Isn't that a ballistic parabolic curve?

As for the lift thing.
Yes, my Emeraude weighs 600kg, lift is produced at 40 knots.
That is the answer, not the equation that Claude Piel, bless his cotton socks, used to get that answer!!

As for the K8 I thought it had the LD of a pair of pliers, and penetration a bit less than me at 77years.

Oh, good - a physics problem.

Let's get rid of all the ideas that 'the momentum' of an aircraft somehow reduces its susceptibility to turbulence. Momentum has no effect on how an aircraft moves given an external force. Mass, on the other hand does (ie inertia). But that's not the same as momentum.

To first order:

1. Take the equation for lift (proportional to v^2, wing area, CL). Again to first order, CL is proportional to angle of attack.
2. In level flight lift = mass * gravity
3. Now consider what happens when your aircraft moves from air that's not going up or down into air that is (eg a thermal). Effectively, the angle of attack changes, and so the lift from the wing changes, accelerating the aircraft vertically. That's your turbulence.

Solve the equations above, and you find that the extra 'g' experienced by the aircraft is proportional to velocity and wing surface area, and inversely proportional to mass - or if you like proportional to velocity and inversely proportional to wing loading.

That all makes intuitive sense. For otherwise identical aircraft - if you fly faster you get more 'g' from turbulence (hence manoeuvring speeds), and if you add weight you get less (at higher speeds, the angle of attack is less, by a factor of V^2; the change in angle of attack due to the thermal is less too, but only by a factor of V). Likewise for aircraft of equivalent weight and speed, the one with the bigger wing is effected more, since it starts off with a lower angle of attack.

Of course, in the real world, the linear assumption about lift and angle of attack doesn't quite hold, but I think it's good enough for this exercise.

As a sanity check, putting a few numbers in the formulae above, suggests that a B747 with a wing area of 525 m^2, max weight of 400,000 kg, flying at 200 kts (100 m/s), has a vertical acceleration due to a vertical thermal about a fifth of my glider (wing area 11.6 m^2, weight 500kg, speed 30 m/s).

Paul

Thank you Paul.
If I had a degree in aerodynamics I'm quite sure that would have made perfect sense.
Unfortunately I don't. I work by numbers, therefore 600 kg times the coefficient of lift equals 600kgcl. Which means nothing of any sense.
If the coefficient of lift had a number, it would help.
600 times 3 = 1800. That's easy.
Next!

The co-efficient of lift is dimensionless.

Lift = 1/2 * density of air (rho) * velocity squared * wing area * Coefficient of lift.

Lift is in Newtons (a force, which has units kg*m/s^2)

density is kg/m^3
velocity squared is (m/s)^2
area is m^2


Multiply those last three together and you get (kg/m^3)*[(m/s)^2]*m^2
or kg*m/s^2 - the same units as lift, so the CL is dimensionless.

If you look up graphs of CL vs angle of attack you find that it's varies with the wing design, but is typically around 1.0 at typical angles of attack (5-10 degrees).

See https://commons.wikimedia.org/wiki/File:LiftCurve.gif

https://commons.wikimedia.org/wiki/File:LiftCurve.gif


Paul

I think you've got to be careful about generalising about big v small...one of the smallest fighters back in the day was the F104 but it generally had a ride that was smooth as glass at low level on a windy day vs. even slightly bigger stuff with slightly lower wing loading.

Thank you Paul.
That makes sense.