Best corner speed

[Convoy]

Like the title says, at what speed is the Fulcrum's best sustained turn rate? 

[SAM77]

From reading the posts by others I have written down 850-950km/h with some limiter kickin in at 870. You'll get a more accurate answer from the experts in here soon.

Edited Tuesday at 05:47 AM by SAM77

[Convoy]

36 minutes ago, SAM77 said:

From reading the posts by others I have written down 850-950km/h with some limiter kickin in at 870. You'll get a more accurate answer from the experts in here soon.

I appreciate it thank you

[AeriaGloria]

17 hours ago, Convoy said:

I appreciate it thank you

I made a video with the best speeds for every regime in the MiG-29, but it needs to be updated 

The problem is that max turn rate is affected by a few things 

-Mach speed, this determines LEF position along with AOA. And the reduction in pitch force as center of pressure moves back with transonic speeds and elevator is less effective 

Mach speed is also affected by temperature, altitude, etc 

So ISA, at low altitude, 850-920 kmh is around your best. 
 

However, things are rarely always ISA or low altitude. 
 

The ARU system is also changing elevator deflection as speed changes, minimum at 870-1200 kmh until altitude increases until the elevator is always max deflection above 9 km altitude. 
 

Thus I have found the most consistent best turn rate band to be precisely around Mach 0.75-85. The higher the altitude the closer to Mach 0.85 you need to be. 
 

It is harder to manage airspeed at a slower turn rate as Mach 0.9 is your best subsonic speed for power, only this speed is transonic and thus limits elevator power as well as ARU limiting elevator deflection at transonic speeds and lower altitudes. Get above 5-9 km and your best turn rate does really become closer to Mach 0.9 as elevator authority stops being an issue. 

 

 

[Convoy]

1 hour ago, AeriaGloria said:

I made a video with the best speeds for every regime in the MiG-29, but it needs to be updated 

The problem is that max turn rate is affected by a few things 

-Mach speed, this determines LEF position along with AOA. And the reduction in pitch force as center of pressure moves back with transonic speeds and elevator is less effective 

Mach speed is also affected by temperature, altitude, etc 

So ISA, at low altitude, 850-920 kmh is around your best. 
 

However, things are rarely always ISA or low altitude. 
 

The ARU system is also changing elevator deflection as speed changes, minimum at 870-1200 kmh until altitude increases until the elevator is always max deflection above 9 km altitude. 
 

Thus I have found the most consistent best turn rate band to be precisely around Mach 0.75-85. The higher the altitude the closer to Mach 0.85 you need to be. 
 

It is harder to manage airspeed at a slower turn rate as Mach 0.9 is your best subsonic speed for power, only this speed is transonic and thus limits elevator power as well as ARU limiting elevator deflection at transonic speeds and lower altitudes. Get above 5-9 km and your best turn rate does really become closer to Mach 0.9 as elevator authority stops being an issue. 

 

 

Great info thank you!

[AeriaGloria]

1 hour ago, Convoy said:

Great info thank you!

Another YouTuber made a video with this chart showing best turn rate at sea level at Mach 0.83-0.85. 
 

However, this is at 9 G and easy to black out. It is often preferred to be slower, and turn rate decreases slowly at Mach 0.75-0.85 at speed decreases becuase below 3 km altitude you don’t need to use full power to maintain ma turn rate at Mach 0.8-0.85. At those altitudes you need to use more and more power until you need full power around Mach 0.75-0.8 and this is very hard to maintain as LEFs come out making it bleed speed much faster. 

[Mateo]

Russian charts of turn rate is not the degrees per second but time of 360 turn. 360 deg in 18 seconds gives 20 degrees per second.

Max sustained turn rate speed is 800 km/h circa 0.7 Mach (close to ground) and that speed decreases with altitude but remeber chart is related to airspeed, and Mach number changes in that relation with altitude. Book title: practical aerodynamics of MiG29

By the way DCS MiG29 seems to be correct, and @AeriaGloria also got the similar figure for maybe a bit different gross weight/altitude.

 

Regards,

Mat

Edited Wednesday at 07:56 PM by Mateo
Strict to original Boyd definition

[Lidozin]

10 hours ago, AeriaGloria said:

Another YouTuber made a video with this chart showing best turn rate at sea level at Mach 0.83-0.85. 
 

However, this is at 9 G and easy to black out. It is often preferred to be slower, and turn rate decreases slowly at Mach 0.75-0.85 at speed decreases becuase below 3 km altitude you don’t need to use full power to maintain ma turn rate at Mach 0.8-0.85. At those altitudes you need to use more and more power until you need full power around Mach 0.75-0.8 and this is very hard to maintain as LEFs come out making it bleed speed much faster. 

The fundamental mistake here is a confusion between how the Ps = 0 line is constructed and how the aircraft may be flown along that line.

The Ps = 0 boundary you are looking at was explicitly obtained at maximum available thrust (full AB). By definition, every point on that line answers a single question: at this speed and load factor, what turn can be sustained when all available thrust is used to exactly balance drag? It does not describe the thrust setting required to fly at some nearby point, nor does it imply that reduced thrust is being used at parts of the boundary.

Interpreting the ability to fly more comfortably at slightly lower speed or reduced thrust as evidence that the Ps = 0 boundary itself was obtained with partial power is therefore incorrect. The boundary is defined by maximum thrust; throttle management inside the envelope is a separate operational consideration and does not alter the physics of how the Ps = 0 line is constructed.
Performing a turn at maximum angular velocity at reduced speed and, consequently, with reduced load factor is achieved not by controlling thrust, but by controlling load factor.

Edited Wednesday at 01:36 PM by Lidozin

[AeriaGloria]

1 hour ago, Lidozin said:

The fundamental mistake here is a confusion between how the Ps = 0 line is constructed and how the aircraft may be flown along that line.

The Ps = 0 boundary you are looking at was explicitly obtained at maximum available thrust (full AB). By definition, every point on that line answers a single question: at this speed and load factor, what turn can be sustained when all available thrust is used to exactly balance drag? It does not describe the thrust setting required to fly at some nearby point, nor does it imply that reduced thrust is being used at parts of the boundary.

Interpreting the ability to fly more comfortably at slightly lower speed or reduced thrust as evidence that the Ps = 0 boundary itself was obtained with partial power is therefore incorrect. The boundary is defined by maximum thrust; throttle management inside the envelope is a separate operational consideration and does not alter the physics of how the Ps = 0 line is constructed.
Performing a turn at maximum angular velocity at reduced speed and, consequently, with reduced load factor is achieved not by controlling thrust, but by controlling load factor.

It is a fact stated in the aerodynamic manual that maximum turn rate at 13000 kg between 0-3000m is reached at less then full afterburner. At the speeds of max sustained turn rate at these low altitudes, Ps0 is positive and going full afterburner would cause acceleration. 

This is also true in DCS, you can try it yourself. On the sustained rate turn graph above, this area is at speeds above 500 knots where the blue “-Ps” area completely disappears and the entire area is orange +Ps. Here you cannot turn enough to decrease speed at full AB and low altitude until higher speeds when elevator deflection is increased by ARU above 1200 kmh.

On the turn rate chart it not as if despite no -Ps area between 500-700 knots that the top is still exactly Ps0, it is slightly +Ps while you are not able to pull enough G to make it Ps0 due to transonic speeds and ARU decreasing elevator authority.  

to quote the aerodynamic manual:

”Full afterburner turn g-forces can be fully realized only at altitudes above 4,000 m. At lower altitudes, the g-forces of maximum thrust turns at low speeds are limited by the maximum permissible angles of attack, and at high speeds by the strength of the structure. In addition, at low altitudes, normal g-forces at full afterburner in the instrument speed range of 750...1,100 km/h significantly exceed the g-forces that the human body can withstand for a long time. Therefore, at these speeds, it is necessary to reduce engine thrust or perform a turn with a sufficiently vigorous acceleration.”

Edited Wednesday at 02:22 PM by AeriaGloria

[Lidozin]

If we return to the diagram that was originally referenced, the maximum speed at which the peak rate of turn (ROT) is achieved is approximately 500 knots. At this speed the sustained-turn load factor reaches its maximum permissible value. The recommendation associated with that point is not to insist on maintaining that speed, but rather - using maximum (afterburner) thrust and modulating angle of attack only - to fly the turn at a slightly lower speed. Doing so allows the aircraft to maintain almost the same ROT while requiring significantly lower g, which is far more sustainable from a physiological standpoint.

The extract from the aerodynamic manual that you quote refers precisely to the realisation of maximum sustained normal load factor. At low altitude, increasing speed beyond roughly 500 knots produces only a marginal increase in ROT, while demanding a disproportionate increase in g, which is entirely consistent with the turning performance characteristics of aircraft in this class. In that sense, the manual and the diagram are fully aligned.

You are also quite correct in noting that above ~500 knots at low altitude it becomes impossible to arrest acceleration by increasing g alone. Once in that regime, thrust must indeed be reduced to prevent the aircraft from “running away” to a much higher steady-state speed in the turn. However, this observation does not apply to the region of the EM diagram where the experimental Ps = 0 points are plotted. Within that envelope, a steady turn at any chosen speed can be established solely by adjusting angle of attack, without the need for thrust modulation.

Finally, to address the original question from the topic starter directly:
Corner speed is the speed at which the maximum instantaneous (non-sustained) rate of turn is achieved. In this case, that speed is approximately 385 knots.

Edited Wednesday at 05:19 PM by Lidozin

[Ironhand]

On 2/3/2026 at 12:28 AM, Convoy said:

Like the title says, at what speed is the Fulcrum's best sustained turn rate? 

Ummm…just a note. Corner speed and sustained turn rate are two different things. The latter is “sustainable”. The former is not but it will temporarily turn you faster than the latter.

Edited Wednesday at 07:35 PM by Ironhand

[Mateo]

I must agree, corner speed is originally defined by John Boyd as EM graph point where max G line intersects stalling line, and this results in max instantaneous turn rate. 

Sorry to confuse with my previous post about max sustained turn rate but author ask exaxtly about max sustained turn rate even the title is corner speed which is defined different.

I must admit at this point I've heard some real pilots use corner speed term in relation to max sustained turn rate, while originally John Boyd defined it differently. I have been also confused so maybe someone will appreciate quick clarification. That unfortunate confusion is actualy popular, not only among sim pilots.

[AeriaGloria]

4 hours ago, Lidozin said:

You are also quite correct in noting that above ~500 knots at low altitude it becomes impossible to arrest acceleration by increasing g alone. Once in that regime, thrust must indeed be reduced to prevent the aircraft from “running away” to a much higher steady-state speed in the turn. However, this observation does not apply to the region of the EM diagram where the experimental Ps = 0 points are plotted. Within that envelope, a steady turn at any chosen speed can be established solely by adjusting angle of attack, without the need for thrust modulation.

Then it’s useless if it doesn’t reflect reality or what we see in game, a Ps0 at these higher speeds would require more then 9-10 G to sustain at sea level in some areas.  
 

Or am I not understanding why you agree it’s true that acceleration can’t be completely stopped with G at certain speeds above 500 knots, then why does the graph in this “regime” show us Ps=0 rather then max you can pull (Ps+)? There is currently no way to increase elevator deflection other then small change by adjusting altimeter, and G decreases to 7 or less at certain points. 
 

This is one reason I wish ED would model the ARU switch. 

Edited Wednesday at 09:55 PM by AeriaGloria

[Lidozin]

The misunderstanding comes from how that part of the EM diagram is being read.

When the text refers to the region below ~500 knots where the experimental Ps = 0 points are plotted, that wording is intentional. For plotting convenience, all Ps contours are simply extended through the 9 g boundary, even though the aircraft cannot actually reach those conditions.

Above roughly 500 knots at sea level, sustaining Ps = 0 would indeed require well in excess of 9–10 g, which is not achievable. That is exactly why acceleration cannot be arrested with g alone in this regime. The diagram does not claim otherwise - the Ps  indication there is theoretical, not attainable.

The only meaningful Ps = 0 data in that low-altitude regime are the experimental points below ~500 knots, where Ps can be trimmed using angle of attack alone. Those points appear again above approximately 670 knots, where a sustained turn becomes possible once more under a different balance of lift, drag, and control authority.

Beyond ~500 knots and below that higher-speed region, the achievable envelope is defined by the maximum usable load factor, not by the extrapolated Ps contours.

So there is no contradiction between the diagram, the manual, and what is observed in DCS - the apparent conflict comes from interpreting the extended Ps lines as achievable flight conditions, which they are not.

[AeriaGloria]

10 minutes ago, Lidozin said:

The misunderstanding comes from how that part of the EM diagram is being read.

When the text refers to the region below ~500 knots where the experimental Ps = 0 points are plotted, that wording is intentional. For plotting convenience, all Ps contours are simply extended through the 9 g boundary, even though the aircraft cannot actually reach those conditions.

Above roughly 500 knots at sea level, sustaining Ps = 0 would indeed require well in excess of 9–10 g, which is not achievable. That is exactly why acceleration cannot be arrested with g alone in this regime. The diagram does not claim otherwise - the Ps  indication there is theoretical, not attainable.

The only meaningful Ps = 0 data in that low-altitude regime are the experimental points below ~500 knots, where Ps can be trimmed using angle of attack alone. Those points appear again above approximately 670 knots, where a sustained turn becomes possible once more under a different balance of lift, drag, and control authority.

Beyond ~500 knots and below that higher-speed region, the achievable envelope is defined by the maximum usable load factor, not by the extrapolated Ps contours.

So there is no contradiction between the diagram, the manual, and what is observed in DCS - the apparent conflict comes from interpreting the extended Ps lines as achievable flight conditions, which they are not.

But the manuals do seem to show this. The sustained turn rate chart is capped at the 9/7 G structural limit and instantaneous G chart shows this limitation, and the turn time chart shows turn time increasing at these speeds drastically.  Or do you mean something else? 

[AeriaGloria]

Perhaps with G limit it’s more like this