Lift Coefficient Questions

[neorich2002]

Good afternoon, I wonder if you could help me please, im trying to find some typical numbers for the following.

1.) By how much (i.e. 30% more etc) does the lift on takeoff generally exceed the weight of the aircraft. Also, when defining the maximum take off weight for an aircraft what is the regulation about providing more thrust than weight (i.e. the mtow is the weight at which the lift generated is a minimum of x% more than mtow, or something to that effect).

2.) What is the value of the coefficient of lift for say a 747-400, firstly with full flaps, then in the cruise (i.e. no flaps).

3.) Taking a 747-400 (please feel free to comment on other aircraft types also) if it lost all 4 engines, what would be its optimum rate of descent to acheive greatest distance of glide, obviously taking account of the speed required to land and not to stall.

Thanks for all your assistance, look forward to receiving some replies soon.

Regards

neorich2002

[barit1]

Originally Posted by neorich2002

Good afternoon, I wonder if you could help me please, im trying to find some typical numbers for the following.
1.) By how much (i.e. 30% more etc) does the lift on takeoff generally exceed the weight of the aircraft. Also, when defining the maximum take off weight for an aircraft what is the regulation about providing more thrust than weight (i.e. the mtow is the weight at which the lift generated is a minimum of x% more than mtow, or something to that effect)...

This is a simple Newton's Third Law problem in the vertical axis. If the plane accelerates from zero ROC (on the runway) to 30 fps ROC (= 1800 fpm) in, say, 10 seconds, the vertical acceleration averages 3 ft/sec/sec.

Then F(lift for vertical acceleration) = M * a. Since the lift required to carry the ship straight & level is M * 32.2 ft/sec/sec, the added lift for acceleration is about 10% of the baseline S&L value.

So the answer is - it all depends on how fast you achieve the rate of climb.

[neorich2002]

Thankyou for your reply, and thanks for putting in the numbers (which was the problem to begin with). What would be the maximum climb rate achievable on a typical commercial airliner, and the minimum time to achieve it?

Thanks Again

Regards

neorich2002

[barit1]

Again, it all depends. I suspect if you work one case (esp. some TOGW, low temperature, near SL), that will be representative of the max. lift for the wing. At any other TOGW, taking that max. lift as constant, the vertical acceleration will be inversely proportional to weight. (a = F / M, right?)

Maybe someone can give us representative numbers.

[neorich2002]

Yes.

Representative numbers would be very helpful though, if anyone could oblige.

Thanks

Regards

neorich2002

[gearpins]

ok for A320.....
TOW...77 tons
ROC......3000f/m...at sea level temp 30dgs (ISA+15)
with all engines gone...best L/Dspeed is a function of wt(wt intonsx2+80)
so TOW=65t best speed=65x2+80=210kts

[neorich2002]

Thanks barit1 and gearpins, so 210kts would be the optimum speed to acheive longest glide distance in the situation you described?

Just to rephrase the end of my question 1:

In determining what the maximum take off weight of an aircraft is, presumably a calculation of maximum lift is done for the aircraft, and the maximum take off weight is then a certain percentage (less than 100%) of that figure for maximum lift, correct?. Anyone have any details about the calculation of mtow?

Thanks

Regards

neorich2002

[chornedsnorkack]

Maximum thrust-to-weight is usually less than 100% on most aircraft... a few fighters are exceptions in that they can actually hover and climb straight up. Most planes are unable to hover and fly and take off by lift.

Typical thrust/weight ratios of transport airplanes at MTOW seem to be in the range of 25 %...35%. Lift is of course over 100 %.

Most transport planes are actually unable to reach the maximum (stalling) lift coefficient on ground, because they would then have a tailstrike. The maximum lift coefficient is thus limited by the tail clearance needed for safety.

I think that when the MTOW is limited by air density (hot and/or high conditions), the certification requirement is for a certain climb slope to be achieved with one engine inoperative. This slope is something like 2,4 % for 1 out of 2 engines, 2,7 % for 2 out of 3 engines, 3,0 % for 3 out of 4 engines. So presumably the thrust/weight ratio has to be the slope required + the drag.

[Wizofoz]

In determining what the maximum take off weight of an aircraft is, presumably a calculation of maximum lift is done for the aircraft, and the maximum take off weight is then a certain percentage (less than 100%) of that figure for maximum lift, correct?.

Yes and no. I think you are under the impression that MTOW is a function of the achievable takeoff performance. The performance limited MTOW is, but that is not what you are asking (I don't think)

When talking about MTOW, we usually mean maximum structural take off weight. That is, the maximum weight the airframe is certified to operate at without breaking. The limiting factor for this will indeed often be the maximum allowed load on the wing.

Transport aircraft are certified to +2/-0G in take off and landing configuration, with a 150% saftey margin. To my knowledge, th highest certified weight for a 737 is 75 000Kg (the -700ER), so it's wing is ceritfied to produce 150 000KG of lift (Lf= L/M). It never does in normal operation as pulling 2g on takeoff would land you in the CPs office very quickley!!

Performance limited MTOW may then be more limiting for a particualr take-off (Short runway, hot-high conditions,obstacle clearence etc.),but this is calculated from factored flight test data. Whilst a component of this will be how fast and to what rate the aircraft achieves climb (which, as you state is a function of how much lift is achieved), it is not quantified as a certification figure. You get what you get and measure the results.

Hope this helps

[bekolblockage]

Apart from the initial acceleration, once the aircraft is established in a steady climb, lift is actually less than weight. It is the vertical component of thrust that balances the forces.

[Mad (Flt) Scientist]

Originally Posted by neorich2002

1.) By how much (i.e. 30% more etc) does the lift on takeoff generally exceed the weight of the aircraft. Also, when defining the maximum take off weight for an aircraft what is the regulation about providing more thrust than weight (i.e. the mtow is the weight at which the lift generated is a minimum of x% more than mtow, or something to that effect).

There's no direct regulation in Part 25 (or equivalent) that sets a minimum % of lift at takeoff.

Things that may provide hints:

Vmu could be taken as the equivalent of stall-speed for an on-ground/in-ground-effect aircraft, since it's the minimum speed you could get airborne at. FAR25.107 states (at amdt 42):

25.107 (e) (1) VR may not be less than--
.....
(iv) A speed that, if the airplane is rotated at its maximum practicable rate, will result in a VLOF of not less than 110 percent of VMU in the allengines-operating condition and not less than 105 percent of VMU determined at the thrust-to-weight ratio corresponding to the one-engine-inoperative condition.

So min Vlof=1.05Vmu (OEI) and so CL@lo=CLmu/(1.05*1.05)

So the answer is that CL at liftoff must have a margin of at least 11% below CLmu.

There's no regulation about minimum thrust, other than that the aircraft must meet the various climb gradients, which imply minimum thrust capabilities dependent upon drag, etc.

[neorich2002]

Thanks guys, very helpfull.

chornedsnorkack:

Most planes are unable to hover and fly and take off by lift.

Do you mean that most planes cannot takeoff solely on the lift generated by the wings, and therefore require some vertical component of thrust also? If so, approximately how much of the lift required for takeoff is actually generated by the wings?

Wizofoz:

The performance limited MTOW is, but that is not what you are asking (I don't think)

When talking about MTOW, we usually mean maximum structural take off weight. That is, the maximum weight the airframe is certified to operate at without breaking. The limiting factor for this will indeed often be the maximum allowed load on the wing.

To be honest i was asking about performance limited mtow, although i didnt know it. I was under the illusion that mtow was defined in terms of maximum lift, and conditions etc.

Am i right to assume that the structural limits of an aircraft far exceed those that the aircraft is capable of producing on takeoff? i.e. you said transport aircraft are certified to +2g, but on takeoff the g's on the wing will be nowhere near 2g right?

So with regard to structural mtow, is this a useful concept for pilots day to day, it would seem not? Since although the wings could take 2g's, it may be not be possible to load them with that much anyway, and hence impossible to generate that much lift, so a plane loaded to structural mtow, may not even be able to take off right?

It would seem like a more useful one would be performance limited mtow (is there a term for this) since the pilot can then deduce from the weight of the aircraft whether it can actually take off, is this correct?

bekolblockage:

It is the vertical component of thrust that balances the forces.

Yes, how about during cruise. I am aware that some aircraft fly with the nose at a few degrees above horizon, but i always thought this was done purely because it was the optimum angle to reduce drag, is this also partly to create a vertical component of thrust, or is a vertical component not required at all in the cruise?

Thanks again.

Regards

neorich2002

[Mad (Flt) Scientist]

chornedsnorkack:

Quote:
Most planes are unable to hover and fly and take off by lift.

There's a comma missing:

Most planes are unable to hover [on thrust alone], and fly and take off by lift.

In other words, the vast majority of the force generated to oppose the weight of an aircraft is aerodynamic lift; thrust generally provides only a small amount of the "anti-gravity".

If you took a typical airliner thrust-weight ratio of 25-35% at max power, and considered a relatively nose-high attitude of 15 degrees, the component of force in the vertical direction due to thrust is sin(15 deg)*(25-35% of tyhe weight) - i.e. about 6-8% of the aircraft weight. In cruise, where the pitch attitude is close to zero, virtually ALL the weight is carried by lift, and all the engines do is overcome drag.

If you assumed for a general case that about 95% of the weight was lift, you'd be fairly close, especially given other factors (like uncertainty of the weight, and trim lift)

Am i right to assume that the structural limits of an aircraft far exceed those that the aircraft is capable of producing on takeoff? i.e. you said transport aircraft are certified to +2g, but on takeoff the g's on the wing will be nowhere near 2g right?

So with regard to structural mtow, is this a useful concept for pilots day to day, it would seem not? Since although the wings could take 2g's, it may be not be possible to load them with that much anyway, and hence impossible to generate that much lift, so a plane loaded to structural mtow, may not even be able to take off right?

It would seem like a more useful one would be performance limited mtow (is there a term for this) since the pilot can then deduce from the weight of the aircraft whether it can actually take off, is this correct?

At a normal takeoff speed (once the plane is safely airborne) it will be flying about 25% faster than stall speed, and will be capable of generating about 1.5'g', so it's true to say that you can't structurally break a plane on takeoff with lift alone. However, the maximum permitted speed in a takeoff configuration (Vfe) will be about 75-100% higer than takeoff speed - say twice the stall speed. Which means that you could in theory generate 4'g' if you went as fast as permitted then pulled as hard as you could. That WILL break something, I'd expect.

MTOW for structural purposes has to cover ALL design manouvres - not just during takeoff, cruise and landing normal operations, but various extreme cases. The extra margin you get in normal operations provides additional safety margin and allows for a decent fatigue life. And, yes, a plane at its structurally certified MTOW can takeoff, certainly on a 'normal' day - otherwise there'd be no point in designing the structure to take the relevant loads!

For simplicity, MTOW (= 'structural MTOW') is the usually quoted value when doing simple comparisons because it's (generally) a single value and is indicative of aircraft performance limits (since most designs will balance out strure vs performance limits to optimise the design). But "MTOW as limited by [brake energy/climb/runway length/whatever]" - there are various performance limits on takeoff weight - is necessarily a complex beast, usually presented on charts in the flight manual as a function of things like temperature, altitude, runway slope, configuration of the aircraft, etc., so it's VERY difficult to choose a single case to take as a datum.

These charts are provided in order that the pilot doesn't have to deduce whether he can takeoff safely - he looks up the conditions of the day, and it'll tell him whether his current weight is ok or not.

[Wizofoz]

So with regard to structural mtow, is this a useful concept for pilots day to day, it would seem not?

Mad Sci covered it but, no, Structural MTOW is most often the limiting factor for takeoff. Only on shorter runways, high climb-out obstacles or especially hot/high conditions will the Performance limit be less than the structural limit.

[chornedsnorkack]

Originally Posted by Mad (Flt) Scientist

At a normal takeoff speed (once the plane is safely airborne) it will be flying about 25% faster than stall speed, and will be capable of generating about 1.5'g', so it's true to say that you can't structurally break a plane on takeoff with lift alone. However, the maximum permitted speed in a takeoff configuration (Vfe) will be about 75-100% higer than takeoff speed - say twice the stall speed. Which means that you could in theory generate 4'g' if you went as fast as permitted then pulled as hard as you could. That WILL break something, I'd expect.

And in theory, you could create 8 G or something if the OEW is, say, 50 % of MTOW and the plane is taking off light.

So, for planes that rotate on takeoff (many donīt because the wing is at a high AoA on takeoff run, either because the tail is low on a short tailwheel or less freqently because although the fuselage is level the wing is at a high angle of incidence):

The maximum allowed speed in takeoff configuration, Vfe, is around, as you say, say, 200 % of stall speed. However, can the plane actually run on ground at Vfe, or are there other speed limits, like tire speed limits?
The normal speed when rotation is attempted is then 125 % of stall speed?
The minimum speed when it is allowed to attempt rotation is 110 % of stall speed, if the plane can stall on ground - usually cannot, then the minimum allowed speed to attempt rotation is accordingly higher as determined by tailstrike clearance?

Airplanes in takeoff configuration are to be able to take +2 g at MTOW... presumably more at less than MTOW. They have to be capable of 1,21 g or more at minimum allowed Vlof... and would be capable of 1,5 g at normal Vlof?

What are the typical g-s actually achieved at a normal liftoff - what would constitute a too abrupt takeoff and what would be regarded as too slow and floating liftoff?

Originally Posted by Mad (Flt) Scientist

MTOW for structural purposes has to cover ALL design manouvres - not just during takeoff, cruise and landing normal operations, but various extreme cases. The extra margin you get in normal operations provides additional safety margin and allows for a decent fatigue life. And, yes, a plane at its structurally certified MTOW can takeoff, certainly on a 'normal' day - otherwise there'd be no point in designing the structure to take the relevant loads!

For simplicity, MTOW (= 'structural MTOW') is the usually quoted value when doing simple comparisons because it's (generally) a single value and is indicative of aircraft performance limits (since most designs will balance out strure vs performance limits to optimise the design).

How is MTOW measured?

While it is usually not allowed, there are plenty of planes alleged to have flown overloaded... And I said usually... Wasnīt it the case that in Alaska, planes are officially permitted to fly overloaded?

[Wizofoz]

So, for planes that rotate on takeoff (many donīt

Even most tailwheel aircraft "Rotate" as the normal takeoff procedure is to lift the tail into a level attitude early in the takeoff run.

However, can the plane actually run on ground at Vfe, or are there other speed limits, like tire speed limits?

you may indeed hit tyre speed first (175kt in my aircraft) but this is a purerly hypothetical question. There is no reason to even contemplate being on the ground at anything close to Vfe. The absolute highest Vr s I ever see (Max weight, improved climb on a long runway) are aroung 158kts.

What are the typical g-s actually achieved at a normal liftoff

At a guess, no more the 1.1G. We aren't fighter pilots. Normal rotation rate is three degrees per second to a nominal target attitude of 15deg, so five seconds of rotation.

How is MTOW measured?

While it is usually not allowed, there are plenty of planes alleged to have flown overloaded... And I said usually... Wasnīt it the case that in Alaska, planes are officially permitted to fly overloaded?

Aircraft are wieghed periodically to determine an empty weight. What ever else you are carrying is added to this (fuel, crew, cargo, Pax) and the result is a Take Off Weight.

There are large saftey margins built into all the calculations, so you can (illegally) exceed MTOW and get away with it "Most" of the time, but you are reducing your saftey margins. There are some circumstances where approval for a flight at higher than usual weight (I know ferry flights are allowed 1.25 of the usual weight) will be given on the basis that the higher risk is accepted and it is a rare occurence. Routine operation at more that MTOW will increase airframe fatique and reduce it's life.

There may be some places (I hadn't heard itm was so in Alaska but it may be) where remote area operation leads to an approval with all of the above taken into account.

[neorich2002]

Thankyou all for your contributions, very helpfull indeed.

Just one more thing with regard to my question 2. Does anyone have a typical value for the coefficient of lift, Cl, which I could use in future when using the lift equation.

Perhaps the value for a 747-400, firstly with 40deg flaps, and secondly in cruise (0deg flaps).

Thanks

Regards

neorich2002

[Wizofoz]

What you have been missing here is one simple thing- Lift=Weight in level, unaccelerated flight. As such there is no one value for Cl in a particular configuration.

If you consider that L=M, and L= Cl*1/2Rho*V^2*A (Where A is wing area), plug ther variables in for a particular weight and speed, you get a value for Cl. Change any value, and you change Cl.

[neorich2002]

Yes, i know L=W in level flight. Therefore given the weight i could calculate Cl.

But say I was doing a calculation to figure out what velocity i would need to be at in order to take off where L is slightly greater than W, say 110% of W. Obviously I know Rho, and also assume i know A, and W, and hence i also know L. I am left with two unknowns in the equation, v and Cl, since I am calculating v, surely i need a value of Cl right?

Regards

neorich2002

[Wizofoz]

I do hear what you're saying, but Cl also varies with Angle of Attach, so there is still no definative value for it. The minimum speed at which an aircraft will theoretically be able to fly is that at which L=W at Clmax, which is to say stall speed (Vs). (The other limit on this may be physical-you may strike the tail before achieving that AofA). This is also refered to a Vmu (V minimun unstick)

I'm not sure what use you are intending to put these calculations to, but the fact is that performance figures are not arrived at in that way. Vs is determined in flight test, things like take-off speeds are calculated using factors of this data.

At no stage does a performance engineer say "The Cl is this or that, so I can achieve 110% of lift at this speed."

So in terms of what you are asking, drag the tail on the ground, and the aircraft will fly when it produces enough lift. This is what they actuall do in flight test!!

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