theory of flight

[virgo]

Remembering basic training................they said that lift is proportional to the square of the speed (Didn't they ?????)
On an ISA day, if I rotate a 747 weighing 200,000 kg to 15 degrees at 110 knots, the wings produce just over 200,000 kg of lift and we fly.
On the same day, the same aeroplane weighing 400,000 kg requires only an extra 67 knots to produce double the lift.

How does this fit into the V squared formula ????

[BOAC]

I reckon you were 11.5kts too fast at rotation:).

[Clarence Oveur]

theory of flight
It's only a theory.

[james ozzie]

I think it fits the formula reasonably well: the heavier weight needs 177kt while the lighter needs 110kt. If you square each of these speeds & then divide the bigger result by the smaller result

(177 x 177)/(110X110)

you get 2.59 which is more than double (is that the point of your question?)
But I guess there are a host of other technical complexities coming into play, what with the rotational momentum of the plane, varying ground effects, varying engine performances etc etc

By this reasoning, you actually only need about 156kt to double the lift from 110kt (is that what boac was alluding to?)

[Old Smokey]

Has it occurred to you that whilst 200 Tonnes is a very light weight for a B747, 400 Tonnes is at or close to maximum for most variants?

Whilst 200 Tonnes can easily be accommodated using the normal V2min speed schedule, it may be necessary to resort to the "Increased Speeds" or "Improved Climb" speed schedule at very high weights, whereby V1/Vr/V2 at 400 Tonnes will be somewhat in excess of the normal speeds formulae. This will account for the apparently high Takeoff speeds at the higher weights.

Regards,

Old Smokey

[Mad (Flt) Scientist]

Also, you don't 'fly' at Vr. You start to rotate. Liftoff occurs at some, higher, speed - depending on the speed and rate you rotate at, the acceleration the aircraft has, and the speed you need to lift off at a given AoA. So you can't necessarily assume that Vr follows the speed-squared relationship.

Consider - even if you didn't rotate, eventually you'd be fast enough (assuming a long enough runway) to lift off at nominally zero AoA. You couldn't correlate that speed with a normal technique at a different weight.

[gearpins]

lets keep it simple,
Assume 1/2.roh.s to be constant and lets give it a value of 1
now V sq is the only variable
Next lets try some values
15 sq=225
increase 15 to say 20 and we have
20sq=400
similarly
150 sq=22500
200sq=40000
so a small increase to v gives us a large increase to the figure on the right.
Hope that explains the puzzle.:)

[Keith.Williams.]

And even if we ignore the realitities of life referred to by OS and MFS the theory also states that stalling speed is proportional to the square root of the weight. So doubling the weight will multiply the stalling speed by a factor or 1.41.

If we assume that VR is affected by weight to the same degree, we can multiply your original VR figure of 110 by 1.41 we get 155.1. This is less than the quoted VR value of 110 + 67 = 177 at 400 000.

This makes your "only an extra 67 knots to produce double the lift" appear to be excessive, rather than surprising small.

[virgo]

Thank you chaps, especially those who expanded the sums to digestable chunks...........I was going backwards and squaring the speed to double the lift (requiring 12,100 knots to get airborne at 400,000 kg - which I KNOW is absolutely not true cos I've been there - 400,000 kg not 12,100 knots !!!)
Obviously the birds haven't worked it out either 'cos I don't think there's a bird that can double its weight and still fly ?

[Pontius's Copilot]

Tum teh tum ... it's here somewhere ... ummmmm ... Yep, got it - a can of worms! Now then, can opener, and here we go!

Let's have some considered opinions from the TechLog pundits on this; and just in case you think it's written by cranks - the authors are David Anderson, Fermi National Accelerator Laboratory, and Scott Eberhardt, formerly of the Department of Aeronautics and Astronautics, University of Washington, now at the Boeing Company.

http://www.allstar.fiu.edu/AERO/airflylvl3.htm

[balsa model]

Not all birds are made equal. Notable examples are:
• A bald eagle was seen carrying a mule deer fawn weighing 6.8 kg (15 lb.) (Martin, 1987).
Bald eagle female weigh approximately 5.8 kg (12.8 lb), males weigh 4.1 kg (9 lb).
• A harpy eagle of South America was recorded carrying a sloth weighing 5.9 kg (13 lb.) (Martin, 1987).
Female Harpy Eagles typically weigh about 7.5 kg (16.5 lb). Males are smaller. Exceptional harpy eagle females have weighed over 9 kg (20 lb).
Also, some birds (I forgot the names) can eat as much as they weigh in one day.
(Ref: Martin, Brian P. World Birds. Enfield, England: Guinness Books, 1987.)

[Pontius's Copilot]

"Also, some birds (I forgot the names) can eat as much as they weight in one day."

You're telling me ... I've met some very skinny birds who can really pack it away, especially if you're paying!

[GOLF_BRAVO_ZULU]

Does nobody use unstick speed anymore? or is my aviation stuck in the last Century?

[chemical alli]

thats when you forget one of your socks leaving in the dark from the first time cc,s hotel room.

[Forkandles]

Not all birds are made equal. Notable examples are:

• A bald eagle was seen carrying a mule deer fawn weighing 6.8 kg (15 lb.) (Martin, 1987).
Bald eagle female weigh approximately 5.8 kg (12.8 lb), males weigh 4.1 kg (9 lb).

• A harpy eagle of South America was recorded carrying a sloth weighing 5.9 kg (13 lb.) (Martin, 1987).
Female Harpy Eagles typically weigh about 7.5 kg (16.5 lb). Males are smaller. Exceptional harpy eagle females have weighed over 9 kg (20 lb).

Also, some birds (I forgot the names) can eat as much as they weight in one day.

(Ref: Martin, Brian P. World Birds. Enfield, England: Guinness Books, 1987.)


An African Swallow, was rumoured to have carried a coconut to England. Definitely not a European Swallow though. :ok:

[balsa model]

Just to, somehow, tie this bird business up with the original post, the takeoffs must have been all vertical (the claws being busy hanging onto prey), so V1, V2, V3, and Vunstick all computed to 0 and fit perfectly in with square and cubic approximation formulas. :)

My reference didn't specify any rate of climb but I would love to see the video.

[sycamore]

Virgo, you should be able to get the basic `clean` stalling speeds for the two weight conditions,as that is the `basic` point for L= W., from your POH.
To relate that to T/O,you must be in the same configuration,vis-a-vis flap/slat settings, otherwise your Cl will be different; nearest approximation for T/O would be the Vmus( minimum unstick speed- rotate early ,scrape the wear block on the tail, until you get airborne), which the FAA/CAA all require for certification..... ; or have I now dug a bit of a hole...............?..Syc..

[virgo]

Thank you gentlemen..............the story develops ?

GBZ, if me memory serves me right, the term "unstick" is only used for tail-wheel aircraft ?

Sycamore, The assumption is the same everything except weight

AAN Having got airborne (lift exceeding weight) I seem to remember that acceleration - and therefore ROC - is a function of thrust compared to drag which will obviously be a much greater margin on the lighter aircraft since drag is a combination of lift (induced drag) and speed (profile drag)

Birds.........I'm sure I saw an Attenborough programme showing Vultures that had eaten so much that their weight was above RTOW and they had to wait for a reduction of OAT in the evening to get airborne. (Having operated in and out of Africa many years ago, I sympathise with their problem !)

[balsa model]

Hm.. Why would you eat so much and why particularily in Africa?

As to the ROC, it varies with the ratio of excess power to weight.

Even if we assume that the climbout speed is pretty much the same for all weights, reducing the question to "what is the excess thrust?", we can't say much more since we don't know whether you've got an Eazy Rocket or a motorglider type of a powerplant. At least I can't simplify it further..