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Finite rotations of a rigid body and the airplane landing-gear problem
Jnl. MechanismsVolume 3, pp.203-208/Pergamon Press1968/Printed in Great Britain
Finite Rotations of a Rigid Body and the Airplane Landing-Gear Problem K. E. Bisshopp* Received 12 February 1 968 Abstract The hinge axis for an airplane landing gear frequently is determined graphically. Alternatively the solution by vectors is more general and can be made with more precision in less time. The methods described here also apply to the deployment from launch configuration to the cruise configuration of antennas and various sensors on a space craft. In addition, the application of vectors to these problems elucidates relationships between rigid body components of spatial configurations. Rodrigues' formula, for which an unfamiliar proof is given, and solutions of the vector equation A × U=B, are the principal techniques on which the analysis is based. A numerical example is included. Zusammenfassung--Endliche Drehungen eines festen K6rpers und das Problem des Flugzeug Landungs-Ger~ites: K. E. Bisshopp. Die Dreh-Achse ffir ein Flugzeug Landungs-Ger6t wird 6fters graphisch bestimmt. Abwechslungsweise ist die LSsung mittels Vektoren allgemeiner und kann genauer und in kfirzerer Zeit durchgef0hrt werden. Die hier beschriebenen Methoden sind auch for den 0bergang vonder in Gang-Setzung Konfiguration in die Reise-Konfiguration von Antennen und verschiedenen Sensoren von Raumschiffen anwendbar. Zus~tzlich kl~rt, die Anwendung von Vektoren zur LSsung dieser Probleme, auf den Zusammenhang zwischen den Komponenten von r~umlichen Konfigurationen. Die Formel von Rodrigues, for welche ein ungew6hnlicher Beweis angegeben ist, und die L6sung der Vektor-Gleichung A x U----B sind die HauptGrundlagen fLir die Analyse. Ein Zahlenbeispiel ist eingeschlossen. Pe3mMe----KoHeqnb~eBpaRIeHHflTBepllaro "lena H I'[po6J1eMa Hoca~oqHaro Yc'rpot~c'rBa CaMoneTa: I<. E. l~amon Oct, Bpa~em~[~rm nocaaoqHaro yCTpO~CTBa caMoneTa qaCTO onpe~eJl~ieTc~[ rpac~bM'lecZH. Pemetme npH iIOMOLL[HBeKTOpOB ~B.rLqe'l"C~6once OSLI~HM H MO~vr 6blTb abmonHeuo TOqHee H S 5 o a e e KOpOTKOe BpeM~L 3,ReCb o l I H C a ~ l e MeTO./~! Tax)ze [IpHMeHHMM ,~J'I.q nepexo~la o r KOH~Itrypaa~a nycxa a xo~ x KoHqbnrypaunHIIOJ1eTa,aHTeHHH pa3HblXCeHcOpOBKOCMHtlecI(HXxopa6nefl.
,~onormnTe.nbno, Hp~IMeHeHHeBeXTOpOaI< pemeHii~o 3TIiXnpo6aeM, abt~[CHgeT3aBHCHMOCTbMexc,zly KOMIIOHeHTaM~ITBcp~aro Tena a HpOCTpaHCTBeHHoflKoH~m"ypaui~. Cl)opMy.rlaPo~pm-eu~, Jzn~[KOTOpO[i npmac~euo rleo6uqH0e ~oxa3aTeJ1t[eTao, a TaloKe pctuemae sem'opHaro yq~aauemu~ A × U = B ~an~eTc~[ OCHOBHO~TeXHHKO~,Ha ZOTOpO~ana.rll,13OCHOBaH. qHc~[eHHMI~ npHMep Tax'~e B~o,~e~.
THE DETERMINATION o f the hinge axis for retracting the l a n d i n g gear o f an a i r p l a n e reduces to the equivalent idealized p r o b l e m o f r o t a t i n g one o f a p a i r o f skew lines a b o u t the other as an axis. F i g u r e 1 shows that the c o n d i t i o n s for existence o f a unique axis are satisfied since any p o i n t Q on the axis o f r o t a t i o n belongs to successive positions A Q B a n d A ' Q B ' o f a rigid body. A m e t h o d o f solution is suggested by observing that the surface generated by r o t a t i o n o f the line A B is a unique h y p e r b o l o i d o f revolution. * Professor of Mechanical Engineering, Rensselaer Polytechnic Institute, Troy, New York.
203
204
/ ~"
•
\
R", /Q,z
.~
Z)'
J
o/ Figure 1. Let A and B be points on the landing gear in its initial position, and A' and B' these same points after rotation of the line AB. Position vectors R,, R 2, R~, and R_~, drawn from an arbitrary origin O, respectively identify the points A, B, A', and B'. From given values of R t, R2, R '1 and R; it is required to find the angle of rotation* 0 < ~ , the direction of the axis n, and one point on this axis. The vectors R ' I - R 1 and R ; - R 2 lie in parallel planes g~ and ~_, respectively, perpendicular to the axis of rotation, hence + x n = (R't
(l)
- R t ) x ( R ~ - R2)-~0.
The ambiguity in the sign of the unit vector n is removed by restricting 0 to angles less than z. This condition is derived from the body-fixed vectors -R t
(2a)
D'=R~-R't
(2b)
D=R
2
by observing that D x n and D ' x n are equal length vectors parallel to the plane zt (or 7z_,). If these vectors are localized at Q' say, (_D'+_D)x
. (D__'--_D) x_n
~
n
Q
Figure 2. Figure 2 shows that [(D' - D) x n] x n = K ( D ' + D) x n
(3)
* The angle 0 is reckoned positive in the sense of a right-hand screw relative to n as shown in Fig. 1.
205
where K is a positive scalar factor. Since D • n = D' • n, Equation (3) reduces to
Knx(D'+D)=D'-D which determines the sign of n so that n x ( D ' + D ) and D ' - D
(4) have the same sense for
O
tD ×.lID' ×oE = c o s _ t D • D ' - ( D • n)", (0<~) ID × n[ 2
(5)
since D ' n = D ' ' n and ]D × n I = ID' × hi. The vectors ( D ' + D ) × n and ( D ' - D ) x n are perpendicular since they are diagonals of a rhombus, Figure 2, and ( D ' + D ) x n bisects the angle 0, therefore
}(D'- D) x n[_ tan0 I(D'+ D) × nI 2"
(6)
On the other hand, Equation (3) shows that K =[[-(D'- D) xn] xnl, I(D'+D) xn I but n is perpendicular to ( D ' - D ) x n, hence I [ ( D ' - D) xn] x.,[= I ( D ' - D) x n l , and K = tan 0/2,
(0 < 0 < •).
(7)
Let Kn = n tan 0/2 = W,
(8)
then Equation (4) becomes W x(D'+D)=D'-D
(9)
which is Rodrigues' formula [1] for any vector fixed in a rigid body when it is rotated about an axis. A direct solution for 0 and n can be obtained by solving Equation (9) for W which gives
w_(D' + D) x ( D ' - D) + 2(D, + D)
ID'+DI
2DxD'
- [ D ' + D[ 2 + 2(D'+ D).
(10)
206
The constant 2 is evaluated by observing from Fig. 1 that R'I - R I is perpendicular to n, hence W • (R'1--R1)=0 and
,
A.--
2D' x D" (R't -R~)
(11)
ID'+ D]Z(D' + D). (R~ -RI)"
The equation for the vector R e from O to a point say Q on the axis of rotation is obtained from Fig. 1 and Rodrigues' formula, Equation (9), thus Q--X=Rt-R e and Q-X'=R~-R e are body-fixed vectors before and after rotation, then
R ~ - R e - ( R ~- R e ) =W x(R~ - R e + R ~ - R e ) . This equation reduces to
W x Re =½['W x (R't + R,)-R't + R , ]
(12)
which has a solution if and only if W" ['W x (R~ + R x ) - R ~ + R t ] = 0 . The triple product vanishes and R~ - R x is perpendicular to W, therefore Equation (12) has the solution R e = {l-W × (R~ + R~) - R~ + R[] x W / W 2 + 2 W / W
(13)
Alternatively,expansion of the right side of this equation, or vector multiplicationon both sides of Equation (l2) by W, yieldsvectors which are parallelto W and they can be neglected so that another form of Equation (13) is =
½[R+
+
n
cot x
R I)]+ 2n.
(14)
Equations (13) and (14) represent the axis of rotation on which there are an infinite number of hinge points corresponding to the choice of 2 which is dictated by restrictions of the airframe. It can readily be verified that the vector RQ, given by Equation (13), is perpendicular to W (or n) when 2= 0 and the problem is solved. Equations (1) and (5) are still valid when 0 = rc even though W becomes infinite, in which case the degenerate form of Equation (13) is Re=½[n xfR xR1)] x n + 2 n ={(R~ + R t ) + 2'n.
(15)
Numerical Example
An application of Equations (1), (4), (5) and (14) is illustrated by the following numerical values taken from the sketch of Fig. 3. Refer to [2].
207 I
o
X~"
~
n.=
'
,Zt2S6)
I" 30
0 x~ Figure 3. R1=66i;
R2=36i;
R]=42j+52k
R~ = - 6.24i + 42j + 81-34k D= -30i;
D ' = -- 6-24i+ 29"34k
R] - R I = - 66i + 42j + 52k R~ - R 2 = - 42.24i + 42j + 81"34k D ' - D = 23-76i + 29.34k Equation (1)
4- x n =
± 66
42
52
1-42-24
42
81-34~
= i(42 x 8 1 . 3 4 - 42 x 52) + j(66 x 8 1 . 3 4 - 52 x 4 2 . 2 4 ) - k(66 x 4 2 - 42 x 42.24) =12311+3172j-998k;
[~n[=3542.
I f the plus sign is used, then n =0.3475i+0.8945j-0.2814k.
208
Equation (4)
n x (D' + D) = 0"3i475 - 36.24
J 0"8945
-0~814
0
29-34
= i(29'34 x 0.8945) + j(36'24 x 0-2814 - 29-34 x 0.3475) + k(36"24 x 0.8945) = 26"25i + 0j + 32.43k = 1' 105(D' - D). From Equation (8) this result shows that tan 0/2 =0"905, hence 0 = 84.3 °. On the other hand Equation (5) gives cos 0 = 187"2 - (10.42) 2 = 0.0994 7909 and again 0 = 84-3 °. The vector to any point Q on the axis of rotation, Equation (14), is
R~-~n=½
6 6 i + 4 2 j + 5 2 k + 1-105
0.3i754
J 0-8945
k -0.2814
- 66
42
52
= 65-2i + 21"22j + 66"7k .
i
If Equation (13) had been used RQ-- 2n = 57.3i + 0"76j + 73.2k
(t5)
R~ - RQ = 7.9i + 20" 5j -- 6"5k = 22.8(0-348 i + 0"900j - 0"285k) "~ 22.8n, which verifies Equations (13) and (14), except for the accuracy lost in the substractions indicated above. When 2 = 0 in Equation (13) R Q = R 0 which is perpendicular to n as shown by the following computation: R o • n=(57.27 ×0"3475+0.76 ×0-8945-73.15 x 0"2814),~0. Parallel solutions of this problem by vectors and the methods of descriptive geometry have proven to be advantageous in developing better visualization of three dimensional relationships and their two dimensional representation, however the vector method lends itself more readily to a computer solution. References [1] Fox E. A. Mechanics, p. 6. Harper and Row (1967); CoE C. J. Theoretical Mechanics, Chap. V. Macmillan (1938). [2l WARNERF. M. and McNEARYM. Applied Descriptive Geometry (5th ed.), pp. 232, 233. McGraw-Hill. (1959).
Finite Rotations of a Rigid Body and the Airplane Landing-Gear Problem K. E. Bisshopp* Received 12 February 1 968 Abstract The hinge axis for an airplane landing gear frequently is determined graphically. Alternatively the solution by vectors is more general and can be made with more precision in less time. The methods described here also apply to the deployment from launch configuration to the cruise configuration of antennas and various sensors on a space craft. In addition, the application of vectors to these problems elucidates relationships between rigid body components of spatial configurations. Rodrigues' formula, for which an unfamiliar proof is given, and solutions of the vector equation A × U=B, are the principal techniques on which the analysis is based. A numerical example is included. Zusammenfassung--Endliche Drehungen eines festen K6rpers und das Problem des Flugzeug Landungs-Ger~ites: K. E. Bisshopp. Die Dreh-Achse ffir ein Flugzeug Landungs-Ger6t wird 6fters graphisch bestimmt. Abwechslungsweise ist die LSsung mittels Vektoren allgemeiner und kann genauer und in kfirzerer Zeit durchgef0hrt werden. Die hier beschriebenen Methoden sind auch for den 0bergang vonder in Gang-Setzung Konfiguration in die Reise-Konfiguration von Antennen und verschiedenen Sensoren von Raumschiffen anwendbar. Zus~tzlich kl~rt, die Anwendung von Vektoren zur LSsung dieser Probleme, auf den Zusammenhang zwischen den Komponenten von r~umlichen Konfigurationen. Die Formel von Rodrigues, for welche ein ungew6hnlicher Beweis angegeben ist, und die L6sung der Vektor-Gleichung A x U----B sind die HauptGrundlagen fLir die Analyse. Ein Zahlenbeispiel ist eingeschlossen. Pe3mMe----KoHeqnb~eBpaRIeHHflTBepllaro "lena H I'[po6J1eMa Hoca~oqHaro Yc'rpot~c'rBa CaMoneTa: I<. E. l~amon Oct, Bpa~em~[~rm nocaaoqHaro yCTpO~CTBa caMoneTa qaCTO onpe~eJl~ieTc~[ rpac~bM'lecZH. Pemetme npH iIOMOLL[HBeKTOpOB ~B.rLqe'l"C~6once OSLI~HM H MO~vr 6blTb abmonHeuo TOqHee H S 5 o a e e KOpOTKOe BpeM~L 3,ReCb o l I H C a ~ l e MeTO./~! Tax)ze [IpHMeHHMM ,~J'I.q nepexo~la o r KOH~Itrypaa~a nycxa a xo~ x KoHqbnrypaunHIIOJ1eTa,aHTeHHH pa3HblXCeHcOpOBKOCMHtlecI(HXxopa6nefl.
,~onormnTe.nbno, Hp~IMeHeHHeBeXTOpOaI< pemeHii~o 3TIiXnpo6aeM, abt~[CHgeT3aBHCHMOCTbMexc,zly KOMIIOHeHTaM~ITBcp~aro Tena a HpOCTpaHCTBeHHoflKoH~m"ypaui~. Cl)opMy.rlaPo~pm-eu~, Jzn~[KOTOpO[i npmac~euo rleo6uqH0e ~oxa3aTeJ1t[eTao, a TaloKe pctuemae sem'opHaro yq~aauemu~ A × U = B ~an~eTc~[ OCHOBHO~TeXHHKO~,Ha ZOTOpO~ana.rll,13OCHOBaH. qHc~[eHHMI~ npHMep Tax'~e B~o,~e~.
THE DETERMINATION o f the hinge axis for retracting the l a n d i n g gear o f an a i r p l a n e reduces to the equivalent idealized p r o b l e m o f r o t a t i n g one o f a p a i r o f skew lines a b o u t the other as an axis. F i g u r e 1 shows that the c o n d i t i o n s for existence o f a unique axis are satisfied since any p o i n t Q on the axis o f r o t a t i o n belongs to successive positions A Q B a n d A ' Q B ' o f a rigid body. A m e t h o d o f solution is suggested by observing that the surface generated by r o t a t i o n o f the line A B is a unique h y p e r b o l o i d o f revolution. * Professor of Mechanical Engineering, Rensselaer Polytechnic Institute, Troy, New York.
203
204
/ ~"
•
\
R", /Q,z
.~
Z)'
J
o/ Figure 1. Let A and B be points on the landing gear in its initial position, and A' and B' these same points after rotation of the line AB. Position vectors R,, R 2, R~, and R_~, drawn from an arbitrary origin O, respectively identify the points A, B, A', and B'. From given values of R t, R2, R '1 and R; it is required to find the angle of rotation* 0 < ~ , the direction of the axis n, and one point on this axis. The vectors R ' I - R 1 and R ; - R 2 lie in parallel planes g~ and ~_, respectively, perpendicular to the axis of rotation, hence + x n = (R't
(l)
- R t ) x ( R ~ - R2)-~0.
The ambiguity in the sign of the unit vector n is removed by restricting 0 to angles less than z. This condition is derived from the body-fixed vectors -R t
(2a)
D'=R~-R't
(2b)
D=R
2
by observing that D x n and D ' x n are equal length vectors parallel to the plane zt (or 7z_,). If these vectors are localized at Q' say, (_D'+_D)x
. (D__'--_D) x_n
~
n
Q
Figure 2. Figure 2 shows that [(D' - D) x n] x n = K ( D ' + D) x n
(3)
* The angle 0 is reckoned positive in the sense of a right-hand screw relative to n as shown in Fig. 1.
205
where K is a positive scalar factor. Since D • n = D' • n, Equation (3) reduces to
Knx(D'+D)=D'-D which determines the sign of n so that n x ( D ' + D ) and D ' - D
(4) have the same sense for
O
tD ×.lID' ×oE = c o s _ t D • D ' - ( D • n)", (0<~) ID × n[ 2
(5)
since D ' n = D ' ' n and ]D × n I = ID' × hi. The vectors ( D ' + D ) × n and ( D ' - D ) x n are perpendicular since they are diagonals of a rhombus, Figure 2, and ( D ' + D ) x n bisects the angle 0, therefore
}(D'- D) x n[_ tan0 I(D'+ D) × nI 2"
(6)
On the other hand, Equation (3) shows that K =[[-(D'- D) xn] xnl, I(D'+D) xn I but n is perpendicular to ( D ' - D ) x n, hence I [ ( D ' - D) xn] x.,[= I ( D ' - D) x n l , and K = tan 0/2,
(0 < 0 < •).
(7)
Let Kn = n tan 0/2 = W,
(8)
then Equation (4) becomes W x(D'+D)=D'-D
(9)
which is Rodrigues' formula [1] for any vector fixed in a rigid body when it is rotated about an axis. A direct solution for 0 and n can be obtained by solving Equation (9) for W which gives
w_(D' + D) x ( D ' - D) + 2(D, + D)
ID'+DI
2DxD'
- [ D ' + D[ 2 + 2(D'+ D).
(10)
206
The constant 2 is evaluated by observing from Fig. 1 that R'I - R I is perpendicular to n, hence W • (R'1--R1)=0 and
,
A.--
2D' x D" (R't -R~)
(11)
ID'+ D]Z(D' + D). (R~ -RI)"
The equation for the vector R e from O to a point say Q on the axis of rotation is obtained from Fig. 1 and Rodrigues' formula, Equation (9), thus Q--X=Rt-R e and Q-X'=R~-R e are body-fixed vectors before and after rotation, then
R ~ - R e - ( R ~- R e ) =W x(R~ - R e + R ~ - R e ) . This equation reduces to
W x Re =½['W x (R't + R,)-R't + R , ]
(12)
which has a solution if and only if W" ['W x (R~ + R x ) - R ~ + R t ] = 0 . The triple product vanishes and R~ - R x is perpendicular to W, therefore Equation (12) has the solution R e = {l-W × (R~ + R~) - R~ + R[] x W / W 2 + 2 W / W
(13)
Alternatively,expansion of the right side of this equation, or vector multiplicationon both sides of Equation (l2) by W, yieldsvectors which are parallelto W and they can be neglected so that another form of Equation (13) is =
½[R+
+
n
cot x
R I)]+ 2n.
(14)
Equations (13) and (14) represent the axis of rotation on which there are an infinite number of hinge points corresponding to the choice of 2 which is dictated by restrictions of the airframe. It can readily be verified that the vector RQ, given by Equation (13), is perpendicular to W (or n) when 2= 0 and the problem is solved. Equations (1) and (5) are still valid when 0 = rc even though W becomes infinite, in which case the degenerate form of Equation (13) is Re=½[n xfR xR1)] x n + 2 n ={(R~ + R t ) + 2'n.
(15)
Numerical Example
An application of Equations (1), (4), (5) and (14) is illustrated by the following numerical values taken from the sketch of Fig. 3. Refer to [2].
207 I
o
X~"
~
n.=
'
,Zt2S6)
I" 30
0 x~ Figure 3. R1=66i;
R2=36i;
R]=42j+52k
R~ = - 6.24i + 42j + 81-34k D= -30i;
D ' = -- 6-24i+ 29"34k
R] - R I = - 66i + 42j + 52k R~ - R 2 = - 42.24i + 42j + 81"34k D ' - D = 23-76i + 29.34k Equation (1)
4- x n =
± 66
42
52
1-42-24
42
81-34~
= i(42 x 8 1 . 3 4 - 42 x 52) + j(66 x 8 1 . 3 4 - 52 x 4 2 . 2 4 ) - k(66 x 4 2 - 42 x 42.24) =12311+3172j-998k;
[~n[=3542.
I f the plus sign is used, then n =0.3475i+0.8945j-0.2814k.
208
Equation (4)
n x (D' + D) = 0"3i475 - 36.24
J 0"8945
-0~814
0
29-34
= i(29'34 x 0.8945) + j(36'24 x 0-2814 - 29-34 x 0.3475) + k(36"24 x 0.8945) = 26"25i + 0j + 32.43k = 1' 105(D' - D). From Equation (8) this result shows that tan 0/2 =0"905, hence 0 = 84.3 °. On the other hand Equation (5) gives cos 0 = 187"2 - (10.42) 2 = 0.0994 7909 and again 0 = 84-3 °. The vector to any point Q on the axis of rotation, Equation (14), is
R~-~n=½
6 6 i + 4 2 j + 5 2 k + 1-105
0.3i754
J 0-8945
k -0.2814
- 66
42
52
= 65-2i + 21"22j + 66"7k .
i
If Equation (13) had been used RQ-- 2n = 57.3i + 0"76j + 73.2k
(t5)
R~ - RQ = 7.9i + 20" 5j -- 6"5k = 22.8(0-348 i + 0"900j - 0"285k) "~ 22.8n, which verifies Equations (13) and (14), except for the accuracy lost in the substractions indicated above. When 2 = 0 in Equation (13) R Q = R 0 which is perpendicular to n as shown by the following computation: R o • n=(57.27 ×0"3475+0.76 ×0-8945-73.15 x 0"2814),~0. Parallel solutions of this problem by vectors and the methods of descriptive geometry have proven to be advantageous in developing better visualization of three dimensional relationships and their two dimensional representation, however the vector method lends itself more readily to a computer solution. References [1] Fox E. A. Mechanics, p. 6. Harper and Row (1967); CoE C. J. Theoretical Mechanics, Chap. V. Macmillan (1938). [2l WARNERF. M. and McNEARYM. Applied Descriptive Geometry (5th ed.), pp. 232, 233. McGraw-Hill. (1959).