free-precession paradox

The free precession picture PhL 3.14.17 For a rigid object that is axially symmetric, the symmetry axis is a principal axis and the other two orthonormal principal axes can be selected in any manner to lie in the plane perpendicular to the symmetry axis. For a pancake, this symmetry axis has the largest principal moment, while for a pencil it has the smallest principal moment. The other two principal moments are equal. If e'3 is the symmetry axis, then I'1 = I'2 and the last of equations (G.8.18) reads 0 = I'3()'3 , so (ω)'3 = constant. The other two equations (G.8.18) are then easily solved. Write them out, ()'1 = (ω)'3 (ω)'2(I'1 - I'3)/ I'1 = Ω'(ω)'2 where Ω' ≡ (ω)'3 (I'1-I'3)/I'1 ()'2 = - (ω)'3 (ω)'1 (I'1 - I'3)/I'1 = - Ω'(ω)'1 (G.8.18) Then ()'1 = Ω'()'2 = - Ω'2 '(ω)'1 Selecting an arbitrary time origin, we can write the solution as (ω)'1 = A sin(Ω't) (ω)'2 = ()'1 / Ω' = Acos(Ω't) . (G.8.20) The Frame S ω vector (showing Frame S' components) is thus doing conical motion: The kinetic energy of this rigid object is (with diagonal I) T = (1/2) ω I ω = (1/2) (ω)'i(I)'ij(ω)'j = (1/2) I'i(ω)'i2 = (1/2)I'1[(ω)'12+ (ω)'22] + (1/2)I'1(ω)'32 = (1/2)I'1 A2 + (1/2)I'1(ω)'32 (G.8.21) If the rigid body has kinetic energy T, then the precession amplitude is given by solving the above for A, A = (2T/I'1) - (ω)'32 (G.8.22) What are the angular momentum components in Frame S' ?? (L)'1 = I'1 (ω)'1 = I'1 A sin(Ω't) (L)'2 = I'1 (ω)'2 = I'1 A cos(Ω't) (L)'3 = I'3 (ω)'3 = I'3 (ω)'3 The first two components are changing in time. But it seems that since N = and N = 0 in Frame S, we should have L = constant. How does this fit with the moving L components above? Just for interest, angular momentum magnitude squared for this rigid body is L2 = (L)'i (L)'i = I'i (ω)'i I'i (ω)'i = I'i2(ω)'i2 = = I'12[(ω)'12+ (ω)'22] + I'32(ω)'32 = I'12A2 + I'32(ω)'32 (G.8.23) and is thus distributed some into the spinning direction and some into the precession directions. Imagine the L vector fixed in Frame S and I make its components be L1 = 0 L2 = 0 L3 = L3 L = L3 What are the Frame S' components of this L ? (L)'1 = L ' = L3 ' = L3 = (sinψsinθ)' + (cosψsinθ)' + (cosθ) ' (G.5.32) ' = sinψsinθ So I conclude that (L)'x = L3sinψsinθ = L3sinψ(t)sinθ = I'1 A sin(Ω't) (L)'y = L3cosψ(t)sinθ = I'1 A cos(Ω't) (L)'z = L3cosθ = I'3 (ω)'3 These equations seem inconsistent. The only solution is to have ψ(t) = Ω't but I thought we had ψ(t) being rotation about the figure axis which is more like ψ(t) = (ω)'3t which is a large number. Let's try brute force to compute the Frame S components of ω . What I have in Frame S' is this (ω)'1 = A sin(Ω't) = sinθsinψ + cosψ (ω)'2 = ()'1 / Ω' = Acos(Ω't) = sinθcosψ - sinψ (ω)'3 = K = cosθ + I know that (ω)' = Rω so ω = R-1(ω)' ωi = (R-1)ij (ω)'j But what is R? Should I use the Euler angles R to relate body frame to space frame? Let's try that. Then R = Rz(-ψ)Rx(-θ)Rz(-φ) R-1 = Rz(φ)Rx(θ)Rz(ψ) Here it is Well, I can just use (G.5.30) I think, so I get ω1 = (cosφcosψ-cosθsinφsinψ) A sin(Ω't) + (-cosφsinψ-cosθsinφcosψ) Acos(Ω't) +sinθsinφ (ω)'3 ω2 = (sinφcosψ + cosθcosφsinψ) A sin(Ω't) + (-sinφsinψ+cosθcosφcosψ) Acos(Ω't) - sinθcosφ (ω)'3 ω3 = sinθsinψ A sin(Ω't) + sinθcosψ Acos(Ω't) + cosθ (ω)'3. How would I show this is conical motion? It just seems a total mess. At is at least moving. If this is supposed to match the picture of Marion on page 392 it seems that if L is in the direction I would expect to see that ω3 = constant. But that seems very unlikely looking at the last line above. Unless ψ(t) = Ω't But how would I justify that? Well go back and add new information: (ω)'1 = A sin(Ω't) = sinθsinψ + cosψ (ω)'2 = ()'1 / Ω' = Acos(Ω't) = sinθcosψ - sinψ (ω)'3 = K = cosθ + So I really have these three equations never yet looked at A sin(Ω't) = sinθsinψ + cosψ A cos(Ω't) = sinθcosψ - sinψ K = cosθ + OK, the first two imply that A2 = [ sinθsinψ + cosψ]2 + [ sinθcosψ - sinψ]2 Cross terms cancel, leaving just the 4 main terms. Then A2 = 2sin2θ + 2 So now I have these two equations A2 = 2sin2θ + 2 K = cosθ + Not going anywhere fast. Go back to the first two A sin(Ω't) = sinθsinψ + cosψ A cos(Ω't) = sinθcosψ - sinψ so sinψA sin(Ω't) = sinθsin2ψ + cosψsinψ cosψA cos(Ω't) = sinθcos2ψ - sinψcosψ Add these to get sinψA sin(Ω't) + cosψA cos(Ω't) = sinθ Do it differently now to cancel the other term cosψA sin(Ω't) = sinθsinψcosψ + cos2ψ -sinψA cos(Ω't) = - sinθsinψcosψ + sin2ψ Then cosψA sin(Ω't) - sinψA cos(Ω't) = So we have at least two new and interesting equations sinψA sin(Ω't) + cosψA cos(Ω't) = sinθ cosψA sin(Ω't) - sinψA cos(Ω't) = Write as Acos(ψ-Ω't) = sinθ Asin(Ω't- ψ) = So I now have four interesting equations Acos(ψ-Ω't) = sinθ Asin(ψ-Ω't) = - A2 = 2sin2θ + 2 K = cosθ + There are four unknowns here: , θ, , so maybe I can somehow solve. I want to show = 0 I think, but not clear how to do that. All four are not independent since third is sum of squares of 1 and 2. Ansatz: Just try to find a solution with = 0. The equations then say Acos(ψ-Ω't) = sinθ Asin(ψ-Ω't) = 0 A2 = 2sin2θ K = cosθ + The 2nd seems to say that ψ = Ω't. The other info left is only this A = sinθ K = cosθ + Ω' Recall that A = (2T/I'1) -K2 not very useful. Instead process Acosθ = sinθcosθ Ksinθ = cosθsinθ + Ω' Acosθ - Ksinθ = -Ω' = - K (I'1-I'3)/I'1 (A/K) cosθ - sinθ = - (I'1-I'3)/I'1 This must be how θ gets determined? Maple at least gives solutions although ugly. Let's now go back to the brute force result: ω1 = (cosφcosψ-cosθsinφsinψ) A sin(Ω't) + (-cosφsinψ-cosθsinφcosψ) Acos(Ω't) +sinθsinφ (ω)'3 ω2 = (sinφcosψ + cosθcosφsinψ) A sin(Ω't) + (-sinφsinψ+cosθcosφcosψ) Acos(Ω't) - sinθcosφ (ω)'3 ω3 = sinθsinψ A sin(Ω't) + sinθcosψ Acos(Ω't) + cosθ (ω)'3. But now set Ω't = ψ to get some simplification maybe, ω1 = (cosφcosψ-cosθsinφsinψ) A sinψ + (-cosφsinψ-cosθsinφcosψ) Acosψ +sinθsinφ K ω2 = (sinφcosψ + cosθcosφsinψ) A sinψ + (-sinφsinψ+cosθcosφcosψ) Acosψ - sinθcosφK ω3 = sinθsinψ A sinψ + sinθcosψ Acosψ + cosθK. Simplify them one at a time: ω1 = A { cosφcosψsinψ-cosθsinφsin2ψ - cosφsinψcosψ-cosθsinφcos2ψ } + sinθsinφ K = A{ -cosθsinφsin2ψ - cosθsinφcos2ψ } + sinθsinφ K = Acosθ{ -sinφsin2ψ - sinφcos2ψ } + sinθsinφ K = -Acosθsinφ + sinθsinφ K = (-Acosθ + Ksinθ )sinφ ok ω2 = (sinφcosψ + cosθcosφsinψ) A sinψ + (-sinφsinψ+cosθcosφcosψ) Acosψ - sinθcosφK = A{sinφsinψcosψ + cosθcosφsin2ψ -sinφsinψcosψ+cosθcosφcos2ψ } - sinθcosφK = A{cosθcosφsin2ψ+cosθcosφcos2ψ } - sinθcosφK = A{cosθcosφ} - sinθcosφK = Acosθcosφ - sinθcosφK = (Acosθ - Ksinθ)cosφ ω3 = sinθsinψ A sinψ + sinθcosψ Acosψ + cosθK = sinθsin2ψ A + sinθcos2ψ A + Kcosθ = Asinθ + Kcosθ Here then are my "brute force" results where I inserted ψ(t) = Ω't : ω1 = (-Acosθ + Ksinθ )sinφ ω2 = ( Acosθ - Ksinθ)cosφ ω3 = Asinθ + Kcosθ Now recall from above that A = sinθ = A/sinθ φ = (A/sinθ) t The results are now ω1 = - (A'cosθ - Ksinθ )sin[(A'/sinθ) t] ω2 = (A'cosθ - Ksinθ)cos[(A'/sinθ) t] ω3 = A'sinθ + Kcosθ This does show precession! The amplitude and frequency seem strange. ω1 = - A sin[Ω t] A = A'cosθ - Ksinθ ω2 = Acos[Ω t] Ω = (A'/sinθ) ω3 = A'sinθ + Kcosθ I think it goes backwards? How can I verify this result somewhere? I think it has to be a textbook somewhere. I have found a few things. going back to my picture Note the angle α. It seems clear that tanα = A'/K sinα = A'/ω Then my result above can be written = A'/sinθ = ωsinα/sinθ φ = (ωsinα/sinθ)t and I do see this in a paper I am looking at so there at least is some kind of confirmation of what I am doing. But he also makes this claim: I can write my body frame stuff as (ω)'1 = ωsinα sin(Ω't) (ω)'2 = ωsinαcos(Ω't) . (G.8.20) (ω)'3 = K How about this idea ω2 = (ω)'i2 = ω2sin2α + K2 which says ω2(1-sin2α) = K2 ω2 cos2α = K Then nice result is K = ω cosα So now I have A' = ωsinα K = ωcosα A'/K' = tanα which I already knew. A'2 + K2 = ω2 obvious from picture, but still new to me. Now my author is claiming that ωsinα = ωy' A' ωcosα = ωz' K He claims also this where is that coming from? These are "intermediate axes" for Euler I think. The ξ' and η' guys. Lξ' = 0 Lη' = Lsinθ Lz' = Lcosθ Looking at my EA picture I think the above are correct. I agree that Lη'/ Lz' = tanθ But he is claiming next that Lη'= I'1 ωη' Lz'= I'3 ωz' He is somehow using those intermediate angles. The right one here seems OK. I get some hints here, but just not very systematic! Go back to ω1 = (-A'cosθ + Ksinθ )sinφ ω2 = ( A'cosθ - Ksinθ)cosφ ω3 = A'sinθ + Kcosθ and use A' = ωsinα K = ωcosα Then these say ω1 = (- ωsinαcosθ + ωcosαsinθ )sinφ ω2 = ( ωsinαcosθ - ωcosαsinθ)cosφ ω3 = ωsinαsinθ + ωcosαcosθ or ω1 = -ω (sinαcosθ + cosαsinθ )sinφ ω2 = ω( sinαcosθ - cosαsinθ)cosφ ω3 = ω(sinαsinθ + cosαcosθ) Now sinαcosθ - cosαsinθ = sin(α-θ) sinαsinθ + cosαcosθ = cos(α-θ) Then have ω1 = -ω sin(α-θ)sinφ φ = (ωsinα/sinθ)t ω2 = ω sin(α-θ)cosφ ω3 = ω cos(α-θ) L = Le3 Getting pretty simple now! It seems like the Frame S cone angle is not θ, but is α-θ. Here is a picture the guy draws, You see that indeed the space cone has half angle θ - α, so confirmation. Not sure what the Big Cone is in this picture. Again the space cone has α-θ as its half angle. What is that other cone?? New idea 3.15.17. What happens if I back map L from Frame S' to Frame S and require it be z-aligned in Frame S? Maybe better to use the forward map which is x' = (cosψcosφ - sinψcosθsinφ) x + (cosψsinφ + sinψcosθcosφ) y + sinψsinθ z y' = (- sinψcosφ - cosψcosθsinφ) x + (-sinψsinφ + cosψcosθcosφ) y + cosψsinθ z z' = sinθsinφ x - sinθcosφ y + cosθ z . (G.5.26) Write as (L')1 = (cosψcosφ - sinψcosθsinφ) L1 + (cosψsinφ + sinψcosθcosφ) L2 + sinψsinθ L3 (L')2 = (- sinψcosφ - cosψcosθsinφ) L1 + (-sinψsinφ + cosψcosθcosφ) L2 + cosψsinθ L3 (L')3 = sinθsinφ L1 - sinθcosφ L2 + cosθ L3 . (G.5.26) Then require that L1 = L2 = 0 in Frame S to get (L')1 = sinψsinθ L3 (L')2 = cosψsinθ L3 (L')3 = cosθ L3. (*) But then we know that (L')1 = I'1(ω)'1 = I'1 ωsinα sin(Ω't) (L')2 = I'1(ω)'2 = I'1 ωsinα cos(Ω't) (L')3 = I'1(ω)'3 = I'3 ω cosα (**) Compare this to (ω)'1 = ωsinα sin(Ω't) (ω)'2 = ωsinαcos(Ω't) . (G.8.20) (ω)'3 = ω cosα So viewed in Frame S', the vectors L and ω are different, but they both precess in Frame S'. Meanwhile, from (*) and (**) we have this possibly new information I'1 ωsinα sin(Ω't) = sinψsinθ L3 = sin(Ω't)sinθ L3 I'1 ωsinα cos(Ω't) = cosψsinθ L3 =cos(Ω't)sinθ L3 I'3 ω cosα = cosθ L3 or // here I assume that ψ = Ω't I'1 ωsinα = sinθ L3 I'1 ωsinα = sinθ L3 I'3 ωcosα = cosθ L3 Divide the last two equations to get (I'1/I'3) tanα = tanθ which compare to the PDF claim So I have finally obtained this key result. It relates the angles θ and α for the first time. I think that θ is the cone half angle for the Frame S precession! This also says θ = constant, something I was looking for! At this point this is what I know about Frame S viewing ω1 = -ω sin(α-θ)sinφ φ = (ωsinα/sinθ)t ω2 = ω sin(α-θ)cosφ ω3 = ω cos(α-θ) L = Le3 I also know that L is same in both frames as a mag, and I already computed, L2 = I'12A'2 + I'32(ω)'32 = (I'1ωsinα)2 + ( I'3ωcosα)2 = a nice expression L2 = ω2 [ (I'1sinα)2 + ( I'3cosα)2 ] so L and ω are locked to each other in this simple manner. Summary of what I now know: 1. α is the cone angle for ω motion viewed in Frame S'. 2. In Frame S' the motion of the ω vector is this (cone angle α) (ω)'1 = ωsinα sin(Ω't) (ω)'2 = ωsinα cos(Ω't) (ω)'3 = ωcosα Given the above, we would compute the cone angle this way tan(cone half angle) = (ω)'1 coefficient / (ω)'3 = ωsinα / ωcosα = tanα and therefore : cone half angle = α 2. In Frame S' the motion of the L vector is this, (L')1 = I'1 ωsinα sin(Ω't) (L')2 = I'1 ωsinα cos(Ω't) (L')3 = I'3 ω cosα tan(cone half angle) = (L')1 coefficient / (L')3 = I'1 ωsinα / I'3 ω cosα = ( I'1/I'3) tan α 3. I have shown also that (I'1/I'3) tanα = tan θ. Combining this with the previous shows that the L precession half angle is just θ !! I think I then know about all three cones in the guy's pictures. Update 3.15.17 I have tried to write some of this up in line in "draft of G.8" but it is a bit messy. I solve for the Euler angle and get what appears to be conical motion of the rotor. But what are my conclusions about the ω and the L vector in Frame S? I know that L is fixed in the z direction, yes. I know that the rotor figure axis is connected to the Euler angles. But what is the ω vector doing? Maybe I just use these results, (ω)x = sinθsinφ + cosφ (ω)y = - sinθcosφ + sinφ (ω)z = cosθ + . // Frame S ok (G.6.4) which I can rewrite as [ using (ω)x = Ω' sinθsin[ (L/I'1)t] (ω)y = - Ω' sinθcos[(L/I'1)t] (ω)z = Ω' cosθ + (L/I'1) . since θ(t) = tan-1([I'1/I'3]tanα) // = constant ψ(t) = Ω't // Ω' = ωcosα (I'1-I'3)/I'1 φ(t) = (L/I'1)t Does the mag of ω work out right or not? It looks a bit fishy, ω2 = Ω'2sin2θ + (Ω' cosθ + (L/I'1) )2 ?? ω2 = Ω'2sin2θ + Ω'2cos2θ + (L/I'1)2 + 2 (L/I'1)Ω' cosθ ?? ω2 = Ω'2 + (L/I'1)2+ 2 (L/I'1)Ω'cosθ ?? I know that cosθ L = I'3ωcosα so I can eliminate that to get ω2 = Ω'2 + (L/I'1)2 + 2 (1/I'1)Ω' I'3ωcosα ?? But I know that Ω' = ωcosα (I'1-I'3)/I'1 So then have ω2 = ω2cos2α (I'1-I'3)2/I'12 + (L/I'1)2 + 2 (I'3/I'1){ωcosα (I'1-I'3)/I'1}ωcosα ?? ω2 = ω2cos2α (I'1-I'3)2/I'12 + (L/I'1)2 + 2 (I'3/I'1)ω2cos2α (I'1-I'3)/I'1 ?? I'12 ω2 = ω2cos2α (I'1-I'3)2 + L2 + 2 I'3ω2cos2α (I'1-I'3) ?? I also know that L2 = ω2 [( I'1sinα)2 +( I'3cosα)2] = ω2 I'12sin2α + ω2 I'32cos2α so put that in as well to get I'12 ω2 = ω2cos2α (I'1-I'3)2 + ω2 I'12sin2α + ω2 I'32cos2α + 2 I'3ω2cos2α (I'1-I'3) ?? I'12 = cos2α (I'1-I'3)2 + I'12sin2α + I'32cos2α + 2 I'3cos2α (I'1-I'3) ?? I'12 = cos2α (I'1-I'3)2 + I'12 - I'12cos2α + I'32cos2α + 2 I'3cos2α (I'1-I'3) ?? 0 = cos2α (I'1-I'3)2 - I'12cos2α + I'32cos2α + 2 I'3cos2α (I'1-I'3) ?? 0 = (I'1-I'3)2 - I'12+ I'32 + 2 I'3(I'1-I'3) ?? 0 = I'12 + I'32 - 2 I'1 I'3 - I'12+ I'32 + 2 I'3I'1 - 2 I'32 ?? 0 = I'32 - 2 I'1 I'3 + I'32 + 2 I'3I'1 - 2 I'32 ?? 0 = - 2 I'1 I'3 + 2 I'3I'1 ?? 0 = 0 YES!! ?? Now go back to (ω)x = Ω' sinθsin[ (L/I'1)t] Ω' = ωcosα (I'1-I'3)/I'1 (ω)y = - Ω' sinθcos[(L/I'1)t] (ω)z = Ω' cosθ + (L/I'1) . I don't like this form much. Go back to x = (cosφcosψ-cosθsinφsinψ) x' + (-cosφsinψ-cosθsinφcosψ) y' +sinθsinφ z' y = (sinφcosψ + cosθcosφsinψ) x' + (-sinφsinψ+cosθcosφcosψ) y' - sinθcosφ z' z = sinθsinψ x' + sinθcosψ y' + cosθ z' . (G.5.30) Apply this to the ω vector to get ω1 = (cosφcosψ-cosθsinφsinψ) (ω)'1+ (-cosφsinψ-cosθsinφcosψ) (ω)'2 + sinθsinφ (ω)'3 ω2 = (sinφcosψ + cosθcosφsinψ) (ω)'1 + (-sinφsinψ+cosθcosφcosψ) (ω)'2 - sinθcosφ (ω)'3 ω3 = sinθsinψ (ω)'1 + sinθcosψ (ω)'2 + cosθ (ω)'3 . Now insert our Frame S' solutions which were (ω)'1 = ωsinα sinψ (ω)'2 = ωsinα cosψ (ω)'3 = ωcosα α = cone half-angle **************************** Our Frame S' solution for (ω)' was (ω)' = To obtain the Frame S components of ω we have Maple compute ω = R-1(ω)' : where W = ω . The resulting vector can be simplified by combining the trig terms: The final result is then ω1 = ω sinφ sin(θ-α) ω2 = - ωcosφ sin(θ-α) ω3 = ωcos(θ-α) Using φ = (L/I'1)t this becomes ω1 = ω sin(θ-α)sin[(L/I'1)t ] ω2 = - ω sin(θ-α)cos[(L/I'1)t ] ω3 = ω cos(θ-α) This set of equations describes the precession in Frame S components of the vector ω at rate φ = (L/I'1)t with a cone half-angle of θ-α. Verify the minus sign: ω sinα cosθcosφ - sinθcosφωcosα = ωcosφ (sinα cosθ - cosαsinθ) = = ωcosφ sin(α-θ) = - ωcosφ sin(θ-α) First line is sinθcosα - sinαcosθ = sin(θ-α) signs are correct! Ω' figure axis ********************** sinθ L = I'1ωsinα ? sin2θ ω2 [( I'1sinα)2 +( I'3cosα)2] = I'12ω2sin2α ? sin2θ [( I'1sinα)2 +( I'3cosα)2] = ( I'1sinα)2 ? sin2θ [1 +( [I'3/I'1]cotα)2] = 1 ? [1 +( [I'3/I'1]cotα)2] = csc2θ = 1 + cot2θ ? ≡ cotθ = [I'2/I'1]cotα yes! same as *** *************** L2 = ω2 ( I'12sin2α + I'32cos2α ) 2T = ω2 ( I'1sin2α + I'3cos2α ) 2TI'1 = ω2 ( I'12sin2α + I'3 I'1cos2α ) 2TI'1 - L2 = ω2 ( I'12sin2α + I'3 I'1cos2α ) - ω2 ( I'12sin2α + I'32cos2α ) = ω2 [ I'3 I'1cos2α - I'32cos2α ] = ω2cos2α I'3(I'1-I'3) which says ω2cos2α = (2TI'1 - L2) / [I'3(I'1-I'3)] Do it again this way L2 = ω2 ( I'12sin2α + I'32cos2α ) 2TI'3 = ω2 ( I'1I'3sin2α + I'32cos2α ) 2TI'3 - L2 = ω2 ( I'1I'3sin2α - I'12sin2α ) = ω2I'1 ( I'3 - I'1)sin2α which says ω2sin2α = (2TI'3 - L2) / I'1( I'3 - I'1) So at this point we have ω2cos2α = (2TI'1 - L2) / [I'3(I'1-I'3)] ω2sin2α = (2TI'3 - L2) / I'1( I'3 - I'1) ω2cos2α = ω2sin2α = Divide to get tan2α = / = (I'3/I'1) ω2 = + As verification, let I'1 = a and I'3 = b so that I then get this result from adding the two equations ω2  = [2T(a+b)-L2]/ab But I started with L2 = ω2 ( a2sin2α +b2cos2α ) 2T = ω2 ( asin2α + bcos2α ) I would then like to show that ω2 = [ω2 ( asin2α + bcos2α )(a+b) - ω2 ( a2sin2α +b2cos2α )] /ab or 1 = [( asin2α + bcos2α )(a+b) - ( a2sin2α +b2cos2α )] /ab or ( asin2α + bcos2α )(a+b) - ( a2sin2α +b2cos2α ) = ab ? sin2α(a2+ab - a2) + cos2α(ab+b2 - b2) = ab? sin2α ab + cos2α ab = ab OK So my solution is ω2sin2α = (2TI'3 - L2) / [I'1( I'3 - I'1)] ω2cos2α = (2TI'1 - L2) / [I'3(I'1-I'3)] tan2α = (2TI'3 - L2) / I'1( I'3 - I'1) // (2TI'1 - L2) / [I'3(I'1-I'3)] = - (2TI'3 - L2) / I'1 // (2TI'1 - L2) / I'3 = - (I'3/I'1) (2TI'3 - L2) / (2TI'1 - L2)