Helical Turns: Part 1, Turn Conditions. W. Premerlani and Peter Hollands, May 1, 2015

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1 elical Turns: Part, Turn Conditions W. Premerlani and Peter ollands, May, 05 This document is art one of a three art document describin the theory and imlementation of helical turn controls. This art summarizes the theoretical foundation for achievin steady, coordinated, helical turns around a vertical axis with a fixed win aircraft. It develos the conditions that a steady coordinated helical turn must satisfy. This technique is a more eneral aroach than is resently used for turn control in MatrixPilot. The new control has the followin features: The technique is related to "virtual ose" in which the 9 elements of the direction cosine matrix are secified for the desired attitude of the aircraft. For the control described in this document, only the 3 elements in the bottom row are secified. This establishes the itch and roll orientation only. Yaw orientation is not secified. owever, yaw rate is secified and controlled. The technique does not assume that the aircraft is nearly level. Controls will work aroriately for any orientation whatsoever. In addition to controllin turn rate, this method will also rovide stabilization for any desired attitude of the aircraft. This method treats all orientations uniformly, by simly secifyin the desired values for the bottom row of the matrix. The new method works well with flyin wins even if the naviation ains are set rather hih. The new method is not subject to fliin an aircraft over if the naviation ain is set aressively. The controls will execute a smooth helical turn around the earth frame vertical axis usin a fly-by-wire (or, if you will, a virtual ose) technique. Earth frame yaw rotation rate and itch attitude are secified either from ilot inuts in fly-by-wire mode, or by the naviation controls in wayoint mode. Deendin on itch attitude, the turn is climbin, descendin, or level. Only the three elements in the bottom row of the direction cosine are used in the comutations, which roceed as follows:. The desired vertical axis turn rate is secified as an inut.. The desired rotation rate in the body frame for the desired earth frame rotation rate and the actual orientation of the aircraft is comuted. The actual value of the bottom row of the direction cosine matrix (DCM) is the actual earth vertical axis as seen in the body frame. Therefore, the desired rotation rate vector in the body frame is the desired earth frame turnin rate multilied by the bottom row of the actual DCM. 3. The values of the bottom row the DCM required to achieve the desired turn rate and itch attitude are comuted from the desired itch of the helix, the desired turn rate, and the airseed. 4. The actual and desired values of the bottom row of the DCM, and the actual and desired rotation rates in the body frame, are used as inuts to a controller for the fliht control surfaces. - -

2 As of this date, fly-by-wire and naviation controls based on this new aroach have been imlemented and tested in MatrixPilot, demonstratin smooth and tiht turns even with the aircraft inverted. In develoin the math, the assumtion is made that the anles of attack and sidesli are zero. This is a very rouh aroximation for the sidesli anle when the commanded turn rate is chanin, but is a reasonable aroximation durin a steady turn, because the action of the controls will roduce a coordinated turn without much sidesli. The actual anle of attack is hardly every zero, but much of the math to follow is valid even when the anle of attack is not zero. The only art that chanes a bit is the relationshi between the itch attitude and rate of climb or descent. The chanes in the math required for a nonzero anle of attack are discussed in a searate document. All vectors are in the UDB coordinate system, rotation rates are in radians er second, counter clockwise is ositive. The desired earth frame vertical axis turn rate in radians er second is secified: ω desired rotation rate around earth frame vertical axis Equation The turn is secified entirely in terms of the rotation rate of the orientation of the aircraft with resect to the earth vertical axis. Deendin on the manitude and direction of the wind, the rotation rate of the course over round vector is not the same as the rotation rate of the aircraft attitude. In the case of fly-by-wire, the ilot will have to comensate for the wind by increasin commanded turn rate when turnin with the wind, and by decreasin commanded turn rate when turnin into the wind. For autonomous naviation, it is ossible to account for the wind eleantly within the framework of the resent theory, but that has not been imlemented and tested yet. As it stands now, naviation control is based on the IMU course over round vector without needin to take into account the wind, so this miht be a moot oint. The lan is to first imlement the technique described in this reort and see how well it works in windy conditions before makin a decision whether or not to imlement modulation of the turn rate taret based on wind seed and air seed. The desired itch is secified as the ratio of the vertical comonent of the air seed vector in the earth frame, divided by the manitude of the earth frame horizontal air seed: desired itch actual itch Equation - -

3 desired actual vertical airseed, vertical airseed horizontal airseed total airseed ositive is uword in earth frame Next, denote the actual bottom row of the DCM as follows: [ r r r ] actual bottom row of DCM R Equation 3 zx zy zz Recall that the bottom row of the DCM is the earth frame vertical axis rojected into the body frame. To simlify the math that is to follow, I will rename the three elements of the bottom row of DCM as the scalers X, Y, and : [ X Y ] R Equation 4 Note that the sum of the squares of X, Y and is equal to. The desired body frame rotation rate vector is related to the actual vertical orientation and the desired earth frame rotation rate around the vertical axis by: [ ω X ω Y ω ] Ω Ω desired rotation rate vector in the body frame Equation 5 Equation 5 can be understood more easily by first considerin a simler case: a steady turn in the horizontal lane. In that case the airseed vector and the acceleration vector are both in the horizontal lane. Because the airseed vector is constant in manitude, the acceleration vector must be erendicular to it. The axis of the turn is erendicular to the lane of the turn. The lane of the turn is horizontal. ence the axis of the turn is vertical in the earth frame. Therefore the rotation rate vector of the aircraft is arallel to the earth vertical axis. ince the bottom row of the DCM is the main of the earth vertical axis to the body frame, in the body frame the rotation rate vector must be roortional to the bottom row of the DCM. It is also ossible to rove that equation 5 must hold durin a steady turn in the horizontal lane. For such a turn, roll and itch attitude do not chane, only yaw chanes. The bottom row of the DCM contain roll and itch information only. Therefore they do not chane durin a steady horizontal turn. The kinematics of the DCM leads to the udate equation in which the time rate of chane of the bottom row of the DCM is equal to the cross roduct of the bottom row of the DCM with the rotation rate vector in the body frame. In order for the result of the cross roduct to be zero, the rotation rate vector in the body frame must be arallel to the bottom row of the DCM

4 Equation 5 has several interestin imlications for a steady turn in which the bottom row of the DCM (the tilt) does not chane. In other words, in order to maintain a iven roll and itch orientation, there must be rotation around all three axes in the aircraft frame of reference. The third () comonent around the vertical axis is obvious, it reresents yaw rotation in the body frame. After all, the lane must yaw in order to turn. The first comonent (X) is a little less obvious, but it does not take too much thouht to realize it must be there. The second (Y) comonent took a little while for the author to believe, but it must be there as well whenever the lane is not level, that is whenever the turn is ascendin or descendin at a constant rate. One imlication of the (Y) comonent of rotation is that if the roll rate feedback term is used with ailerons, the resent MatrixPilot controls introduce a small variation in roll attitude durin an ascendin or descendin turn. That is because the roll rate feedback will bias the roll rate toward zero, even thouh it is required for the turn. As a result, the controls will tend to roll into an ascendin turn, and roll out of a descendin turn. There are other imlications of equation 5 for fixed win aircraft that do not have rudder, elevator, and ailerons. For aircraft that do have all three control surfaces, it is ossible to achieve all three elements of the desired rotation rate. The rudder can be used for the majority of yaw rate in the body frame, elevator for itch rate, and aileron for roll rate. But how is it ossible to satisfy equation 5 if rudder control is not used, if rudder is missin (flyin win), or if there are no ailerons? For lanes with no ailerons, the win enerally has dihedral, so yawin the aircraft also causes it to roll. owever, there are only two derees of freedom as a result. As a consequence, there is a definite relationshi amon X, Y, and durin a turn. Another interestin situation arises if the rudder is not used for control. The question arises, what causes the aircraft to achieve yaw rotation? The answer is, there is some amount of side sli durin the turn, creatin an airflow erendicular to the rudder. You miht arue that the elevator hels with the turn, but a little thouht will show in the body frame, the elevator mainly creates (X) axis rotation, not (). Of course, for very tiht turns, () oes to zero, and in that case, there is no yaw rate in the body frame. Finally, the question arises for flyin wins, what causes them to turn? In other words, what causes the requisite yaw rate in the body frame? ome flyin wins have vertical fins that act as fixed rudders. For flyin wins without any vertical fins, it is the author's seculation that the key is the fact that the center of ravity is forward of the center of ressure, so ravity and aerodynamic forces can enerate a yaw torque. The math for comutin the orientation required to accomlish the turn is develoed next. The oal is to determine an orientation that roduces an acceleration in the horizontal lane of the earth frame of reference, erendicular to the fuselae of the aircraft. In the other two erendicular directions, the oal is to balance the forces of lift, thrust, and ravity

5 tart with the equations for the horizontal acceleration associated with the turn rate. There are several equivalent exressions: a a ω acceleration in the ω rotation rate around earth frame κ horizontal comonent of κ curvature of the turn earth frame horizontal lane vertical axis the airseed in the earth frame Equation 6 Both actual and desired values will be considered. Equation 6 resents us with three obvious methods for controllin the turn. We may secify the desired acceleration, the desired rotation rate, or the desired curvature. A reasonable way to control the turns is by secifyin the desired turn rate. In that case, the desired horizontal acceleration is iven by: a a ω ω desired turn rate air seed Y required lateral acceleration to achieve the turn Equation 7 Note the square root factor that reduces the acceleration that would otherwise be required to achieve the turn, due to the itch. To see how the factor arises, it is helful to understand that the second column of the DCM is the fuselae direction in the earth frame, with the first two elements indicatin the horizontal comonents. Next, we need to comute the attitude of the aircraft that, taken in conjunction with the turn rate, will result in the desired lateral acceleration, with acceleration in the other two erendicular directions equal to zero. In order to achieve that, certain values of lift, dra, and thrust will be required in the body frame. The anle of attack of the win will automatically adjust to roduce the needed lift. The seed control will adjust the throttle to balance thrust and dra such that there is no (or little) chane in airseed. It can be shown that the lift and net thrust needed to balance ravity are iven by: Lift Thrust Dra, Equation 8 mass mass - 5 -

6 Y, acceleration of ravity The several stes required to o from equation 7 to equation 8 are summarized in Aendix. The lift and dra forces are erendicular in the body frame. The vertical and horizontal forces are erendicular in the earth frame. Therefore the square of the earth frame horizontal force is equal to the square of the lift, lus the square of the thrust, minus the square of ravity. That leads to the followin relationshi: X ω Equation 9 The several stes required to o from equation 8 to equation 9 are summarized in Aendix. Equation 9 leads to the followin relationshis that the elements of the bottom row of the DCM must satisfy to achieve the desired turn: X ω Equation 0 Y Equation 0 can be imlemented in a few stes. The followin is the first ste: ω 0 [ ] R X Y Equation ω In other words, the first ste comutes roll only. tart with the roll arameter laced in the X element, lace in the element, and normalize. This will result in a first ass result with Y equal to zero, and the sum of the squares of X and elements equal to. Then, insert the desired value of itch in the Y element, and renormalize. The ratio of X/ will be reserved, and the second art of equation 0 will be satisfied: [ ] [ X ] X Y R Equation ( ) Equation alies for riht side u fliht. Inverted fliht can easily be achieved by simly fliin the sins of the first and last elements in the vector

7 Aendix : Derivation of equation 8 This section summarizes the math leadin u to equation 8. It assumed anle of attack is zero, and that the controls erform coordinated turns so side sli is zero. Therefore the velocity vector in the earth frame is the velocity manitude times the second column of the direction cosine matrix. In other words, the velocity vector is arallel to the body frame y axis. It is not assumed that the acceleration alon the earth frame vertical axis is zero. It is not assumed the roeller thrust vector is alined with the body frame y axis, it may be anled u or down. For convenience we define the itch as the ratio of vertical to horizontal velocity, with ositive values for ascent: actual itch Equation A. actual earth frame vertical airseed earth frame Also note that: horizontal airseed total airseed Equation A. Next, denote the actual bottom row of the DCM as follows: [ r r r ] actual bottom row of DCM R Equation A.3 zx zy zz Recall that the bottom row of the DCM is the earth frame vertical axis rojected into the body frame. To simlify the math that is to follow, we will rename the three elements of the bottom row of DCM as the scalers X, Y, and : [ X Y ] R Equation A.4 Note that the sum of the squares of X, Y and is equal to. Y can be comuted from as follows: - 7 -

8 Y Equation A.5 imilarly, can be comuted from Y: Y Equation A.6 Y The startin oint of the analysis is to comute the time rate of chane of the airseed. In the most eneral case, airseed is not constant. The rate of chane is most easily comuted in the body frame alon the body frame y axis, in which case the net force alon that axis is the rojection of the roeller thrust alon that axis, minus the dra force, minus the rojection of ravity force alon the body frame y axis: m m & Thrust Dra Equation A.7 m total mass & time rate of chane of airseed ravity Thrust rojection of roeller thrust alon fuselae Dra rojection of dra alon fuselae In order to determine how much lift is required in the body frame axis, or in other words, the force enerated by the win that is erendicular to the win, we need to analyze the acceleration of the lane alon the earth vertical axis. This is obtained by multilyin equation A.7 by the rojection factor /sqrt(*) from equation A. to obtain: m m & ( Thrust Dra) Equation A.8 Now, mass time the time rate of chane of earth vertical seed is equal to the sum of the earth frame vertical forces. There are three forces to consider and roject onto the earth frame vertical axis: body frame lift, body frame thrust-dra, and earth frame ravity. In the earth frame, the rojection of body frame lift onto the vertical axis is times the lift. The rojection of thrust-dra is thrust-dra times /sqrt(*). Gravity rojects urely alon the earth vertical axis. This leads to a net force alon the earth vertical axis iven by: - 8 -

9 F F Lift net force alon ( Thrust Dra) the earth frame vertical axis Lift net force erendicular to the win m Equation A.9 Note that lift includes all forces that are erendicular to the win in the body frame, includin win forces and elevator forces, and the contribution from the roeller if its axis is not arallel to the fuselae. ettin equation A.8 equal to equation A.9 roduces: m Equation A.0 ( Thrust Dra) Lift ( Thrust Dra) m Note that the thrust-dra terms are the same on both sides of equation A.0, so they can be removed, roducin: m Lift m Equation A. Equation A. can be rearraned to: m Lift m m m ( ) m Equation A. The final ste is to divide equation A. by and by m to roduce: It can be shown that the lift and net thrust needed to balance ravity are iven by: Lift m Equation A.3 Note that we have not made any assumtions reardin thrust and dra

10 Aendix : Derivation of equation 9 This document summarizes the math leadin u to equation 9. It assumed anle of attack is zero, and that the controls erform coordinated turns so side sli is zero. Therefore the velocity vector in the earth frame is the velocity manitude times the second column of the direction cosine matrix. In other words, the velocity vector is arallel to the body frame y axis. It is not assumed that the acceleration alon the earth frame vertical axis is zero. It is not assumed the roeller thrust vector is alined with the body frame y axis, it may be anled u or down. For convenience we define the itch as the ratio of vertical to horizontal velocity, with ositive values for ascent: actual itch Equation A. actual earth frame vertical airseed earth frame Also note that: horizontal airseed total airseed Equation A. Next, denote the actual bottom row of the DCM as follows: [ r r r ] actual bottom row of DCM R Equation A.3 zx zy zz Recall that the bottom row of the DCM is the earth frame vertical axis rojected into the body frame. To simlify the math that is to follow, we will rename the three elements of the bottom row of DCM as the scalers X, Y, and : [ X Y ] R Equation A.4 Note that the sum of the squares of X, Y and is equal to. Y can be comuted from as follows: - 0 -

11 Y Equation A.5 imilarly, can be comuted from Y: Y Equation A.6 Y The startin oint of the math that roduced equation 9 are the followin equations: m m & Thrust Dra Equation A.7 m total mass & time rate of chane of airseed ravity Thrust rojection of roeller thrust alon fuselae Dra rojection of dra alon fuselae Lift m a Equation A.8 ω a required lateral acceleration to achieve the turn ω desired turn rate Equation A.9 It is ossible to continue with the most eneral assumtions of equation A.7, but to simlify the math, take the time rate of chane of the airseed to be zero. In that case, equation A.7 becomes: m Thrust Dra Equation A.0 Next, reconize that a rotational transformation reserves vector manitudes. Therefore, the manitude of the net aerodynamic force vector on the aircraft is the same in both earth and body frame of reference. In the body frame, there are two erendicular comonents, lift and dra. In the earth frame, there are two erendicular comonents, a horizontal force required to accelerate horizontally, and a vertical force to oose ravity. Therefore the square of the horizontal earth force must be equal to the sum of the squares of two body frame forces, minus the square of ravity. This roduces equation A.: ω Equation A. ( ) - -

12 - - Equation A. can be rearraned slihtly as: ( ) ω Equation A. Equation A. can be simlified a little by cancelin two of the terms: ω Equation A.3 Next take advantae of the followin identity: Y Equation A.4 Use equation A.4 to substitute for one of the terms inside the square bracket on the riht: [ ] Y ω Equation A.5 Now, it is identically true that: Y X Equation A.6 Therefore: Y X Equation A.7 ubstitute A.7 into A.5 to roduce: X ω Equation A.8 Take the square root of both sides, takin care with the sin relationshi between X and turnin rate, to obtain: X ω Equation A.9

xy plane was set to coincide with the dorsal surface of the pronotum. (4) The locust coordinate

xy plane was set to coincide with the dorsal surface of the pronotum. (4) The locust coordinate 4 5 6 7 8 9 0 4 5 6 7 8 9 0 Sulementary text Coordinate systems and kinematic analysis. Four coordinate systems were defined in order to analyse data obtained from the video (Fig. S): () the video coordinate