A320 loss of all pitot vs Mach
Only half a speed-brake
Thread Starter
A320 loss of all pitot vs Mach
Hi,
recently in sim we'd practiced loss of all airspeed, simulated as total blockage of pitot sensors in cruise. However, the Mach indication remained available. I suppose, that if such failure would happen on a real aircraft, there is no way Mach info could be retained. True or false?
thanks,
FD (the un-real)
recently in sim we'd practiced loss of all airspeed, simulated as total blockage of pitot sensors in cruise. However, the Mach indication remained available. I suppose, that if such failure would happen on a real aircraft, there is no way Mach info could be retained. True or false?
thanks,
FD (the un-real)
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Macn number
Hi,
First of all I must point out that I'm not in any position to speak authoritatively on this subject (as a mere 'modestly' experienced A320 F/O :-) )
However, my two cents worth is that the Mach no. indication would/should not be available in this scenario given that Mach no. is an expression of the aircrafts IAS(CAS I imagine, to be pedantic) relative to the local speed of sound.
So if you deprive the Air data computers of (all) the pitot information then it can't know its airspeed hence it can't can't know how close it is to the local speed of sound hence it can't compute or display the Mach no.
I stand to be corrected of course! :-)
I'm also very interested to hear the replies of the more learned contributors.
Regards,
Emerald Flyer.
First of all I must point out that I'm not in any position to speak authoritatively on this subject (as a mere 'modestly' experienced A320 F/O :-) )
However, my two cents worth is that the Mach no. indication would/should not be available in this scenario given that Mach no. is an expression of the aircrafts IAS(CAS I imagine, to be pedantic) relative to the local speed of sound.
So if you deprive the Air data computers of (all) the pitot information then it can't know its airspeed hence it can't can't know how close it is to the local speed of sound hence it can't compute or display the Mach no.
I stand to be corrected of course! :-)
I'm also very interested to hear the replies of the more learned contributors.
Regards,
Emerald Flyer.
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also no bus-driver....
...but Mach needs also on temperature. so if pitot is blocked and rosemount probe (if found on airbus?) still working it could at least indicate something, but in no way reliable
...but Mach needs also on temperature. so if pitot is blocked and rosemount probe (if found on airbus?) still working it could at least indicate something, but in no way reliable
Only half a speed-brake
Thread Starter
Yes, the indicated Mach showed values I would normally expect. Also, the TAS was available from ND, and that also makes no sense to me. Indeed, the sim used is rather aged.
One other thing comes to mind, real A/C will have a tendency to pitch continiously in cruise with A/P off, is this behaviour correctly replicated in the sim you use? Ours does not provide the high alt pitch-up and the onset of the (no airspeed) problem is then easily deal with by basically not touching anything. In real life, this is not an option I would say. Keep in mind that in ALTN law you cannot trim the pitch tendency away.
FD (the un-real)
One other thing comes to mind, real A/C will have a tendency to pitch continiously in cruise with A/P off, is this behaviour correctly replicated in the sim you use? Ours does not provide the high alt pitch-up and the onset of the (no airspeed) problem is then easily deal with by basically not touching anything. In real life, this is not an option I would say. Keep in mind that in ALTN law you cannot trim the pitch tendency away.
FD (the un-real)
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but Mach needs also on temperature
Moderator
but Mach needs also on temperature
M depends only on CAS and pressure height. No OAT input required. Plenty of older threads with the relevant equations quoted if you so desire.
mach is needed to display SAT
stagnation temperature rise depends on M. So M is an input to convert TAT to SAT (with whatever recovery coefficient is relevant to the installation). Likewise plenty of threads with the equations
real A/C will have a tendency to pitch continuously in cruise with A/P off,
you are probably thinking of the phugoid. Bit of a roller coaster motion which is only a minor nuisance to the pilot. With the A/P height lock in the phugoid degenerates to a similar period speed oscillation unless the A/T takes care of it. Some aircraft have a pronounced phugoid, others are quite benign.
M depends only on CAS and pressure height. No OAT input required. Plenty of older threads with the relevant equations quoted if you so desire.
mach is needed to display SAT
stagnation temperature rise depends on M. So M is an input to convert TAT to SAT (with whatever recovery coefficient is relevant to the installation). Likewise plenty of threads with the equations
real A/C will have a tendency to pitch continuously in cruise with A/P off,
you are probably thinking of the phugoid. Bit of a roller coaster motion which is only a minor nuisance to the pilot. With the A/P height lock in the phugoid degenerates to a similar period speed oscillation unless the A/T takes care of it. Some aircraft have a pronounced phugoid, others are quite benign.
Moderator
It does NOT depend on IAS but on TAS M = TAS / a
.. respectfully suggest you might do a bit more homework. Your equation is quite correct but the normal calculation for M works on CAS and Hp, I suggest. No point using less accurate data when you have immediate inputs.
Example thread plucked out of the air is this one.
.. respectfully suggest you might do a bit more homework. Your equation is quite correct but the normal calculation for M works on CAS and Hp, I suggest. No point using less accurate data when you have immediate inputs.
Example thread plucked out of the air is this one.
Only half a speed-brake
Thread Starter
real A/C will have a tendency to pitch continuously in cruise with A/P off,
you are probably thinking of the phugoid. Bit of a roller coaster motion which is only a minor nuisance to the pilot. With the A/P height lock in the phugoid degenerates to a similar period speed oscillation unless the A/T takes care of it. Some aircraft have a pronounced phugoid, others are quite benign.
you are probably thinking of the phugoid. Bit of a roller coaster motion which is only a minor nuisance to the pilot. With the A/P height lock in the phugoid degenerates to a similar period speed oscillation unless the A/T takes care of it. Some aircraft have a pronounced phugoid, others are quite benign.
Basically, AB FBW uses a "1g" reference for pitch control and any forward/back stick movement is then translated as "load factor" demand. Problem is that the normal acceleration pertaining to level flight is not of a constant value. When stationary on ground, one gets a constant gravity force opposed by constant centrifugal force (Earth's rotation) and normal acceleration is determined. In altitude, the gravity force is different (smaller) compared to surface value due to increased radius. Any forward tangential velocity in the direction of Earth’s rotation will increase the perceived orbiting speed and centrifugal force. Extreme case is high altitude flight over equator on 090 true track (yes, very similar to AF447). It all boils down to one thing - aircraft probably has s hard-wired "level flight 1g" constant but at level flight under described conditions the real g value is less. As the aircraft measures that g is less than expected for level flight, the FBW will attempt to compensate and pull up to achieve the hard wired value.
My small trial (of course properly approved by ATC in a NON RVSM airspace) was in a A320 cca 30°N on an easterly track. At 300 ft above assigned level -500 fpm trajectory was manually commanded and stick released. Expected behaviour of the FBW would be to keep the trajectory, however probably due to logic described, the FBW performed a constant positive load manoeuvre with trajectory low at 100 ft below assigned FL and within 30 sec from stick release we hit 300 ft above assigned again, this time with +700 fpm. My observation is, that in high altitude and easterly track a small constant forward input on stick is required to maintain level flight.
I wonder if this behaviour is available from some newish simulators or not, ours certainly cannot replicate it.
FD (the un-real)
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It is an interesting theory.
But I think that if normal acceleration due to the actually curved path of a level flight was noticeable (even eastbound along the ecuator), it would be corrected for by the computers (either ADIRUs or ELACs, SECs). I think the ADIRUs aldready take it into account in order to display attitude correctly.
Otherwise, all what they explain in the FCOMs would be rubish. The airplane would never be path stable, as they claim.
But maybe you have a point, because what the FCOMs explain about the roll axis, for instance, is rubish, in my opinion. So why would it not enter in phugoids? I would really love to do some "flight testing"...
Mach number cannot be correctly displayed if pitots are blocked, but if you have IAS readings (incorrect) you will have TAS and Mach readings, too (incorrect, naturally). And I guess that the two "frozen" ADRs rejected the good one? So you did see all readings (incorrect) without warnings nor red flags.
When studiying UNRELIABLE SPEED procedure, it seems that the memory items "when safety of flight is affected" are just for shortly after take off, or during final approach or go around. But they are equally valid at cruising level when you are half asleep. Disconnecting the A/THR is vital but, for some reason, is the most frequently missed item when doing this drill.
Regarding to the mach measurement, I know the equations and relationships among TAS, EAS, FL, OAT, Mach... But I don't remember/understand well how Mach is computed if IAS has a compressibility error proportional to the mach number itself. I guess that in the computations, somehow this effect is "gone out" of the equations?
Cheers
But I think that if normal acceleration due to the actually curved path of a level flight was noticeable (even eastbound along the ecuator), it would be corrected for by the computers (either ADIRUs or ELACs, SECs). I think the ADIRUs aldready take it into account in order to display attitude correctly.
Otherwise, all what they explain in the FCOMs would be rubish. The airplane would never be path stable, as they claim.
But maybe you have a point, because what the FCOMs explain about the roll axis, for instance, is rubish, in my opinion. So why would it not enter in phugoids? I would really love to do some "flight testing"...
Mach number cannot be correctly displayed if pitots are blocked, but if you have IAS readings (incorrect) you will have TAS and Mach readings, too (incorrect, naturally). And I guess that the two "frozen" ADRs rejected the good one? So you did see all readings (incorrect) without warnings nor red flags.
When studiying UNRELIABLE SPEED procedure, it seems that the memory items "when safety of flight is affected" are just for shortly after take off, or during final approach or go around. But they are equally valid at cruising level when you are half asleep. Disconnecting the A/THR is vital but, for some reason, is the most frequently missed item when doing this drill.
Regarding to the mach measurement, I know the equations and relationships among TAS, EAS, FL, OAT, Mach... But I don't remember/understand well how Mach is computed if IAS has a compressibility error proportional to the mach number itself. I guess that in the computations, somehow this effect is "gone out" of the equations?
Cheers
Only half a speed-brake
Thread Starter
Mach: Sure, Mach and TAS display with all three pitot channels blocked had been a simulator imperfection.
Phugoid: I would expect that the airframe would manifest phugoids, Dutch roll, etc. in a way comparable to other conventional designs. In direct law that is [to non AB world: movement on the stick commands movement of the control surfaces]. Once in normal or alternate [stick commands the trajectory and flight control computers calculate what kind of control movement is required to achieve so] I would guess that you would not get any of these natural behaviours as FCCs filter the unwanted effects out.
1g This is a can of worms. I would again, hazard a guess, that AB does not want to explain all the true details behind FWB logic in FCOM as it is a) too complex for pilot to understand b) in turn might freak you out and lose your trust in the system c) reveal too much know-how.
Doc 7488 Manual of the Standard Atmosphere:
It is known that gravity is a vectorial summation of the gravitational attraction and the centrifugal force induced by the earth's rotation; it is therefore a complex function of a latitude and a radial distance from the earth's centre and the expression for acceleration due to gravity is generally awkward and unpractical for use. However, the acceleration g may be obtained with sufficient accuracy for the purpose of this standard atmosphere by formally neglecting centrifugal acceleration and using only Newton's gravitational law. NOTE FD: For standard g the centrifugal force cannot be neglected (effect of 0,0337 m*s-2 at equator). However, this paragraph further develops the principle of geopotential altitude and changes of g with increasing height. Because the difference in centrifugal force and associated acc is miniscule with height (see my calculation 1 below), to determine the total change of g with height the delta(Fcentripetal) is being ignored.
....
Doc 7488
h=0 m ---- g = 9,8067 m*s-2
h=11000 m ---- g = 9,7728 m*s-2
See, delta g between surface and 36000 ft is 0,0339 ms-2.
conclusion A The altitude plays a role.
On the other hand, I've done some calculations and this is what I got:
-1) the difference of centripetal acc of a stationary object (still subject to earth's rotation) on surface compared to tethered object 11000 m above = 0,000058 m*s-2: conclusion B The effect of change in centripetal acceleration due increased height has only 0,17% effect on total g change with altitude; negligible.
-2) the difference in centripetal acc of a tethered object (still subject to earth's rotation) 11000 m above surface compared to moving object at same height travelling 500 kt in a direction identical to earth's rotation = 0,000000037 m*s-2: conclusion C1 The effect of increased tangential velocity (calculated at altitude) on g compared to total g change with altitude, would only be +0,00011%; negligible.
-3) the difference in centripetal acc of a tethered object (still subject to earth's rotation) 11000 m above surface compared to moving object at same height travelling 500 kt in a direction opposite to earth's rotation = -0,000000037 m*s-2: conclusion C2 The effect of increased tangential velocity (calculated at altitude) on g compared to total g change with altitude, would only be -0,00011%; negligible.
To summarize:
(i) Normal g changes with altitude due to increased distance from earth's centre.
(ii) Normal g changes with altitude due to difference in orbiting speed; compared to (i) above, this delta g is inconsequential for aircraft.
Based on (ii), while eastbound trajectory provides the theoretical maximum delta g, the speed and direction of flight has no relevant impact.
My points:
If FWB had a hard-wired "g" value, at altitude it would naturally seek a pitch up to increase load to the hard-wired value.
The designers surely must have been aware of this. In order to provide compensation algorithm, you would need altitude information thus linking F/CTL system to altimetry and make it vulnerable to possibly incorrect data. I do not know squat about aircraft design and system engineering, yet this does not seem like a clever thing to do.
Contrary to what I had written in previous post, the FCOM1 clearly states [1.27.20 p2 A320] that manual trim will disengage automatic p.t.
Questions:
Is it true that FBW will attempt to pitch up in altitude due to different normal g?
If so, does your simulator replicate this phenomenon?
Yours,
FD (the un-real)
Microburst2002:
I was told on my instrument class that the curved path is compensated using the Shuler's pendulum principle Pendulum - Wikipedia, the free encyclopedia, or algorithm providing the same in strap-down, laser ring gyro systems. Also I fell that NAV (IRU) and F/CTL (FAC, ELAC, SEC) could be independent. True, however, that loss of all 3 IRs puts you to Direct Law. There's one line on FCOM1.27.20 p2 saying ... the system maintains 1 g in pitch (corrected for pitch attitude)... With no pitch, no g correction, no normal law.
Phugoid: I would expect that the airframe would manifest phugoids, Dutch roll, etc. in a way comparable to other conventional designs. In direct law that is [to non AB world: movement on the stick commands movement of the control surfaces]. Once in normal or alternate [stick commands the trajectory and flight control computers calculate what kind of control movement is required to achieve so] I would guess that you would not get any of these natural behaviours as FCCs filter the unwanted effects out.
1g This is a can of worms. I would again, hazard a guess, that AB does not want to explain all the true details behind FWB logic in FCOM as it is a) too complex for pilot to understand b) in turn might freak you out and lose your trust in the system c) reveal too much know-how.
Doc 7488 Manual of the Standard Atmosphere:
It is known that gravity is a vectorial summation of the gravitational attraction and the centrifugal force induced by the earth's rotation; it is therefore a complex function of a latitude and a radial distance from the earth's centre and the expression for acceleration due to gravity is generally awkward and unpractical for use. However, the acceleration g may be obtained with sufficient accuracy for the purpose of this standard atmosphere by formally neglecting centrifugal acceleration and using only Newton's gravitational law. NOTE FD: For standard g the centrifugal force cannot be neglected (effect of 0,0337 m*s-2 at equator). However, this paragraph further develops the principle of geopotential altitude and changes of g with increasing height. Because the difference in centrifugal force and associated acc is miniscule with height (see my calculation 1 below), to determine the total change of g with height the delta(Fcentripetal) is being ignored.
....
Doc 7488
h=0 m ---- g = 9,8067 m*s-2
h=11000 m ---- g = 9,7728 m*s-2
See, delta g between surface and 36000 ft is 0,0339 ms-2.
conclusion A The altitude plays a role.
On the other hand, I've done some calculations and this is what I got:
-1) the difference of centripetal acc of a stationary object (still subject to earth's rotation) on surface compared to tethered object 11000 m above = 0,000058 m*s-2: conclusion B The effect of change in centripetal acceleration due increased height has only 0,17% effect on total g change with altitude; negligible.
-2) the difference in centripetal acc of a tethered object (still subject to earth's rotation) 11000 m above surface compared to moving object at same height travelling 500 kt in a direction identical to earth's rotation = 0,000000037 m*s-2: conclusion C1 The effect of increased tangential velocity (calculated at altitude) on g compared to total g change with altitude, would only be +0,00011%; negligible.
-3) the difference in centripetal acc of a tethered object (still subject to earth's rotation) 11000 m above surface compared to moving object at same height travelling 500 kt in a direction opposite to earth's rotation = -0,000000037 m*s-2: conclusion C2 The effect of increased tangential velocity (calculated at altitude) on g compared to total g change with altitude, would only be -0,00011%; negligible.
To summarize:
(i) Normal g changes with altitude due to increased distance from earth's centre.
(ii) Normal g changes with altitude due to difference in orbiting speed; compared to (i) above, this delta g is inconsequential for aircraft.
Based on (ii), while eastbound trajectory provides the theoretical maximum delta g, the speed and direction of flight has no relevant impact.
My points:
If FWB had a hard-wired "g" value, at altitude it would naturally seek a pitch up to increase load to the hard-wired value.
The designers surely must have been aware of this. In order to provide compensation algorithm, you would need altitude information thus linking F/CTL system to altimetry and make it vulnerable to possibly incorrect data. I do not know squat about aircraft design and system engineering, yet this does not seem like a clever thing to do.
Contrary to what I had written in previous post, the FCOM1 clearly states [1.27.20 p2 A320] that manual trim will disengage automatic p.t.
Questions:
Is it true that FBW will attempt to pitch up in altitude due to different normal g?
If so, does your simulator replicate this phenomenon?
Yours,
FD (the un-real)
Microburst2002:
I was told on my instrument class that the curved path is compensated using the Shuler's pendulum principle Pendulum - Wikipedia, the free encyclopedia, or algorithm providing the same in strap-down, laser ring gyro systems. Also I fell that NAV (IRU) and F/CTL (FAC, ELAC, SEC) could be independent. True, however, that loss of all 3 IRs puts you to Direct Law. There's one line on FCOM1.27.20 p2 saying ... the system maintains 1 g in pitch (corrected for pitch attitude)... With no pitch, no g correction, no normal law.
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Hi FD,
Similarly I was doing some maths - and calculated that the difference in "g" between being parked on the ground on the Equator (but travelling at 40 thousand Km in 24 hours relative to space) and travelling at 1,000 Km per hour in the same rotational direction as Earth, produced an additional delta g of 0.0019. So Altitude + tangential velocity produces a total delta g of 0.0358.
Using S = UT + 1/2 Accel T squared)
Over 1 minute this would result in a vertical height change of:
0 + 0.5 * 0.0358 * 60 * 60 = 64 meters.
So there may be a very small delta g difference at high Altitude and speed, but 64 meters deviation over 1 minute of stick free flight is hardly significant.
Similarly I was doing some maths - and calculated that the difference in "g" between being parked on the ground on the Equator (but travelling at 40 thousand Km in 24 hours relative to space) and travelling at 1,000 Km per hour in the same rotational direction as Earth, produced an additional delta g of 0.0019. So Altitude + tangential velocity produces a total delta g of 0.0358.
Using S = UT + 1/2 Accel T squared)
Over 1 minute this would result in a vertical height change of:
0 + 0.5 * 0.0358 * 60 * 60 = 64 meters.
So there may be a very small delta g difference at high Altitude and speed, but 64 meters deviation over 1 minute of stick free flight is hardly significant.
Only half a speed-brake
Thread Starter
I checked my spreadsheet again and the numbers do not seem right, neither mine or yours.
Given:
Earth radius R=6356766 m [ICAO Doc]
Earth's rotation 360 deg / 24 hrs
(angular velocity) 2*PI/ (24*60*60) rad/s
tang. spd v=2*PI*r[circumference] / (24*60*60) [time] m/s
crz alt h=11000 m
Formulas:
centripetal acceleration
a(c) = (v^2)/r
Calculated:
surface centripetal acceleration on equator
a(cS) = (v^2)/R = 0,33617737 m/s2
alt (height 11 km) centripetal acceleration over equator
a(cA) = ((2*pi*(R+11000)/(24*60*60))^2)/(R+11000) = 0,33617795 m/s2
crz eastbound centripetal acceleration (h=11km, trk=090, gs=1000 km/h=278 m/s) over EQ
a(cE) = ((2*pi*(R+11000)/(24*60*60)+278)^2)/(R+11000) = 0,08615724 m/s2
crz westbound centripetal acceleration (h=11km, trk=270, gs=1000 km/h=278 m/s) over EQ
a(cE) = ((2*pi*(R+11000)/(24*60*60)+278)^2)/(R+11000) = 0,00535496 m/s2
Assumption:
The ICAO sea level g = 9,8067 m/s2 includes the centripetal effect of earth's rotation.
The ICAO 11 km g = 9,7728 m/s2 neglectes the difference of centripetal effect which is calculated from above to be 0,000000058 m/s2.
The difference of centripetal effect with additional speed 1000 km/h at altitude is compared to surface stationary is 0,052539503 m/s2. This is the nubmer I believe we both got very incorrect the first time.
So, just as you summed up the difference between sea level stationary g and g required for FL360 equatorial westbound level flight with 550 kt GS is 0,0336 [ICAO alt difference] + 0,0525 [speed in crz difference] = delta g = 0,0861 m/s2 = compensating acceleration.
Until further mistakes are discovered, using your perfect example that gives:
With 10 sec reference V/S 170 fpm and altitude deviation 14 ft.
With 30 sec reference V/S 510 fpm and altitude deviation 127 ft.
With 1 min reference V/S 1015 fpm and altitude deviation 510 ft.
With 2 min reference V/S 2030 fpm and altitude deviation 2030 ft.
With 3 min reference V/S 3050 fpm and altitude deviation 4580 ft.
For a FCS that you were trained keeps level flight with free stick, that is a difference. Indeed, the FCOM says 'maintains 1 g compensated for pitch'.
FD (the un-real)
Given:
Earth radius R=6356766 m [ICAO Doc]
Earth's rotation 360 deg / 24 hrs
(angular velocity) 2*PI/ (24*60*60) rad/s
tang. spd v=2*PI*r[circumference] / (24*60*60) [time] m/s
crz alt h=11000 m
Formulas:
centripetal acceleration
a(c) = (v^2)/r
Calculated:
surface centripetal acceleration on equator
a(cS) = (v^2)/R = 0,33617737 m/s2
alt (height 11 km) centripetal acceleration over equator
a(cA) = ((2*pi*(R+11000)/(24*60*60))^2)/(R+11000) = 0,33617795 m/s2
crz eastbound centripetal acceleration (h=11km, trk=090, gs=1000 km/h=278 m/s) over EQ
a(cE) = ((2*pi*(R+11000)/(24*60*60)+278)^2)/(R+11000) = 0,08615724 m/s2
crz westbound centripetal acceleration (h=11km, trk=270, gs=1000 km/h=278 m/s) over EQ
a(cE) = ((2*pi*(R+11000)/(24*60*60)+278)^2)/(R+11000) = 0,00535496 m/s2
Assumption:
The ICAO sea level g = 9,8067 m/s2 includes the centripetal effect of earth's rotation.
The ICAO 11 km g = 9,7728 m/s2 neglectes the difference of centripetal effect which is calculated from above to be 0,000000058 m/s2.
The difference of centripetal effect with additional speed 1000 km/h at altitude is compared to surface stationary is 0,052539503 m/s2. This is the nubmer I believe we both got very incorrect the first time.
So, just as you summed up the difference between sea level stationary g and g required for FL360 equatorial westbound level flight with 550 kt GS is 0,0336 [ICAO alt difference] + 0,0525 [speed in crz difference] = delta g = 0,0861 m/s2 = compensating acceleration.
Until further mistakes are discovered, using your perfect example that gives:
With 10 sec reference V/S 170 fpm and altitude deviation 14 ft.
With 30 sec reference V/S 510 fpm and altitude deviation 127 ft.
With 1 min reference V/S 1015 fpm and altitude deviation 510 ft.
With 2 min reference V/S 2030 fpm and altitude deviation 2030 ft.
With 3 min reference V/S 3050 fpm and altitude deviation 4580 ft.
For a FCS that you were trained keeps level flight with free stick, that is a difference. Indeed, the FCOM says 'maintains 1 g compensated for pitch'.
FD (the un-real)
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Hi
The goal of the FCS, in normal law, is to be "path stable". So "stick free", which has not much sense here, the airplane should maintain flight path angle. In the case of level flight, FPA=0º. But it is in fact a curved path right? So I conclude that the system does not keep an inertial path, but a path linked to earth, linked to gravity.
I think the ADIRUs use schuller pendulum (i have never fully understood it) to help calculating the gravity vector. So the calculated accelerations, velocities and position are not corrupted be a small component of gravity.
But the effect of the centripetal force is a different matter, even independent of gravity. It just depends on the radius of turn and the square of the inertial velocity. I guess that the FCS uses IR (rather than ADR) altitude, true heading and GS information to correct the 1g value due to this effect, and thus maintain "curved stable paths".
I did not read well your post. So you have all pitot blocked. What happened? If the three of them give the same incorrect values, the failures would go undetected by the ADRs?. Was this the case or did they fail one by one?
Regarding stick free in direct law... Do the surfaces float of remain fixed in the position corresponding to the sidestick angle?
The 1g compensated for pitch... This means that the system calculates actual g forces or the component of the g forces at right angles with the airplane path or the component at right angles with the longitudinal axis?
The goal of the FCS, in normal law, is to be "path stable". So "stick free", which has not much sense here, the airplane should maintain flight path angle. In the case of level flight, FPA=0º. But it is in fact a curved path right? So I conclude that the system does not keep an inertial path, but a path linked to earth, linked to gravity.
I think the ADIRUs use schuller pendulum (i have never fully understood it) to help calculating the gravity vector. So the calculated accelerations, velocities and position are not corrupted be a small component of gravity.
But the effect of the centripetal force is a different matter, even independent of gravity. It just depends on the radius of turn and the square of the inertial velocity. I guess that the FCS uses IR (rather than ADR) altitude, true heading and GS information to correct the 1g value due to this effect, and thus maintain "curved stable paths".
I did not read well your post. So you have all pitot blocked. What happened? If the three of them give the same incorrect values, the failures would go undetected by the ADRs?. Was this the case or did they fail one by one?
Regarding stick free in direct law... Do the surfaces float of remain fixed in the position corresponding to the sidestick angle?
The 1g compensated for pitch... This means that the system calculates actual g forces or the component of the g forces at right angles with the airplane path or the component at right angles with the longitudinal axis?
Only half a speed-brake
Thread Starter
My understanding is that IRs have no altitude data.
It was a LOFT sim scenario, first pitot 3 was blocked in climb (stby ASI overreads gibberish, ADR3 fault), then, en-route, loss of speed on ADR1+2, no speed info information whatsoever. Managed according to Unreliable Airspeed tables from QRH, done easier and more precise than I would have had expected.
By "stick free" I ment 'manual flight with stick released' in the scenario given and sim used we did not encounter DCT law at all. If so, as you well know, the control surfaces are controlled conventially in proportion to side stick movement.
FD (the un-real)
I did not read well your post. So you have all pitot blocked. What happened? If the three of them give the same incorrect values, the failures would go undetected by the ADRs?. Was this the case or did they fail one by one?
Regarding stick free in direct law... Do the surfaces float of remain fixed in the position corresponding to the sidestick angle
FD (the un-real)
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perhaps a bit off subject but I read that all lockheed U2 spyplanes had a yaw string at the bottom of pilots front view in case the instruments failed, so why not have some simple external spring indicator that is pushed by the wind to give a rough indication of airspeed? there have been too many blocked pitot head crashes
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recently in sim we'd practiced loss of all airspeed, simulated as total blockage of pitot sensors in cruise. However, the Mach indication remained available. I suppose, that if such failure would happen on a real aircraft, there is no way Mach info could be retained. True or false?