0316gadget.htm updated 29jul2008

MY 11 TRISECTION APPROXIMATION METHODS Pg. 16

TRISECTION GADGET

FIG. 3

DESCRIPTION:

1.	Most arms in FIG. 3 are hollow square tubing that fit inside each

	other with a slip fit. Ends are built up for proper height or width.

2.	The left end of arm A is hinged at pin 1 to arm F and welded to
	Pg. 17
	arm B at 45^o; the middle is hinged to arm H through hinge 5;

	and the right end is welded to pin 3, see FIG. 3 above.

	NOTE:

	Hinges are 1 or 3 sections of round tubing, with a rod or pin inside,

	welded to square tubing; while 'pins 1, 2 & 3' are sharp tipped rods

	extending » 0.5" below model, see VIEW "A - A" in FIG. 3, above.

3.	Arm B is welded securely to arm A at 45^o and hinged to arm J

	by hinge 4, as shown in FIG. 3 and in VIEW "A-A" above.

4.	Arm C holds a slider arm D and is welded to arm J at one end and

	hinged to arm I at hinge 0 at the other end, see FIG. 3 above.

5.	Arm D slides inside arm C at one end and is welded and hinged at

	pin 2 to slider E at the other end, as shown in FIG. 3 above.

6.	Slider E slides freely over arm F and is hinged at pin 2 to arm D.
	Pg. 18
7.	Arm F is hinged (1) to arm A at one end and slides inside slider E,

	midway, and hinged (6) to slider G at the other end, see FIG. 3.

8.	Slider G is hinged (6) to arm F and freely slides over arm H.

9.	Arm H is hinged (5) to arm A at one end, slides through slider G,

	midway, and is hinged (7) to arm I at the other end, see FIG. 3.

10.	Arm I is hinged (8) and built up to fit arm C at one end and is

	hinged (7) to arm H at the other end , see FIG. 3 above.

11.	Arm J is hinged at pin 4 to arm B at one end and welded to arm

	C, along its centre line, at the other end, see FIG. 3 above.

12.	Arms (J + C), I, H create a machine linkage or a parallelogram,

	to keep arms C, I, H parallel to each other, see FIG. 4 below.

13.	At the extreme right end of arm A is pin 3 which is located 3 units

	away from pin 1, see FIG. 4 below.
	Pg. 19
	NOTE

	The distance between pin 1 and pin 4 is equal to 0.8965755

	units and pins' Ð413 = 45^o , explained below.

14.	Together, the pins 1, 2 & 3 form the angle being trisected:

	at pin 1 (pins' Ð213 = 3Æ) and the resultant trisection angle at

	pin 3 (pins' Ð231 » Æ) for all angles of 3Æ, from 0º to 45º, see

	FIG. 3, above, and FIG. 5b, below.

NOTES FOR FIG. 4 AND FIG. 5b, BELOW:

1.	FIG. 4 , below, is FIG. 3, above, with all hinges or pins and their

	connecting centerlines reproduced exactly.

2.	FIG. 5b , below, is FIG. 4, below, with pins 1, 2 & 3 reproduced

	for all angles.

3.	By inspecting FIG. 5b , below, one can see that the gadget might
	Pg. 20
	be more accurate for ÐFOA (pin Ð213 = 3Æ) larger than 45º

	wherethe drawing approaches a straighter line.

4.	Mathematically the hinge (pin 4) attaching arm C to arm B in

	FIG. 3 can be a 'loose fit'.

5.	Arm B could have been welded further to the right or drilled thru to

	expose the hinge and pin as long as the distance between pin 1

	and pin 4 remains 0.8965755 units ie. pins' Ð413 = 45º.

Pg. 21

FIG. 4

CONSTRUCTION:

1.	Draw any suitable straight line OA = 3 units and the straight line
1.
	OB so that ÐBOA = 3Æ (0^o < 3Æ £ 45^o), see FIG. 4, above.

2.	Locate a suitable pt D on the line OD'B so that ÐD'OA = 3Æ.

3.	With a radius length OD' and centre pt O draw an arc across line

	OC'A at the intersection pt C' just below pt D', see FIG. 4 above.
	Pg. 22
4.	Join pt C' and D' with a straight line and extend a suitable

	distance to pt E, see FIG. 4 for guidance.

5.	Draw the line OF (pt F is hinge 4), so that ÐFOA = 45^o and

	length OF = 0.8965755 units, as shown in FIG. 4, above.

6.	Draw the straight line EFDC parallel to the straight line E'D'C'

	thus DO = OC through parallelism.

	NOTE:

	Now pts O, D, & A are pins 1, 2, & 3 respectively, see FIG. 4.

	Pins' Ð213 = ÐBOA = 3Æ and pins' Ð231 = ÐDAO = Æ

	Pts D & C are on the lines ODD'B & OCC'A respectively.

	Pts C', D' & E' are not shown in FIG. 3 or FIG. 4, above; since

	they represented MATH only.

7.	Draw the centerline JG of the mechanical 'arm A', see FIG. 3
	Pg. 23
	above, parallel to and below line OA, as shown in FIG. 4 above.

	NOTE:

	Mathematicallly, pt J could be pt O with pt G a suitable distance

	on line OGA; however, the idea is to prove the gadget in FIG. 3.

8.	Similarly draw a suitable centerline line JH of 'arm F', see FIG. 3,

	above, parallel to and above the line OB, with the same offset

	as JG and call their intersection pt J, as shown in FIG. 4, above.

	NOTE:

	ÐHJG = ÐDOA = 3Æ by parallel construction, see FIG. 4 above.

9.	As in step 3 above; with a suitable radius of length JG and centre

	pt J draw an arc across the lines JGA and JH, produced, so that

	JG = JH as shown in FIG. 4, above.

10.	Join pts G and H with a suraight line.
	Pg. 24
	NOTE:

	line HG is parallel to line EDC, by construction and parallelism.

11.	Take any suitable pt E on line CDFE; with radius length FE and

	center pt G draw an arc to intersect line GHI produced, at pt I.

12.	Join pts E and I and pts F and G with a straight line.

	NOTE: in FIG. 4

	Figure FGIE is a parallelogram, by construction and parallelism.

	and rotates about stationary pts F(4) and G(5) - a MACHINE LINKAGE.

13.	Drop a perpendicular from pt F to the meet line OKCA at pt

	K and extend line FK up a suitable distance to pt L, as

	shown in FIG. 4, above.

14.	Similarly drop a perpendicular from pt D to meet the line OKMCA

	at pt M.

15.	Thus Ð D'OA = 3Æ and Ð OAD' = Æ' for all positions of pt D' on
	Pg. 25
	the line EFD'C - explained later.

16.	Thus Ð D'AO = Æ trisects Ð D'OA = 3Æ, as required.

PROOF:

1.	In DFOA: in FIG. 4

	Since FK/OK = tan45^o = 1.

	and FK/KA = tan15^o = 2 - 3^½ = 0.2679492. [from CRC TABLES]

	then 3 = OK + KA = FK*((1/tan45^o) + (1/tan15^o)),

	thus length OK = FK = 3/((1/tan45^o) + (1/tan15^o)) = 0.6339746 units.

	NOTE:

	DM = 3/((1/tan33^o) + (1/tan11^o)) = 0.44880 = pi/7 = p/7.

	Can one parallel that; it still meeds a 22 or a 695 NOVA!

	See http://www.geocities.com/klaus_vanv/nick.htm#nova

	thus length *FO = 0.63397462^½ = 0.8965755** units.
	Pg. 26
	used in construction, as shown in FIG. 3 and FIG. 4, above.

2.	In DDOA:

	Let ÐDOA = 3Æ = 36^o and since OD = OC, by construction,

	then ÐODC = ÐOCD = (180^o - 36^o)/2 = 72^o; [isosceles D DOC]

	thus ÐKFC = ÐMDC = 90^o - ÐOCD = 90^o - 72^o = 18^o; [parallel]

	And thus KC = FKtan18^o = 0.63397460.3240167 = 0.2059908 units.

	therefore OC = OD = OK + KC = 0.6339746 + 0.2059908 = 0.8399655 units.

	thus DM = ODsin36^o = Ocsin36^o = 0.8399655*0.7265425 = 0.4937193 units.

	thus length OM = DMtan36^o = 0.49371930.7265425 = 0.6795464 units.

	and length MA = 3 - OM = 3 - 0.6795464 = 2.3204536 units.

	therfore, ÐOAD = atan(OM/MA) = 0.4937193/2.3204536 = 12.01161^o

	An ERROR of 100(12 - ÐOAD)/12 = 0.0967800%* - well within limits.

	Thus ÐOAD = Æ = 12.0^o trisects ÐAOD = 3Æ = 36.0^o, as required.

Pg. 27

MECHANICAL PROOF:

1.	In DDOC:

	The radii OD = OC make D DOC isosceles and thus ÐODC = ÐOCD =

	= (180^o - ÐDOA)/2 = (2i - 3Æ)/2 = 90^o - 3Æ/2 = i - 3Æ/2

2.	In DDCA:

	Since ÐDCA is the complement of ÐDCO, therefore

	ÐDCA = 180^o - ÐDCO = 2i - ÐOCD = i + 3Æ/2 and

	ÐCDA = 180^o - (i + 3Æ/2) - Æ = i - 5Æ/2.

3.	In DOFC:

	ÐOFC = 180^o - 45^o - ÐOCD = 2i - i/2 - (i - 3Æ/2) = (i + 3Æ)/2.

4.	In DCFK:

	ÐCFK = 180^o- 90^o - ÐOCD = 2i - i - (i - 3Æ/2) = 3Æ/2

	- ie. the bisector of ÐDOC = 3Æ. [CONSTRUCTION: 2, & 3., Pg. 21]

	One could have used this angle to get pt D directly.
	Pg. 28
5.	In DHJG & DDOC:

	DHJG is congruent to DDOC; since the sides OD = OC are parallel

	to HJ = GJ = R respectively - making them both isosceles.

	Thus ÐJHG = ÐODC = i - 3Æ/2 and thus the bases (HG & DC) of

	both Ds, or lines IHG and EFDC, are parallel at all times.

6.	In figure EFGI:

	Since EF = IG, by construction, and since EF and IG are parallel

	makes figure EFGI parallelogram and a RHOMBUS.

	Thus EI = FG and are always parallel to each other, see FIG. 4.

	NOTE: pts F & G are fixed pts with hinges 4 & 5, respectively.

7.	Thus Ð DAO = Æ trisects Ð DOA = 3Æ, as required.

Pg. 29

FORMULAS: (FIG. 4, above)

	In DOFK & DOFA:

	With length OA = 3 units and OK = FK of a 'right angled' D of 45^o

	tanÐFAO = tan15^o = 2 - 3^½ = FK/KA [from CRC TABLES]

	OA = OK + KA = 3 = OK + OK/(2 - 3^½)

	OK = 3(2 - 3^½)/(3 - 3^½) = 0.6339746 units.

	*OF = 2^½OK = 0.63397461.4142136 = 0.8965755** units

	Since sin15º = 2^½(1 - 3^½) = FK/AF = OK/FA [from CRC TABLES]

	FA = OK/sin15^o = 3(2 - 3^½)/[(3 - 3^½)2^½(1 - 3^½)] = 6^½ = 2.44949 units

	OD = OC = OK + KC = OK + FK/tan(i - 3Æ/2)

	OD = OK + OKtan(3Æ/2) = 0.6339746[1 +* tan(3Æ/2)]

	Let OM = ODcos3Æ = 3 - MA = 3 - x

	x = OM - 3 = 3 - ODcos3Æ

	Let DM = ODsin3Æ = y
	Pg. 30
	tanÐOAD = y/x = DM/MA = DM/(3 - OM) = ODsin3Æ/(3 - ODcos3Æ)

	ÐOAD = atan(y/x) = atan[sin3Æ/(3/OD - cos3Æ)]

	= atan[sin3Æ/(3/(0.6339746*(1 + tan[3Æ/2])) - cos3Æ)].

CALCULATIONS FOR ERROR TABLE 1:

	Let 3Æ = ÐDOA

	Æ = 3Æ/3 = ÐDOA/3

	b = ÐOAD = atan[sin3Æ/(3/(0.6339746(1 + tan[3Æ/2*])) - cos3Æ)]

	% ERROR = (Æ - *b)/Æ100**

	giving ERROR TABLE 1 for 3 different lengths of OA below:

Pg. 31

Pg. 33

TRUE LENGTH TABLE 2: (for length OA)

	Let 3Æ = ÐDOA and Æ = 3Æ/3 = ÐDOA/3 = ÐDAO

	OM = ODcos3Æ and DM = ODsin3Æ

	KC = FKtan(3Æ/2) = OKtan(3Æ/2)

	OD = OC = OK + KC = OK + OKtan(3Æ/2) = OK[1 +* tan(3Æ/2)]

	MA = DM/tanÆ = ODsin3Æ/tanÆ

	OA = OM + MA = 0.6339746[1 + tan(3Æ/2)](cos3Æ + sin3Æ/tanÆ)

	giving TRUE LENGTH TABLE 2 for OA, for every degree, below:

Pg. 34

Pg. 35

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MY 11 TRISECTION APPROXIMATION METHODS Pg. 16

TRISECTION GADGET

FIG. 3

DESCRIPTION:

NOTE:

NOTE

NOTES FOR FIG. 4 AND FIG. 5b, BELOW:

FIG. 4

CONSTRUCTION:

NOTE:

NOTE:

NOTE:

NOTE: in FIG. 4

PROOF:

NOTE:

MECHANICAL PROOF:

FORMULAS: (FIG. 4, above)

CALCULATIONS FOR ERROR TABLE 1:

TRUE LENGTH TABLE 2: (for length OA)