G-Load Calculation Issue

[Shortround6]

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?

Note enough data/information.

Please look at the chart/s posted.

The Spitfire had a range of turns.
at 20 seconds for a 360 degree turn the Spit could fly at 150mph at around 2.3-2.4 Gs and fly just under a 700ft radius. It could also fly at a much higher speed over a wider radius and also complete the turn in 20 seconds.

Did the Russians test have a standard turn radius/diameter?
Was the Russian turn limited by needing to hold altitude or climb a certain amount while turning?

If we restrict the Spitfire to holding altitude and not diving the the best turn speed (least time) would be about 160mph and 19 seconds for the 360 degree turn. If you give the Spitfire 24 seconds for the turn you can fly a very tight circle at close to 130mph or you could fly a much wider circle at around 230mph in the same 24 seconds.

I believe (could be wrong) the Russian turn test was at 1000 meters (3280ft) which is not going compare well with 12,000ft (3657 meters)

Later fighters could turn better than early fighters from a practical standpoint because they had more thrust and could hold speed/turn/altitude better. The early fighter can turn as tight or tighter but it is going to loose altitude quicker. Look at the chart again. The Spitfire I could manage a 6 G turn at 350mph, it just had to have it's nose pointed down at just under 19 degrees to do it. Which, if I have done the math right and it may be way off) means the plane dropped just over 3000ft in that 17 second turn.

So yeah you could pull three 360 degree turns in just under minute flying at 350mph. You just need 9,000ft of altitude to do it. There is a reason that most dogfights/furballs ended at low altitude.

 

[special ed]

Aaah... if I only understood all this math, I would have become the world's greatest fighter pilot. Oh well, Que Sera.

 

[GregP]

Hi Shortround6!

Since you so nicely posted the graphs for the Spitfire and the Bf 109, I thought I'd post them again with some gridlines to make interpolation a bit clearer.

The gridlines aren't perfect, but they are very close. Thanks for the neat charts!

 

[Wild_Bill_Kelso]

I want to see if I can make a computer simulation of this. I have some business app software here that I use for work, not ideal for this kind of calculation but I think I could use it to crunch all these numbers. I'll see if I can get something set up.

 

[GregP]

Very interesting that, at 250 mph, the Spitfire can only turn level at just under 3 g. Anything more and the path is descending.

The Bf 109, at 250 mph, can only turn level at 2.4 g or so. Anything more and the path is descending.

Easy to see the Spitfire can out-turn the Bf 109 ... at 12,000 feet, anyway.

The Spitfire in the chart is a very early model, with only 1,050 hp, so it's a Spitfire I.

And these charts are at 12,000 feet! That probably means that at 30,000 feet +, they could only do very gentle turns without dropping fairly rapidly.

Of note, the 1 g stall speed is about 88 mph. So, the 3 g stall speed SHOULD be 152.4 mph, but the chart shows 191 mph or so. That makes me wonder fairly heavily about the chart since the prediction and fact are too far apart for comfort. I figured it would be a bit higher than ideal, but not by 40 mph!

 

[Ivan1GFP]

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



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.

Hello Wild Bill Kelso,
Yes, Wing Loading and the amount of lift needed is what determines the angle of attack.
As an example if each aircraft were attempting a 3G turn, The A6M which weighs maybe 5000 pounds needs to pull enough AoA for its wing to support 15,000 pounds. The Thunderbolt which weighs perhaps 16,000 pounds needs to pull enough AoA for its wing to support 48,000 pounds. The wing of the Thunderbolt is only about 25% bigger, so it has to pull a lot more angle to generate that much lift.

You are absolutely right, extra thrust can be used to offset the energy bleed, but the amount of thrust available was generally not even close to the amount needed to offset the amount of energy lost in a tight turn.

 

[Ivan1GFP]

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.

Hi GregP,
You misread my post. I didn't state that the BANK Angle was different. I stated that the the Angle of Attack to fly the same arc was different.

The point was that when the two aircraft enter the turn at 200 MPH, both are flying exactly the same arc. (And same bank angle) The P-47 needs to use a higher AoA to make the same turn because it has a higher wing loading. After perhaps 180 degrees of turn, the P-47 will have bled off a bit more speed than A6M. That was the only point being made.

To be honest, I am not all that certain the P-47 could even pull a 4G turn at that speed.

- Ivan.

 

[Ivan1GFP]

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...

If I understand you correctly then Yes.
Basically you are adding two right angle vectors. One is horizontal (the turn), and one is vertical (gravity). To do this, you use the Pythagorean theorem.

Now keep in mind that the measure that folks here are using for the turn is what our poor Ant is experiencing inside the Ping Pong Ball (aeroplane). In this case it was 1.414G.You can figure out the changes for a 3G turn (Hypotenuse is 3, Gravity remains 1 of course) and so forth.
The Bank Angle isn't that hard to figure out either. In our example, it is the angle of the string compared with vertical.

 

[Ivan1GFP]

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

As GregP pointed out, any aeroplane making a turn will pull the same amount of G.
The difference is that the biplane and the A6M Zero are fairly light.
Most wings stall at pretty close to the same angle.
With the light aeroplane, you have two choices. You can make the same turn as the heavy aeroplane by not pulling as hard (using lower Angle of Attack on the wings) because you don't need as much lift. - OR -
You can pull as hard as possible at which point, your wing is generating more lift in proportion to the weight of your aeroplane. The heavy aeroplane may generate the same amount of lift but won't turn as tight because their plane is heavier.

 

[Ivan1GFP]

Very interesting that, at 250 mph, the Spitfire can only turn level at just under 3 g. Anything more and the path is descending.

The Bf 109, at 250 mph, can only turn level at 2.4 g or so. Anything more and the path is descending.

Easy to see the Spitfire can out-turn the Bf 109 ... at 12,000 feet, anyway.

The Spitfire in the chart is a very early model, with only 1,050 hp, so it's a Spitfire I.

And these charts are at 12,000 feet! That probably means that at 30,000 feet +, they could only do very gentle turns without dropping fairly rapidly.

Of note, the 1 g stall speed is about 88 mph. So, the 3 g stall speed SHOULD be 152.4 mph, but the chart shows 191 mph or so. That makes me wonder fairly heavily about the chart since the prediction and fact are too far apart for comfort. I figured it would be a bit higher than ideal, but not by 40 mph!

IAS versus TAS?

 

[Wild_Bill_Kelso]

Ok now I'm thoroughly confused again

 

[GregP]

Hello Wild Bill Kelso,
Yes, Wing Loading and the amount of lift needed is what determines the angle of attack.
As an example if each aircraft were attempting a 3G turn, The A6M which weighs maybe 5000 pounds needs to pull enough AoA for its wing to support 15,000 pounds. The Thunderbolt which weighs perhaps 16,000 pounds needs to pull enough AoA for its wing to support 48,000 pounds. The wing of the Thunderbolt is only about 25% bigger, so it has to pull a lot more angle to generate that much lift.

You are absolutely right, extra thrust can be used to offset the energy bleed, but the amount of thrust available was generally not even close to the amount needed to offset the amount of energy lost in a tight turn.

Sorry, Ivan! Didn't mean to mis-interpret your post. Seems like it was in regular English. I just misread it. Mea Culpa.

About the IAS versus TAS, if you look at the high end, it HAS to be TAS.

I wish all this stuff was labeled on the charts. Perhaps it is in some accompanying text and it just wasn't posted.

 

[Dimlee]

I believe (could be wrong) the Russian turn test was at 1000 meters (3280ft) which is not going compare well with 12,000ft (3657 meters)

Soviet tests were done at various altitudes but the turn time at 1,000 m was the shortest, so it's safe to assume that the best turn times mentioned in the literature were achieved at 1,000 m.

 

[Wild_Bill_Kelso]

Does anyone have a link to the literature (in Russian or English) which explains the context of the turn times tests? I often see the results posted but rarely with any of the primary source data associated with it.

 

[GregP]

Hi WIld Bill,

I've seen some compilations in the airwar.ru page and a other Russian websites, but they were all a text listing, not a primary source. I found one with some source data, but could never find any of the sources listed.

When I found it, I put it into Excel. That doesn't make it any more primary than the source I found it in, but I attach it here for you. The source website is at the top of both tabs with data in them. So, no primary sources here, but at least some data. I'd offer some explanatory text about the turn times, but there was none. I am left feeling good about finding the data, but no so good about the validity of same.

Cheers.

 

[Wild_Bill_Kelso]

Gotcha, thanks. I figure there has to be some kind of trail of records back to the original document(s). I'll keep looking.

 

[GregP]

I've been looking for years.

If you find it, please post where you found it!

Cheers.

 

[Zipper730]

If I understand you correctly then Yes.
Basically you are adding two right angle vectors. One is horizontal (the turn), and one is vertical (gravity). To do this, you use the Pythagorean theorem.

Can you draw this out? Or can somebody?

Some things are better visually explained...

 

[pbehn]

Can you draw this out? Or can somebody?

Some things are better visually explained...

Are you really asking someone to draw a right angled triangle?
a) Is the lift vector, always 1 in level flight.
b) Is the turning vector, so whatever "g" value you are discussing.
c) Is the resultant vector combining the two.

If you imagine c) is a piece of string keeping a toy plane circling the ceiling , the theoretical banking angle of it is at 90 degrees to the hypotenuse

 

[GregP]

Here's my attempt at it. Suppose you bank at a 30° angle. See below.

Arbitrarily, I choose 1.155 g, for no other reason than to avoid a round number. So, 1 / 1.155 = .86580, and cos-1 (.86580) = 30°.

1.00 g can be found by 1.155 g * cos (30°) and the .5774 g can be found by 1.155 * sin (30°).

Edit: The vertical g-load will always be 1 g in a level turn. If it was less, you'd descend. If it was more, you'd climb. End Edit.

You have 1 g vertical and .5774 g horizontal. Square both, add them up, and take the square root. Your butt feels 1.155 g down through the seat.