missing info MiG-29A (Maneurability performance)

[AeriaGloria]

1 hour ago, Kuky said:

Well, ED should take into account that some countries have tuned down their engines so to make them last longer, I believe DDR did that, also after being integrated into NATO they also got less engane performance due to different fuel used. Si if ED used data from such aircraft they (again - referringto modellingof SPO) modelled downgraded and not as per original specs, aircraft.

It’s a switch in the wheel well. Easily set to full for exercises according to past German pilots 

[Lidozin]

12 hours ago, Logan54 said:

Specific excess power comparson game (blue+red) vs real data (green)

   The core problem is that a near-vertical climb at constant speed destroys the fundamental assumptions behind SEP. In this regime the aircraft is no longer in a 1g condition; lift and normal load factor collapse toward zero, and induced drag essentially disappears. That immediately breaks equivalence with the standard SEP definition, which is rooted in 1g flight and is directly comparable to acceleration in level flight. You are no longer measuring excess power under the same aerodynamic conditions, so the result is not “SEP with error bars,” it is a different physical regime altogether. At that point the comparison is invalid in principle, not just in accuracy.

   On top of that, the results become even less trustworthy for practical reasons. The altitude band from 0 to 1000 m is traversed extremely quickly in a high-energy fighter, so you are trying to infer a local quantity from a very short, highly integrated maneuver. The transition into the climb inevitably introduces significant normal load, further corrupting the energy balance right where you want clean data. And once established in a constant-speed, high-pitch climb, lift and g are almost absent, so any pilot input needed to “hold the regime” shows up as fluctuations in load factor and energy partitioning, directly increasing measurement error. Even without invoking second-order, the method fails at the level of basic physics.

   The methodology of determining SEP via vertical climb rate at constant airspeed is only more or less valid for aircraft with low thrust-to-weight ratio, where the flight path angle remains small and the maneuver can reasonably be treated as a 1g condition. In that regime, lift is still providing nearly all of the weight support, induced drag remains present, and the energy balance is close to that of level flight with a small vertical component. Under those assumptions, vertical speed can be used as a rough proxy for specific excess power.

   Once thrust-to-weight becomes high, the climb angle grows large, normal load factor departs significantly from 1, and the aircraft transitions into a thrust-supported climb rather than a lift-supported one. At that point the relationship between vertical speed and SEP breaks down, because induced drag collapses and excess power is no longer being expended primarily against weight in a 1g framework. The method does not merely lose accuracy; it ceases to represent the same physical quantity.

   

I really enjoy conducting tests on virtual aircraft; they are far safer and significantly cheaper than testing real ones, and I believe that in this thread my results would be more than appropriate.

 

 

 

 

[DummyCatz]

To calculate 1g SEP (Vy) you can simply measure the level flight acceleration. Take note of the True Airspeed (TAS), and use the equation as I did in the below thread, where I confirmed discrepancies in SEP of both DCS F-16 and F-18:

For example, take note of TAS (in m/s) at several seconds apart, like how much of TAS you gained (dV) in seconds (dt), using slow motion and align the time accurately. This will be your dV/dt.

SEP = (V/g)(dV/dt)

V is just your TAS, in m/s, and g is 9.81 m/s^2. So the SEP will be in m/s too.

BTW, isn't the 330m/s come from a configuration of 2 x R-60 and 2 x APU-470 pylon with a weight of 13000kg, instead of 2 x R-27 and 2 x R-73?

 

 

On 1/6/2026 at 5:49 PM, Logan54 said:

Some Hornet performance data, that can be useful:  (page 30)  https://www.gao.gov/assets/nsiad-96-98.pdf

In DCS with this loadout (2*AIM-9+2*AIM-120C) I had perfect timing for level acceleration (for 5 000 ft, 20 000ft and 35 000ft), but STR in DCS better
20.8-21 deg/s at sea level (430kts) that biger then 19.2 deg/s
and 13.3 deg/s at 15.000ft (0.73 M) that biger then 12.3 deg/s
I also tried to check speed bleed during hard turn from 430 kts, I slowed down to 250kts in 8s, but this is only (430-250)/8=22.5 kts/s but should be 54 kts/s, this means that ingame Hornet hold speed.
 

 

You're welcome to do further flight testing in this F18 bleed rate thread:

 

Edited January 9 by DummyCatz

[Logan54]

5 часов назад, DummyCatz сказал:

SEP = (V/g)(dV/dt)

Thank you for this idea, I checked and Ps really 330 at sea level.
I m sory for my weird calculating in the past. Seems like I thought that Ps just vertical speed but its not.

[Kefa]

According to the Basic Employment Manual, the USAF simply defines Ps as 'rate of energy loss or gain during a particular maneuver.'  And this particular maneuver as according to the -1-1 is defined as either climb/dive or turning (Pg D8-5)

   

I am guessing that the Mig-29's 330ms at sl  (about 65,000fpm)  is probably better than nearly all the teen jets.  Only with the F-16 manual, do we have actual figures for +/-Ps from the jets manufacturer; and for the Block 50/52 versions.  The F-14A/B manual has some.  And the Hornet's Ps diagrams are not even in its performance supplement.   Its turn figures are in a totally different booklet, which requires a security clearance, as which none here have.  Flxtrm and Eidetics are not the official NATOPS.  The only public official NATOPS figures are scant in number and in the GAO document.  According to real life pilots (USAF/USN/GAF/RCAF),  the Mig-29's energy addition rates are superior to the F-18 (any), F-14 (any), F-15 (any), and nearly all Blocks of the F-16.  

 

[IvanK]

Ps=V(T-D)/W

[DummyCatz]

Full form is Ps = dh/dt + (V/g)(dV/dt) = V(T-D)/W

But be careful, T and D are not in the same frame of reference.

Ref Page 21 of https://dspace.mit.edu/bitstream/handle/1721.1/36378/16-885JFall2003/NR/rdonlyres/Aeronautics-and-Astronautics/16-885JFall2003/1A2F9B8F-B307-4461-83AE-9CCD658DBD50/0/aircraft_murman.pdf

[Logan54]

Se=dh/dt+(v/g)(dv/dt) seems working good at sea level, but not shows real things higher because this is not real aircraft, no real engine, no real drag and even not real air... 
Developer using script that accelerate aircraft in programmed dots, so between them there is some kind of timer to the next dot.
I waste some time with this math, but I undertood something. Ingame jet acceleration is kind of "boost-delay-boost" sequence.
Ps in the sec 1 and sec 2 could be very different even if it should be the same. Calculating with combining 2 seconds as 1 Ps (to decrease pulsating) also can not show real things.
Officially I can not say this is real Ps of ingame 29, because I calculated Ps only near dots like 0.5M, 0.6M,..etc but I`ve not calculate Ps between them, so it mostly wasted time imo.
I will test 29 acceleration with timer later.

Edited January 10 by Logan54

[BIGNEWY]

Please include the track replays from your test. 

thank you 

[Logan54]

2 часа назад, BIGNEWY сказал:

Please include the track replays from your test. 

thank you 

Pitch up with 0.9M+
As you can see there is no acceleration there. And only 285 m/s.
Not close this thread please. I think I have to do more stuff like this.
Thank you.

Tacview-20260110-144242-DCS.zip.acmi 29 vert_test .trk

[Lidozin]

5 hours ago, Logan54 said:

Btw, I think need to check vertical 90° acceleration with all available data from aerodynamic book.
From this source I know that engine thrust during flight at 0.9M=10000 kgs (kilogram-strength), so total thrust of 2 engine would be 20000 kgs.

Drag index (Cx) for 0.9M=0.025 for clean airplane.
Cx for 2*APU-470=0.00075
Cx for 2*R-60M=0.002
Total Cx=0.025+0.00075+0.002=0,02775

Step 1: Determination of atmospheric parameters and velocity at an altitude of H=1000 m (ISA), the air density ρ =1,1117 kg/m^3, the speed of sound a=336.43 m/s.
The true speed (V) for M=0.9:
V=0.9*336.43=302.79 m/s

Step 2: Calculation of velocity pressure (q) and Drag force (D):
q=(1,1117*302,79^2)/2=50950,5 Pascals

Drag force  in Newtons:
D=Cx*q*S, where S-wing area=38.056 m^2
D=0.2775*50950.5*38.056=53 811.5 H
Conversion to kilogram-strength:
Dkgs=5 3811.5 / 9.80665=5 487.2 kgs

Step 3: Calculate the acceleration in the vertical set (90°) In vertical flight, the aircraft is affected by thrust (T), drag (D) and weight (W).Total force (F):
F=T-D-W
In kgs: Fkgs=20 000-5 487.2-13 000=1 512.8 kgs
Convert the excess force to Newtons to calculate the acceleration:  Fh=1512.8*9.80665=14835 H
Acceleration (a):
a=14835/13000≈1.141 m/s^2

Step 4: Calculation of Specific excess power (Ps) (characterizes the energy potential of the aircraft (possible rate of climb)):
Ps=(T-D)*V/W=(20 000-5 487.2)*302.79/13000≈338m/s

Answer: The aircraft accelerates vertically because the available thrust (20,000 kgf) is greater than the sum of weight and drag (18487.2 kgf). The acceleration is 1.14 m/s2, and the theoretical rate of climb (Ps) is 338 m/s.

Let me check how much vertical speed gived DCS during 90°climb:
Vs=17036/60=283.9 m/s, its 54 m/s less then should be.

What is it? Additional drag or not enough thrust? Or both?
Thank you.
 

 

null

[Logan54]

41 минуту назад, Lidozin сказал:

 

null

What is it?

[Lidozin]

It's a tombstone for your entire test.

 

Even if the simulator integrates the equations of motion with a small time step and the thrust values you are using (10 000 kgf per engine at 1 km and M 0.9) are indeed “available thrust” from the documentation, the comparison you are attempting is still fundamentally invalid because the measured regime does not correspond to the definition of Ps you are trying to verify.

The reference Ps data you are using are explicitly defined for 1g, steady, horizontal flight. That is a very specific physical condition: lift equals weight, induced drag is present at its normal 1g level, the flight path angle is zero, and the aircraft is in a locally steady energy state. Ps in that context is a local energetic property of a single point (V, h, ny = 1).

What is shown in the telemetry screenshot is something completely different. With G ≈ 6.5 and AoA ≈ 6–7°, the aircraft is in a highly loaded maneuver. This is not a “vertical steady climb” in the energetic sense, but a transient pull-up where a large fraction of the available energy is being spent on generating lift and rotating the velocity vector. In such a condition, induced drag increases dramatically (proportional to CL²), and the force balance along the trajectory cannot be reduced to the simple T − D − W expression derived for 1g reference conditions.

Because of this, the lower observed vertical speed does not indicate missing thrust or excessive parasite drag in the model. It is exactly what you should expect when you compare a 1g reference quantity to data taken from a high-g, non-stationary maneuver. You are not measuring the same physical quantity.

There is also a conceptual mismatch in comparing Ps directly to vertical speed. Even in a vertical trajectory, Ps is not simply dh/dt; by definition it also includes the kinetic energy term (V/g)·dV/dt. Unless the motion is both steady and correctly constrained, Ps and vertical speed will not coincide numerically, even if the model were perfect.

The correct conclusion is therefore not “DCS has extra drag” or “engine thrust is insufficient.” The correct conclusion is that this test has no diagnostic value for Ps at all. Vertical pull-ups or near-vertical accelerations cannot be used to validate Ps data that are defined for 1g horizontal flight. This is not a matter of tuning or refining the method; it is a methodological error at the level of experiment design.

If the goal is to compare with reference Ps data, the only meaningful approach is a controlled 1g horizontal test with fixed mass, configuration, and engine regime, and Ps extracted from properly smoothed acceleration data. Anything outside that regime simply measures something else.

 

 

 

[Logan54]

1 час назад, Lidozin сказал:

In short: the diagram is about what is energetically possible in steady flight, not about what can be “caught” during a pull-up near the ground.

I honestly think we are talking on different languages. I`ve calculated that 29 should have 338 m/s vertical climb performance during it;
I loaded all posible docs for check vertical climb performance.
I said about sustained climb rate
I honestly not understand your point. There is altitude on the left side, that is 0 m, yes, for 1000m it would be another, but this jet have more weight if you noticed.
You cycling around my old screenshoot, and seems not hear me.
The final result doesn't depend on my G`s on screenshoot. I have no idea how to explain, sorry. You use some kind of your own opinion without any official confirmation.
I m really don`t know how react on this.
Could you explain your owb opinion via phisycs or any scientific statement? I`m really not understand what you actually talking about (in physix way).
Steady rate climb with 338m/s possible with vertical climb if your engine` T/W= or more then 1. As I already said, btw I think you not hear me.
Thank you.

Edited January 11 by Logan54

[DummyCatz]

22 hours ago, Logan54 said:

Se=dh/dt+(v/g)(dv/dt) seems working good at sea level, but not shows real things higher because this is not real aircraft, no real engine, no real drag and even not real air... 
Developer using script that accelerate aircraft in programmed dots, so between them there is some kind of timer to the next dot.
I waste some time with this math, but I undertood something. Ingame jet acceleration is kind of "boost-delay-boost" sequence.
Ps in the sec 1 and sec 2 could be very different even if it should be the same. Calculating with combining 2 seconds as 1 Ps (to decrease pulsating) also can not show real things.
Officially I can not say this is real Ps of ingame 29, because I calculated Ps only near dots like 0.5M, 0.6M,..etc but I`ve not calculate Ps between them, so it mostly wasted time imo.
I will test 29 acceleration with timer later.

I think you can continue to smooth out the level flight acceleration data and calculate Ps from the smoothed data. Just need to make sure your data and trackfile is there to be submitted as a bug report.

[Lidozin]

9 hours ago, Logan54 said:

I honestly think we are talking on different languages. I`ve calculated that 29 should have 338 m/s vertical climb performance during it;
I loaded all posible docs for check vertical climb performance.
I said about sustained climb rate
I honestly not understand your point. There is altitude on the left side, that is 0 m, yes, for 1000m it would be another, but this jet have more weight if you noticed.
You cycling around my old screenshoot, and seems not hear me.
The final result doesn't depend on my G`s on screenshoot. I have no idea how to explain, sorry. You use some kind of your own opinion without any official confirmation.
I m really don`t know how react on this.
Could you explain your owb opinion via phisycs or any scientific statement? I`m really not understand what you actually talking about (in physix way).
Steady rate climb with 338m/s possible with vertical climb if your engine` T/W= or more then 1. As I already said, btw I think you not hear me.
Thank you.

You are missing one basic thing: drag depends on lift.

Your screenshot shows G = 6.5.
That means the wing is producing 6.5× weight in lift.

For your own inputs (H≈1000 m, M=0.9, m=13000 kg, S=38.056 m²):

Required lift coefficient:
CL ≈ 0.43

From the trimmed polar for CL ≈ 0.43:
CD ≈ 0.045, not 0.025.

Now the numbers:

Dynamic pressure term:
q·S ≈ 1,939,000 N

Drag:
D = q·S·CD ≈ 87,300 N ≈ 8,900 kgf

You assumed:
D ≈ 5,500 kgf (using CD0 only)

That error alone is +3,400 kgf of drag.

With your own thrust and weight:
T = 20,000 kgf
W = 13,000 kgf

Available force along the flight path:
T − D ≈ 11,100 kgf

This is less than weight.

So:

•  steady vertical climb at constant speed is impossible
•  acceleration along the vertical is negative
•  your 338 m/s result cannot exist in this regime

Correct SEP for these conditions:
Ps ≈ 259 m/s, not 338 m/s

Your mistake is simple:
you used CD0 (lightly loaded flight)
while flying at 6.5 g (highly loaded flight).

I am quite confident you will see those 338 m/s at 1000 m, once the aircraft is already fully established in a steady unloaded vertical climb, which, unfortunately, is not a regime that can be entered or sustained at low altitude.

[Logan54]

1 час назад, Lidozin сказал:

you used CD0 (lightly loaded flight)

At first you said that vertical test illegal, now you trying to deal with it
I really can not understand one thing.
Why don`t you do your own test with this set up?
Load out: 2*R-60+2*APU, GW=13 000kg, standard pressure, +15°
Show me how to have 338m/s with ingame MiG-29, please.
Thank you.

 

8 часов назад, DummyCatz сказал:

I think you can continue to smooth out the level flight acceleration data and calculate Ps from the smoothed data. Just need to make sure your data and trackfile is there to be submitted as a bug report.

For doing it in right way, need to calculate every second of the test. I have not so much free time for this. And output data not so stable for doing any statement. Also I can not control aircraft without trimming, so testing aircraft flying up and down near the target altitude, this thing also not let to increase measurement accuracy. Altitude hold not helps. Thank you for the idea anyway, it was new experience for me.

Edited January 11 by Logan54

[Lidozin]

You’re still missing the same point: I am not “trying to make the vertical test work.” I am explaining why it cannot validate neither the 1g nor 0g reference Ps data, and why your particular vertical maneuver (with high G) cannot be compared to that diagram.

I’m not going to “show 338 m/s in-game” for one simple reason: 338 m/s is not a promise of what you can demonstrate at 0–1000 m by pulling to vertical at M 0.9. It is a steady-state performance value that assumes the aircraft is already established in the correct regime (steady straight-line climb, constant airspeed, and the corresponding force balance). Your test does not establish that regime, and at low altitude you do not have the vertical space to transition into it without a pull-up and the associated induced drag.

So the request “do the same setup and show 338 m/s” is exactly the methodological error: you are asking for a low-altitude transient maneuver to reproduce a steady-state performance point. That is not how flight test or performance validation works.

If you want this to be testable, provide a track where at about 1000 m altitude the aircraft holds, for several seconds:

•  M = 0.90 (constant)
•  flight-path angle near vertical (constant)
•  normal load factor close to cos γ (not 6.5g)
•  stable AoA / no pull-up

Then we can talk about whether the measured steady ROC matches the document. Without those conditions, you are not testing the diagram — you are testing your pull-up technique.

[Logan54]

53 минуты назад, Lidozin сказал:

You’re still missing the same point

You still missing there is no acceleration. LonG=1.0, but it should be more via T/W ratio more then 1.
You can also to do more math and try to explain anything you want, but what the point?

Edited January 11 by Logan54

[Lidozin]

First of all, I want to genuinely acknowledge your persistence. You kept refining the setup, reduced G, and as a result you pushed the discrepancy down to just a few percent. That is exactly how basic flight-dynamics axioms are usually rediscovered in practice.

Now let’s go through your latest snapshot step by step, using the same approach you prefer, but with all inputs made consistent.

From your screenshot and assumptions:

• Altitude H ≈ 1435 m
• TAS ≈ 1084 km/h
• Normal load factor ny ≈ 1.9
• Gross weight = 13,000 kg
• Wing area S = 38.056 m²

Step 1) Convert speed
1084 km/h = 1084 / 3.6 ≈ 301.1 m/s

Step 2) Atmosphere at 1435 m (ISA)
Air density ρ ≈ 1.065 kg/m³

Step 3) Dynamic pressure
q = 0.5 · ρ · V²
q ≈ 0.5 · 1.065 · (301.1)² ≈ 48,277 Pa

Step 4) Required lift coefficient at ny = 1.9
Lift = ny · weight, therefore:
CL = (ny · m · g) / (q · S)

CL ≈ (1.9 · 13,000 · 9.81) / (48,277 · 38.056) ≈ 0.132

This point is crucial: CL ≈ 0.13 is small, so induced drag is small, and using CD ≈ CD0 is justified here. That is why your vertical speed increased once you reduced G.

Step 5) Drag at this point
First compute qS:
qS ≈ 48,277 · 38.056 ≈ 1,837,227 N

Drag:
D = qS · CD ≈ 1,837,227 · 0.025 ≈ 45,931 N
D ≈ 45,931 / 9.80665 ≈ 4,684 kgf

Step 6) Thrust consistent with altitude (4-point interpolation as the function is nonlinear)

Given thrust per engine (kgf):

• 0 km: 10,200

• 1 km: 9,250

• 3 km: 7,900

• 5 km: 6,250

Using cubic interpolation through all four points, thrust at H = 1.435 km is:
T ≈ 8,919 kgf per engine

Total thrust (two engines):
T_total ≈ 17,838 kgf

Step 7) Specific excess power (same formula you use)
Ps = (T − D) · V / W

T − D ≈ 17,838 − 4,684 = 13,154 kgf

Ps ≈ 13,154 · 301.1 / 13,000 ≈ 304.7 m/s

Step Compare with what you measured
Your screenshot shows:
VS = 18,015 m/min ≈ 300.3 m/s

Difference:
304.7 − 300.3 ≈ 4.4 m/s
Relative error ≈ 1.5%

Final conclusion

1. Your earlier 338 m/s mismatch was not caused by “hidden drag” or “wrong engines.” It was mainly a regime issue: high ny forces high CL, which increases induced drag, so CD is no longer 0.025.

2. In this newer test you reduced ny to ~1.9. That reduced CL to ~0.13, made CD ≈ CD0 valid, and the numbers immediately converged.

3. With thrust corrected for altitude using a higher-order interpolation, predicted Ps ≈ 305 m/s and measured VS ≈ 300 m/s differ by only ~1–2%.

That is well within 5%, and in practice very good agreement — especially given that the maneuver is still not a perfectly textbook steady climb.
The rounding of the longitudinal load factor to one decimal place alone can introduce an uncertainty of about ±15 m/s, i.e. roughly 5% at a 300 m/s climb rate, which is several times larger than the observed ~1.5% (≈4–5 m/s) discrepancy.

In other words: once the assumptions behind the equations are respected, the discrepancy essentially disappears.


As a cross-check, the same result can be obtained much faster and with far less effort by flying at the same altitude and airspeed with a load factor close to 1g and then computing PsP_sPs from the standard energy equations; the values will agree within normal measurement uncertainty.

Edited January 11 by Lidozin

[Logan54]

12 часов назад, Lidozin сказал:

First of all, I want to genuinely acknowledge your persistence.

First of all, Thrust not 8 919, but ≈9 500 according to the table data. So total thrust not 17 838 but 19 000 kgf.

Step 1: Determination of atmospheric parameters and velocity at an altitude of H=1435 m:
Temp: 288.15-(0.0065*1435)=278.82K (+5.67°C)
Pressure: P=101325*(278.82/288.15)^525588≈85 488 Pa
Air density: ρ=P/(278.05*T)=85488/(287.05*278.82)≈1.0681 kg/m^3
Speed of sound (a): a=20.05*√278.82≈334.8 m/s
True airspeed (V): for 0.9M V=0.9*334.8=301.32 m/s

Step 2: Calculation of velocity pressure (q) and Drag force (D):
q=0.5*ρ*V^2=0.5*1.0681*301.32^2≈48489.2 Pa
D=Cx*q*S=48489.2*38.056≈51211.7 H
in kgf: Dkgf=51211.79/9.80665≈5221 kgf

Step 3: Calculate the acceleration in the vertical set (90°) In vertical flight, the aircraft is affected by thrust (T), drag (D) and weight (W).Total force (F):
T≈9 500*2=19000kgf
W=13 000kg*9.80665m/s^2=127486.5 H
F=T-D-W=19000-5221-13 000=779 kgf

Acceleration (a): a=F/m=779/13 000≈0.588 m/s^2
Lets find LonG:
LonG=(T-D)/W=(19 000-5221)/13 000=13 799/13 000≈1.06
I honestly can not interpolate G and Alpha during this, because there was not sustained parameters (pull down), but you see difference between our calculations.

I will do 1 more thrust test: ground speed up from 0 km/h to 0.2M (flaps up). Thrust for 0.2M is 7 700kgf*2=15 400kgf total force (accordind to the table data).
I have 15 224 kg GW MiG-29 and lets take a look on LonG during first seconds (full AB, with chocks installed).
Lets calculate theoretical LonG:

ρ=1.225 kg/m^3
V=0.18*340.3≈61,25 (around 220 km/h, that is close to 216 on the screenshoot)
q=0.5*ρ*V^2=0.5*1.225*61.25^2≈2294.6 Pa
D=0.02775*2294.6*38.056≈247.1 kgf (I used Cx for 0.9M, that is much more then at 0.18M)
Summary drag force(Dtotal)=D+Ffric, where μ is coefficient of friction
Ffric=μ*W=0.225*15224≈380.6 kgf
Dtotal=247,1+380.6≈627.7 kgf
Calculate theoretical LonG: 
LonG=(T-Dtotal)/W=(15400-627.7)/15224=14772.3/15224≈0.969≈0.97

Let`s find Cx for LonG=0.7:
Faccel=LonG*m=0.7*15 224=10 656.8 kgf
Dtotal=T-Faccel=15400=10 656.8=4743.2 kgf
Subtract the rolling friction force Ffric=μ*W with μ=0.025 (its standard for DCS):
Ffric=15224*0.025=380.6 kgf
Clean aerodynamic drag:
D=4743.2-380.6=4362.6 kgf or 42783.1 H
Calculating q: 
V=0.18*340.3≈61.25 m/s
q=0.5*ρ*V^2=0.5*1.225*61.25^2≈2298 Pa
Calculating Cx:
D=Cx*q*S, where S=38.056 m^2
Cx=D/(q*S)=42783.1/(2298*38.056)=4278.1/87452.7≈0.4892

This is additional drag or not enough thrust. Appreciate you try to find my mistakes.

Edited January 12 by Logan54

[Logan54]

1 more level acceleration test at sea level, that shows sustained less power during 0.6M..0.95M and 2 power drops (0.65M and 0.9M). 0.9M Drop is around 100m/s that is really can decrease climb performance during 0.9M Climb.

Edited January 12 by Logan54

[BIGNEWY]

The team have checked and it is within tolerances. 

Some things to note: 

you need to use thrust diagram from the PA to obtain actual thrust for 1400 m and 0.9 M and you seem to be using some CD estimation instead of using direct numbers from PA. 

thank you 

 

[Logan54]

14 минут назад, BIGNEWY сказал:

The team have checked and it is within tolerances. 

Some things to note: 

you need to use thrust diagram from the PA to obtain actual thrust for 1400 m and 0.9 M and you seem to be using some CD estimation instead of using direct numbers from PA. 

thank you 

 

At this picture there is PA data, and for all 3 flight tests there is significant power drop at 0.9M. I m using only Practical Aerodynamic book data (yellow line). No idea why this power drop is normal for team.
Thank you.

[Kuky]

3 hours ago, BIGNEWY said:

The team have checked and it is within tolerances. 

Some things to note: 

you need to use thrust diagram from the PA to obtain actual thrust for 1400 m and 0.9 M and you seem to be using some CD estimation instead of using direct numbers from PA. 

thank you 

 

Within tolerance? Tolerance is say under 1% deviations, here it looks like 10 to 20+% which is extreme deviation.

Edited January 12 by Kuky