seats, MTOW, and design range is available. In cases where one has access to MSP(max. structural payload), the following metric is probably a better measure:

.......................(Max. structural payload/MTOW) X (MSP range/1000)

788..................(102,500/502,500) X (5,438/1,000) = 1.11

332HGW..........(112,788/524,700) X (4,784/1,000) = 1.03

The above metric is better at capturing the cargo capability of A332 as it shows B788 is better than A332HGW by 7.7%, and not 10% using only passenger payload and design range.

I should add that the above metrics are no match for a complete cost benefit analysis which accounts for fuel prices, revenue potential of passenger seats and cargo, and acquisition costs. However, they can be a quick and rough approximation that capture, fairly accurately, the rank order for competing aircraft.

Thanks LAXDESI, your first comparison where you normalise the range directly is much better then my way over the fuel burn as you then have to assume OEW. Must admit I have not checked it 100% yet for snags but it seems watertight.

Anyway it is elegant, the second MSP/MTOW*range is also very good, you need a bit more data but it gives a more complete "total payload ratio" picture. Glad you found the thread .

Anyway it is elegant, the second MSP/MTOW*range is also very good, you need a bit more data but it gives a more complete "total payload ratio" picture. Glad you found the thread .

Non French in France

Quoting ferpe (Reply 12):Glad you found the thread |

Me too, and thanks for exploring new metrics to evaluate competing aircraft. Let me know if the metrics I have presented don't perform well for certain comparisons. Perhaps we can modify it to better capture missing variables.

One of these days we should start a thread with rule of thumb metrics that have proven themselves.

Applying the MPP metric to B744ER and 773ER:

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B744ER............(87,360/910,000) X (7,670/1,000) = 0.73

B773ER............(76,650/775,000) X (7,930/1,000) = 0.78

B773ER is nearly 7% better. I suspect B773ER looks even better using MSP metric.

Applying the MPP metric to B748 and 773ER:

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B748................(98,070/975,000) X (8,000/1,000) = 0.80

B773ER...........(76,650/775,000) X (7,930/1,000) = 0.78

748 is about 2.5% better than 773ER.

Applying the MPP metric to B748 and 773ER(10-abreast 395 seat NG):

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B748................(98,070/975,000) X (8,000/1,000) = 0.80

B773ER(NG).....(82,950/775,000) X (7,930/1,000) = 0.85

773ER(NG) is likely to better than B748 by nearly 6%.

Going by history and the difference of 7% in MPP metric between 773ER and 744ER, 773ER(10-abreast NG) could nearly kill B748's prospects.

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B744ER............(87,360/910,000) X (7,670/1,000) = 0.73

B773ER............(76,650/775,000) X (7,930/1,000) = 0.78

B773ER is nearly 7% better. I suspect B773ER looks even better using MSP metric.

Applying the MPP metric to B748 and 773ER:

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B748................(98,070/975,000) X (8,000/1,000) = 0.80

B773ER...........(76,650/775,000) X (7,930/1,000) = 0.78

748 is about 2.5% better than 773ER.

Applying the MPP metric to B748 and 773ER(10-abreast 395 seat NG):

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B748................(98,070/975,000) X (8,000/1,000) = 0.80

B773ER(NG).....(82,950/775,000) X (7,930/1,000) = 0.85

773ER(NG) is likely to better than B748 by nearly 6%.

Going by history and the difference of 7% in MPP metric between 773ER and 744ER, 773ER(10-abreast NG) could nearly kill B748's prospects.

@LAXDESI

So I have figured a bit, unfortunately I don't think a version of the efficiency index with range included is totally clean, here is why:

Initial index:

This divides the usefull weight transported with the total weight to do that, the index is thus dimensionless. When I introduced a correction for unequal transport distance for the usefull weight transported I did this using the fact that the fuel weight that is consumed for the A/C flying a further distance could as well be traded as added usefull weight carried the same distance for all compared A/C. This is a valid assumption as the frame (within the MZFW) does not distinguish between payload or fuel weight. The corrected index has the drawback that you need to define OEW somehow.

Your index:

This multiplies the dimensionless Payload/MTOW with range/1000 (with the dimension nm), we therefore have an index with the dimension range at the end. This is not OK I think. I would like to be proven wrong however as your index is more elegant .

So I have figured a bit, unfortunately I don't think a version of the efficiency index with range included is totally clean, here is why:

Initial index:

This divides the usefull weight transported with the total weight to do that, the index is thus dimensionless. When I introduced a correction for unequal transport distance for the usefull weight transported I did this using the fact that the fuel weight that is consumed for the A/C flying a further distance could as well be traded as added usefull weight carried the same distance for all compared A/C. This is a valid assumption as the frame (within the MZFW) does not distinguish between payload or fuel weight. The corrected index has the drawback that you need to define OEW somehow.

Your index:

This multiplies the dimensionless Payload/MTOW with range/1000 (with the dimension nm), we therefore have an index with the dimension range at the end. This is not OK I think. I would like to be proven wrong however as your index is more elegant .

Non French in France

Quoting ferpe (Reply 15):Your index:
This multiplies the dimensionless Payload/MTOW with range/1000 (with the dimension nm), we therefore have an index with the dimension range at the end. This is not OK I think. I would like to be proven wrong however as your index is more elegant . |

Not sure I fully understand your point about the index being dimensionless. Staying with your terminology, a dimensionless index(ratio) normalised by design range is still a dimensionless index(ratio).

However, as I have said before:

Quoting LAXDESI (Reply 11):I should add that the above metrics are no match for a complete cost benefit analysis which accounts for fuel prices, revenue potential of passenger seats and cargo, and acquisition costs. However, they can be a quick and rough approximation that capture, fairly accurately, the rank order for competing aircraft. |

Applying the MPP metric to B789 and A359:

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B789(280 seats)............(58,800/553,000) X (8,000/1,000) = 0.85

A359(314 seats)............(65,940/590,800) X (8,100/1,000) = 0.90

A359 is nearly 6% better. However, A359 is nearly 22% more expensive at list. Furthermore, I am using the old range of 8,000 nm for 789(with 290 passengers) before the bump in MTOW.

In summary, one can further normalise the MPP metric by list price to gauge the competitiveness of two different aircraft.

Applying the MPP metric to B789, A359, and B772ER:

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B789(280 seats)............(58,800/553,000) X (8,000/1,000) = 0.85

A359(314 seats)............(65,940/590,800) X (8,100/1,000) = 0.90

B77E(301 seats)............(63,210/656,000) X (7,725/1,000) = 0.75

A359 is nearly 6% better than B789, but 22% more expensive at list.

A359 is 20% better than B77E, but 15% more expensive at list..

B789 is nearly 13% better than B77E, and 6% cheaper at list!

It will be interesting to see which of the two aircraft, B789 and A350, garners a larger share of 300 seat market. I would expect 789 to garner a larger share of the 300 seat medium haul market(A333), and A359 to garner a larger share of long haul market(B772E). However, entry of 787-10 would change the picture for the 300 seat medium haul market.

.......................(Max. passenger payload/MTOW) X (Design range/1000)

B789(280 seats)............(58,800/553,000) X (8,000/1,000) = 0.85

A359(314 seats)............(65,940/590,800) X (8,100/1,000) = 0.90

B77E(301 seats)............(63,210/656,000) X (7,725/1,000) = 0.75

A359 is nearly 6% better than B789, but 22% more expensive at list.

A359 is 20% better than B77E, but 15% more expensive at list..

B789 is nearly 13% better than B77E, and 6% cheaper at list!

It will be interesting to see which of the two aircraft, B789 and A350, garners a larger share of 300 seat market. I would expect 789 to garner a larger share of the 300 seat medium haul market(A333), and A359 to garner a larger share of long haul market(B772E). However, entry of 787-10 would change the picture for the 300 seat medium haul market.

This may have already been discussed...if so, I seem to have missed it.

How do you eliminate the variable of more fuel efficient engines on the newer craft when calculating the efficiency of the construction material?

How do you eliminate the variable of more fuel efficient engines on the newer craft when calculating the efficiency of the construction material?

What the...?

Quoting joecanuck (Reply 18):How do you eliminate the variable of more fuel efficient engines on the newer craft when calculating the efficiency of the construction material? |

Hi Joe, we are not calculating the efficiency of the construction material (perhaps the heading is a bit misleading), we are doing ratios for total aircraft:

- modern or old (7810 vs 763ER or 333)

- mostly alu or with up to 50% CFRP

The new fuel efficient engines is the very factor that threw me off a bit initially as I thought the OEW to MTOW ratio should be lower then (say 45-47% vs. the typical 48-52% for older frames) but they were not, the 788 will land at 50% when it has finished it's diet. Astuteman brought the key, the engines consumes less fuel as well, this makes the OEW compared to payload + fuel be proportionally larger as the payload stays the same but fuel weight for a given range goes down. The whole aircraft (=MTOW) for a given payload and range is smaller however, this is your gain with CFRP.

@LAXDESI, your range factor does not cater for differences in SFC as you multiply directly with the range, if you go via the fuel weight you will include the difference in SFC (better SFC means less kg fuel consumed for the range diff).

Non French in France

Quoting ferpe (Reply 19):
Hi Joe, we are not calculating the efficiency of the construction material (perhaps the heading is a bit misleading), we are doing ratios for total aircraft: |

Ah...I get it. Makes sense now. Interesting thread. Thanks.

What the...?

I have now worked a bit more on these simple rule calculations and ratios one can do with the rather sparse data that we have on e.g. 787 and 350.

In the above analysis the compensation method (to get a class of frames equalised to the same range e.g. 8000nm) is to course, one need to use the FINAL fuel burn instead of the average one to be more close to reality. One can also multiply the fuel per nm with speed which gives the more commonly used kg/hr fuel burn.

Using PianoX it seems modern aircraft start the cruise segments burning about 110% of average and finish with about 88% and their average cruise speed is about 480 kts for M 0,85 frames.

Using these improvements and update data on the frames one gets (calling the ratio the more commly used Payload fraction):

______mPnC range_____Comp kg____Payl fract%____Fuel burn kg/h

788..........7730................2375.................9,1...............4900

789..........8150...............-1421 ..............11,2...............5100

359..........8100............... -992................11,5...............5400

351..........8400...............-4434...............12,3...............6050

77W........7930.................1041................9,9...............8100

and for the midrange bunch:

______mPnC range_____Comp kg____Payl fract%____Fuel burn kg/h

7810........6900...............-9584...............16,0...............5800

333..........5850................1734 ..............11,1...............6180

767..........5975..................252...............11,0...............5300

Once again all caveats apply.

As can be seen the back of the envelope method gives a reasonable fuel burn prediction compared to data we have from Zeke et al, it is sensitive to wrongly estimate OEW for 787 and 350 but only to a degre of about 150kg/hr wrong per 1% wrongly projected OEW/MTOW ratio.

Given that we have no data this is better then nothing

[Edited 2011-06-29 03:59:18]

In the above analysis the compensation method (to get a class of frames equalised to the same range e.g. 8000nm) is to course, one need to use the FINAL fuel burn instead of the average one to be more close to reality. One can also multiply the fuel per nm with speed which gives the more commonly used kg/hr fuel burn.

Using PianoX it seems modern aircraft start the cruise segments burning about 110% of average and finish with about 88% and their average cruise speed is about 480 kts for M 0,85 frames.

Using these improvements and update data on the frames one gets (calling the ratio the more commly used Payload fraction):

______mPnC range_____Comp kg____Payl fract%____Fuel burn kg/h

788..........7730................2375.................9,1...............4900

789..........8150...............-1421 ..............11,2...............5100

359..........8100............... -992................11,5...............5400

351..........8400...............-4434...............12,3...............6050

77W........7930.................1041................9,9...............8100

and for the midrange bunch:

______mPnC range_____Comp kg____Payl fract%____Fuel burn kg/h

7810........6900...............-9584...............16,0...............5800

333..........5850................1734 ..............11,1...............6180

767..........5975..................252...............11,0...............5300

Once again all caveats apply.

As can be seen the back of the envelope method gives a reasonable fuel burn prediction compared to data we have from Zeke et al, it is sensitive to wrongly estimate OEW for 787 and 350 but only to a degre of about 150kg/hr wrong per 1% wrongly projected OEW/MTOW ratio.

Given that we have no data this is better then nothing

[Edited 2011-06-29 03:59:18]

Non French in France

Quoting ferpe (Reply 21): have now worked a bit more on these simple rule calculations and ratios one can do with the rather sparse data that we have on e.g. 787 and 350. |

Nice work. Could you provide numbers for 772ER and A332. I am curious to see the difference between 772ER and A333 as well as A332 and B788.

Quoting ferpe (Reply 19):@LAXDESI, your range factor does not cater for differences in SFC as you multiply directly with the range, if you go via the fuel weight you will include the difference in SFC (better SFC means less kg fuel consumed for the range diff). |

It seems to me that the range figure incorporate the differences in sfc, as everything else being equal the lower sfc engines will carry the passengers farther. Applying my MPP metric to revised A350-10 specs:

.......................Max. passenger payload/MTOW) X (Design range/1000)

A350-10............(73,500/679,000) X (8400/1,000) = 0.91

B773ER............(76,650/775,000) X (7,930/1,000) = 0.78

The simple MPP metric shows the revised A350-10 to be about 17% better than B77W. Clearly, it is not going to be as good as more comprehensive analysis, but it does provide a useful first order approximation when more data is not available.

Quoting LAXDESI (Reply 22):Could you provide numbers for 772ER and A332. |

Will do, just want to check the data before putting it in. Will also once again think about your comments re MPP.

If you compare the equalised payload fractions of 77W (9,9%) and 3510 (12,3%) the 3510 has a 24% higher fraction then the 77W. Further the 3510 has a 25% lower fuel burn then the 77W (6050/8100).

[Edited 2011-06-29 13:13:56]

Non French in France

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