MY 11 TRISECTION APPROXIMATION METHODS         Pg. 90

 

SAIL SURFER TRISECTION

 

tri17c.gif
FIG. 17

 

CONSTRUCTION:

 
  1.  Draw any suitable straight line OA and draw a quarter arc with
   
   center pt O and radius R = OA = 1 unit, see FIG. 17 above.
    
  2.  Draw a line OB = R = 1 unit so that ÐBOA = 3Æ, see FIG. 17.
    
  3.  Drop a perpendicular from pt B to meet line OCA at pt C, as
    Pg. 91
   shown in FIG. 17. [OC/OB = OC/R = OC/1 = OC = a = cos3Æ]
    
  4.  Divide line CDA so that lengths CD/CA = 0.48
   
  

NOTE:

   
   see FACTOR TABLE below for ÐBOA = 3Æ = 75°.[ac = 0.52]
   
   I used ac = 0.48 to induce enough of an error to display all
   
   my figures FIG. 17 through FIG. 23.
    
  4.  a. With center pt C (pt a) and a suitable 100 unit radius just
   
  4.  larger than CA draw an arc to cut a perpendicular, extended
   
  4.  up from pt A to the intersection pt b on line acb, see FIG. 17.
   
  4.  b. Join pts a & b with a straight line, as shown in FIG. 17.
   
  4.  c. Locate pt c on line acb, so that ac = 48 units.
   
  4.  d. Drop a perpendicular from pt c to meet line ODA at pt D and
   
  4.  extend line Dc up to cut arc BEA at pt E.
    Pg. 92
  4.  [OD/OE=OD/R=OD/1=OD=cos2b & 2b = ÐEOA = cos-1OD & b'= ÐBOE = (3Æ - 2b)]
   
  4.  e. In similar 'right angled' DCDc & DCAb:
   
  4.  ÐbCA is common, thus sides CD/CA = ac/ab = 48/100 = 0.48
   
  5.  Duplicate ÐBOE = b' = 3Æ - 2b on line OA:
   
  5.  a. with center pt A and radius length BE draw an arc to cut
   
  5.  thearc BEFA at pt F above pt A, as shown in FIG. 17 above.
    
  5.  b. Join pts F & O with a straight line, see FIG. 17.
    
  5.  [Equal arcs and chords subtend the same angle at the center of the circle - theorem]
    
  5.  thus ÐBOE = ÐFOA = b'.
    
  6.  Drop a perpendicular from pt F to meet line OHA at pt H.
   
  4.  [OH/OF = OH/R = OH/1 = OH = cosb']
    
  7.  With center pt O and radius OH draw an arc to cut line OHA
   
   at pt H and line OIR at pt I, as shown in FIG. 17 above.
     Pg. 93
  8.  Drop a perpendicular from pt I to meet the line OJA at pt J.
    
  9.  Joint pts I & J with a straight line, as shown in FIG. 17 above.
   
  9.  [OJ/OI = OJ/OH = OJ/cosb' = cosb' & OJ = cos2b'] 
    
10.  Bisect the line DGA at pt G so that DG = GA = DA/2:
    
10.  a.With center pts D & A and a radius length just larger than
   
10.  DA draw arcs to to intersect each other twice.
   
10.  b. Join the 2 intersecting pts with a straight line and call the
   
10.  intersection pt with line DGA pt G.   [OG=OD+DG=cos2b+(1-cos2b)/2]
 

NOTE ERROR GJ ¹ 0 IN FIG. 17:

 
i.  cos2Æ = 2cos2Æ - 1 & 1 - cos2Æ = 2 - 2cos2Æ       [from CRC TABLES]
   
   cos2Æ + (1 - cos2Æ)/2 = cos2Æ               [IDEAL]
   
   == my 'FOCUSING EQUATION' used throughout this method!
   
   ie. OD + DG + GJ = OJ = 1 - JA               [IDEAL: GJ = 0]
    Pg. 94
ii.  with ERROR GJ:
   
   OG + GJ = cos2b + (1 - cos2b)/2 + GJ = cos2b'
   
   OG + GJ = 1 - JA = OJ = cos2b' = cos2(3Æ - 2b)
   
   ie. OD + DG - OJ = -GJ = ±e ¹ 0, as shown in FIG. 17 above.
   
   Length GJ = ±e is the present 'error value' to be corrected.
   
iii.  OG+GJ/3 = cos2b+(1 - cos2b)/2+GJ/3 » cos2Æ
   
   This method continues to decrease the 'error value' = ±e
   
   with FIG. 18 below.
   
iv.  FIG. 18 is a blow up of FIG. 17 at area GJ.
   
v.  ÐBOE = ÐFOA = b' is still not equal to Æ, see FIG. 17 above.
   
vi.  Let's continue with this closer approximation of GJ/3, as in
   
   FIG. 18 below.
  Pg. 95

FIG 18 GIF
FIG. 18

 
11. Divide the line GKJ so that the ratio GK/GJ = 1/3.
    
11. a. Similar to step 4.
    
11. b. With center pt G (pt d) and a 3 unit radius, just larger than
   
11. length GJ, draw an arc to cut line Jel at e, see FIG. 18.
    Pg. 96
11. c. Join pts d & e with a straight line, as shown in FIG. 18.
   
11. d. Locate pt f on line dfe so that df = 1 unit and df/de = 1/3
   
11. e. Drop a perpendicular from pt f to meet line GKJ at pt K.
   
11. f. In similar 'right angled' DGKf & DGJe:
   
11. ÐeGK is common, thus sides GK/GJ = df/de = 1/3.
   
12. Draw a circle with pt K as center and radius KA = LK = r to cut
   
  an arc across line OLKA at pts A & L, see FIG. 18/19.
   
13. At pt L draw a perpendicular LM to meet arc BMA at pt M.
   
  4.  [OL/OM=OL/R=OL/1=OL=cos2b'' & 2b'' = ÐMOA = cos-1OL & b'' = ÐBOM = (3Æ - 2b')]
    
14. Duplicate ÐBOM on line OA.
    
  a. Similar to step 5.:
   
   b. With center pt A and a radius length BM draw an arc to cut
   
   arc BMMA at pt N above pt A, as shown in FIG. 19 below.
  Pg. 97
tri19c.gif


FIG. 19

 
14. c. Join pts N & O with a straight line, see FIG. 19 above.
    
  5.  [Equal arcs and chords subtend the same angle at the center of the circle - theorem]
    
  thus ÐNOA = ÐBOM = b'''.
    
15. Drop a perpendicular from pt N to meet line OPA at pt P.
    Pg. 98
  [OP/ON = OP/R = OP/1 = OP = cosb''' = cos(3Æ - 2b'')]
    
16. With center pt O and radius OP draw an arc PQ to cut line
   
   OQN at pt Q, as shown in FIG. 19 above.
    
17. Drop a perpendicular from pt Q to meet line ORPA at pt R.
   
18. Joint pts Q & R with a straight line, as shown in FIG. 19 above.
   
  [OR/OQ = OR/OP = OR/cosb''' = cosb''' & OR = cos2b''' = cos2(3Æ - 2b'')]
 

NOTE ERROR KR ¹ 0 IN FIG. 19:

 
i.  Using my 'FOCUSING EQUATION':
   
  cos2Æ + (1 - cos2Æ)/2 = cos2Æ               [IDEAL]
   
   ie. OL + LK + KR = OR = 1 - RA               [IDEAL: KR = 0]
   
ii.  with ERROR KR:
   
   OL + LK + KR = cos2b'' + (1 - cos2b'')/2 + KR = cos2b'''
   
   OK + KR = 1 - RA = OR = cos2b''' = cos2(3Æ - 2b'')
    Pg. 99
  ie. OL + LK - OR = - KR = ±e ¹ 0, as shown in FIG. 19 above.
   
   Length KR = ±e is the present 'error value' to be corrected.
   
   When I used ac = 0.5 in FIG. 17 the KR = ±e error here
   
   was already nil.
   
iii. Assuming: OL + LK + 2KR/3 » cos2Æ
   
  ie. cos2b'' + (1 - cos2b)/2 + 2KR/3 » cos2Æ
   
   This method continues to decrease the 'error value' = ±e
   
   with FIG. 20 below.
   
iv.  FIG. 20 is a blow up of FIG. 19 at area KR = ±e.
   
v.  ÐNOA = ÐBOM = b''' is still not equal to Æ, see FIG. 19 above.
   
vi.  Let's continue with this closer approximation of 2KR/3,
   
   as inFIG. 20 below.
  Pg. 100
tri20c.gif


FIG. 20

 
19. Divide the line KSR so that the ratio KS/KR = 2/3.
    
19. a. Similar to steps 4. & 11..
    
19. b.With center pt K (pt g) and a suitable 3 unit radius, just larger
   
19. than length KR, draw an arc to cut line RiQ at pt i.
   
19. c. Join pts g & i with a straight line, as shown in FIG. 20, above.
   
19. d. Locate pt h on line ghi so that gh = 2 units.
   
19. e. Drop a perpendicular from pt h to meet line KSR at pt S.
   
19. f. In similar 'right angled' DKSh & DKRi:
  Pg. 101
tri21c.gif


FIG. 21

 
19. f. ÐiKR is common, thus sides KS/KR = gh/gi = 2/3.
    
20. With pt S as center draw a circle with radius SA to cut an arc
   
   across line OTSA at pt T to the left of pt S, see FIG. 21
   
   above.   [ TA = 2SA = 2TS = 2r ]
     Pg. 102
21. At pt T draw a perpendicular TU to meet arc BUA at pt U,
   
   see FIG. 21.   [OT/OU=OT/R=OT/1=OT=cos2biv & 2biv = ÐUOA = cos-1OT]
    
22. To bisect ÐMOU:
    
22. a. With pts M & U as center draw circles with a suitable radius
   
22. just larger than MU to cut each other at pt V just above U as
    
22. shown in FIG. 22 below.
    
22. b. Join pts O & V with a straight line, as shown in FIG. 22/23.
   
      [ ÐVOA = (ÐMOA + ÐUOA)/2 = (2b'' + 2biv)/2 & ÐBOV = Æ]
 
tri22c.gif


FIG. 22

  Pg. 103

NOTE:

 
Line OV agrees with the actual Ð2Æ mark which was carried along
 
(as an AutoCAD14 Drawing image), as shown in most of the
 
drawings above.
 
tri23c.gif


FIG. 23

  Pg. 104
23. Thus ÐBOA = 3Æ is trisected by ÐBOV = Æ as required,
   
  see Fig.23 above.

NOTE:

 
In the program formula [ see FACTOR TABLE below ] M & U
 
are reversed, with a much larger separation, and bisected to give
 
a similar accurate ÐBOV = Æ. However, the computer can repeat
 
lines 1 to 21 until the 'error value' = 0!
 

PROOF:

 
1. In 'right angled' D OCB:
   
  ÐBOA = 3Æ (75o) and OB = OA = R = 1 unit  [ see FIG. 17]
   
2. Let n = OC/OB = OC/R = OC/1 = OC = cos3Æ [cos75o = 0.2588190]
   
3. cos2Æ = 2cos2Æ - 1 or cos2Æ = 1 - 2sin2Æ  [from CRC TABLES]
   
  cos2Æ + (1 - cos2Æ)/2 = cos2Æ = cos2(3Æ - 2Æ)  [IDEAL]
    Pg. 105
  cos2z + (1 - cos2z)/2 + error = cos2z' = cos2(3Æ - 2z) [ƹz¹z']
   
  - used as the 'FOCUSING EQUATION'.    [ see FIG. 18/20].
   
4. Let Error Factor 'ac/ab' = CD/CA =
   
  = (cos2Æ - cos3Æ)/(1 - cos3Æ) = (cos2Æ - n)/(1 - n)
   
      [ = (cos50o - cos75o)/(1 - cos75o) = (0.6427876 - 0.2588190)/(1 - 0.2588190) = 0.518490 ]
   
  ie. ac/ab = 0.52 adjusted high, see 3Æ = 75o in FACTOR TABLE below.
   
5. In 'right angled' DODE:    [see FIG. 17]
   
  Let OD/DE = OD/R = OD/1 = OD = OC + CD = OC + (ac/ab)CA
   
  = n +.52CA = cos2b  [ = n + .52(1 - n) = .2588190 + .52(1 - .2588190) = 0.6442331 ]
   
  Let 2b = ÐEOD = cos-1OD  [ = cos-1(0.6442331) = 40.3917o = 40o 53' 30.480'' ]
   
  Let b' = ÐEOB = ÐFOA = 3Æ - ÐEOD = 3Æ - 2b  [ = 25.1082o ]
   
6. Let 2y1 = DA = 1 - OD = 1 - cos2b = 2DG = 2GA  [ = 0.355767 ]
   
7. OH/OF = OH/1 = cosÐFOA = cosb' = cos(3Æ - 2b) [=0.9065081]
  Pg. 106
tri24c.gif


FIG. 24  [ compare with Fig. 18/19 ]

 
7. OJ/OI = OJ/OH = OJ/cosb' = cosb' or OJ = cos2b'  [ = 0.819949 ]
   
  Let x1 = JA = 1 - OJ = 1 - cos2b' = 1 - cos2(3Æ - 2b)  [=0.180055]
   
  1st. ERROR: GJ = GA - JA = y1 - x1 = ±e ¹ 0  [ see Fig. 18/24 ]
   
      [ e = 0.177883 - 0.180055 = -2.171695E-3 ]

NOTE:

 
The above graphical method 'Error Factor' 0.48 (75o) gives
 
good display results but the computer formula method is accuracy
 
with y1 = 0.177883 which is out a mere 0.000717 from Y.
  Pg. 107
ie. X = cos2Æ = Y = (1 - cos2Æ)/2 = (1 - OD')/2 = 0.1786.
 
The 'Error Factor' is doctered to give a larger OD' giving a
 
larger ÐBOE = ÐFOA = b' > Æ and a smaller y1 < Y = X < x1.
 
cos2b + (1 - cos2b)/2 + error = cos2b  [ my 'FOCUSING EQUATION']
 
or 1 - 2Y' + Y' + e = 1 - X' or e = -X'/ Y'
 
ie. Y' is inversely proportional to X'.
 
8. KA = KJ + JA = LK = LA/2  [ see Fig. 18/19/24 ]
   
  KA = JA + 2GJ/3 = JA -2(JA - GA)/3  [ see Fig. 24 ]
   
  Let y2 = KA = x1 + 2(y1 - x1)/3  [ = x1 + 2(y1 - x1)/3 = 0.178607 ]
   
9. Let 2b'' = ÐMOA = cos-1(1 - LA) = cos-1(1 - 2y2)  [see FIG.19]
   
  [ = 50.00017o = 50o 0' 0.62'' ]
   
10. Let b''' = ÐBOM = ÐNOA = 3Æ - ÐMOA = 3Æ - 2b''  [=24o59'59.38'']
    Pg. 108
11. OP/R = OP/1 = OP = cosb'''     [ = 0.906391 ]
   
  OR/OQ = OR/OP = OR/cosb''' = cosb'''
   
  or OR = cos2b'''     [ = 0.82140 ]
   
12. Let x2 = RA = 1 - OR = 1 - cos2b'''     [ = 0.17860 ]
   
  2nd. ERROR: KR = KA - RA = y2 - x2 = ±e ¹ 0  [ see Fig. 25 ]
 
tri25c.gif


FIG. 25  [ compare with Fig. 20 ]

 

NOTE:

 
In the '1st. ERROR' the 'error value': -e is negative while in the
 
'2nd. ERROR' the 'error value': +e is positivetive; also y2 > Y > x2.
  Pg. 109
This negative cross-over property allows the computer formula to
 
home in on the solution like a spider web; since, x1 always over
 
shoots y1 (ie. x1 > y1), y2 always overshoots x2 (ie. y2 > x2),
 
and etc.; which did not happen with the graphic solution.
 
13. SA = SR + RA = TS = TA/2    [ see FIG. 20/21/25]
   
  Let y3 = SA = RA + KR/3 = RA + (KA - RA)/3
   
  Let y3 = TA/2 = x2 + (y2 - x2)/3  [ = x2 + (y2 - x2)/3 = 0.178605]
   
  NOTE: y3 = y2 = Y = 0.17860 to 5 decimal places already.
   
14. ÐUOA = cos-1(1 - TA) = cos-1(1 - 2y3) = 2biv  [ see FIG. 26]
   
  [ = 49.99982o = 49o 59' 59.3683'' ]
   
15. ÐBOV = ÐBOA - (ÐMOA + ÐUOA)/2 = Æ       [ see FIG. 26]
   
  ÐBOV = 3Æ - [cos-1(1 - 2y2) + cos-1(1 - 2y3)]/2
  Pg. 110
tri26c.gif


FIG. 26  [ compare with Fig. 23 ]

 
15. ÐBOV = 3Æ - (2b'' + 2biv)/2, as shown in FIG. 26 above.
   
   [ = 75o - (50.00017o + 49.99982o)/2 = 25.000001o = 25o 0' 0.0069'' ]
   
16. ÐBOA = 3Æ is trisected by ÐBOV = Æ, as required.
  Pg. 111

COMPUTER FORMULAS FOR ERROR FACTOR TABLE:

 
1.  By construction or reading a protractor 3Æ = 75o
   
2.  n = cos3Æ = cos75o = 0.2500190.  [R = 1]
   
3.  cos2b = OD = n + Error Factor*(1 - n).
   
 

NOTE:

   
  See 'Error Factor Table' below for 3Æ = 3o - 180o and
   
  'Error Factors' 0.55 - 0.25; and 'Error Factor' =0.52 for 75o
   
  OD = n + 0.52(1 - n) = 0.6442331
   
4.  2b = cos-1OD = 49.0918o
   
5.  b' = 3Æ - 2b = 75o - 2b = 25.1082o
   
6.  y1 = (1 - OD)/2 = 0.1770035
   
7.  cosb' = 0.9055081
   
  cos2b' = 0.819949
   
8.  x1 = 1 - cosb' = 0.180055
    Pg. 112
9.  y2 = x1 + 2(y1 - x1)/3 = 0.178607
   
10.  cos2b'' = 1 - y2 = 0.642786
   
  2b'' = cos-1(1 - y2) = 50.00017o
   
11.  b''' = 3Æ - 2b'' = 24.99983o
   
  cosb''' = 0.906391
   
  cos2b''' = 0.82140
   
12. x2 = 1 - cos2b''' = 0.17860
   
13. y3 = x2 + (y2 - x2)/3 = 0.178605
   
14. cos2biv = 1 - 2y3 = 0.642790
   
15. 2biv = cos-1(1 - y3) = 49.99982o
   
16. 2Æ' = (2b' - 2biv)/2 = 49.999995o
   
  or Æ' = 3Æ - (2b' - 2biv)/2 = 25.000001o = 25o 0' 0.0069''
    Pg. 113
 

NOTE:

   
i.  Accutacy is dependant on how far the program is iterated; for
   
  instance: bvii = Æ always.
   
ii. The table shows the optimum 'Error Factors', 0.55 to 0.25,
   
  for each angle from 3Æ = 3o to 180o respectively.
   
iii.  Blue text in the table below displays % ERROR = 100(Æ - Æ')/Æ
   
iv.  This table looks like a SAILSURFER going at full speed.
   
v.  only best ERROR for Error Factor values are shown.
 
Sail Surfer   SAILSURFER  
 

table5c.gif
FACTOR TABLE V


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