F6F Stall Speeds/Flaps

[dtango]

Check out dtangos data. The climb rates are all over the board. If there is any climb or decent rate the aircraft is not in a 1g state.

You can be in a climb or dive at 1g.  An unaccelerated steady climb is at 1g.

[jamdive]

I see your missunderstanding, but no, 1g is one gravity or a neutral state, where the lift component has the same force as gravity. Im at 1g just sitting in my chair. If you change any portion of the lift component your not at 1g anymore. A 0g state is anything in free fall. The same thing also applies to the thrust and drag components. Look at it this way. If your climbing at 1500 ft. per minute you are actually accelerating away from the force of gravity at 1500 ft. per minute.

So in your test, if you put the plane in a climb your actually increasing the stall speed do to the fact that your changing the airplanes 1g state. If your stall speed is 75mph and your plane is climbing at 4g's(for ease of this example lets say), your stall speed would increase to the square root of 4, which is 2 times your 1g stall speed of 75mph. The plane will stall at 150 mph.

In the case of a 1g stall in level flight, whats really going on is the slower the airspeed the more the wing has to change its angle of attack to create the lift nessesary to counter the gravitation force its opposing. Eventually the flow of air will separate from the airfoil and that is the stall. Your not stalling the plane, your stalling the wing. Dont confuse the angle of attack with the planes attitude (compared to the horizon). The planes attitude doesnt even factor. You can have a combination of the four vectors at various speeds and create the same effect. For example. Fly strait then turn the plane on its side. It can remain at that altitude but now your lift component is being maintained by the rudder and fusalage not the wings. The planes attitude is now perpendicular to the horizon, no effect on the 1g state. The plane is currently on its side flying level, now pull the stick back, now your changing the thrust and drag components but not lift and gravity. The airfoil doesnt know the difference and will stall the same as if it where horizontal.

Hope this helps.

[Baumer]

Not to sound overly contrite, but 1G is 1G no matter what your orientation is. If your accelerometer is aligned to measure in the direction of motion and it registers 1G and is not changing then you are not accelerating.

[jamdive]

Not to sound overly contrite, but 1G is 1G no matter what your orientation is. If your accelerometer is aligned to measure in the direction of motion and it registers 1G and is not changing then you are not accelerating.

Yep, you are absolutely correct. Im making more of this that I should be. Im trying to make the point that his climb rate is not constant in any of the tests that where done. Try the test while maintaining a 0 rate of climb for accuracy. If im flying level getting close to a stall, if I start to climb (start jerking the stick) I am changing the dynamic. I'm going to pull +1g's and that is going to affect the stall speed. The same is true if I let the plane decend. Im creating and accelerated stall not a normal stall.

Normal stall speeds are always determined from strait and level flight.
Just trying to help out.

[dtango]

Yep, you are absolutely correct. Im making more of this that I should be. Im trying to make the point that his climb rate is not constant in any of the tests that where done. Try the test while maintaining a 0 rate of climb for accuracy. If im flying level getting close to a stall, if I start to climb (start jerking the stick) I am changing the dynamic. I'm going to pull +1g's and that is going to affect the stall speed. The same is true if I let the plane decend. Im creating and accelerated stall not a normal stall.

Normal stall speeds are always determined from strait and level flight.
Just trying to help out.

#1 They weren't my tests.  They were tests done by Pyro.

#2 I'm not the one that's misunderstanding it.  It's you .  You can absolutely be at 1g in a climb or a dive.  Infact for normal steady climbs that's assumed.  Maybe someone else has the energy to explain it to help you work through the kink in your physics.  I can't think of a pithy way of helping you work through your thinking except to say that if you were at a load factor greater or less than 1g then your flight path would be curved and not straight.

#3 "Normal" stall speeds are determined a various number of ways, not just level flight.

Cheers!
Tango

[hitech]

#2 I'm not the one that's misunderstanding it.  It's you Smiley.  You can absolutely be at 1g in a climb or a dive.  Infact for normal steady climbs that's assumed.  Maybe someone else has the energy to explain it to help you work through the kink in your physics.  I can't think of a pithy way of helping you work through your thinking except to say that if you were at a load factor greater or less than 1g then your flight path would be curved and not straight.

Not sure what you thinking is today dtango, I'm surprised you seem to be  missing this one. I may be wrong but I believe load factor is lift / weight of plane. IF this is the case then in any non accelerated flight, I.E. same speed and not turning load factor is 1 only in level flight. Flying straight up or down it is 0, and more specifically it is the cosine of the climb or decent angle not accounting for any engine incidence or AOA's.


HiTech

[Stoney]

May be semantics, but load factor is a dimensionless "g" that isn't supposed to be used interchangeably with the gravitational "g", right?

[dtango]

Not sure what you thinking is today dtango, I'm surprised you seem to be  missing this one. I may be wrong but I believe load factor is lift / weight of plane. IF this is the case then in any non accelerated flight, I.E. same speed and not turning load factor is 1 only in level flight. Flying straight up or down it is 0, and more specifically it is the cosine of the climb or decent angle not accounting for any engine incidence or AOA's.


HiTech

Well, I'm full of surprises I suppose , sometimes wrong ones!  Yes you're right HT.  I mixed up load factor with the fact that in an unaccelerated climb or dive forces normal (perpendicular) to the flight path cancel each other out just like they do in level flight.  So my original response about 1g climbs & dives to jamdive was incorrect.

What I meant to say is that in an unaccelerated climb or dive there is no net acceleration normal to the aircrafts flight path which means lift = weight * cosine (climb_angle).  In other words the amount of lift equals the amount of weight normal to the flight path just like it is in level flight.  The stall tests performed in the F6F video in this thread appear to be in this climbing flight mode and thus because of the effect of climb angle gamma could explain why the stall speeds quoted are as low as they are.

Cheers,
Tango

[bozon]

May be semantics, but load factor is a dimensionless "g" that isn't supposed to be used interchangeably with the gravitational "g", right?

The dimensionless "G" (lift/weight) is usually used with a capital. Lower case "g" is usually used for the gravity acceleration.

p.s.
How does a G meter on a plane actually work? I guess it measures acceleration in a given axis set to be aligned with the lift.

In that case, when pointing at an angle "a" above the horizon, the acceleration in the measured axis is:
L/m-g*cos(a)
and the G load it will show is:
G=L/(m*g)+(1-cos(a))
Since in a stead climb L=mg*cos(a):
G=cos(a)+(1-cos(a)) = 1.

So the G meter in that case will show 1.

[Stoney]

This is another excerpt that may or may not have any bearing:

Deceleration rate has a pronounced affect on lift coefficient. Changes to the flow pattern within 25 chord lengths of an airfoil have been shown to produce significant nonsteady flow effects. The lift producing flow around the airfoil (vorticity) does not change instantaneously. During rapid decelerations the wing continues to produce lift for some finite time after the airspeed has decreased below the steady state stall speed. The measured stall speed for these conditions is lower than the steady state stall speed. For this reason, a deceleration rate not to exceed 1/2 kn/s normally is specified when determining steady state stall speed for performance guarantees.

Could be that during the stall speed testing, this wasn't being accounted for?

[hitech]

The dimensionless "G" (lift/weight) is usually used with a capital. Lower case "g" is usually used for the gravity acceleration.

p.s.
How does a G meter on a plane actually work? I guess it measures acceleration in a given axis set to be aligned with the lift.

In that case, when pointing at an angle "a" above the horizon, the acceleration in the measured axis is:
L/m-g*cos(a)

and the G load it will show is:

G=L/(m*g)+(1-cos(a))
Since in a stead climb L=mg*cos(a):
G=cos(a)+(1-cos(a)) = 1.

So the G meter in that case will show 1.

Didn't run threw your math (sorry lazy today), but you must be missing something, if the meter is positioned to measure force in the lift axis, then straight up the meter would show zero G regardless of acceleration. If you wish I can test it some time this week in the RV.

With a quick scan it appears you are measuring earth relative instead of object relative? I.E. you are saying the G meter measures mass/gravity?


HiTech

[bozon]

Didn't run threw your math (sorry lazy today), but you must be missing something, if the meter is positioned to measure force in the lift axis, then straight up the meter would show zero G regardless of acceleration. If you wish I can test it some time this week in the RV.

With a quick scan it appears you are measuring earth relative instead of object relative? I.E. you are saying the G meter measures mass/gravity?


HiTech

Ignore my previous post, that is an error... (sorry, I cannot remove it now)

I forgot to add the earth's gravity along the tube which is not 1g but g*cos(a). This cancel the g*cos(a) in the net acceleration of the plane frame of reference L/m-g*cos(a) and all that is left is G=L/(m*g) aways. For no net accelerations (L/m=g*cos(a)) it means G=cos(a) of the climb angle.

The simplest device I can think of is just a spring measuring the weight of a ball that can slide along a tube. Zero G would be the 0 point of the spring. 1G would be the point of spring load equals m*g of the ball. 2G is 2*m*g, etc. In a level flight, the tube is vertical to the ground and the ball weighs m*g against the spring. In a free fall, or a vertical climb (horizontal tube) the ball will apply no force on the spring and the device will read 0G.

I have no idea if this is how it works in practice.

[jamdive]

What ww2 aircraft had an actual g-meter? I don't ever remember actually seeing one in any cockpit photos or museum exhibits.

[save]

No G meter in ww2 planes.

However you feel G forces in a plane flying it , you don't in a sim without motion

[Widewing]

p.s.
How does a G meter on a plane actually work? I guess it measures acceleration in a given axis set to be aligned with the lift.

Although everything is digital today and MEMS accelerometers are commonly used, back in the 40s and 50s, they used strain gauge accelerometers. Essentially, a beam with a mass on the end of it. If the accelerometer (or aircraft) is dropping at the rate of gravity (32.2 ft/sec/sec) the relative g loading goes to zero and is indicated as such on the g meter (often referred to as "zero g, absolute"). Pilots will "unload" their aircraft to zero g to increase acceleration.