A/C Speed/ "Apparent Weight"

[strake]

A very idle question that probably has little relevance but, I'm hot and bored....

If, hypothetically, one were able to use the runway as a weighing device, is it fair to assume that the "apparent" weight of the aircraft will decrease in direct proportion to the speed/lift. In other words, if a fully laden 747-400 weighs X tonnes on commencement of roll, will it weigh X/2 tonnes (apparent) at exactly half rotate speed and so on?
If so, is the reverse true for landing?

[G-ALAN]

yes the apparent weight of the aircraft would decrease but not by X/2 at half the rotate speed that you suggest. This is because even at rotate speed you still have some weight bearing down on the runway until you rotate which in turn increases the angle of attack of the wing which increases lift and lifts a/c into the air. On landing remember on airliners the spoilers come up and get rid of most of the lift over the wing, planting the a/c firmly on the ground. On light aircraft with no spoilers it can be a problem braking in slippery conditions because the aicraft is quite light after landing as you rightly suggest. Hope this has helped

[Genghis the Engineer]

Yes, the weight on wheels will decrease by the amount of lift.

The amount of lift is hard to estimate, since it's a function of airspeed and angle of attack, modified by the fact that the wing is functioning in ground effect. A back of envelope sum says that it'll reduce by something around the weight multiplied by the square root of (speed/stalling speed), but that value could be as much as 50% out and I wouldn't hang my hat on it.

G

[Keith.Williams.]

The situation described in this question is probably too complex to come up with any reasonably simply and accurate reply.

If however we simplify the question into "How is lift related to airspeed if angle of attack is constant?" then the answer is a bit easier.

The lift equations is L = Cl 1/2 Rho Vsquared, where V is the TAS.

If angle of attack is constant then Cl is constant. And if altitude, temperature and humidity are constant then so is Rho (if we ignore compressibility).

Taking the (now constant Cl, and 1/2Rho) out of the equation gives:

Lift is proportional to TAS squared.

So if we double the TAS we get 4 times the lift. Or to put it another way if lift equals weight at a given speed, then lift will be 1/4 of the weight at 1/2 of that speed.

The above argument is of course a simplification in that it ignores the complications of such things as upwash, downwash and effective angle of attack.

[bookworm]

The amount of lift is hard to estimate, since it's a function of airspeed and angle of attack, modified by the fact that the wing is functioning in ground effect. A back of envelope sum says that it'll reduce by something around the weight multiplied by the square root of (speed/stalling speed), but that value could be as much as 50% out and I wouldn't hang my hat on it.

That estimate would be based on the assumption that the AOA with the aircraft rolling on all three wheels is equal to the critical (stalling) AOA. That would be an extreme case and would make the aircraft very difficult to land on the mains. As G-ALAN suggests, because the aircraft has to rotate at Vr, the AOA while rolling is less than the critical AOA.

Some aircraft (and I'm ignoring "conventional" undercarriage here!) sit at a higher AOA than others. The Twin Comanche was designed to allow the passengers to enplane without the use of a step, hence it has a very nose-up resting attitude. In my experience, this makes it somewhat trying in crosswinds, as you have very little weight left on the wheels late in the take-off roll.

Others -- and it tends to be the high wing ones, though I'm not sure I know why -- have a very low lift coefficient while rolling along the runway, and have to be hauled fairly vigorously into the air.

[Genghis the Engineer]

You are of-course correct, except that available lift will increase significantly in ground effect. As I said, it's a gross approximation.

G