I love the BMS flight model
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@VDK partly correct, by applying rudder you just remove the portion of the lift that is provided already by the vertical fin. if you push more rudder you will start climbing.
The aircraft is behaving almost like a bullet
A bullet doesnt have wings, though it still flies straight for a couple hundred meters before dropping
Edit: added some bits to the first phrase to be more clear
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@frapes45 said in I love the BMS flight model:
There was a video of a f16 pilot talking about how to perform knife edge and he didnt mention pushing the stick.
I don’t know in what circumstances he said that, but it is comum that pilots in podcasts and interviews don’t get much “technical” because of the vast majority of the audience is lay.
@frapes45 said in I love the BMS flight model:
but it doesnt mean the aircraft would change heading cause the stabilisers vector also changes
I don’t get your point here… because the stabilisers lift are in the same direction of the wings (and i understand that you are referring to the horizontal stabilisers).
Also, the F-16 has a “lifting body”, that will push even more the aircraft in the direction of the bank
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@frapes45 said in I love the BMS flight model:
you just remove the portion of the lift that is provided already
What portion?
@frapes45 said in I love the BMS flight model:
if you push more rudder you will start climbing.
Yep
@frapes45 said in I love the BMS flight model:
The aircraft is behaving almost like a bullet
A bullet doesnt have wings, though it still flies straight for a couple hundred meters before dropping
Sure, i agree with you.
I think the difference between the two is that the F16 can maintain speed (and with that, can maintain the “straight” flight), and the bullet will always lose speed as soon as it comes out of the barrel.
Anyway, interesting conversation my friend.
Greetings from Brazil
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@VDK said in I love the BMS flight model:
I don’t get your point here… because the stabilisers lift are in the same direction of the wings (and i understand that you are referring to the horizontal stabilisers).
Also, the F-16 has a “lifting body”, that will push even more the aircraft in the direction of the bank
Thats correct. But more like pushes the whole aircraft rather than turn.
Thats why I said some degrees of change is realistic and you can see that in the video as he changes like 4 degrees of initial heading.I cant seem to find the video but there was also an other in flight one, were pilot also said clearly he only uses rudder during knife edge.
I have tried the Lockheed martin own simulator at a recent event. The knife edge behavior is more tamed and much easier to attain. You are free to believe me or doubt me.
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Reading through this thread is both fascinating and amusing. From an outsider looking in, it’s like Einstein arguing with Hawking, over the the 9th decimal of the cosmological constant, with the majority of us not aware of any difference between the two. I commend the efforts that both sides are striving to make BMS into the most realistic flight model.
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@spotdott I was was thinking exactly the same. I have some knowledge of hydrodynamics and can follow some of what they say, but not all. It is a fascinating discussion indeed.
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Guys you are doing a mess of something extremely simple
The FLCS pitch command receives Nz acceleration from the Z axis
Without pilot inputs the FLCS commands (1-Nz)
So in level flight without pilot input the commanded pitch is 1-1 =0
When you bank , there is no more acceleration on the Zaxis therefore commanded pitch is 1-0=1 G , therefore the aircraft automatically commands pitch to maintain 1G , therefore the aircraft’s starts pitching and turn automatically
So during a knife edge pass you have to push stick to maintain heading , there is absolutly no way otherwise and we shouldn’t worry about a pilot just omitting to say it in an interview
This is exactly the same phenomenon that when you pitch the f16 , as you all know you need to push stick to maintain constant pitch angle see it continues to pitch by itself (as described in hart manœuvres documents)
As usual , do t forget in all your analysis that FLCS is heavily changing the rules here , you can’t applied non flcs aircraft reasoning
For instance of course without flcs the f16 wouldn’t need a stick push during knife edge pass
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@Mav-jp said in I love the BMS flight model:
Guys you are doing a mess of something extremely simple
The FLCS pitch command receives Nz acceleration from the Z axis
Without pilot inputs the FLCS commands (1-Nz)
So in level flight without pilot input the commanded pitch is 1-1 =0
When you bank , there is no more acceleration on the Zaxis therefore commanded pitch is 1-0=1 G , therefore the aircraft automatically commands pitch to maintain 1G , therefore the aircraft’s starts pitching and turn automatically
So during a knife edge pass you have to push stick to maintain heading , there is absolutly no way otherwise and we shouldn’t worry about a pilot just omitting to say it in an interview
This is exactly the same phenomenon that when you pitch the f16 , as you all know you need to push stick to maintain constant pitch angle see it continues to pitch by itself (as described in hart manœuvres documents)
As usual , do t forget in all your analysis that FLCS is heavily changing the rules here , you can’t applied non flcs aircraft reasoning
For instance of course without flcs the f16 wouldn’t need a stick push during knife edge pass
The facts you mention are indeed valid in the given theory.
However, there is still a question, to be answered
When you bank 90 degrees from straight flying at a given speed, are you or are you not still at 1G?
You actually said it yourself flcs trys to maintain 1g 1-0nz is still 1g why would the flcs command pitch then?
The evidence in the particular video that the pilot doesnt actually push the stick down is actually right in front of our eyes.
The aircraft changes slight heading (4degrees) during knife edge towards the banking side. That would be not the case if he indeed pushed the stick. -
@frapes45 said in I love the BMS flight model:
This is exactly the same phenomenon that when you pitch the f16 , as you all know you need to push stick to maintain constant pitch angle see it continues to pitch by itself (as described in hart manœuvres documents)
This is correct , indeed you need to pitch down during climb. But that is not the whole picture
The pitch down is required due to the increase of alpha and the unstable nature design of the viper.
There is not such an extended change on alpha when performing knife edge
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@frapes45 said in I love the BMS flight model:
@Mav-jp said in I love the BMS flight model:
Guys you are doing a mess of something extremely simple
The FLCS pitch command receives Nz acceleration from the Z axis
Without pilot inputs the FLCS commands (1-Nz)
So in level flight without pilot input the commanded pitch is 1-1 =0
When you bank , there is no more acceleration on the Zaxis therefore commanded pitch is 1-0=1 G , therefore the aircraft automatically commands pitch to maintain 1G , therefore the aircraft’s starts pitching and turn automatically
So during a knife edge pass you have to push stick to maintain heading , there is absolutly no way otherwise and we shouldn’t worry about a pilot just omitting to say it in an interview
This is exactly the same phenomenon that when you pitch the f16 , as you all know you need to push stick to maintain constant pitch angle see it continues to pitch by itself (as described in hart manœuvres documents)
As usual , do t forget in all your analysis that FLCS is heavily changing the rules here , you can’t applied non flcs aircraft reasoning
For instance of course without flcs the f16 wouldn’t need a stick push during knife edge pass
The facts you mention are indeed valid in the given theory.
However, there is still a question, to be answered
When you bank 90 degrees from straight flying at a given speed, are you or are you not still at 1G?
You actually said it yourself flcs trys to maintain 1g 1-0nz is still 1g why would the flcs command pitch then?
The evidence in the particular video that the pilot doesnt actually push the stick down is actually right in front of our eyes.
The aircraft changes slight heading (4degrees) during knife edge towards the banking side. That would be not the case if he indeed pushed the stick.The accelerometer measures the acceleration on the vertical Axis
So when flying level the accelerometer measures 1G ( gravity)
When you bank 90 deg the gravity is no more measures on the Z axis of the aircraft and therefore the accelerometer outputs 0 . ….
when the pilot or the pilot starts to pitch then the accelerator starts measuring on the Zaxis
This is the same phenomenon when you pitch wings level, the Gravity measures is Gxcos(theta) and therefore the accelerometer measures a value lower than 1
As FLCS G commends is 1- Gxcos(theta) , the aircraft is then pitching up by itself
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@frapes45 said in I love the BMS flight model:
@frapes45 said in I love the BMS flight model:
This is exactly the same phenomenon that when you pitch the f16 , as you all know you need to push stick to maintain constant pitch angle see it continues to pitch by itself (as described in hart manœuvres documents)
This is correct , indeed you need to pitch down during climb. But that is not the whole picture
The pitch down is required due to the increase of alpha and the unstable nature design of the viper.
There is not such an extended change on alpha when performing knife edge
@frapes45 said in I love the BMS flight model:
@frapes45 said in I love the BMS flight model:
This is exactly the same phenomenon that when you pitch the f16 , as you all know you need to push stick to maintain constant pitch angle see it continues to pitch by itself (as described in hart manœuvres documents)
This is correct , indeed you need to pitch down during climb. But that is not the whole picture
The pitch down is required due to the increase of alpha and the unstable nature design of the viper.
There is not such an extended change on alpha when performing knife edge
That is absolutly INCORRECT , the pitching up is Not due to the instability because the Aoa feedback handles it.
As a proof : fly Mach 1.3 and pitch 40 deg up, your aoa will still be VERY small though the aircraft will want to pitch up by itself
This is only due to Flcs NZ feedback
Guys please , you cannot apply standard flight mechanics theory on a FLCS aircraft, you need to study and understand the FLCS before drawing any analysis
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@Mav-jp said in I love the BMS flight model:
Gxcos(theta)
Exactly Gxcos(theta) is only on pitch axis. It doesnt change that much in a knife edge maneuver
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@Mav-jp said in I love the BMS flight model:
Guys please , you cannot apply standard flight mechanics theory on a FLCS aircraft, you need to study and understand the FLCS before drawing any analysis
Look, I am certain you know and understand the FLCS better than me and better than perhaps all of us.
I am not engaging in a fact fight.
I commented about the knife edge at first place because of personal experience I had in a recent event at Lockheeds martin very own full flight simulator.
In fact I tried a bunch of other maneuvers
It was exactly like the video. No heading change more than 4-5degrees without pushing the stick down.
Feel free to question if I am valid in what I am saying or not. I just pointed out something that I think might be not exactly accurately perfected in bms.
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@Mav-jp You have been most helpful in my other fm questions. Peace !
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@frapes45 said in I love the BMS flight model:
@Mav-jp said in I love the BMS flight model:
Gxcos(theta)
Exactly Gxcos(theta) is only on pitch axis. It doesnt change that much in a knife edge maneuver
Jeez the accelerometer measures on the Z axis
When you bank 90 deg it measures 0 from gravity
NZ = GxCos(theta)xcos(Phi)
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@Mav-jp cos (phi) ? Isnt φ the roll axis?
It would be very interesting if it actually accounts for roll angle.
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@frapes45 said in I love the BMS flight model:
@Mav-jp cos (phi) ? Isnt φ the roll axis?
It would be very interesting if it actually accounts for roll angle.
Okay please read nasa TP1538
https://www.cs.odu.edu/~mln/ltrs-pdfs/NASA-79-tp1538.pdfPage 2 , définition of An
And page 37 appendix B which gives you how this is computed
As you can see it takes both pitch and roll angles into account (with accelerations on all axis for dynamic )
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@Mav-jp not only that, An is also dependant on pitch acceleration among other things
φ is the euler angle for roll.
It is also not as simple as Z axis reading 0g as you mentioned initially.
Dont want to sound rude or insulting at all.
I will keep looking. Obviousely I am not in a position to know if and how the bms flcs is precision engineered.
Kind regards
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@frapes45 said in I love the BMS flight model:
@Mav-jp not only that, An is also dependant on pitch acceleration among other things
φ is the euler angle for roll.
It is also not as simple as Z axis reading 0g as you mentioned initially.
Dont want to sound rude or insulting at all.
I will keep looking. Obviousely I am not in a position to know if and how the bms flcs is precision engineered.
Kind regards
Thank you I think I know what the Euler angles are
BMS flight models includes An that is described in the nasa TP1538 , including all rates p,q and Wdot
My comments earlier was a simplified approach to make you understand the effect of pitch and bank on the flcs response which explains why the FLCS commands pitch up when banked at 90deg
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@frapes45 said in I love the BMS flight model:
@Mav-jp not only that, An is also dependant on pitch acceleration among other things
φ is the euler angle for roll.
It is also not as simple as Z axis reading 0g as you mentioned initially.
Dont want to sound rude or insulting at all.
I will keep looking. Obviousely I am not in a position to know if and how the bms flcs is precision engineered.
Kind regards
BMs flcs is not engineered, it’s a copy paste of the real