Steep turns and stall speed

BCBC

New Member
Hi all,

Several pilot friends and myself have been kicking this one around. If you're in a steep turn and holding altitude, this requires considerable back pressure on the yoke/stick, which increases g-loading (and stall speed), yes? But if you're in a steep turn and descending, you're not increasing back pressure on the yoke/stick and therefore not increasing the g-loading. Which means you're not increasing the stall speed. So...stall speed is higher in a 45 degree bank constant altitude turn than in a 45 degree descending turn.

Any thoughts? :banghead:

Thanks everyone!
 

subpilot

Squawking 7600
That is true. If you bank into a steep turn but do not pull back on the yoke to maintain altitude then you will maintain a 1G load on the wings and thus stall speed should remain the same. I am sure a technical explanation will soon follow my non-technical answer :)
 
R

Roger, Roger

Guest
Not quite true. As long as your rate of climb or descent is static, your wing will be producing the same amount of lift. It comes back to basic physics.
F=MA.
However, you are correct that by not pulling back on the stick you will keep stall speed the same. In that situation, not only would you be descending, but your rate of descent would be increasing, because the net upward lift is less than the mass of the airplane. Therefore gravity takes over and the airplane accelerates downward.
 

MidlifeFlyer

Well-Known Member
In that situation, not only would you be descending, but your rate of descent would be increasing, because the net upward lift is less than the mass of the airplane. Therefore gravity takes over and the airplane accelerates downward.
Yep. I think the issue is -steady state=, not =level, climbing or descending=.

Leave out the steep turn for a moment. In terms of the VSI, if the VSI is =constant=, we're at 1 G, whether level, climbing or descending. If the VSI is =changing=, an upward change is an increase in load, a downward change a decrease in load.

The key to understanding this is to understand that there is a difference between a descent in which the =rate= of descent is the =same= (say 500 fpm) and a descent in which the rate of descent is =changing= (say 50 fpm the 1st second, 100 fpm the second second, 150 fpm the third second, etc)

That's easy to demonstrate in an airplane in flight. Push down aggressively on the yoke and you'll feel the negative G (decreased load); let the airplane settle in a 500 fpm descent and, once established, the G will go back to 1.

The steep turn adds load. But it adds load to the load that otherwise exists.

If the load factor is 1 - whether level or in a stable 500 fpm descent, you are increasing the load (and you =are= pulling back on the stick in order to maintain that 500 fpm descent)

If, OTOH, the load factor is, say -1, because you are in a descent that is constantly increasing, then the load factor is less than 1 and the steep turn will tend to neutralize it.

There's actually a maneuver that was in the private PTS for a while a few years ago that demonstrates this: the emergency descent. The first step in the procedure (after reducing power) was to aggressively push forward on the stick to start the descent and add a 45° bank to neutralize the negative Gs. Then, once you stabilized the descent rate, you needed to wither decrease the bank or increase the back pressure.

I'm sure tgrayson will step in with a better, more technical answer.
 

tgrayson

New Member
Which means you're not increasing the stall speed. So...stall speed is higher in a 45 degree bank constant altitude turn than in a 45 degree descending turn.
Not significantly. If you roll into a 45 degree turn, and don't increase the backpressure, you won't immediately experience an increase in load factor. However, the airplane will accelerate to a higher speed to restore the vertical component of lift = weight. As the speed increases, the load factor will increase too and so will your stall speed.

Anyone who has done a steep spiral for the Commercial SEL rating knows this is true.

There is a slight reduction in load factor for a given bank angle in a descent, but it's due to the fact that some of the lift the airplane needs is supplied by a component of drag. For a 45 degree bank, it's pretty trivial. You can calculate the load factor in a descent by this formula:
n = cos(descent angle)/cos(bank angle)
For a 45 degree bank, and a descent angle of say, 10 degrees, the load factor is
=cos(10)/cos(45)
=.985/.707
= 1.393
A level turn load factor for a 45 degree turn is
=cos(0)/cos(45)
=1/.707
=1.414
So the difference is 1.414 - 1.393 = .021. Not much.

Thus an aircraft with a wings-level stall speed of 50 knots will have a stall speed in a level steep turn of
new stall speed = sqrt(1.414) * 50
= 59 knots.
In a descending turn, the stall speed will be
new stall speed = sqrt(1.393)*50
= 59 knots
In other words, the small difference is lost during rounding.
 

BCBC

New Member
Hi all. Thanks for the replies. Not sure I get this:

"If you roll into a 45 degree turn, and don't increase the backpressure, you won't immediately experience an increase in load factor. However, the airplane will accelerate to a higher speed to restore the vertical component of lift = weight. As the speed increases, the load factor will increase too and so will your stall speed."

Why would the stall speed increase just because airspeed increases? Load factor increases as a result of increased angle of attack (i.e., back pressure), not as a result of an increase in airspeed. In a gliding turn with no increase in back pressure, load factor (and therefore stall speed) would remain the same, correct?
 

tgrayson

New Member
Load factor increases as a result of increased angle of attack (i.e., back pressure), not as a result of an increase in airspeed.
Not so. Load factor is the ratio of Lift to Weight; anything that increases lift increases the load factor. Do you know the lift equation? Lift is a function of AOA and airspeed. At a given AOA, with an increase in airspeed, lift increases and so does load factor.

In a gliding turn with no increase in back pressure, load factor (and therefore stall speed) would remain the same, correct?
No. Go try it yourself. Roll into a 60 degree bank and hold it. The load factor will build rapidly as the airspeed increases. The airspeed should end up about 40% higher, once it stabilizes.
 
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