kill(all)$

load(lrats)$

geometry :trajectory of ballistic missile, part 2  , 'version 1 '
peter.vlasschaert@gmail.com,22/06/2018

Construction fig 1
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combination : paint ( = color),geogebra(=geometry),maxima( = symbol ,ψ,η,λ)
Geogebra :  https://www.geogebra.org/classic  (online soft : save  ellipsegeo4.png) ,easy to use
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point      :   F   on the ellipse  (= burn-out  point) 
point      :   l    on the ellipse  (= re-entry   point)
point      :   0   center of earth ( = focus from the ellipse)
point      :   B   other focus from the ellipse
point      :   M  divide angles around  point F  ( = η _bo : red  ,brown)
distance :   | MF | = h
distance :   |0B   | = distance between focuses, 0 = center of the earth
distance :   |0F   | = distance between center of earth and the burn-out point = r_bo
distance :   |FB  | = r_bo' = r_boo
distance :   |LF  | = local horizon
distance :   |FE  | = vector 'v_bo'  tangent to the ellipse.(line trough F and E = tangent line)
distance     |KF |  =  gives 'right angle with vector ' = v_bo '
definition : ellipse  => |0F|+|FB| = 2*a , a = semi major axis, b= semi minor axis
distance   :     |0F|+|FB| = r_bo + r_bo '  = 2*a            
triangle         : | B0F | ,include two 'sub triangles ' = {pp,qq}
triangle         :  pp = |BMF| right triangle 
triangle         :  qq = |FM0| right triangle
sub triangle  : divide = distance between two sub triangles = | MF | = h
angle      : between( M0F) =ψ/2
angle      : between(MBF)  = λ
angle      : between (BF0)  = 2*η_bo
angle      : between( LFE)  = η_bo = flight path angle.

? How to find  'h' ,triangle qq ' = right triangle'
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p1 : sin(ψ/2) =  h/r_bo;
p2: solve(p1,h);
p3: p2[1];

?  How to find 'λ' , λ =λ(η_bo,ψ)
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1e ) definition (general triangle) : sum three angles :  'triangle         : | B0F |  = π'
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p4:λ+2*η_bo +ψ/2 = %pi;

2e)  λ : ?
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p5:solve(p4,λ);

p6:ratexpand(p5);
p7:p6[1];

? How to find  'h' ,triangle pp ' = right triangle'
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1e)  find  sin( λ)  ,triangle pp
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rem : r_bo ' = r_boo

p8:sin(λ) = h/r_boo;

2e)  h : ?
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p9:solve(p8,h);
p10:p9[1];

3e) use 'λ' see above 'sum of triangles'
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p11: lratsubst(p6,p10);

from both ,sub triangles ={pp,qq},we find 'h'
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rem :  h = h

p12:rhs(p3) =rhs(p11);

p13::p12/r_boo;

How to removes ,r_bo and r_boo from 'p13'  
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rem : with use  Δ = Δ b_o ,from first part geometry.
rem : because point F ( = missile,point of burn-out)

1e)  definition ellipse : 'see above fig 1'
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=> remove : r_boo

p14:r_boo+r_bo = 2*a;
p15:solve(p14,r_boo);
p16:p15[1];

2e)  from part 1 of geometry ,relation  a=a(Δbo,r_bo)
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p17: a=r_bo/(2-Δbo);

3e) combine  1e) and 2e)
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p18:lratsubst([p17],p16)$
p19:ratexpand(p18);

How to remove : r_bo and r_boo from 'p13'
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p20:lratsubst([p19],p13);

p21:expand(p20);

p22:factor(p21);

To use  formula 'p21' => to find   ' η_bo = flight path angle '
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given : missile                     : Δr_bo < 1
            range free flight      :  ψ
=> value on rhs 'p21' 
result : η_bo = flight path angle
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p23:solve(p21,η_bo);
p24:p23[1];

reason :' two angles' with same value sine '=n' =>  lhs(p21)
reason : see goniometric circle 'fig 2'
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trajectory : small value => low trajectory 
trajectory : high  value => high trajectory
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=>  for given :  Δr_bo Δbo =Δ(r_bo,v_bo),see first part 'geometry'
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rem : 
max range : Δr_bo < 1  =>  π < range, high and low trajectory
max range : Δr_bo > 1  => + and - values η_bo 
                   1e) neg values : penetrate the earth (= low  trajectory)
                   2e) pos values  : good                       (= high trajectory)
                   =>  π >  range ,exit possible trajectory over north pole
(= south pole), result will be connected to place of launch  a missile)
rem : low and high are to target
                     Δr_bo  =1 => not of use.
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