another terrible ATPL PoF question
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Exo,
The TR has no controlled flapping, its thrust is reasonably assumed to be perpendicular to its disk, and its disk is reasonably assumed to be perpendicular to the helos vertical axis. Thus it is shown differently from the main rotor thrust which is the subject of a great deal of controlled flapping via cyclic.
The roll angle changes with weight because it changes with the power drawn by the MR, which requires more TR thrust to offset the torque. The roll angle for most helos is 1.5 degrees greater in HOGE than in HIGE!
The issue we face here is that we are trying to teach Statics (a branch of mechanics for engineers) to intelligent people who have not had it, and to do so through one (albeit complex) scenario. It is asking an awful lot.
I could try this for you:
No rotorcraft designer believes you. Furthermore, any instructor who said the TR was located to somehow achieve a hover roll angle correction if full of crap, since no designer would spend one pound of weight to appease a pilot!
The TR has no controlled flapping, its thrust is reasonably assumed to be perpendicular to its disk, and its disk is reasonably assumed to be perpendicular to the helos vertical axis. Thus it is shown differently from the main rotor thrust which is the subject of a great deal of controlled flapping via cyclic.
The roll angle changes with weight because it changes with the power drawn by the MR, which requires more TR thrust to offset the torque. The roll angle for most helos is 1.5 degrees greater in HOGE than in HIGE!
The issue we face here is that we are trying to teach Statics (a branch of mechanics for engineers) to intelligent people who have not had it, and to do so through one (albeit complex) scenario. It is asking an awful lot.
I could try this for you:
No rotorcraft designer believes you. Furthermore, any instructor who said the TR was located to somehow achieve a hover roll angle correction if full of crap, since no designer would spend one pound of weight to appease a pilot!
Lots of points to ponder, Lifting - I was wondering about those links in your posts for a while, then finally decided to click on them - great stuff!
More thinking required for me - even if we do talk about the helicopter rolling around its cg, I'm finding it hard to get past thinking of the two force couples idea - one between lift at the head and weight, and the other between side force at the head and tail rotor thrust.
I can follow the rolling moments around the cg idea OK, but the other way still makes sense to me, too.
Looking back at some of the posts here, it seems that both ways come up with very similar answers - for example, looking at Nick's picture there where the higher tail rotor gets the machine to straighten up and fly right, we could say that tail rotor thrust's further up from the cg creating a bigger moment to make that happen, or alternatively that it's moved closer to the level of main rotor and therefore there's less of a pro-roll couple, as the filthy rotor-height-comparing lobby would say!
More thinking required for me - even if we do talk about the helicopter rolling around its cg, I'm finding it hard to get past thinking of the two force couples idea - one between lift at the head and weight, and the other between side force at the head and tail rotor thrust.
I can follow the rolling moments around the cg idea OK, but the other way still makes sense to me, too.
Looking back at some of the posts here, it seems that both ways come up with very similar answers - for example, looking at Nick's picture there where the higher tail rotor gets the machine to straighten up and fly right, we could say that tail rotor thrust's further up from the cg creating a bigger moment to make that happen, or alternatively that it's moved closer to the level of main rotor and therefore there's less of a pro-roll couple, as the filthy rotor-height-comparing lobby would say!
Been away for very long time. See you are talking about the subject of my moniker.
Important for designers. Boring and ultimately useless and irrelevant for drivers.
Now I know why I went away!
Important for designers. Boring and ultimately useless and irrelevant for drivers.
Now I know why I went away!
Passion Flying Hobby Science Sponsor Work
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Numbers for the R44 Raven I
Did some calculations for a R44-I by means of my scientifique simulator,
for the amateurs of numbers.
Case 1.
Full fuel, 1 pilot (80kg), total TOW 890 kg ISA sea level, OGE
MR-sideways banking (to left down from pilots perspective) = 2.55 °
Helicopter roll (to left down from pilots perspective) = 1.25°
so relative left banking of MR wrt Body = 1.20°
Interesting data :
vertical forces on body 214 N downward, so total MR trust 8945 N
lateral forces on body 1N to the right (neglectable, also the resulting moment)
TR net trust 395 N
pitch up 3.3°, effectively lowering tail rotor to 1.63 m below MR
Observation: even with pilot on righthand side heli banks to the left
Case 2.
153 lit fuel, 4 POB each 75 kg, total TOW 1080 kg, ISA sea level, OGE
MR-sideways banking (left down from pilots perspective) = 2.63 °
Helicopter roll (to left down from pilots perspective) = 0.73 °,
so relative left banking of MR wrt Body = 1.90°
Interesting data :
vertical forces on body 261 N downward , so total MR trust 10856 N
lateral forces on body 1N to the right (neglectable, also the resulting moment)
TR net trust 499 N
pitch down 1°, effectively putting tail rotor at 1.3 m below MR
Observation : putting more POB, and shifting CoG to the left is overruled,
to a large extent, by much small arm of TR wrt MR, resulting in smaller roll of 0.73° instead of 1.25°.
d3
for the amateurs of numbers.
Case 1.
Full fuel, 1 pilot (80kg), total TOW 890 kg ISA sea level, OGE
MR-sideways banking (to left down from pilots perspective) = 2.55 °
Helicopter roll (to left down from pilots perspective) = 1.25°
so relative left banking of MR wrt Body = 1.20°
Interesting data :
vertical forces on body 214 N downward, so total MR trust 8945 N
lateral forces on body 1N to the right (neglectable, also the resulting moment)
TR net trust 395 N
pitch up 3.3°, effectively lowering tail rotor to 1.63 m below MR
Observation: even with pilot on righthand side heli banks to the left
Case 2.
153 lit fuel, 4 POB each 75 kg, total TOW 1080 kg, ISA sea level, OGE
MR-sideways banking (left down from pilots perspective) = 2.63 °
Helicopter roll (to left down from pilots perspective) = 0.73 °,
so relative left banking of MR wrt Body = 1.90°
Interesting data :
vertical forces on body 261 N downward , so total MR trust 10856 N
lateral forces on body 1N to the right (neglectable, also the resulting moment)
TR net trust 499 N
pitch down 1°, effectively putting tail rotor at 1.3 m below MR
Observation : putting more POB, and shifting CoG to the left is overruled,
to a large extent, by much small arm of TR wrt MR, resulting in smaller roll of 0.73° instead of 1.25°.
d3
delta3 - observation on 2 - you have raised the TR with respect to the c of g which is why the roll angle is less. You have also ignored the change in the vertical position of the c of g by adding extra pax.
Read Nick's posts, it is NOT the relationship between the heights of main and tail rotors, it IS the distance of main and tail rotors from the c of g.
Read Nick's posts, it is NOT the relationship between the heights of main and tail rotors, it IS the distance of main and tail rotors from the c of g.
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choice of reference point
Crab, you seem to read things in my post that are not there.
SAFETY NOTICE : For math-allergists, skip this post….
1. Model is a full 3-d dynamic model. It is a very extensive mathematical set of differential equations, involving multiple rotating frames.
2. The result of the model is the presented data : so I did not put the tail rotor anywhere, the model did, calculating for instance the rotor blade elements centimeter per centimeter, degree per degree as they rotate until the desired attitude is achieved (so billions of calculations)
3. At the risk of presenting a different opinion, my mathematical experience with respect to the CoG is:
-> for static equilibrium calculation, it does not matter where you put the origin, can be hub, can be CoG. Personally I use the hub, because it is a fixed coordinate, whereas the CoG changes. But one has to be complete in the setup and take all forces and arms into consideration. In that respect all loads where put at the exact 3-D positions (also the passengers, fuel etc). Looking to Helo-practise, I note that different constructors also use different reference points for CoG, including the nose of the Heli.
-> for dynamic equations : the so-called rigid body dynamics around the CoG is the only way to go, because expressing the differential equations not in the CoG become very difficult, read impossible.
4. Some math about the static equilibrium : forces and moments need to be zero
-> Force = 0 => sum(Fi)=0, where Fi are the different 3D-force vectors, such as TR trust, MR trust, aerodynamic forces on body, tail fin, horizontal stabilizer, gravity acting on the heli, fuel and passengers.
-> Moment = 0 => sum(Fi X Li) = 0, where Li are the respective 3D-arms, and X is the cross product of vectors
Changing the reference point to say a position Lo creates the transformation Li = Li' + Lo
sum(Fi X Li) = sum(Fi X Li') + sum(Fi X Lo) = sum(Fi X Li') + sum(Fi) X Lo = sum(Fi X Li'), because at equilibrium sum(Fi)=0. In simple terms : in equilibrium the equations remain quite independent of the choice of the reference point
5. The R44 is theatering, so no moment is acting at the rotor hub. For ridgid/excentiric rotors this moment needs to be put at the rotor hub location. It typically will depend on relative rotor bankings.
d3
SAFETY NOTICE : For math-allergists, skip this post….
1. Model is a full 3-d dynamic model. It is a very extensive mathematical set of differential equations, involving multiple rotating frames.
2. The result of the model is the presented data : so I did not put the tail rotor anywhere, the model did, calculating for instance the rotor blade elements centimeter per centimeter, degree per degree as they rotate until the desired attitude is achieved (so billions of calculations)
3. At the risk of presenting a different opinion, my mathematical experience with respect to the CoG is:
-> for static equilibrium calculation, it does not matter where you put the origin, can be hub, can be CoG. Personally I use the hub, because it is a fixed coordinate, whereas the CoG changes. But one has to be complete in the setup and take all forces and arms into consideration. In that respect all loads where put at the exact 3-D positions (also the passengers, fuel etc). Looking to Helo-practise, I note that different constructors also use different reference points for CoG, including the nose of the Heli.
-> for dynamic equations : the so-called rigid body dynamics around the CoG is the only way to go, because expressing the differential equations not in the CoG become very difficult, read impossible.
4. Some math about the static equilibrium : forces and moments need to be zero
-> Force = 0 => sum(Fi)=0, where Fi are the different 3D-force vectors, such as TR trust, MR trust, aerodynamic forces on body, tail fin, horizontal stabilizer, gravity acting on the heli, fuel and passengers.
-> Moment = 0 => sum(Fi X Li) = 0, where Li are the respective 3D-arms, and X is the cross product of vectors
Changing the reference point to say a position Lo creates the transformation Li = Li' + Lo
sum(Fi X Li) = sum(Fi X Li') + sum(Fi X Lo) = sum(Fi X Li') + sum(Fi) X Lo = sum(Fi X Li'), because at equilibrium sum(Fi)=0. In simple terms : in equilibrium the equations remain quite independent of the choice of the reference point
5. The R44 is theatering, so no moment is acting at the rotor hub. For ridgid/excentiric rotors this moment needs to be put at the rotor hub location. It typically will depend on relative rotor bankings.
d3
D3 this was the paragraph I was commenting on
'Observation : putting more POB, and shifting CoG to the left is overruled,
to a large extent, by much small arm of TR wrt MR, resulting in smaller roll of 0.73° instead of 1.25°'
And my point was that you were talking about the postition of the TR with respect to the MR which, if you have read many of the previous posts, you will know has no bearing on the roll angle in the hover.
As for your last post, I'm afraid I didn't understand most of it except your reference to a c of g forward of the nose. It doesn't mean that is where the c of g is - it is a datum point that will always give a positive ie aft of the datum position.
If the datum is under the rotor head it means having positive and negative (fwd of the datum) sums to do when you do your loading calculations which is hardly arduous but in an effort to keep it simple for pilots (which I like) some manufacturers build their RFM loading diagrams about a datum outside the aircraft.
'Observation : putting more POB, and shifting CoG to the left is overruled,
to a large extent, by much small arm of TR wrt MR, resulting in smaller roll of 0.73° instead of 1.25°'
And my point was that you were talking about the postition of the TR with respect to the MR which, if you have read many of the previous posts, you will know has no bearing on the roll angle in the hover.
As for your last post, I'm afraid I didn't understand most of it except your reference to a c of g forward of the nose. It doesn't mean that is where the c of g is - it is a datum point that will always give a positive ie aft of the datum position.
If the datum is under the rotor head it means having positive and negative (fwd of the datum) sums to do when you do your loading calculations which is hardly arduous but in an effort to keep it simple for pilots (which I like) some manufacturers build their RFM loading diagrams about a datum outside the aircraft.
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.... Never ask an engineer (let alone a mathematician) ?? ...
That is what I also see in Belgian exams. Many of the questions are ambigous because the full context is either ommitted and/or misjudged.
Never easy to develop good questions. Probably takes a good mix of theoreticians and practitioners.
d3
That is what I also see in Belgian exams. Many of the questions are ambigous because the full context is either ommitted and/or misjudged.
Never easy to develop good questions. Probably takes a good mix of theoreticians and practitioners.
d3
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Attempt to Reconciliate the CoG and Hub reference view
For illustration I let the simulator generate some plots (rear view). This to have real data and not artistic impressions. Making the vectors visible was not an easy task. I had to fool around with camera viewing angles and alpha transparancy, but I hope it does the job. On my PC I can freely rotate the views, but I don't know how to post that.
1. Pilot case (same data as earlier post)
> Yellow downward arrows are gravity forces of Body,Pilot and 2 fuel tanks
> Red downward arrow is the combined gravity force, acting out of CoG (the starting point of this forces defines the CoG). From the picture one can see that the total gravity forces are to the right of the Body gravity force, illustrating the right shift of gravity by the asymetric loading.
> Both of the above gravity forces are also shifted to the right of the hub because of the counter clockwise roll of the Heli of 1.35° (hub is the reference point in the plot)
> MR is clearly banked (2.55°) and coned
> the offset measured from the top of the MR trust to the vertical (measured horizontally) is equal to the TR-trust (red arrow close to the TR, difficult to see)
Same case before take off
2. 4 POB case
One can hopefully distinguish the gravity forces of the pilot and 3 passengers and the resulting total gravity force defining again the CoG (at the root of the red downward arrow). The resulting gravity force is now almost centered because of the symmetric loading (slight offset still possible due to fuel loading).
Discussion of influence of TR height
Approach via Hub reference
Hub = blue cross, same rear view, (sorry for my mediocer free hand drawing)
> MR-trust generates per definition no moment, because it acts in the reference point
> TR generates a counter clockwise moment around Hub
> Gravity generates a clockwise moment around Hub equal to TR moment. Total gravity force is offset to the right of the Hub in part because of assymetric loading, in part because of left roll.
From this viewpoint we can make the following reasoning :
> the deaper the tail rotor, the greater the counter clockwise moment it generates around the Hub.
> the opposing moment by the gravity forces around the Hub must increase to maintain equilibrium, so CoG must shift to the right, equivalent to a left roll.
Approach via CoG reference
CoG = blue cross, same rear view
> MR generates a counter clockwise moment around the CoG
> TR generates a clock wise moment around the CoG
> Gravity by choice of reference point generates no moment
From this viewpoint we can make the following reasoning :
> the deaper the tail rotor, the closer it comes to the CoG, the less the clockwise roll moment it generates around the CoG.
> the opposing moment of the MR-trust generates around the CoG must also decrease to maintain equilibrium.
> This is achieved by shifting the CoG to the right (closer to the extended line of the MR-trust), equivalent to a left roll.
For Nick :
> implicite in the above arguments is that the MR trust does not work through the CoG because of other moments which it has to counter act.
> no aero forces on the tail fin, because in the model interaction with tail rotor is ignored (only net TR-trust needed is calculated in hover)
> no aero forces on stabiliser, because induced velocity is assumed to 'miss' the stabiliser in steady hover.
Hope that helps
d3
1. Pilot case (same data as earlier post)
> Yellow downward arrows are gravity forces of Body,Pilot and 2 fuel tanks
> Red downward arrow is the combined gravity force, acting out of CoG (the starting point of this forces defines the CoG). From the picture one can see that the total gravity forces are to the right of the Body gravity force, illustrating the right shift of gravity by the asymetric loading.
> Both of the above gravity forces are also shifted to the right of the hub because of the counter clockwise roll of the Heli of 1.35° (hub is the reference point in the plot)
> MR is clearly banked (2.55°) and coned
> the offset measured from the top of the MR trust to the vertical (measured horizontally) is equal to the TR-trust (red arrow close to the TR, difficult to see)
Same case before take off
2. 4 POB case
One can hopefully distinguish the gravity forces of the pilot and 3 passengers and the resulting total gravity force defining again the CoG (at the root of the red downward arrow). The resulting gravity force is now almost centered because of the symmetric loading (slight offset still possible due to fuel loading).
Discussion of influence of TR height
Approach via Hub reference
Hub = blue cross, same rear view, (sorry for my mediocer free hand drawing)
> MR-trust generates per definition no moment, because it acts in the reference point
> TR generates a counter clockwise moment around Hub
> Gravity generates a clockwise moment around Hub equal to TR moment. Total gravity force is offset to the right of the Hub in part because of assymetric loading, in part because of left roll.
From this viewpoint we can make the following reasoning :
> the deaper the tail rotor, the greater the counter clockwise moment it generates around the Hub.
> the opposing moment by the gravity forces around the Hub must increase to maintain equilibrium, so CoG must shift to the right, equivalent to a left roll.
Approach via CoG reference
CoG = blue cross, same rear view
> MR generates a counter clockwise moment around the CoG
> TR generates a clock wise moment around the CoG
> Gravity by choice of reference point generates no moment
From this viewpoint we can make the following reasoning :
> the deaper the tail rotor, the closer it comes to the CoG, the less the clockwise roll moment it generates around the CoG.
> the opposing moment of the MR-trust generates around the CoG must also decrease to maintain equilibrium.
> This is achieved by shifting the CoG to the right (closer to the extended line of the MR-trust), equivalent to a left roll.
For Nick :
> implicite in the above arguments is that the MR trust does not work through the CoG because of other moments which it has to counter act.
> no aero forces on the tail fin, because in the model interaction with tail rotor is ignored (only net TR-trust needed is calculated in hover)
> no aero forces on stabiliser, because induced velocity is assumed to 'miss' the stabiliser in steady hover.
Hope that helps
d3
Last edited by delta3; 21st Oct 2006 at 19:41.
