G-Load Calculation Issue

Wild_Bill_Kelso · Jun 21, 2022

GregP said:
Actually no. If you are in a level turn at a certain speed and g-load, then you have enough lift to sustain the turn at that speed. If your wing cannot supply the lift required due to one of the factors you mention above, then you will stall and cannot sustain the turn you are trying to make. The stall speed is the level-flight stall speed multiplied by the square root of the g-load.

Ah ok. So the lift effect of the wings for a turn is basically determined by the stall speed.

GregP said:
So, say you have a level flight stall speed of 90 mph (or 90 knots, etc.) and say you want to turn at 2 g. The square root of 2 is 1.414 and 1.414 time the stall speed is 127.26 mph. So, you cannot make a level turn at ANY speed below 127.26 mph because you will stall first. But, at 130 mpg and 2 g, you CAN turn and the turn radius will be 652 feet using: r = (V^2)/(g tan [bank angle]), and at 2.0 g, the bank angle is 60°. It takes 10.7 seconds to complete a 180° turn and the rate of turn is 17° per second.

Thanks this is very helpful.

GregP said:
You can get these number by knowing the turn radius, calculating the circumference, and then take half of that distance and calculate how long it take to fly that distance at 130 mph. Turn rate is 180° / time to turn, in degrees per second.

Sorry if I went too far into the explanation. Math is fun for me.

Cheers.

No very much appreciated. So I guess the other factor that is relevant is power vs. drag and weight, which determines how fast you lose speed in a turn?

Wild_Bill_Kelso · Jun 21, 2022

So slats and combat flaps do reduce the stall speed and that is how they help sustain sharper turns

Engineman · Jun 21, 2022

GregP said:
OK, I can buy that.

So, if the centripetal acceleration is 1.000 g and the actual g-load on the aircraft is 1.414 g, AND if the speed is 250 mph, then the turn radius is 4,180 feet. A 180° turn can be completed in 35.8 seconds, and the bank angle is 45° (well, 44.991° anyway). That isn't much of a combat turn. The time to complete a 180° turn gets worse as the speed goes up.

Edit:
Just as an aside, everyone in here who is instrument-rated knows a standard-rate turn is a 30° turn. Supposedly, it takes one minute to complete a 180° turn at standard rate. In my formulas, it doesn't quite work out that way. A 30° banks works out to a g-load of 1.155 g, and a 180° turn takes 61.9 seconds. To get to exactly 60 seconds for a 180° turn, the bank angle needs to be 30.9°. But, on a standard turn and bank indicator, you can't READ down to the tenth of a degree, so the "standard" was made into 30° since you can't tell the difference between the two numbers while in the cockpit. In the airplanes I used to fly, a standard turn and bank indicator had a mark every 10°, and shading to the outside edge of the 30° mark was right about 1 minute for a 180° turn.

Well,
A Rate 1 turn is, as you say, 180 degrees (course reversal) in one minute. However, the bank angle required to achieve that is variable, depending on TAS. Instrument rated pilots should know that the required procedural turn should be made at 25 degrees of bank or that required for a Rate 1 turn, whichever is the lower. To make it safe, procedures use the 25 degrees at the max published speed for the procedure as the max radius of turn, as lower speeds will achieve a tighter radius.

Eng

Engineman · Jun 21, 2022

Wild_Bill_Kelso said:
So slats and combat flaps do reduce the stall speed and that is how they help sustain sharper turns

Boy, that is complicated.
Generally, Slats and Flaps are designed to improve the coefficient of lift of a high speed low-drag wing for slower speeds, and the stall speed is also reduced. However, the lift/drag ratio of a wing can be very speed critical, so that the use of high lift devices at high speeds is actually reducing the energy rapidly. Similarly, not using the high lift devices at low speeds can
bleed energy through high buffet approaching the stall. In the days/aircraft without high lift devices, there was the need for aircraft to keep high speed when manoeuvring and stay away from high drag/ high alpha, (high AoA).
Of course, you could compare a Spitfire with a EE Lightning, the Spitfire could turn well inside but, it could not win a fight.

Eng

Wild_Bill_Kelso · Jun 21, 2022

Engineman said:
Boy, that is complicated.
Generally, Slats and Flaps are designed to improve the coefficient of lift of a high speed low-drag wing for slower speeds, and the stall speed is also reduced. However, the lift/drag ratio of a wing can be very speed critical, so that the use of high lift devices at high speeds is actually reducing the energy rapidly. Similarly, not using the high lift devices at low speeds can
bleed energy through high buffet approaching the stall. In the days/aircraft without high lift devices, there was the need for aircraft to keep high speed when manoeuvring and stay away from high drag/ high alpha, (high AoA).
Of course, you could compare a Spitfire with a EE Lightning, the Spitfire could turn well inside but, it could not win a fight.

Eng

Of course. Flaps and slats impose a drag penalty, in other words, and also can't even be used generally speaking at very high speeds. They more or less help you 'ride the stall'

GregP · Jun 21, 2022

Wild_Bill_Kelso said:
Ah ok. So the lift effect of the wings for a turn is basically determined by the stall speed.

Thanks this is very helpful.

No very much appreciated. So I guess the other factor that is relevant is power vs. drag and weight, which determines how fast you lose speed in a turn?

Actually, no. The stall speed simply determines the lowest speed at which you can turn at whatever g-force you decide on. The highest speed is determined by excess power and angle of attack. Most WWII fighters would be stalled at more than about 18° angle of attack or a bit less. In between the stall speed and the highest speed at which you can corner, the engine has enough power to pull you through a level turn.

Shortround6 · Jun 21, 2022

Wild_Bill_Kelso said:
So slats and combat flaps do reduce the stall speed and that is how they help sustain sharper turns

Kind of.

Note that they are talking about full span flaps or slats. Not a few feet in front of the ailerons.

Also note the angle of attack. If your wing is pulling a 10 degree or more angle of attack you need lots of thrust (F-15 or F-16 type thrust) or you are not going to sustain sharper turns for very long.

The slats and combat flaps reduce the stall speed by allowing the wing to operate at a higher angle of attack without stalling.

The famous slats of the 109 probably affected about 1/3 of the wing area. granted the prop wash may have been degrading some of the wing root area but the slats on the 109 didn't change the coefficient of lift of the entire wing all that much. The slats did mean the pilot still had alerion control just above stall so he was less likely to crash.

GregP · Jun 21, 2022

Wild_Bill_Kelso said:
So slats and combat flaps do reduce the stall speed and that is how they help sustain sharper turns

Yes, they simply can turn at a slightly lower speed, meaning shorter turn radius, for the g-force maintained.

Cheers.

Ivan1GFP · Jun 22, 2022

Wild_Bill_Kelso said:
Of course. Flaps and slats impose a drag penalty, in other words, and also can't even be used generally speaking at very high speeds. They more or less help you 'ride the stall'

Let's use another example that might make more sense..... or add to the confusion.
Let's say you have a lightweight Japanese A6M2 Type Zero fighter and the big heavy P-47 Thunderbolt flying against each other. Both are going 200 MPH and enter a 4G turn.
They will both fly the same path, but the A6M doesn't need to pull the same angle of attack on its wings to make the turn as the P-47 does. The P-47 will bleed off more speed. It will reach its accelerated stall speed sooner and have to drop out of the turn. It can drop its stall speed a bit with a bit of flaps (Combat Flaps) but then it bleeds off speed even faster though it can probably turn for a bit longer.
The difference gets even worse when you compare the biplane against the P-47.

Ivan1GFP · Jun 22, 2022

Zipper730 said:
.......

I would have thought it would have been centrifugal acceleration, but you're describing 1G+1G right? I'm curious how the square root gets in there

Let's say you take an Ant and put it inside a Ping Pong ball and tie a string to the ball.
Now you take the string and whirl the ball around.
If you get the speed of the ball just right, the angle of the string will be at 45 degrees to the horizon.
In order to get that exact angle, the centripetal acceleration of the ball around the circle will be exactly 1G.
Gravity will of course be 1G.
The tension on the string will be a sum of the vectors which are two vectors of 1G at right angles to each other and will result in Square Root of (1 Squared Plus 1 Squared). Thus Square Root of 2.
The Ant inside will experience 1.414G and wonder what is going on.

- Ivan.

Wild_Bill_Kelso · Jun 22, 2022

Ivan1GFP said:
Let's use another example that might make more sense..... or add to the confusion.
Let's say you have a lightweight Japanese A6M2 Type Zero fighter and the big heavy P-47 Thunderbolt flying against each other. Both are going 200 MPH and enter a 4G turn.
They will both fly the same path, but the A6M doesn't need to pull the same angle of attack on its wings to make the turn as the P-47 does.

Ok yes that is confusing. What determines the angle of attack exactly then? Does wing loading come into it here?

Ivan1GFP said:
The P-47 will bleed off more speed. It will reach its accelerated stall speed sooner and have to drop out of the turn. It can drop its stall speed a bit with a bit of flaps (Combat Flaps) but then it bleeds off speed even faster though it can probably turn for a bit longer.
The difference gets even worse when you compare the biplane against the P-47.

The other factor would be the power though as well right? If you had a higher power to weight ratio your speed wouldn't drop off as much.

GregP · Jun 22, 2022

Ivan1GFP said:
Let's use another example that might make more sense..... or add to the confusion.
Let's say you have a lightweight Japanese A6M2 Type Zero fighter and the big heavy P-47 Thunderbolt flying against each other. Both are going 200 MPH and enter a 4G turn.
They will both fly the same path, but the A6M doesn't need to pull the same angle of attack on its wings to make the turn as the P-47 does. The P-47 will bleed off more speed. It will reach its accelerated stall speed sooner and have to drop out of the turn. It can drop its stall speed a bit with a bit of flaps (Combat Flaps) but then it bleeds off speed even faster though it can probably turn for a bit longer.
The difference gets even worse when you compare the biplane against the P-47.

Very definitely NOT TRUE Ivan. For a level turn, the angle of bank depends ONLY on the velocity, the g-force, and angle of bank. But the angle of bank depends only on the g-force. So, the real dependency in a 4 g level turn is only the velocity. The Zero can turn tighter than the P-47 only because his stall speed is lower, so the Zero can either turn at 4 g slower than the P-47 can OR the Zero can pull more g-force before it stalls.

The stall speed changes as follows. Stall at some g load = (stall at 1 g) * square root (new g load).

An A6M-2 stalls at around 60 mph gear and flaps down. At 4 g in the same configuration, the stall speed will be 120 mph.

A P-47 stalls at around 100 mph, gear and flaps down. At 4 g in the same configuration, the stall speed will be 200 mph.

So, at any speed above 120 mph and below 200 mph., the Zero can make a 4 g turn and the P-47 simply cannot because it stalls first.

Alternately, suppose the fight is at 230 mph. The P-47 will stall at 5.29 g and will still be below the normal (8 g) and the ultimate g limit (12 g). The Zero, on the other hand COULD turn at up to 14.69 g before it stalls, but the normal g-limit is 6 and the ultimate limit is 12. So, the Zero pilot could pull anywhere between 6 and 12 g without expecting to break the wings and avoid the P-47 easily. If he stays at 6 g, he won't even damage the aircraft and the P-47 cannot turn with him.

But if both airplanes are at 230 mph and both are pulling 4 g in a level turn, they are both at the same bank angle (75.5°) and both have the same turn radius (913 feet).

The real strength of the P-47 is that the normal g limit is higher (8 versus 6) so, at high speeds, the P-47 can turn at 8 g without damage. The Zero, on the other hand, is constrained by the 6 g limit unless he doesn't mind damaging the airplane.

Wild_Bill_Kelso · Jun 22, 2022

But doesn't a zero or a biplane turn sharper even at low speed and without pulling very high Gs?

GregP · Jun 22, 2022

Wild_Bill_Kelso said:
But doesn't a zero or a biplane turn sharper even at low speed and without pulling very high Gs?

No. In any level turn at some defined g-load, and speed, the angle of bank and turn radii are the same. It's just that the biplane usually stalls at a slower speed than a monoplane, so it can make a defined turn at a lower speed than the monoplane because the monoplane stalls first. This all depends upon the speed and the g-load.

In the case of the P-47 above (100 mph stall), the Zero (60 mph stall) can fly at any airspeed between 60 mph and 100 mph, but the P-47 cannot because it stalls at 100 mph. So, he keeps his speed high, separates, climbs, and makes another pass.

Most, but by no means all, biplanes have their upper wing at a slightly higher angle of attack than the lower wing. This is so the upper wing will always stall first and keep airflow over the lower wing to maintain roll control. The center of lift moves aft and the nose drops, as it should to recover flight after the stall. Of course, this assumes ailerons on the lower wing of the fighter.

Wild_Bill_Kelso · Jun 22, 2022

But the radius of the turn circle also depends on the speed right?

GregP · Jun 22, 2022

Yes. Depends on speed, g-load, and bank angle ... and bank angle depends on g-load only.

These are for level turns only. That is, turns at a constant altitude. It gets VERY complicated for turns NOT in level flight since you have to take gravity into account as speed is gained and/or lost. It gets even worse if the power setting changes in the turn.

Zipper730 · Jun 22, 2022

Ivan1GFP said:
Let's say you take an Ant and put it inside a Ping Pong ball and tie a string to the ball.
Now you take the string and whirl the ball around.
If you get the speed of the ball just right, the angle of the string will be at 45 degrees to the horizon.
In order to get that exact angle, the centripetal acceleration of the ball around the circle will be exactly 1G.
Gravity will of course be 1G.
The tension on the string will be a sum of the vectors which are two vectors of 1G at right angles to each other and will result in Square Root of (1 Squared Plus 1 Squared). Thus Square Root of 2.
The Ant inside will experience 1.414G and wonder what is going on.

- Ivan.

I think I sort of get where you're going... the square of the numbers seems to do more with something like the pythagorean theorem than the square of the change in velocity then?

Math is not my strong-point...

GregP · Jun 22, 2022

Bank angle = arc cos (1 / g-load) = cos ^-1 (1 / g-load).

So, if you are in a level 3-g turn, the bank angle = 70.5° = arc cos (1 / 3). Actually 70.52878139°. Close enough to 70.5° for me.

Shortround6 · Jun 22, 2022

Please see : http://www.spitfireperformance.com/spit109turn.gif

Read the notes. like only good at 12,000ft.
These are calculations.
All so note that the two diagrams are not drawn to the same scale or rather that they are drawn with different start and end points. Like Spitfire starts at 0mph and ends at 450mph and the 109 starts at 50mph and ends at 500mph.

Now note that both the Spitfire and the 109 IF they are flying at 225mph true and pulling 3Gs they will have a radius of 1200ft and take just about 23 seconds to complete a full turn and they will have the same bank angle.

EXCEPT the Spitfire might be able to sustain that turn and speed (if everything is going well and the pilot is good) while the 109 cannot. The 109 has got to descend with it's nose pointed down at about 5 degrees in order to trade altitude for speed in order to maintain the speed and turn rate.

Do not complete turn at altitude, do not maintain turn, do not collect 200 dollars on Monopoly turn board.

Please note that adding power to the aircraft (Like a Spit V with a Merlin 45) will change the climb line and bring more of the chart under the climb line curve. Also note that a heavier airplane (Like the Spitfire V) will move the stall line a bit a right. Different altitudes (air density) can move the chart/s up or down a bit.

Also note that the Spitfire cannot pull a 6 g turn at 225mph. It is right at the stall boundary at 225mph at 5 g.

There is a lot of "stuff" going on in the turn between the lift, drag, thrust and the coefficient of lift and other things.

It can be calculated using enough data, But it is well above my pay grade (ability) to do so.

I am guessing that such calculations (charts) were done for quite a number of aircraft but not published outside of the engineering departments.

Wild_Bill_Kelso · Jun 22, 2022

Can you translate to the various figures in the Russian turn times test data, which seem to range from around 17 to 23 or 24 seconds to execute a turn?

G-Load Calculation Issue

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