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Some real information, or as close to real as we're likely to see in publicly available sources:

 

A TsAGI report apparently states that the base line Su-27(S?) accelerates from 600km/h to 1,100km/h in 15 seconds, at 1,000m altitude, at an all-up starting mass of 18,920kg, and this equates to an average acceleration of 9.25m/s^2.

 

I can't find an actual acceleration graph or the TsAGI report mentioned in the source, so the above should all be taken with a pinch of salt until it can be verified with actual documents. Obviously due to non-linear drag variation & other factors adjusting the figures for different altitudes & different speeds is far from simple, and since it's been 17 years since I last solved a differential equation I'm not about to try or do the integration to find the theoretical thrust value for that matter.

 

It's good being back to useful conversations. Thank you Darkfire! So, if that plane's version should have similar performances to the Su-27's version in DCS, that acceleration would relate to almost 1G, better said 0.9432Gs along the X axis as an average acceleration between those 2 speed ranges at that altitude.

 

By simple math: 18,920kg * 9.25 / 9.80665 = 17846 kgf or 39350 lbf is the resultant total propulsive force (Thrust - Drag), which turns out that the plane you've talked about accelerates relative to a total propulsive force of about 71.4% of 27560lbf (which is the manufacturer's max thrust rating). But that also includes the drag. If the drag would be eliminated, it's logic that the resultant thrust per engine would be greater than 71.4%. How much? Depends on drag! Eliminating the drag (which is relatively high at those speeds) in the given example, the Thrust / Weight ratio will certainly go above 1, resulting in higher engine thrust than we can see with DCS Su-27.

 

 

So how bad can the AL-31 actually be then? Let's leave it to the devs for an answer.

When you can't prove something with words, let the maths do the talking.

I have an insatiable passion for helping simulated aircraft fly realistically!

Sincerely, your correct flight model simulation advisor!

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I respectfully disagree with GGTharos about the loss of thrust with an installed engine. I have no numbers for the Su-27 and F-15 specifically, but looking at other airframes this number is less 20-15% percent. For example, the F-14 approaches 13% of a difference between installed and uninstalled thrust. SR-71 with 4.5%, A-10 with 1.8%. But all these numbers are pointless since they are different airframes.

 

But I don't think the loss of thrust is that intense. I wouldn't use those numbers as reference, you need to find out the exact difference (installed and uninstalled thrust values), and then use that to prove your point. A general rule of thumb isn't going to help you here. Measuring the thrust in the game is nearly impossible due to the lack of tools.

 

Also what was your speed when testing this out? Forward speed (in this case vertical speed) affects thrust too.

 

I haven't checked the tracks yet, I'll take a look at them later.

 

Discussing this kind of topic is really difficult though, as I said, due to the lack of proper tools to measure thrust, drag and other things. We simply don't know the thrust and drag values. So that's why I believe that flying the aircraft vertically (obtaining altitude and forward speed, which affect thrust dramatically) isn't a good way to find out the installed thrust of the DCS Su-27. And saying you're smarter than everybody here isn't going to help either, that's quite rude, to say the least... Again, I think your posts lack evidence, documents, and real proofs. Not one cares about thinks, guesses or whatever.

 

I totally agree with you, honestly, at whatever you said, it's just perfect, and yes, I act rude sometimes, but according to the fact that all the jetfighters in DCS receive almost a flat 15% maximum thrust reduction (85% of the thrust rating told by manufacturers), what Tharos said is reflected in DCS, except for the Mirage which hasn't received any engine thrust tables fixes from RAZBAM yet which proves to have 0% thrust reduction on that plane (at 9400kg, the Mirage-2000 stands on it's tail in DCS) and the Su-27's engines in this discussion which seems to suffer the highest thrust decay ever.

 

Indeed, as you say and also Tharos did as I remember, the resultant maximum engine thrust output primarily varies due to the aerodynamics of the plane's inlets designs and also secondarily due to the shape of the aircraft (but in smaller proportions), but maybe Tharos was correctly giving that amount for these DCS fighters.

 

 

Regards!

When you can't prove something with words, let the maths do the talking.

I have an insatiable passion for helping simulated aircraft fly realistically!

Sincerely, your correct flight model simulation advisor!

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That would not give an accurate figure since the mass constantly changes due to fuel usage. It would be something like F = (Ms - dm/dt) * a where Ms is the starting mass and dm/dt is the change in mass due to fuel drain. The calculation is also altitude, humidity and temperature dependant.

 

 

Yes, but maybe I forgot to tell: with unlimited fuel checked (hopefully bug free now) and in ISA conditions at sea level only no more than 500m high to not string too far from the expected results. The only bad thing is the drag which will increase the calculated engine thrust error as speed builds up (reason why the first method of tail standing is a faster approach), thus if I want to use the 2nd law of Newton I should only do trial tests with short time intervals and near a zero lift alpha (to reduce the drag as much as possible). If the engine's thrust would also increase with the same percentage as the drag increases, then it would be compared as not having drag while the thrust is constant, but I personally believe that the thrust increases slower than the drag builds up, at least for these military fighters inlets with rectangular shapes, because the pressure recovery isn't good enough at lower speeds.


Edited by Maverick Su-35S

When you can't prove something with words, let the maths do the talking.

I have an insatiable passion for helping simulated aircraft fly realistically!

Sincerely, your correct flight model simulation advisor!

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Some real information, or as close to real as we're likely to see in publicly available sources:

 

A TsAGI report apparently states that the base line Su-27(S?) accelerates from 600km/h to 1,100km/h in 15 seconds, at 1,000m altitude, at an all-up starting mass of 18,920kg, and this equates to an average acceleration of 9.25m/s^2.

 

I can't find an actual acceleration graph or the TsAGI report mentioned in the source, so the above should all be taken with a pinch of salt until it can be verified with actual documents. Obviously due to non-linear drag variation & other factors adjusting the figures for different altitudes & different speeds is far from simple, and since it's been 17 years since I last solved a differential equation I'm not about to try ;) or do the integration to find the theoretical thrust value for that matter.

 

 

I have just spent 5 minutes to test that acceleration (it is so damn hard to keep this thing level...)

 

See attached track:

00:10 600 km/h

00:26 1100 km/h

 

 

AC mass was 19000 kg. This gives 16 sec instead of 15, but keep in mind, that there were lots of control inputs by me, creating additional drag, so that 1 second is on me...

 

 

Have a nice day!

Su-27 horizontal acceleration.trk

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@HWasp

Easiest way to do that is with autopilot. Then you shouldn't have much unnecessary control deflection. I dunno how the AP handles full burner, though, as I'm already level and trimmed when using it.

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I have tried using the autopilot, being lazy as I am, but it was not able to hold altitude at all, so there was no other way, but to do it by hand.

 

Which autopilot mode did you try? LAlt-3 (wings level) does not seem to hold altitude but LAlt-4 (barometric altitude hold) works pretty well.

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There's an easier way to check engine thrust. A 1G acceleration is 9.8 m/s^2, or about 35 km/h per second (22 MPH per sec.)

 

 

Load the aircraft lightly with no external stores, set autopilot to straight and level, fly at a low airspeed to minimize the effect of drag, and hit the gas.

If the plane gains about 35 km/h per second, then it's thrust is about equal to its weight.

 

 

Seems about right to me, any opinions?

 

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There's an easier way to check engine thrust. A 1G acceleration is 9.8 m/s^2, or about 35 km/h per second (22 MPH per sec.)

 

 

Load the aircraft lightly with no external stores, set autopilot to straight and level, fly at a low airspeed to minimize the effect of drag, and hit the gas.

If the plane gains about 35 km/h per second, then it's thrust is about equal to its weight.

 

 

Seems about right to me, any opinions?

 

AD

 

 

Sorry, that is not correct at all. :) You need to calculate with drag, especially at low speeds (induced drag).

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There's an easier way to check engine thrust. A 1G acceleration is 9.8 m/s^2, or about 35 km/h per second (22 MPH per sec.)

 

 

Load the aircraft lightly with no external stores, set autopilot to straight and level, fly at a low airspeed to minimize the effect of drag, and hit the gas.

If the plane gains about 35 km/h per second, then it's thrust is about equal to its weight.

 

 

Seems about right to me, any opinions?

 

AD

 

That is not a good way to calculate thrust since drag is included in the equation. And Drag can be seen as the opposite of thrust, so if the aircraft isn't matching the charts, we don't know if the drag is too high or the thrust is too low.

 

Flying the aircraft at low speeds would impose induced drag (as said above), which is the drag due to lift. In order to maintain a level flight, the aircraft (or the AP) must increase the AoA, which induces drag. Flaps will be extended, slats and so on...

 

Best way to measure thrust as described here, is by using the thrust formula Ms (V2 - V1) / g where Ms is the mass flow, V1 being the entry speed, V2 being the exhaust velocity and G being gravity obviously, but getting these data from the sim is impossible without proper tools or asking to Yo-Yo... that's why those discussions about the TF34 thrust were always stagnated by the lack of info.

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I see good efforts to calculate a simulated parameter by using real world physics.

 

Whatever your conclusion is, it means nothing until ED confirm the same real world physics apply to the Sim.

 

It is easier to Ask ED to confirm the thrust reduction or the simulated thrust curve, than proofing it based on unknown physics.

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  • 2 years later...

 

On 9/8/2018 at 7:20 PM, Maverick Su-35S said:

 

 

Yes, but maybe I forgot to tell: with unlimited fuel checked (hopefully bug free now) and in ISA conditions at sea level only no more than 500m high to not string too far from the expected results. The only bad thing is the drag which will increase the calculated engine thrust error as speed builds up (reason why the first method of tail standing is a faster approach), thus if I want to use the 2nd law of Newton I should only do trial tests with short time intervals and near a zero lift alpha (to reduce the drag as much as possible). If the engine's thrust would also increase with the same percentage as the drag increases, then it would be compared as not having drag while the thrust is constant, but I personally believe that the thrust increases slower than the drag builds up, at least for these military fighters inlets with rectangular shapes, because the pressure recovery isn't good enough at lower speeds.

Hi, i also had such a feeling, that su-27 is underpowerd. I did quite simple test: I let su-27 accelerate from brakes at full power to speed of 120 kts and measured the travelled distance. Then i switched off engines and after droping back to 120 kts I again measured the distance to full stop - this showed me the average friction and drag force. Then I did it again without afterburner. 
You are right my calculation showes, that the thrust at full AB is around 161kN (70%) and 102kN without (73%).
I did the tests on OMDW airport (4,500m long, elev. 170ft, no wind, 15°C)

I also think, the the fuel consumption has to be tuned.  The dcs manual states range of 3,530km with 4 missiles. Other sources like "Famous Russian Aircraft Sukhoi Su-27" from Yefim Gordon states even more (probably without weapons).
In my best affords I´am hardly able to reach 3,000km without weapons with landing fuel below 500kg.

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Thrust is lower than bench rating when the engines are installed on the aircraft and at low speed.  So, your test showed something that is expected.

 

And how is Yefim's book an authoritative source for any numbers?  Does he mention the flight profile used to achieve this range?


Edited by GGTharos
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  • 2 weeks later...
On 9/7/2018 at 11:26 PM, DarkFire said:

 

That would not give an accurate figure since the mass constantly changes due to fuel usage. It would be something like F = (Ms - dm/dt) * a where Ms is the starting mass and dm/dt is the change in mass due to fuel drain. The calculation is also altitude, humidity and temperature dependant.

 

I think this can be "locked" by using infinite fuel setting in DCS.
Looks like I was late to reply. Please go ahead an disregard my post.


Edited by Cmptohocah

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5 hours ago, GGTharos said:

That's great, but what is it?  On the bottom is looks like mach number, what are the units and things being measured on the vertical axis?

I believe that the upper left scale, with 40, 60, 80 marks, is in kN (kilo-Newtons) - "кН".
Very bottom scale should be in Mach.
Bottom left says kilograms over something that looks like hour - "ч". Maybe fuel consumption? "кг/Нч"?

I feel like an investigator, looking at this picture.


Edited by Cmptohocah
Added info.

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That would make sense and it's a great graph.   You could compute straight and level acceleration from this, though you have to factor drag in.

 

It appears obvious from the graph though that below M0.5 you're in the hole.  Not terribly surprising.


Edited by GGTharos

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2 hours ago, Cmptohocah said:

I believe that the upper left scale, with 40, 60, 80 marks, is in kN (kilo-Newtons) - "кН".
Very bottom scale should be in Mach.
Bottom left says kilograms over something that looks like hour - "ч". Maybe fuel consumption? "кг/Нч"?

I feel like an investigator, looking at this picture.

 

Doing this on my phone, so no Cyrillic. Concerning the info on lower left: Cud is the abbreviation for specific fuel consumption—the ratio of hour-long fuel consumption to thrust. So those numbers are amount of fuel required under specific conditions to create 1N thrust. Unit Cud-kg/(N/h).

 

Google was my friend, in this instance.


Edited by Ironhand

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Верхняя шкала Тяга в килоньютонах, внизу  скорость в махах, а также температура на входе  в входе в двигатель, а также удельный расход топлива

20 часов назад GGTharos сказал:

В этом есть смысл, и это отличный график. Из этого вы можете вычислить ускорение по прямому и по уровню, хотя вы должны учитывать сопротивление.

 

Однако из графика, что ниже M0,5, вы находитесь в яме. Неудивительно.

 

яма образует из-за того, что скорость полета самолета растет быстрее, чем расход воздуха через двигатель


Edited by BBCRF

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4 hours ago, Cmptohocah said:

Can anyone translate this to English ^^^

The upper scale is Thrust in kilonewtons; the lower, speed in Mach, as well as the inlet temperature in the engine inlet; as well as the specific fuel consumption. The hole forms due to the fact that the aircraft's flight speed increases faster than the air flow through the engine.


Edited by Ironhand
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Ok, so I decided to go the other way and determine the drag force on a Su-27 with clean configuration (no flaps, no weapons, engines on idle).
Disclaimer: there are number of simplifications in this model, so please bare that in mind.

Here is the theory:

Free falling body in a fluid (vertical drop) has two forces acting on it:

1. gravitation force -  G = m x g (pointing down)

2. drag force - Fd (pointing up)

 

The instant the body is dropped its drag force is 0 (drag force increases with speed and initial vertical speed is 0 at the moment of drop) and the only force acting on it is G. This causes the body to accelerate, but as the object picks up speed the drag force increases and continues increasing until the force of gravity and the drag force equalize. At this moment the vertical acceleration is 0 and the body has reached its terminal velocity - vt.

 

Now for some math:

Drag force equals to Fd = 0.5 * Rho * A *Cd * v^2, where:

  • Rho - air density [kg/m^3]
  • A - frontal area of the object exposed to the fluid [m^2]
  • Cd - coefficient of drag
  • v^2 - speed of the object squared [m/s]

For the simplicity let's introduce a helper element K, which equals to: K = Rho * A * Cd, so now the drag force becomes Fd = K * v^2

 

From Newton's second law we get the sum of force to be: m * g - Fd = m * a.


Experiment:
I took an Su-27 without any stores and let it drop from an altitude, straight down vertically until it reached its terminal velocity. It turns out that its terminal velocity, at around sea level, is somewhere around 1470km/h.

Here are the flight parameters:
m = 22987kg (mass)
H0 = 16326m (initial height)
H1 = 0 (terminal height - sea level)

/\T = 57s (flight time)

v0 = 257km/h (initial velocity)

v1 = 1470km/h (terminal velocity)

AoA ~ 0.4deg (average angle of attack)

G ~ 0.2 (load when reached terminal velocity)
Alpha ~ 87deg (average pitch angle - negative)

Since at terminal velocity the following holds true:

m * g - Fd = m * a

and since there is no acceleration a = 0,

we get the drag force to be Fd = m * g = 22987kg * 9.81m/s^2  = 225502.47N ~ 225.5kN
Since we know the speed and the force we can get the value of K = Fd / v^2 = 225502.47 * (408.3)^2 = 1.35267
This means that at sea level and standard (DCS) temperature and pressure we can calculate the drag coefficient as we know the value of Rho (air density) and A (frontal area of the Su-27).
Now, I don't know the value of the area so perhaps someone can help out with this one.
 


Edited by Cmptohocah

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Какой коэффициент лобового сопротивления ты используешь при нулевой подъемной силе Cx0 или общий Cx ?

и разложи правильно систему сил

image002.png

Тяга двигателя для Су-27 будет проецироваться на две оси т.к двигатель отклонен от Строительной оси самолета на несколько градусов

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