3
$\begingroup$

Let's say I am flying an airplane that has a weight of 100,000 kg with a 1G load factor at normal steady flight. I have helium lift assist that is providing 50,000 kg (50%) worth of buoyant lift force upwards. Therefore, my aerodynamic lift has to provide 50,000 kg (50%) of upwards lift to remain in flight.

If I go into, let's say, a 60 degree bank maneuver and induce a 2G load factor to the airplane (so now the perceived weight goes up to 200,000 kg), does the buoyant lift force scale in some manner with the load factor, or is the lift force completely static and continue to provide 50,000 kg worth of positive lift regardless of the induced load factor? In other words, does the aerodynamic lift have to provide 150,000 kg of positive lifting force during a 2G maneuver to keep the plane airborne?

| improve this question | |
$\endgroup$
  • 2
    $\begingroup$ As an aside to this hypothetical question, 50,000kg of helium buoyancy is a Zeppelin Graf sized envelope which is unlikely to ever achieve a 60 degree/2g bank. $\endgroup$ – Pilothead Sep 7 '18 at 4:20
  • 1
    $\begingroup$ Pilothead, as human beings, we are vulnerable to confusing the unprecedented with the improbable. $\endgroup$ – Jason Oct 29 '18 at 22:08
4
$\begingroup$

It would probably be easier to think of it as a 50,000kg aircraft with a "normal" wing. This untruth will make calculating accelerations and turn rate incorrect (because the actual mass will make it more sluggish). But the wing at all times has to provide 50,000kg of vertical lift for the craft to not accelerate downward.

Then just like a regular plane, the wing has to double the (aerodynamic) lift at 60 degrees. This allows it to have 50,000kg vertically and the craft doesn't descend.

If I go into, lets say, a 60 degree bank maneuver and induce a 2G load factor to the airplane

I don't prefer this language. The load factor is the result of the lift. If you had a wing that couldn't provide increased lift (it was already at the point of stalling), then banking wouldn't increase the load factor. It would either stall the wing or the plane would start accelerating downward. Saying the load factor has increased suggests to me that the lift has already been accomplished.

does the buoyant lift force scale in some manner with the load factor, or is the lift force completely static and continue to provide 50,000 kg worth of positive lift regardless of the induced load factor?

The buoyant force is constant and always vertical. Instead of coming from the relative wind, it comes from the vertical pressure gradient in the atmosphere. So in the simple analysis, it isn't affected by banking. But since the gravitational pull on the airplane is also constant and always vertical, you can just combine them to understand what the aerodynamic wing is required to do.

| improve this answer | |
$\endgroup$
2
$\begingroup$

Aerostatic lift is always vertical, and doesn't contribute centripetal force in a turn. If you're in a 2G 60 degee turn, the total force in the vertical direction is equal to the aerostatic buoyancy plus the total aerodynamic lift (pointed 60 degrees from vertical) times cos(60 degrees). The centripetal force is just aerodynamic and is equal to the total aerodynamic lift times sin(60 degrees). To be in a 60 degree 2G turn you would have to be accelerating aggressively upward--for example, turning at 60 degrees while pulling up through the bottom of a dive.

To determine the bank angle required to maintain a 2G level turn with half of the aircraft's weight offset by buyancy, we have two equations and two unknowns:

Unkowns:

theta # Bank angle

lift # Expressed as acceleration, pointed theta radians from vertical

Equations:

lift*cos(theta) + 0.5G of aerostatic buoyancy = 1G # Required for a level turn

2G = sqrt((lift*cos(theta) + 0.5G)^2 + (lift*sin(theta))^2) # Setting the total acceleration equal to 2G.

Now we solve:

=> lift*cos(theta) = 0.5G

=> (lift*cos(theta) + 0.5G)^2 = 1G

=> 2G = sqrt(1G + (lift*sin(theta))^2)

=> 3G = lift^2*sin^2(theta)

=> 0.5G = 0.289*lift*sin(theta)

=> lift*cos(theta) = 0.289*lift*sin(theta)

=> cos(theta) = 0.289*sin(theta)

=> theta = atan(0.289)

=> theta = 1.29

=> lift*cos(theta) = 0.5G

=> lift*cos(1.29) = 0.5G

=> lift = 1.803G

1.29 radians is the same as as a 74 degree bank angle.

At 60 degrees in a level turn...

0.5G = lift*cos(60 degrees)

=> lift = 1G

Total G load = sqrt((1G*cos(60 degrees) + 0.5G)^2 + (1G*sin(60 degrees))^2)

Total G load = 1.32G

At 60 degrees and 2G total acceleration...

2G = sqrt((lift*cos(60 degrees) + 0.5G)^2 + (lift*sin(60 degrees))^2)

=> 4G = (lift*cos(60 degrees) + 0.5G)^2 + (lift*sin(60 degrees))^2

=> lift = 1.703G

Upward acceleration = 1.703*cos(60 degrees) + 0.5G

Upward acceleration = 2.203G

Net upward acceleration = 1.203G.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 60 degree turn is a standard training maneuver and is accomplished by rolling the plane into a 60 degree bank (you now have 0.5 g vertical lift) $\endgroup$ – Robert DiGiovanni Sep 9 '18 at 9:31
  • $\begingroup$ Pulling the elevator to 2g now gives you 1 g of lift. On the real world, one can also feed in some rudder, pointing the nose skyward. That, along with the upward pointing prop, also helps vertical lift. The elevator is now also turning the plane through the circle. One can control altitude by adding power, or rolling to a steeper or shallower bank. $\endgroup$ – Robert DiGiovanni Sep 9 '18 at 9:42
  • $\begingroup$ Critical here is sufficient airspeed to avoid accelerated stall but not to exceed the structural limits of the aircraft. $\endgroup$ – Robert DiGiovanni Sep 9 '18 at 9:47
  • $\begingroup$ @RobertDiGiovanni thanks for the comments. The aerodynamic "lift" in my answer is actually the net aerodynamic force, so perhaps it would have been clearer to call it that. In other words, it is the combination of drag, thrust, and aerodynamic forces from the wings, fuselage, and empennage. What do you think could be improved in my answer? $\endgroup$ – Alec Martin Sep 10 '18 at 21:23
  • 1
    $\begingroup$ Certainly nothing wrong with fine tuning a response. Especially after reading and learning from others on this fine forum. After further consideration, I would divide my answer into actual lift required in the light and heavy case, and also the "percieved" weight which may translate into structural loading. $\endgroup$ – Robert DiGiovanni Sep 11 '18 at 8:20
1
$\begingroup$

You may have answered your own question by saying "percieved" weight. Note that a 2G turn does not increase the weight of the plane there for does not increase the amount of VERTICAL lift required to hold it in the air. 2G is what you perceive or feel in the plane. The amount of vertical lift required will always be the weight of the aircraft minus the lifting gas contribution. The remainder must be provided by the wing/engine.

So what must be determined is whether or not your plane is capable of producing enough total lift to generate 50,000 kg of vertical lift at a 60 degree bank. This would be however close to your max Clift you dared to venture before stalling. Another solution would be to use more helium, or reduce the angle of your turn. For recreational flying, trying to hold altitude in a sustained 60 degree turn may not be worth pursuing for long. You may be spilling fuel from your wings.

If you still want to do it, remove the helium and see if the wings will lift 100,000 kg on their own in level flight. At 60 degree bank, if my math is correct, will give 50,000 kg vertical lift.

Bob

As a post script, this question is applicable to comparing weight vs stall speed. The "light" plane needs less AOA than the "heavy" at a given speed. There for, the heavy must either pull more AOA to hold altitude (risking stall) or increase speed (limited by safe maneuvering speed).

What this translates into is the "light" would have less drag, which would save fuel on a long trip. One might think of filling a Super Guppy with helium to find out!

| improve this answer | |
$\endgroup$
1
$\begingroup$

Short answer is “no”. The buoyant force is due to a completely different principle, and is always the same amount, pointing up. No scaling.

By the way, for your partially buoyant aircraft, if you wanted to do a 2G level turn, you would need more than 60 degree AOB.

A lot of times, if you think things through to an extreme, it will help you understand the in-between cases.

Imagine, your airplane has so much helium that it is neutrally buoyant (or is flying in a fluid that is dense enough...think submarine). Then what angle of bank would you need for a level turn? You’d need a 90 degree AOB for any level turn.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.