CGI - Possible configuration of the Boeing F-47 NGAD

Does this post suggest that BOP's gull-wing was for stability rather than stealth? Due to a lack of fly by wire in the demonstrator.

Hydroman said:
They did BOP on a tight budget and with the gull-wing configuration, the aircraft had mechanical flight control system, no hydraulic fly-by-wire/artificial stability, would have cost a lot more. Even with the gull-wing configuration, it could be a hand full for the pilot in certain flight regimes. But BOP demonstrated and proved out what they wanted to accomplished so it was a successful program.

The gull-wing shape could have compromised LO possibly but LO is not my area of expertise.

As an example, there was no way Tacit Blue could have flown without a FBW flight control system, it would be the Tacit Tumbler, not good, same goes for the YF-23, plus from what I was told, Tacit Blue was unstable in all 3 axis. You can add F-16 and F-117 to the list as well which could not fly with a pure mechanical control system even with hydromechanical assist.
Click to expand...
 
Page 12 is where I last saw this thread. Now on page 17 or 18...

VTOLicious said:
Latest iteration with tandem bays... Each bay may also accommodate up to two AGM-158 JASSM ;)
Click to expand...
While I've been pushing for that as a "don't be stupid, make your bays bigger than they absolutely have to be per contract so that you can put bigger booms in there" I don't believe that the F-47 has bays that big.



dark sidius said:
This model ask questions what it could be F-47 ?
View attachment 773549
Click to expand...
F-23 EMD?
 
targetpractice said:
Does this post suggest that BOP's gull-wing was for stability rather than stealth? Due to a lack of fly by wire in the demonstrator.
Click to expand...
Remember, the BOP flight control system was purely mechanical and from what I understand, no hydraulics, it was a little bit of a handful to fly but that's OK for a demonstrator program which is primarily testing LO. There was even stabilization devices installed on or near the main gear to make control easier with gear down. There can be a compromise and balance using a gull-wing for stabilization and LO, just have to have the right design combination. If BOP was going to be a production mission aircraft, it would have a full authority, electrohydraulic FBW flight control system and the gull-wing would have probably gone away.
 
Hydroman said:
Remember, the BOP flight control system was purely mechanical and from what I understand, no hydraulics, it was a little bit of a handful to fly but that's OK for a demonstrator program which is primarily testing LO. There was even stabilization devices installed on or near the main gear to make control easier with gear down. There can be a compromise and balance using a gull-wing for stabilization and LO, just have to have the right design combination. If BOP was going to be a production mission aircraft, it would have a full authority, electrohydraulic FBW flight control system and the gull-wing would have probably gone away.
Click to expand...
Some of the stuff I read speculated that the gull wings was part of the visual stealth process that was being tested by the BoP. Yehoudi Lights development or some such.
 
targetpractice said:
Does this post suggest that BOP's gull-wing was for stability rather than stealth? Due to a lack of fly by wire in the demonstrator.
Click to expand...
Yes.
The Boeing BoP is nose heavy and basically a dart, therefore extremely instable. The wings being placed far in the back it needed them to be dihedral for making it stable. And since it's become so "stable" it no longer needed the fast reaction time and numerous corrections per second that a FBW system offers.
Simplified wing arrangement for illustration:

wings.png
Scott Kenny said:
While I've been pushing for that as a "don't be stupid, make your bays bigger than they absolutely have to be per contract so that you can put bigger booms in there" I don't believe that the F-47 has bays that big.
Click to expand...
Well, one of the japanese research design had tandem bays.
From the looks of it the FCAS/SCAF seems to go for wider instead.
Wider increases cross section so tandem seems better to keep the same cross section or for a slight increase if engines are moved sideway. Since more fuel is needed, hence, lengthening was going to happen anyway it would be a minor addition on top
For the same drag reasons ships tend to be made get longer rather than wider. We'll see soon enough.
 
Seragina said:
Yes.
The Boeing BoP is nose heavy and basically a dart, therefore extremely instable. The wings being placed far in the back it needed them to be dihedral for making it stable. And since it's become so "stable" it no longer needed the fast reaction time and numerous corrections per second that a FBW system offers.
Simplified wing arrangement for illustration:

View attachment 774472
Click to expand...
Gotcha, makes sense now.



Seragina said:
Well, one of the japanese research design had tandem bays.
From the looks of it the FCAS/SCAF seems to go for wider instead.
Wider increases cross section so tandem seems better to keep the same cross section or for a slight increase if engines are moved sideway. Since more fuel is needed, hence, lengthening was going to happen anyway it would be a minor addition on top
For the same drag reasons ships tend to be made get longer rather than wider. We'll see soon enough.
Click to expand...
Yeah, I've been arguing back and forth mentally between 2 bays a little narrower than the F-22 in tandem, or 2 bays that are collectively a lot wider than the F-22 in parallel. The A-12 used 2 bays in parallel, but I suspect that the F/A-XX will be using 2 bays in tandem to avoid the width problem.
 
Seragina said:
Yes.
The Boeing BoP is nose heavy and basically a dart, therefore extremely instable. The wings being placed far in the back it needed them to be dihedral for making it stable. And since it's become so "stable" it no longer needed the fast reaction time and numerous corrections per second that a FBW system offers.
Click to expand...
If in fact the BOP was nose heavy, it would be extremely stable. By putting the wings so aft, it would be even more stable - i.e. - the wings would pull the neutral point further aft, thus causing the longitudinal stability to go further positive. Actually, the forebody configuration was very de-stablizing. To stabilize, the wings needed to be place well aft considering there was no artificial stability.

The gull wing was designed with directional stability in mind (among several other issues also being considered). Remember, the initial flights were done with a temporary ventral/centerline vertical surface. The boys were hedging their bets whether the gull would be sufficient.
 
djfawcett said:
If in fact the BOP was nose heavy, it would be extremely stable. By putting the wings so aft, it would be even more stable - i.e. - the wings would pull the neutral point further aft, thus causing the longitudinal stability to go further positive. Actually, the forebody configuration was very de-stablizing. To stabilize, the wings needed to be place well aft considering there was no artificial stability.

The gull wing was designed with directional stability in mind (among several other issues also being considered). Remember, the initial flights were done with a temporary ventral/centerline vertical surface. The boys were hedging their bets whether the gull would be sufficient.
Click to expand...
It seems we have a misunderstanding of definitions here.
Nose heavy here means the CG is far in front of the aerodynamic center which is determined by the wing mid chord. Your definition of stability refers in terms of aerodynamic tumbling. I'm refering to controlability. An example of nose heavy is said to be the F16. In this case it had a large tail, fins and angled elevators to counter.
 
Seragina said:
It seems we have a misunderstanding of definitions here.
Nose heavy here means the CG is far in front of the aerodynamic center which is determined by the wing mid chord. Your definition of stability refers in terms of aerodynamic tumbling. I'm refering to controlability. An example of nose heavy is said to be the F16. In this case it had a large tail, fins and angled elevators to counter.
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I believe you are a bit confused. You are attempting to describe stability in non-conventional terms. I really have no idea what you are trying to describe when you say "aerodynamic tumbling"- I am guessing forward somersaulting. Additionally, aerodynamic center for a given aircraft is not solely determined by the quarter chord of the wing. The forebody and aft body effects, which can be dramatic, also are part and partial to the aerodynamic center. When you say "Nose heavy here means the CG is far in front of the aerodynamic center", you are describing a positively stable aircraft by definition.

Concerning the F-16, you are totally incorrect. The F-16 is not nose heavy, it is effectively tail heavy, thus the requirement for artificial stability with fly-by-wire flight controls. If you could pick-up the F-16, flight loaded. at the CG (think Jolly Green Giant - lol) to determine how the aircraft balances, it would fall on its tail. Described in simple terms, while in level flight, the horizontal is not trying to lift the nose, it is actually pitching the nose down. Think of the horizontal as a lifting surface. This is a very simple explanation. There are other factors involved, but the above is the basics.
 
Hello.
I have a question: If the F47 is indeed the aircraft they showed us, what is on its tail that they went to such lengths to hide, making the dihedral wings necessary to make it more stable?
The Chinese didn't need to incorporate dihedrals into their tailless designs, so the absence of tails wouldn't be a deciding factor in incorporating a dihedral.
I'm asking this question for the more knowledgeable people on the forum.
Greetings and thanks!
 
Rodrigo Avella said:
Hello.
I have a question: If the F47 is indeed the aircraft they showed us, what is on its tail that they went to such lengths to hide, making the dihedral wings necessary to make it more stable?
The Chinese didn't need to incorporate dihedrals into their tailless designs, so the absence of tails wouldn't be a deciding factor in incorporating a dihedral.
I'm asking this question for the more knowledgeable people on the forum.
Greetings and thanks!
Click to expand...
Hello *
May be new exhaust system....
 
Rodrigo Avella said:
Hello.
I have a question: If the F47 is indeed the aircraft they showed us, what is on its tail that they went to such lengths to hide, making the dihedral wings necessary to make it more stable?
The Chinese didn't need to incorporate dihedrals into their tailless designs, so the absence of tails wouldn't be a deciding factor in incorporating a dihedral.
I'm asking this question for the more knowledgeable people on the forum.
Greetings and thanks!
Click to expand...
Hey! I won't say that hiding the aft section of the aircraft is for the sake of aerodynamic breakthrough, but more so for INFOSEC. I feel that in this day and age that IW is as important as any kind of warfare. Hide the back of the bird, keep the adversary guessing. Let's take what we do here as an example, we speculate for fun, but for others, that speculation is for intelligence gathering. How much resources does a foe waste chasing vapor cones?
 
FlyGuy369 said:
Hey! I won't say that hiding the aft section of the aircraft is for the sake of aerodynamic breakthrough, but more so for INFOSEC. I feel that in this day and age that IW is as important as any kind of warfare. Hide the back of the bird, keep the adversary guessing. Let's take what we do here as an example, we speculate for fun, but for others, that speculation is for intelligence gathering. How much resources does a foe waste chasing vapor cones?
Click to expand...
Hello! I understand what you're saying, but I still don't understand why they need a dihedral wing.
From what I've seen with the Chinese models and the studies conducted by American companies, it's not necessary to maintain the stability of a tailless aircraft. It wasn't even something explored in the FATE and ICE studies, where multiple varieties of controls for tailless aircraft were explored. We only saw it in the Bird of Prey, and as they said here, the dihedral had its reason for being.
So I want to understand the reason for the dihedral wing for the F47, and the only thing I can find is that there's something about its rear end (which is the only part we don't know what it looks like) that's "so crazy" and "so revolutionary" that they need a dihedral wing to compensate for the lack of stability of an aircraft that will have the most advanced software in the world to control it.
Greetings, and by the way, I love debating these topics!
 
I think the dihedral appearance is a play on the photo for distortion to be misleading, even the cockpit and nose wheel are distorted too. Look st the contrast adjustment. The nosewheel was placed over something else, and the canopy was enlarged to throw off scale.
 

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Rodrigo Avella said:
Hello.
I have a question: If the F47 is indeed the aircraft they showed us, what is on its tail that they went to such lengths to hide, making the dihedral wings necessary to make it more stable?
The Chinese didn't need to incorporate dihedrals into their tailless designs, so the absence of tails wouldn't be a deciding factor in incorporating a dihedral.
I'm asking this question for the more knowledgeable people on the forum.
Greetings and thanks!
Click to expand...

I'm very much a vibes based guy, and would not pretend to be an expert on anything. My process has been to try and come up with the most obvious thing possible, then make it look a bit sexy. Also I wanted to build something in Blender.

When I first saw the mocks my initial thought was that both canards and dihedral was misinformation - it is a very long way from the accepted understanding of what a platform like this should be. However, both images are consistent, and there is a lot of stuff that is purposely hidden, so why lie about this? Historically, these teases have turned out to be true representations, and I think we have to start by assuming the same here. If not, without further evidence, we're going to drive ourselves mad.

With that in mind, some ideas:
  1. Maybe the dihedral is an optical illusion due to the v weird, nearly orthographic camera perspective. That would mean the wings are swept in a v strange way...
  2. Maybe this forum's conclusions about the BOP dihedral were incorrect and there is some stealth or aerodynamic advantage in this wing shape. I vaguely remember an earlier USAF quote mentioned something about interesting new aerodynamics. To be honest, this feels like most likely conclusion.
  3. I'd not really considered that there is something else hidden in the image, but it is a super intriguing thought! I wish I had more imagination!
 
targetpractice said:
I'm very much a vibes based guy, and would not pretend to be an expert on anything. My process has been to try and come up with the most obvious thing possible, then make it look a bit sexy. Also I wanted to build something in Blender.

When I first saw the mocks my initial thought was that both canards and dihedral was misinformation - it is a very long way from the accepted understanding of what a platform like this should be. However, both images are consistent, and there is a lot of stuff that is purposely hidden, so why lie about this? Historically, these teases have turned out to be true representations, and I think we have to start by assuming the same here. If not, without further evidence, we're going to drive ourselves mad.

With that in mind, some ideas:
  1. Maybe the dihedral is an optical illusion due to the v weird, nearly orthographic camera perspective. That would mean the wings are swept in a v strange way...
  2. Maybe this forum's conclusions about the BOP dihedral were incorrect and there is some stealth or aerodynamic advantage in this wing shape. I vaguely remember an earlier USAF quote mentioned something about interesting new aerodynamics. To be honest, this feels like most likely conclusion.
  3. I'd not really considered that there is something else hidden in the image, but it is a super intriguing thought! I wish I had more imagination!
Click to expand...
I don't think so, the Face View in the hangar show there is a positive dihedral.
 
If we can understand F22 intake & F119 engine, we can guess & corelate F-47 intake & engine.

F119Doctor said:
The rated air mass flow (AMF) of an engine will be at Mil Power at sea level (14.7 psi) standard day (59F) conditions behind a zero loss inlet bellmouth - engine inlet conditions (station 2) will be Pt2 = 14.7 psi, Tt2=59F.

When the engine is running at or below it's flat rated inlet temperature, it will be flowing "corrected AMF" at Mil and AB power. The actual AMF is corrected by ratio of actual Pt2 / 14.7.

Let's have a theoretical engine with a rated AMF of 100 pps. At sea level, zero Mn with a bellmouth inlet, actual AMF will be 100 pps. If you moved the test cell up to 20K ft, the inlet conditions would be 6.75 psi / -12F. Actual airflow would 100pps x 6.75 / 14.7 = 45.9pps. Gross thrust would 45.9% of rated thrust.

If you put that engine into a aircraft and accelerated to 0.9Mn at 20K ft, the inlet conditions increase to 11.42 psi Pt2, 60F Tt2. Actual airflow would be 100 pps x 11.42 / 14.7 = 77.7 pps

Go back down to sea level at M0.9, and the inlet conditions are 24.86 Pt2, 143F Tt2. If the engine was able to run up to it full rated airflow at that inlet temperature, the actual airflow would 100 pps x 24.86 / 14.7 = 169 pps.
Click to expand...

Is there any online calculator for Pt2 & T2 based on aircraft speed, altitude (ambient temp. T0 & press. Pt0), intake area, inlet area?

So confirming this by example - Does this mean that that F110-GE-132's quoted 124.7 Kg/s is tested value on a ramp at 100% mil power, at sea level & standard conditions 14.7 psi/1atm & 59 F / 15 C / 288.15 K?

1750261868630.png

If we combine these calculations with F119 engine, then
- RATED 138.6 Kg/s air passes through it at sea level & standard conditions, stationary engine & produces dry thrust 116 KN.
- Moving engine to 20K ft. but stationary, reduces AMF & thrust by (6.75/14.7), so dry thrust = 53.27 KN, AMF = 63.64 Kg/s.
- Accelerating engine to Mach 0.9 at 20K ft, changes AMF & thrust by (11.42/14.7), so dry thrust = 90.12 KN, AMF = 107.67 Kg/s.
- Coming down at sea level, Mach 0.9, changes AMF & thrust by (24.86/14.7), so dry thrust = 196.17 KN dry & AMF = 234.39 kg/s. Is this thrust & AMF value true or possible for F119?

The F-22 would achieve Mach 1.8 supercruise at mid-altitude, let's consider 35K ft, where atmospheric pressure is 3.53 psi & temp. is -65.6 F / -54.2 C / 218.9 K. But IDK how to calculate Pt2 & T2.

Can you please tell us what would be Pt2 & T2 & hence the AMF & dry thrust at 35K ft & Mach 1.8?
Thanks.

1750263969649.jpeg

If i put above calculations in a table, it looks like following :
Bcoz air speed before inlet is unknown, so volume is unknown, air density (AMF/vol.) unknown.

1750264653883.png

F-22 with F119 engine as reference​
Altitude​
Aircraft speed​
P0 Atmospheric pressure​
Pt2 inlet pressure​
Atmospheric air density​
Inlet air density​
Atmosphere Air temperature​
Tt2 Inlet air temperature​
Throttle​
AMF entering inlet​
AMF value​
Thrust formula​
Thrust value​
Max Air speed before inlet​
At sea level​
Zero
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
1.22500 Kg/m3​
?
59 F / 15 C / 288.15 K​
59 F / 15 C / 288.15 K​
100%/MIL power​
AMF.sea RATED​
138.6 Kg/s​
RATED Thrust​
116 KN RATED​
Mach 0.5/0.6/0.7/0.8/0.9​
At sea level​
Zero
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
1.22500 Kg/m3​
?
59 F / 15 C / 288.15 K​
59 F / 15 C / 288.15 K​
Afterburner​
AMF.sea RATED​
138.6 Kg/s​
RATED Thrust​
156 KN RATED​
Mach 0.5/0.6/0.7/0.8/0.9​
At sea level​
Mach 0.9
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
24.86 psi / 1.69 atm / 1.71 bar / 1.74 kg/cm2 / 171.4 kpa​
1.22500 Kg/m3​
?
59 F / 15 C / 288.15 K​
143 F / 61.6 C / 334.8 K​
100%/MIL power​
AMF.sea X (24.86/14.7)​
234.39 Kg/s​
Rated Thrust X (24.86/14.7)​
196.17 KN gross​
Mach 0.5/0.6/0.7/0.8/0.9​
20,000 ft.​
Zero
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
0.652694 kg/m3​
?
-12.26 F / -24.58 C / 248.56 K​
-12.26 F / -24.58 C / 248.56 K​
100%/MIL power​
AMF.sea X (6.75/14.7)​
63.64 Kg/s​
Rated Thrust X (6.75/14.7)​
53.27 KN gross​
Mach 0.5/0.6/0.7/0.8/0.9​
20,000 ft.​
Zero
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
0.652694 kg/m3​
?
-12.26 F / -24.58 C / 248.56 K​
-12.26 F / -24.58 C / 248.56 K​
Afterburner​
AMF.sea X (6.75/14.7)​
63.64 Kg/s​
Rated Thrust X (6.75/14.7)​
71.63 KN gross​
Mach 0.5/0.6/0.7/0.8/0.9​
20,000 ft.​
Mach 0.9
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
11.42 psi / 0.77 atm / 0.78 bar / 0.8 kg/cm2 / 78.73 kpa​
0.652694 kg/m3​
?
-12.26 F / -24.58 C / 248.56 K​
60 F / 15.5 C / 288.7 K​
100%/MIL power​
AMF.sea X (11.42/14.7)​
107.67 Kg/s​
Rated Thrust X (11.42/14.7)​
90.12 KN​
Mach 0.5/0.6/0.7/0.8/0.9​



F119Doctor said:
The inlet does not adjust to get the engine its rated airflow. If it has moving parts, they are there to make the inlet work to provide the highest ram pressure at the lowest temperature and distortion at the engine face. At very low speeds, the aircraft inlet can be a flow restriction to the engine because the inlet throat is smaller than the engine diameter. Because of this, you sometimes see things like blow-in doors (early B747 engines) or moveable lower lips (Eurofighter) to increase the actual airflow and thrust at very low speeds. At supersonic speeds, the movable ramps, spikes, and bleeds are there to position the shocks for the best ram compression while maintaining stable flow thru the inlet.
Click to expand...
Yeah i said & showed this already.


F119Doctor said:
Back to the flat rating statement - most engines are controlled to a flat rated inlet temperature. They usually don't include the flat rated temperature in the engine specifications unless that temperature is significantly higher than standard day 59F. Compressor rotor speed is corrected by temperature: Corrected RPM = Indicated RPM / Square Root (Absolute Tt2 / 519R). Constant Corrected RPM = Constant Airflow. Below the flat rated inlet temperature, the engine will run at a constant Corrected RPM and corrected airflow. As the inlet temperature drops, the indicated RPM and turbine temperatures also get lower while the airflow and thrust remain relatively constant. Conversely, indicated RPM and turbine temperatures increase with increasing inlet temperatures. Above the flat rated inlet temperature, the engine runs into a rotor speed and / or turbine temperature limits and will run at the limit. Actual RPM stays constant, but the corrected RPM and airflow decrease along with thrust. In the SR-71, running at its inlet temperature limit of 800F (1260R) with the engine at 100% indicated RPM is only at 64% corrected RPM, which is only a little above Idle in terms of thrust and airflow.
Click to expand...

Corrected RPM, Actual RPM, Rated RPM, Indicated RPM, this bounced off my head. o_O
Can you provide some link or video to understand this RPM formula?

I had some diagram by which fortunately i got that "519 R" means 519 RANKINE, a temp. unit which i didn't know.

So it turns out that 59 F = 519 R approx. = 15 C = 288.15 K.

When i put your formula in Google search then it doesn't give straight same result.
A YT video appears in result:

View: https://www.youtube.com/watch?v=RkEC-2MlsJ0


It says as you stated "Corrected speed is that speed at which the inlet temp. is that of ambient temp. at sea level, standard day i.e. 288.15 K."
It reduced the N1 RPM relation to Temp. as - when T0 increased to T1 then (N1/N0)^2 = T1/T0

1750269559221.jpeg
1750269569238.jpeg

So d=1meter, N0 = Rev/sec = 60*M*sqrt(Y*R*T0)/3.14
Re-labelling for real scenario -
N1 R/sec = 60*M*sqrt(Y*R*T2)/3.14
where M - Mach # before fan inlet, but we don't know its value - M 0.5/0/6/0.7/0.8????
Y = Specific heat ratio of ambient air, diffuser, fan = 1.4
R = Gas constant = 286 J/Kg/K or m2/s2/K
T2 = Fan inlet temp. = ???? F-22 has convergent-divergent diffuser duct so the T2 & P2 are more than atmospheric P0 & T0.

But how to get T2 & P2?

F119 engine diameter = 100 cm = 1m
max theoretical fan tip linear velocity = Mach 1 = 343 m/s
max theoretical RPM = 60*343 / (2*3.14*0.5) = 6554.14 RPM = 109.23 RPS

N1 = 60*M*sqrt(Y*R*T2)/3.14
6554.14 = 60*1*sqrt(1.4*286*T2)/3.14
Max T2 before inlet = 293.82 K = 20.67 C = 69.2 F = 528.87 R
which is little above sea level Std. temp. 59 F = 519 R approx. = 15 C = 288.15 K
But obviously the engine won't run its tip at Mach-1.

So let's consider tip speed at Mach 0.8, a safe gap & T2 same = 293.82 K.
the N1 RPM= 60*0.8*sqrt(1.4*286*293.82)/3.14
N1 RPM = 5243.23 Rev/min. This would be, or close to, SAFE 100% MIL power RPM of F119 engine.

I hope these calculations are correct.
But still IDK how to calculate Pt2, T2 based on aircraft speed, altitude (ambient temp. T0 & press. Pt0), intake area, inlet area?
And bcoz air speed before inlet is unknown, so volume is unknown, air density (AMF/vol.) unknown.
 
Scott Kenny said:
Page 12 is where I last saw this thread. Now on page 17 or 18...

While I've been pushing for that as a "don't be stupid, make your bays bigger than they absolutely have to be per contract so that you can put bigger booms in there" I don't believe that the F-47 has bays that big.
Click to expand...
Actually it's not that big. My aim was to make the bay(s) as compact as possible.

I came to the conclusion that a side by side arrangement (e.g. F-22) cannot provide the efficient use of space and versatility like this layout. It enables carrying of outsized A/G ordnance without having to inflate the bay specifically for this puropse.

In other words, each bay accepts five AMRAAM sized weapons (stacked), or two 2000 lbs JDAM sized weapons (side by side), in the same box. Due to the longitudinal stacking of AMRAAMs this bay would also accept future LRAAM, which most likely exceed AIM-120/260 in length. And it could fit JSOW (4,1 m), JASSM (4,27 m), or future expendable CCA of similar size as well.

NGDA_400_IWB_038.png
 
Last edited:
@VTOLicious

I think your bay is too compact when you include more real life issues:
  • Clearance angles from weapons after release from the end of the ram extension e.g. as per below diagram. This will particularly impact your roof mounted AMRAAMs clearance to the door mounted weapons.
  • Actual release of the top mounted weapons from the bay. At the end of the ejection stroke they are still a long way inside the bay. They might not make it through the boundary layer into the free stream.
  • A different type of launcher with longer stroke would help solve the above two issues but add depth to the bay.
  • Ejection force path for the door mounted weapons. The forces are significant. For F-35 they are routed through the hinge directly whereas here there is a very indirect path to the small swan neck hinge.
Including these factors results in a much larger bay
 

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@red admiral, thank you for your feedback.

Ejection envelope is indeed an area which requires attention, and I actually considered it in the design.
I've based my assumptions on Boeing's Enclosed Weapons Pod design (US 9,180,967 B2), which shows a similar arrangement and thight stacking of missiles... Fig. 11B, 12A/B are about ejection envelope considerations discussed in the patent.
Screenshot 2025-06-19 202950.png
 
VTOLicious said:
Ejection envelope is indeed an area which requires attention, and I actually considered it in the design.
Click to expand...
Do the doors on your bay open to a similarly large angle to get the clearance? Probably some negatives around opening times, max speed with doors open, and some other aspects.

The other method for compact carriage is a more traditional bay with sequential launch and some mechanical mechanism to move the weapons.

Both approaches have negatives around the range of store types and combinations of stores. And how to load the weapons without employing children with small arms and hands :)
 
Low IQ Techie said:
If we can understand F22 intake & F119 engine, we can guess & corelate F-47 intake & engine.



Is there any online calculator for Pt2 & T2 based on aircraft speed, altitude (ambient temp. T0 & press. Pt0), intake area, inlet area?

So confirming this by example - Does this mean that that F110-GE-132's quoted 124.7 Kg/s is tested value on a ramp at 100% mil power, at sea level & standard conditions 14.7 psi/1atm & 59 F / 15 C / 288.15 K?

View attachment 774882

If we combine these calculations with F119 engine, then
- RATED 138.6 Kg/s air passes through it at sea level & standard conditions, stationary engine & produces dry thrust 116 KN.
- Moving engine to 20K ft. but stationary, reduces AMF & thrust by (6.75/14.7), so dry thrust = 53.27 KN, AMF = 63.64 Kg/s.
- Accelerating engine to Mach 0.9 at 20K ft, changes AMF & thrust by (11.42/14.7), so dry thrust = 90.12 KN, AMF = 107.67 Kg/s.
- Coming down at sea level, Mach 0.9, changes AMF & thrust by (24.86/14.7), so dry thrust = 196.17 KN dry & AMF = 234.39 kg/s. Is this thrust & AMF value true or possible for F119?

The F-22 would achieve Mach 1.8 supercruise at mid-altitude, let's consider 35K ft, where atmospheric pressure is 3.53 psi & temp. is -65.6 F / -54.2 C / 218.9 K. But IDK how to calculate Pt2 & T2.

Can you please tell us what would be Pt2 & T2 & hence the AMF & dry thrust at 35K ft & Mach 1.8?
Thanks.

View attachment 774888

If i put above calculations in a table, it looks like following :
Bcoz air speed before inlet is unknown, so volume is unknown, air density (AMF/vol.) unknown.

View attachment 774890

F-22 with F119 engine as reference​
Altitude​
Aircraft speed​
P0 Atmospheric pressure​
Pt2 inlet pressure​
Atmospheric air density​
Inlet air density​
Atmosphere Air temperature​
Tt2 Inlet air temperature​
Throttle​
AMF entering inlet​
AMF value​
Thrust formula​
Thrust value​
Max Air speed before inlet​
At sea level​
Zero
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
1.22500 Kg/m3​
?
59 F / 15 C / 288.15 K​
59 F / 15 C / 288.15 K​
100%/MIL power​
AMF.sea RATED​
138.6 Kg/s​
RATED Thrust​
116 KN RATED​
Mach 0.5/0.6/0.7/0.8/0.9​
At sea level​
Zero
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
1.22500 Kg/m3​
?
59 F / 15 C / 288.15 K​
59 F / 15 C / 288.15 K​
Afterburner​
AMF.sea RATED​
138.6 Kg/s​
RATED Thrust​
156 KN RATED​
Mach 0.5/0.6/0.7/0.8/0.9​
At sea level​
Mach 0.9
14.7 psi / 1 atm / 1.01 bar / 1.03 Kg/cm2 / 101.35 kpa​
24.86 psi / 1.69 atm / 1.71 bar / 1.74 kg/cm2 / 171.4 kpa​
1.22500 Kg/m3​
?
59 F / 15 C / 288.15 K​
143 F / 61.6 C / 334.8 K​
100%/MIL power​
AMF.sea X (24.86/14.7)​
234.39 Kg/s​
Rated Thrust X (24.86/14.7)​
196.17 KN gross​
Mach 0.5/0.6/0.7/0.8/0.9​
20,000 ft.​
Zero
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
0.652694 kg/m3​
?
-12.26 F / -24.58 C / 248.56 K​
-12.26 F / -24.58 C / 248.56 K​
100%/MIL power​
AMF.sea X (6.75/14.7)​
63.64 Kg/s​
Rated Thrust X (6.75/14.7)​
53.27 KN gross​
Mach 0.5/0.6/0.7/0.8/0.9​
20,000 ft.​
Zero
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
0.652694 kg/m3​
?
-12.26 F / -24.58 C / 248.56 K​
-12.26 F / -24.58 C / 248.56 K​
Afterburner​
AMF.sea X (6.75/14.7)​
63.64 Kg/s​
Rated Thrust X (6.75/14.7)​
71.63 KN gross​
Mach 0.5/0.6/0.7/0.8/0.9​
20,000 ft.​
Mach 0.9
6.75 psi / 0.459 atm / 0.465 bar / 0.474 Kg/cm2 / 46.53 kpa​
11.42 psi / 0.77 atm / 0.78 bar / 0.8 kg/cm2 / 78.73 kpa​
0.652694 kg/m3​
?
-12.26 F / -24.58 C / 248.56 K​
60 F / 15.5 C / 288.7 K​
100%/MIL power​
AMF.sea X (11.42/14.7)​
107.67 Kg/s​
Rated Thrust X (11.42/14.7)​
90.12 KN​
Mach 0.5/0.6/0.7/0.8/0.9​




Yeah i said & showed this already.




Corrected RPM, Actual RPM, Rated RPM, Indicated RPM, this bounced off my head. o_O
Can you provide some link or video to understand this RPM formula?

I had some diagram by which fortunately i got that "519 R" means 519 RANKINE, a temp. unit which i didn't know.

So it turns out that 59 F = 519 R approx. = 15 C = 288.15 K.

When i put your formula in Google search then it doesn't give straight same result.
A YT video appears in result:

View: https://www.youtube.com/watch?v=RkEC-2MlsJ0


It says as you stated "Corrected speed is that speed at which the inlet temp. is that of ambient temp. at sea level, standard day i.e. 288.15 K."
It reduced the N1 RPM relation to Temp. as - when T0 increased to T1 then (N1/N0)^2 = T1/T0

View attachment 774902
View attachment 774903

So d=1meter, N0 = Rev/sec = 60*M*sqrt(Y*R*T0)/3.14
Re-labelling for real scenario -
N1 R/sec = 60*M*sqrt(Y*R*T2)/3.14
where M - Mach # before fan inlet, but we don't know its value - M 0.5/0/6/0.7/0.8????
Y = Specific heat ratio of ambient air, diffuser, fan = 1.4
R = Gas constant = 286 J/Kg/K or m2/s2/K
T2 = Fan inlet temp. = ???? F-22 has convergent-divergent diffuser duct so the T2 & P2 are more than atmospheric P0 & T0.

But how to get T2 & P2?

F119 engine diameter = 100 cm = 1m
max theoretical fan tip linear velocity = Mach 1 = 343 m/s
max theoretical RPM = 60*343 / (2*3.14*0.5) = 6554.14 RPM = 109.23 RPS

N1 = 60*M*sqrt(Y*R*T2)/3.14
6554.14 = 60*1*sqrt(1.4*286*T2)/3.14
Max T2 before inlet = 293.82 K = 20.67 C = 69.2 F = 528.87 R
which is little above sea level Std. temp. 59 F = 519 R approx. = 15 C = 288.15 K
But obviously the engine won't run its tip at Mach-1.

So let's consider tip speed at Mach 0.8, a safe gap & T2 same = 293.82 K.
the N1 RPM= 60*0.8*sqrt(1.4*286*293.82)/3.14
N1 RPM = 5243.23 Rev/min. This would be, or close to, SAFE 100% MIL power RPM of F119 engine.

I hope these calculations are correct.
But still IDK how to calculate Pt2, T2 based on aircraft speed, altitude (ambient temp. T0 & press. Pt0), intake area, inlet area?
And bcoz air speed before inlet is unknown, so volume is unknown, air density (AMF/vol.) unknown.
Click to expand...
I would recommend downloading a copy of the Pratt & Whitney Aeronautical Vest Pocket Handbook. https://www.scribd.com/document/279...-Handbook-Pratt-Whitney-22nd-Edition-Sep-1991

In the Handbook there are lots of formulas on jet engine and rocket engine performance to keep you busy for a month, at least. In addition, there is a table show showing standard ram recovery values of PT2 and TT2 at various Mn and altitudes from see level static to M3 / 80k ft. These values are based on a design Mil Spec and may represent the best case, with all inlet designs being less than optimum. Real world inlets will probably have lower ram recovery and higher temperatures when operating outside their design Mn.

I can’t answer all of your questions, but the inlet duct Mn at the engine face is usually in the 0.6 to 0.8 range. While the airflow inside the inlet duct must be subsonic, the speed of the fan blade tips is not limited to M1. The J91 was the first P&W engine with a transonic compressor, with the blade tips running beyond M1. The J58 was originally an 80% scaled down version of the J91, and the first two stages of the J91 compressor were used in place of the first three stages of the J57 to create the TF33/JT3D turbofan engine. From what I understand, most turbofan engines are designed with the first stage blade tips moving thru the inlet air at approximately M1.4.

I will also caution you that F119 engine airflow, pressure ratio, rotor speeds, turbine inlet temperatures have never been publicly released. Using Chinese guesses will just make your guesses just that more inaccurate.

The F110-132 airflow would be that value at sea level, zero Mn, 15C on the test cell with a zero loss bellmouth. When installed, the actual airflow will be less due to the aircraft inlet restriction bringing the inlet pressure below the 14.7 psi standard number. Once the aircraft starts moving with the airflow aligned with the intake, you are probably back up to the 14.7 psi around M 0.3, and the increasing with additional Mn. But, the faster you go, the higher the inlet temperature. In addition, the installed engine is supplying bleed air and gearbox horsepower to the airframe, both of which increase TIT and move it closer to the limit. The effect is relatively small at low altitudes, but become a larger and larger percentage impact as the aircraft goes higher and slower.
 
VTOLicious said:
Actually it's not that big. My aim was to make the bay(s) as compact as possible.

I came to the conclusion that a side by side arrangement (e.g. F-22) cannot provide the efficient use of space and versatility like this layout. It enables carrying of outsized A/G ordnance without having to inflate the bay specifically for this puropse.

In other words, each bay accepts five AMRAAM sized weapons (stacked), ot two 2000 lbs JDAM sized weapons (side by side), in the same box. Due to the longitudinal stacking of AMRAAMs this bay would also accept future LRAAM, which most likely exceed AIM-120/260 in length. And it could fit JSOW (4,1 m), JASSM (4,27 m), or future expendable CCA of similar size as well.

View attachment 775109
Click to expand...
Okay, that's a more likely size for sure. I'm still slightly concerned about ejection clearance, can you do a picture of the AMRAAM loaded bay with doors open and compare that to the EWP?

Also, JASSMs (and AARGM-ERs) are significantly wider than 2000lb JDAMs, ~25" versus ~19". Plus you need room for the bomb techs to reach around the ordnance to lock it onto lugs, tighten bolts, and connect data umbilicals.
 
Scott Kenny said:
*yoink!*

@Low IQ Techie I assume that you also have a copy of Gas Turbine Engine Aerothermodynamics (originally written by Sir Whittle)?
Click to expand...
Not really bro, never heard his name also, LOL! :D I'm IT engineer, not aero-engineer. Otherwise i would have already got all the answers.
But i'm trying my best to ultimately simplify things in a table for everybody to understand easily.
 
red admiral said:
Do the doors on your bay open to a similarly large angle to get the clearance? Probably some negatives around opening times, max speed with doors open, and some other aspects.

The other method for compact carriage is a more traditional bay with sequential launch and some mechanical mechanism to move the weapons.

Both approaches have negatives around the range of store types and combinations of stores. And how to load the weapons without employing children with small arms and hands :)
Click to expand...

Angle of ejection envelope shown in the drawing is 20°. Door mounted missiles would be launched first. These mounts are also the ones that could rail-launch short-range missiles (or eject small stuff, e.g. SDB).

Screenshot 2025-06-19 202710.png
 
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Scott Kenny said:
Okay, that's a more likely size for sure. I'm still slightly concerned about ejection clearance, can you do a picture of the AMRAAM loaded bay with doors open and compare that to the EWP?
Click to expand...
See post #678

Also, JASSMs (and AARGM-ERs) are significantly wider than 2000lb JDAMs, ~25" versus ~19". Plus you need room for the bomb techs to reach around the ordnance to lock it onto lugs, tighten bolts, and connect data umbilicals.
Click to expand...
I never said two of those would fit side by side ;)

Anyways, the more interesting option is probably the accommodation of future Expendable CCA / Remote Carriers.
 
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