scraps 1_14_17

Scraps 1.14.17 It has taken a LONG TIME to get this point on 12.16.16, Finally we get some oscillation which involves ω and involves a factor of 3. What did my PDF say about this? Here from p 124, So they are saying ωlibration = ω and that agrees with my equation, hurray!!! So you have to ignore the fictitious force to get this result! What is the condition for doing that? I show above that φ' = π/2 would do it, the planar case only! What is the Ref 2 above? ****************************************** Maple Problem: Enter eq1 and eq2, eliminate z as usual, and see if Maple can do a numerical solution. Do this in "satellite in Cartesions v5.mws". I am able to get it to fly, but the odeplot runs out of memory. Here is my code: Are these correctly entered? Yes they are. Next, here is how to eliminate z: Next we have This all looks good. Now we come to the dsolve call: There are only two equations, functions unknown are x(t) and y(t), and there are the inits. I start m1 at location x = 0 and y = 0 and no velocity. But then I ask if for f(0.1) and it locks up! So maybe I can try some other methods now. I started with classical and it hangs on f(0.1). I try rfk45 and get this My error was that I had rp1 and r1p both present. Now it works better. and I get finally some kind of semi-reasonable (perhaps) plot, It starts at the right cusp and goes round the loop and repeat Here is just less than one loop I think as it starts to move, there is Coriolis pushing it to the right. This is I think a big breakthrough in my Cartesian satellite plotting efforts, even though I don't understand the result and I may have lots of math errors still. But it is a reasonable motion! This would be my "out of plane" motion, and I do expect a deviation. Let's try x(0) = 0 and y(0) = 1, then I expect to get the "in plane" motion: Here it stays in the x = 0 plane and swings back and forth. This looks VERY good, I will now take a break, it is 2:30 PM. Now let's try for a circular path? ******************* r'12 = r12 + b2 + 2bz = b2 [ 1 + 2(z/b) + (r1/b)2] r'1-2 = b-2 [ 1 + 2(z/b) + (r1/b)2]-1 r'1-3 = b-3 [ 1 + 2(z/b) + (r1/b)2]-3/2 x = [2(z/b) + (r1/b)2] Now use (1+x)n = 1 + nx + n(n-1)/2 x2 + .... (1+x)-3/2 = 1 +(-3/2) x + (-3/2) (-5/2)/2 x2 + .... (1+[2(z/b) + (r1/b)2])-3/2 = 1 +(-3/2) [2(z/b) + (r1/b)2] + (-3/2) (-5/2)/2 [2(z/b) + (r1/b)2]2 + ... = 1 +(-3/2) 2(z/b) + (-3/2) (r1/b)2 + (15/8) * 4(z/b)2 = 1 -3(z/b) -(3/2) (r1/b)2 + (15/2)(z/b)2 so (b/r'1)3 = 1 -3(z/b) -(3/2) (r1/b)2 + (15/2)(z/b)2 and A ≡ (ω2b3/r'13) ≈ ω2[1 -3(z/b) -(3/2) (r1/b)2 + (15/2)(z/b)2] This is accurate for A through second order!! With full A I get this With first order approx A I get this With second order approx A I get this There must be something wrong with my approximation for A!!! r1p := r1^2+b^2+2*b*z(t); Yes indeed, this is the wrong expression for r1p. I am now keeping "the next term" for the moment. Then z - x = (ω2b3/r'13)bx - ω2(b+z)x + 2ωx = ω2[1 - 3(z/b) - (3/2)(r1/b)2]bx - ω2(b+z)x + 2ωx = [ω2 - 3ω2(z/b) - (3/2)ω2(r1/b)2]bx - ω2(b+z)x + 2ωx = ω2bx - 3ω2(z/b)bx - (3/2)ω2(r1/b)2bx - ω2(b+z)x + 2ωx = ω2bx - 3ω2zx - (3/2)ω2(r12/b)x - ω2bx - ω2zx + 2ωx = - 4ω2zx - (3/2)ω2(r12/b)x + 2ωx = - 4ω2zx + 2ωx - (3/2)ω2(r12/b)x So there \ ****************88 I am messing in v5 of the mws (OINLY!). With the full-bore A I get this With the approximate A I get this which is just the start of the previous curve. Why are these so very different? I thought my approximation for A was pretty good, but here it makes a huge difference in a trajectory! v6: