palgia
24th Oct 2004, 09:04
Hello everyone,
my question refers to power-on stalls in a small piston single like a C-172.
Why should the the inclinometer ball be set slightly right of center in order for the stall to break straight ahead with wings level? It's usually 1/2 ball-width to the right.
In many cases, if you try doing it with the ball centered, the stall breaks to the right.
I've heard/read several explanations, but none make too much sense to me.
One of them goes like this:
P-factor causes center of thrust to be offset to the right of the aircraft longitudinal axis. That force times the distance from the longitudinal axis creates a yawing moment to the left.
Applying right rudder moves the ball to the center but causes the aircraft to sideslip to the left (just like a twin with an inop left engine flown wings level, ball centered). The aircraft is therefore not flying straight through the air (sideslipping) and for this reason the stall breaks to the right.
Leaving the ball slightly right of center cancels out this effect.
The above explanation does not make sense to me for the following reasons:
1. The arm from the center of thrust and the aicraft longitudinal axis is so small that the resulting yawing moment, if any, is not even remotely comparable to the situation of a twin with an engine inop.
2. Even if we assume there is enough yawing and that the aircraft IS sideslipping to the left, it would be the LEFT wing to drop first, since it's the upwind wing that operates at a higher AOA in a slip (isn't that how the stabillizing dihedral effect works?)
3. If the aircraft is in fact sideslipping to the left as if it had asymmetrical thrust, ONLY A BANK towards the right can compensate for the sideslip and bring the aicraft to zero-sideslip. Reducing right rudder pressure would not eliminate the sideslip. The only thing it would do is move the ball to the right, thus giving the "illusion" of zero sideslip since in a zero sideslip situation where the aircraft is banked to the right, the ball would also swing to the right of center due to gravity. But these two are totally different effects that happen for different reasons.
Another explanation claims that the ball should be centered for a power-on stall, but because the left turning tendencies decrease after the stall, most people induce the left break on the stall because they continue to push the right rudder with excessive pressure after the stall. For this reason, they recommend entering the stall with the ball slightly to the right of center.
This explanation is more convincing, but I have experienced cases where the ball was centered, but the aircraft was breaking to the right even before, or just at the same instant, that the nose was dropping. If we look at the left turning tendencies, why should their intensity drop right AT the stall, before the nose has dropped any significant amount?
What about turning power-on stalls? It looks like leaving the ball slightly right of center is the best way to break the stall straight ahead.
Thanks for your replies.
palgia
my question refers to power-on stalls in a small piston single like a C-172.
Why should the the inclinometer ball be set slightly right of center in order for the stall to break straight ahead with wings level? It's usually 1/2 ball-width to the right.
In many cases, if you try doing it with the ball centered, the stall breaks to the right.
I've heard/read several explanations, but none make too much sense to me.
One of them goes like this:
P-factor causes center of thrust to be offset to the right of the aircraft longitudinal axis. That force times the distance from the longitudinal axis creates a yawing moment to the left.
Applying right rudder moves the ball to the center but causes the aircraft to sideslip to the left (just like a twin with an inop left engine flown wings level, ball centered). The aircraft is therefore not flying straight through the air (sideslipping) and for this reason the stall breaks to the right.
Leaving the ball slightly right of center cancels out this effect.
The above explanation does not make sense to me for the following reasons:
1. The arm from the center of thrust and the aicraft longitudinal axis is so small that the resulting yawing moment, if any, is not even remotely comparable to the situation of a twin with an engine inop.
2. Even if we assume there is enough yawing and that the aircraft IS sideslipping to the left, it would be the LEFT wing to drop first, since it's the upwind wing that operates at a higher AOA in a slip (isn't that how the stabillizing dihedral effect works?)
3. If the aircraft is in fact sideslipping to the left as if it had asymmetrical thrust, ONLY A BANK towards the right can compensate for the sideslip and bring the aicraft to zero-sideslip. Reducing right rudder pressure would not eliminate the sideslip. The only thing it would do is move the ball to the right, thus giving the "illusion" of zero sideslip since in a zero sideslip situation where the aircraft is banked to the right, the ball would also swing to the right of center due to gravity. But these two are totally different effects that happen for different reasons.
Another explanation claims that the ball should be centered for a power-on stall, but because the left turning tendencies decrease after the stall, most people induce the left break on the stall because they continue to push the right rudder with excessive pressure after the stall. For this reason, they recommend entering the stall with the ball slightly to the right of center.
This explanation is more convincing, but I have experienced cases where the ball was centered, but the aircraft was breaking to the right even before, or just at the same instant, that the nose was dropping. If we look at the left turning tendencies, why should their intensity drop right AT the stall, before the nose has dropped any significant amount?
What about turning power-on stalls? It looks like leaving the ball slightly right of center is the best way to break the stall straight ahead.
Thanks for your replies.
palgia