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hawker111

G Stall Line Numbers - do they change when an aircraft's weight changes?

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When an aircraft's weight increases, do the G stall line numbers increase, and when an aircraft's weight decreases, do the G stall line numbers decrease?

Here are the G stall line numbers for the F-105D Thunderchief, obtained from the V-G diagram contained in the aircraft's flight manual. These numbers are for a clean aircraft:

1G: 160
2G: 220
3G: 275
4G: 320
5G: 360
6G: 400
7G: 440
8G: 470
9G: 500

Do these numbers increase when the aircraft is heavier, and decrease when the aircraft is lighter?

For example, if the F-105D was really heavy, loaded down with external ordnance, would it still be able to achieve 7Gs at 440 KCAS, or would it have to be going faster in order to have the lift required to pull a 7G turn? Would it still be able to take off at 160 KCAS, or would it have to be going faster that that? Could it still pull 4Gs once it reached 320 KCAS, or would it need to be going faster?

Thanks,

hawker111

 

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Short answer: Yes.
If your weight goes up, your required lift goes up. That means (all else being equal) you'll have to go faster to achieve the same amount of lift/weight relationship.

Determining "g" is nothing but dividing your lift available by your current weight. If "lift = weight", we'll get 1g and so on.
"G available" is your lift available (F_L = q*cl*S) divided by your current weight (F_W = m*g).

That only works that easily "right side up" (for any other case, you'll have to figure out the components of your lift and weight verctors according to your cartesian system first), but it's a good starter for understanding what "g" actually is.

It's nothing but an acceleration based on unequal forces (lift and weight, thrust and drag) acting on your airplane.
"9g" means that the wings are providing enough lift to carry 9 times your current weight, or that your airplane accelerates at nine times the earth's free-fall acceleration (hence the term "g") into the lift-direction.
All is relative :smile:

_______
q = 0.5*air-density*v^2
cl = max available lift-coefficient at current Mach
S = effective wing-area

Edited by Toryu

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Toryu,

If I knew what the F-105D's exact weight is when it could pull Gs at these airspeeds, would I be able to enter in a formula in Excel to give me the new airspeeds as I change the weight?

1G: 160 KCAS
2G: 220 KCAS
3G: 275 KCAS
4G: 320 KCAS
5G: 360 KCAS
6G: 400 KCAS
7G: 440 KCAS
8G: 470 KCAS
9G: 500 KCAS

Thanks very much for your help,

hawker111
 

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Yes. Just figure out your CLmax for each airspeed (it varies with Mach, but assume it to be constant for a first approximation) and apply it to your Excel table.

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Sure. I'll do it in SI units, though.

1) We know that lift equals weight. Weight is mass times earth's gravitational acceleration. L = W = m*g
2) We know that L = 0.5*rho*v^2 * CL * S   ==>>  CL = L / (0.5*rho*v^2*S)

m = your mass in [kg]
g = earth's average gravitational acceleration, which is 9.81m/s^2
rho = air's density, which under ISA-conditions is approximately 1.225kg/m^3
v = airspeed in [m/s]
CL = what we're looking for, dimensionless
S = your reference wing-area in [m^2]

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