### Re: Updated: Qantas saying it is almost ready to select Project Sunrise aircraft

Posted:

**Tue May 14, 2019 6:52 am**Eyad89 wrote:MoKa777 wrote:

You did the calculations for the A350 and 777X a few posts up but may I ask you to break-down the calculation for the 787 effective span? Please.

I just followed the rule of thumb that a raked wingtip adds 80% of the wingtip's length to the physical wingspan, and I roughly assumed a 787's wingtip is 6m on both sides, so its effective wingspan would be 54.12m + 4.8m = 58.92m

Again, those are rough estimations only, and a wind tunnel testing can confirm it.MoKa777 wrote:May you please also elabortae a little bit more on the span loading.

Span loading is the complete spanwise pressure distribution of the wingspan, and it can be calculated by dividing weight by span. Before I elaborate further here, I will quickly go through the principle of lift, as understanding it is crucial for grasping the concept of span loading and induced drag.

As air accelerates around the wing, the pressure below the wing will be higher than the pressure above the wing. This difference in pressure generates the lift force that pushes the wings upwards. Now that the pressure at the lower surface is higher than the pressure at the upper surface, what happens at the tips of the wings when the air from the lower and upper side meet? the air will move from the area of the higher pressure to the area of the lower pressure (fluids always flow from higher to lower pressure) in this fashion:

This circular motion of air is caused by the pressure difference between the upper and lower sides of the wings, and that pressure difference is caused by lift. This circular air motion causes changes in the air speed and direction in a way that air circulates around the entire span of the wing, and that simply what is called lift-induced drag. It is the energy consumed in this circular motion, and it is an inherent part of the lift force, as you cannot create lift without having this drag. You can only reduce it.

How to reduce this disturbing motion of air (lift-induced drag)? Reduce span loading. Bring down weight or increase the effective wingspan.

In order to reduce the impact of this circular air motion, you have to make it slower or weaker. Remember this entire motion is generated by the pressure difference between the two sides of the wing, so reducing this pressure difference will reduce the circular motion. How can you reduce the pressure difference? Reduce lift. And how can you do that? Reduce weight.

Another way to make this circular motion slower is to increase its effective wingspan. This will increase spanload distribution for the wing to make the circular air motion around the wing wider and more optimal.

(note: speed and density affect this induced drag as well).

Thank you!

This explains things very well for me. I really appreciate it.

If I may add the calculation for the posters who may still find it a bit strange,

From the Boeing 777X ACAP, the folding part of the raked tip is 3.47m on each side - 6.94m combined.

The basic wing span - excluding the folding part - is 64.82m.

80% of the folding raked tip is 2.776m on each side - 5.552m combined.

64.82m + 5.552m = 70.372m effective span

Is this correct?

About the 787, my rough estimation using the scale in the 787 ACAP is that the raked section of each wing is +-3.5m.

That gives 53.12m for the basic wing.

80% of the raked tip is 2.8m on each side - 5.6m combined.

53.12m + 5.6m = 58.72m effective span

For the A350 it seems a bit more complicated. Correct me if I am wrong.

From pages 31 and 39 of the A350-900/-1000 ACAP, the "sharklets" have a vertical height of 2.43m each.

On pages 27 and 29, the horizontal length of each "sharklet" is 2.88m.but from the start of each one to the tip measure between 5.18m and 5.27m horizontally if the "curl" is taken into account.

Which one is the preferred measurement for the A350 when calculating (roughly) effective span?