Envelopes generated by coupler-bar line segments in James

Watt’s four-bar linkages

Peter Scales Grant Keady,University of Western Australia

January 18, 2000

1 Introduction

This report was largely prepared while working towards the two papers [KSF, SK]. Theengineering motivation is in [KSF].

For pointers to Maple and Mathematica code – both that due to others and itemsrelated to this document and to [KSF, SK], see:http://maths.uwa.edu.au/~keady/MechEng/FourBar/index.html

1.1 Computational matters

1.1.1 Computer animations

There are animations of linkages in Maple, Mathematica, etc.. For animations in otherpackages, see, for example,http://www.math.ntnu.edu.tw/~jcchuan/demo/gear/machine.html

http://forum.swarthmore.edu/sketchpad/mechlink.html

http://forum.swarthmore.edu/sketchpad/java_linkages.html

The last is a Java variant of animations previously done in Geometer’s Sketchpad, andpresumably similar code is around for Cabri-Geometre and its graphing calculator im-plementations e.g. for Texas Instrument calculators like TI-89 and TI-92. The ‘WorkingModel’ package, widely used in engineering education, has also been used for such anima-tions. Maple code with the animation is at the “Applications” pages at maplesoft.com.The Mathematica code on MathSource, wolfram.com, is more oriented to finding ‘couplercurves’. (Coupler curves are treated in § 1.3 below.) For links, see the URI above.

1.1.2 CAS

There is more mathematics - of various kinds - associated with the problem than isevident in the numerical work of the kind needed for animations, etc.. There is a vastliterature on linkages, this literature ranging from popular high-school level items as in

1

[CR], through to mechanical engineering books of various levels both of mathematics andof specialisation, e.g. [Dij, HN]. In this study there is one less common question - findingenvelopes of the coupler bars which arose in connection with shaping/polishing (see also[KSF] - , but much of the study concerns checking of relations in the older literature.Groebner bases proved to be a useful tool. The code - in both Maple and MathematicaComputer Algebra Systems (CAS) - is collected in an Appendix and is available on theWeb. See the URL given before.

1.2 Notation

We consider linkages which, in their initial configuration, form a symmetric trapeziumABCD, these points occuring as one traverses the trapezium in a clockwise direction.Points A : (−a, 0) and B : (a, 0) are fixed. C and D move on circles centred on B and A

respectively, both with radii of b. The length of bar CD is 2c. The bar BC will be calledthe crank, CD the coupler, DA the follower.

Rough diagram. To be replaced by eps figure.

D-----------------C

/ \

/ \

/ \

A------------O------------B

Figure 1: The four bar linkage. Upper figure is when the linkage is in its initial symmetricposition – an isosceles trapezoid. Lower figure is the linkage in another position.

The angles, α, β, γ are the angles formed with the horizontal by bars DA, CB andDC at A, B and C respectively. Thus angle OAD is α: angle OBC is (π − β).

We will be concerned with the envelope of the lines defined by CD as the linkagemoves through different configurations. For circular segments on the coupler bar, see[KSF].

1.3 Locus of the midpoint of the coupler bar

We will be concerned with the envelope of the lines defined by CD as the linkage movesthrough different configurations.

2

Generalising this concern to one about points on the line CD, we begin by consideringthe locus of the midpoint E of CD:

xE =1

2(xC + xD) , yE =

1

2(yC + yD) .

We begin with this item as the locus is well known and is called Watt’s curve. The nameis in honour of James Watt of steam-engines who considered it. See [Lo] p162, [Sh] p179,andhttp://www-groups.dcs.st-and.ac.uk/~history//Curves/Watts.html

Watt’s curve is

pE := x2E(x2

E + y2E + PE)2 + y2

E(x2E + y2

E +QE)2 − REy2E = 0, (1.1)

where

PE = c2 − b2 − a2,

QE = c2 − b2 + a2,

RE = 4c2a2 = Q2E − P

2E + 4a2b2 .

(The alternative Cartesian form given in [Sh] is equivalent to (1.2) below.) pE is ahomogeneous polynomial of degree 6 in x, y, a, b and c and only even powers in thesevariables occur in pE . (There might have been a case for using RE = 2ac rather than REabove. Then pE would be a homogeneous polynomial of degree 3 in x2, y2, PE , QE , RE.

This requires some patience to derive by hand – though it is not hard. See [Dij]pp246-248. Equation (1.1) is readily checked, using Groebner bases, by the CAS codegiven in the Appendix.

It is easy to show (see (1.2)) that pE > 0 for x2E + y2

E = r2E > b2, which implies that

the Watt’s curve lies entirely within the circular disk centred at the origin and of radiusb.

Remark. A curious little identity (for which, as yet, we have no use) is

1

xE

∂pE∂xE

−1

yE

∂pE∂yE

+ 8a2(x2E + y2

E − b2)

= 0 .

It is also possible to describe the coupler curve in polar coordinates:

xE = rE cos(θ) , yE = rE sin(θ) .

Substituting these into equation (1.1) and rearranging gives

pE/r2E = (r2

E + PE)2 + 4a2 sin(θ)2(r2E − b

2)

= 0 where PE = c2 − a2 − b2 . (1.2)

This is a quadratic in r2E with the following properties:

The product of the roots is positive.The discriminant is positive if and only if | cos(θ)| < c/a.If the discriminant is positive, then it can be shown that the roots satisfy r2

E < b2.Equation (1.2) rearranges in various ways, e.g.

sin(θ) = ±1

2a

√b2 − r2E −

(a2 − c2)√b2 − r2

E

.

3

This last equation is valid provided the absolute value of the right-hand side is less thanor equal to one. Another way of writing equation (1.2) is as the polar equation

r2E = b2 −

(a sin(θ)±

√c2 − a2 cos(θ)2

)2

. (1.3)

(Assuming c 6= a gives complex values from the inner square-root for some range of θ.)In the ‘granite polishing’ application, motions when the mechanism rocks about θ = π/2are of interest. At θ = π/2, equation (1.3) becomes

r2E = b2 − (a± c)2,

and the symmetric trapezoid has r2E = b2 − (a− c)2. If b < (a+ c) this is the only value

for which rE, corresponding to θ = π/2, is real.

In other applications it might be useful to define also

tE = tan

(θ

2

)so xE = rE

2tE1 + t2E

, yE = rE1− t2E1 + t2E

.

For the case c = a, the sextic (1.1) factors into a product, one factor of which cor-responds to a circle of radius b. When c = a = b/

√2, the other factor corresponds to a

Bernouilli lemniscate. See [Lo], p114.

For another item on Bernouilli lemniscates, see:http://www.best.com/~xah/SpecialPlaneCurves_dir/specialPlaneCurves.html

There may also be some useful forms of parametric description of Watt’s curve. Insome contexts it is natural to think of the curve being described as the crank moves, i.e.as B moves around in its circle. xE and yE are determined by the angular positions αand β of C and D. α is related to β as in equation (2.2), and through this one can findrelations to determine xE(β) and yE(β). In one approach one finds a quadratic equationfor xE not containing yE , and then, having solved for xE, yE is determined from

(xE − a− b cos(β))2 + (yE − sin(β))2 − c2 = 0.

The expressions involved are rather elaborate. There are, however, some simple identitiesarising in this problem. For example, in §2.2 we will see a simple relationship between θand the angles, α, β with the horizontal at A, B:

tan(θ) = tan

(α + β

2

).

This identity is useful in finding a relation (2.4) between tan(θ) and β which, used with(1.3), yields a parametric description (xE(β), yE(β)).

Remark. Bipolar coordinates might also prove to be of some use in this problem. Denotethe length from A to E by rAE and that from B to E by rBE. We find

pE =(2 rAE

2rBE2 −

(b2 − 2 c2

) (rAE

2 + rBE2)

+ 2 b2a2 − 8 c2a2 + 2 c4)×(

rAE2 + rBE

2 − 2 b2 − 2 a2)/4 +

b2(−a2 + c2

)2.

4

On putting c = a = b/√

2, this bipolar form of pE factors to(2 rBE rAE + b2

)(2 rBE rAE − b

2)(rAE

2 + rBE2 − 3 b2

)/8 = 0,

and the factor rAErBE − b2/2 corresponds to the Bernouilli lemniscate.

1.4 Locus of a general point on the coupler bar

We now consider a general point K on CD:

xK = (1− k)xC + kxD , yK = (1− k)yC + kyD .

Its locus can be found (easily using Groebner bases in a CAS) as pK = 0 where, withak = (2k − 1)a and ck = (2k− 1)c,

pK := (xK − ak)2 (x2K + y2

K + PK + 2akxK)2 + y2K(x2

K + y2K +QK + 2akxK)2

−RKy2K + 16akk(1− k)(c2 − a2)φK = 0, (1.4)

φK :=(x3K + xKy

2K + akx

2K − (c2

k + a2 + b2 − c2)xK + ak(b2 − a2

k))

where

PK = PE +(2a2

k − c2k

),

QK = QE − c2k ,

RK = RE − 4a2k

(c2 + 4k(1− k)a2

).

1.5 Series expansion for Watt’s curve about θ = π/2

In the symmetric trapezoid configuration we have θ = π/2, xE = 0 and y2E = b2−(a−c)2.

Later in this document we will wish to consider an envelope when the mechanismmoves a little about θ = π/2.

Different parametrisations are possible. The choice of parametrisation might helpreduce the algebra. At present our favorite choice for parameter is

tO = tan(π

4−θ

2) =

1− tE1 + tE

, so tE = tan(θ

2) =

1− tO1 + tO

.

1.5.1 Expansion in terms of xE

Watt’s curve is a sextic in both xE and yE but only even powers of both occur. Thus itis a cubic in both xE2 = x2

E and yE2 = y2E . There are, of course, explicit formulae for

solving cubics ‘by radicals’, but in this application the formulae appear to be messy.

There are several approaches to finding the coefficients ηj in the power series

yE2(xE2) =(b2 − (a− c)2

)+ η1xE2 + η2x

2E2 + . . . .

Define yE2,0 =(b2 − (a− c)2

).

5

One method (which appears to be faster in CAS than the series operations availablein the CAS is to arrange the Watt’s curve polynomial p so that one can iterate. Seek

yE2 = yE2,0 + ρ , ρ =∑j=1

ρjxjE2 ,

and substitute this into Watt’s polynomial

−Aρ = BxE2 + f(xE2, ρ) ,

where f is higher degree of smallness than the other terms.

A = −4ac(−c+ a− b)(−c+ a+ b) , B = 4a(b2c+ 3ac2 − 3a2c+ a3 − c3) ,

so ρ1 = −B/A. We find

ρ2 = −b2c+ (a− c)3

c (b2 − (a− c)2).

1.5.2 Expansion in terms of tO

The polar equation (1.3) suggests (as borne out in calculation) that expansions aboutθ = π/2 will be easier than expansions for yE(xE) about xE = 0.

In our implementation of the calculations we chose to use tO as the parameter, andseek series expansions about tO = 0.

We find:

x2E ∼ 4

(b2 − (a− c)2

)tO

2 + 8

(−b2c+ 5 ca2 − 4 ac2 + c3 − 2 a3

)c

t4O + . . .

y2E ∼

(b2 − (a− c)2

)− 4

(b2c+ (a− c)3

)c

t2O +

4

(2 b2c3 − 14 c3a2 + 9 ac4− 2 c5 + 8 a3c2 − a5

)c3

t4O . . . .

2 Relations involving α, β, θ

2.1 α, β

As defined near Figure 1, Angle BAD is denoted by α: angle ABC is π − β.

Expanding out (xC − xD)2 + (yC − yD)2 − 4c2 = 0, and dividing by 4 gives

2ab (cos(β)− cos(α)) + b2 − 2c2 + 2a2 − b2 cos(β − α) = 0. (2.1)

This leads directly to the relationship between tA = tan(α/2) and tB = tan(β/2):

((a+b)2−c2+t2B(a−c)(a+c))t2A−2b2tAtB−c2+a2 +t2B(−a+b−c)(−a+b+c) = 0 . (2.2)

This is a quadratic equation determining tA from tB .

Various representations of the states of the linkage are possible. They form curvesof allowed (α, β) values on a torus: they form curves of allowed (tA, tB) values in the(tA, tB)-plane.

6

2.2 tan(θ) = tan(α+β2

)

Related geometrical results are presented in [SK].

The result (2.3) of this section concerns the geometry of quadrilaterals which haveone pair of opposite sides equal. The result concerns the line OE joining the midpointsof the other sides.

Is this θ = (α + β)/2 Euclidean geometry result a named theorem and, if so, what isits name?

The coordinates of E are

xE =b

2(cos(α) + cos(β)) , yE =

b

2(sin(α) + sin(β)) .

Hence

tan(θ) =yExE

=sin(α) + sin(β)

cos(α) + cos(β),

=2tA(1 + t2B) + 2tB(1 + t2A)

(1− t2A)(1 + t2B) + (1− t2B)(1 + t2A),

=tA + tB1− tAtB

,

= tan

(α+ β

2

). (2.3)

With τ = tan(θ), tA = (τ − tB)/(1 + τtB). Using this to eliminate tA from equa-tion (2.2), we obtain a quadratic equation determining tan(θ) from tB .

−((−a+ b+ c)

(tB

2 + 1)− 2b

)((c− b+ a)

(tB

2 + 1)

+ 2b)τ2 +

4 b tB((b− a)

(tB

2 + 1)− 2b

)τ +((

a2 − c2) (tB

2 + 1)2

+ 4b2t2B

)= 0 (2.4)

An alternative form for equation (2.4) is

−((−a+ b+ c) tB

2 + c− b− a)(

(c− b+ a) tB2 + c+ b+ a

)τ2 +

4 btB(tB

2 (b− a)− a − b)τ +(

−c2 + a2)tB

4 +(4 b2 + 2 a2 − 2 c2

)tB

2 − c2 + a2 = 0

3 The direction of the coupler line CD

On p247 of [Dij], the equation (1.1) pE = 0 for Watt’s curve is derived starting fromelementary trigonometry. Let the direction of line CD be denoted by (cos(γ), sin(γ)).The steps used in [Dij] p247 include establishing the identities, valid for cd 6= 0,

cos(γ) = −1

2ac

(x2E + y2

E +QE), (3.1)

sin(γ) =xE

2acyE

(x2E + y2

E + PE), (3.2)

7

(Squaring and adding gives equation (1.1), pE = 0.)

This leads to an equation for the slope m of CD:

m = −xEyE

(x2E + y2

E + PE)(

x2E + y2

E +QE) .

The remainder of §3 is largely concerned with more elaborate polynomial relationsinvolving xE , yE and m The relation immediately above is degree 1 in m: those beloware degree 2.

3.1 Quadratics in m(θ) and/or m(xE, yE)

Simple trigonometry gives

xE = (a+ b cos(β)) + c cos(γ) , yE = b sin(β) + c sin(γ) . (3.3)

Eliminating β from equations (3.3) gives

(rE cos(θ)− a− c cos(γ))2 + (rE sin(θ)− c sin(γ))2 = b2

which expands to give an equation which is a quadratic in tC :

r2E − 2rEa cos(θ) − 2rEc cos(γ) cos(θ)−

2rEc sin(γ) sin(θ) + 2ac cos(γ) + a2 + c2 − b2 = 0 (3.4)

where rE is given in terms of θ by equation (1.3). The preceding equation (3.4) can bewritten in terms of (xE, yE) and tC as

2c(a− xE)(1− t2C)− 4yEctC +((xE − a)2 + y2

E + c2 − b2)

(1 + t2C) = 0

which, on collecting terms in tC is((xE − a+ c)2 + y2

E − b2)t2C − 4yEctC +

((xE − a− c)

2 + y2E − b

2)

= 0 .

Define

S0 = −(xE −a)2−y2E , S2 = −4(S0 + (b+ c)2)(S0 + (b− c)2) , and SE = ±

√S2 .

The quadratic can be solved for tC

tC =1

2

4yEc+ SE(xE − a+ c)2 + y2

E − b2.

From this we find

m = tan(γ) =2tC

(1− t2C)

= −1

2

(S0 + b2 − c2

)SE − 8 yE c

2 (xE − a)((xE − a)2 + (yE − c)

2 − b2) (

(xE − a)2 + (yE + c)2 − b2) . (3.5)

8

For the curvature calculations which follow, we will also need some derivatives of m.In this connection it may be helpful to define

pm =(

(xE − a)2 + (yE − c)2 − b2

) ((xE − a)2 + (yE + c)2 − b2

)m2 +

8c2(xE − a)yEm+((xE − a− c)

2 + yE2 − b2

)((xE − a+ c)2 + yE

2 − b2). (3.6)

m satisfies pm = 0. This is readily verified using Groebner basis techniques. See AppendixA. What is shown in Appendix A is that if one sets m = (yC − yE)/(xC − xE) in pm andmultiplies by (xC − xE)2, then the polynomial in xE , yE , xC, yC so formed has normalform zero.

3.2 Quadratics in m(β)

Here is an alternative to the treatment given above in the subsection m(θ). As before,denote by m the slope of the coupler bar CD.

A calculation gives

m =yD − yCxD − xC

=b(sin(α)− sin(β))

b(cos(α)− cos(β))− 2a

= −b (−tB + tA) (1− tB tA)

a (1 + tA2) (1 + tB

2) + b (−tB + tA) (tA + tB )

This is a rational function but involves both tA and tB. Equation (3.9) involving tB only,but involving square roots, appears to be more useful in the envelope calculation whichfollows.

Denote by γ the angle between the horizontal and CD (measured in the same way asα the angle between the horizontal and AD, and β the angle between the horizontal andBC). m = tan(γ). Simple trigonometry gives

xD = −a + b cos(α) = (a+ b cos(β)) + 2c cos(γ)

yD = b sin(α) = b sin(β) + 2c sin(γ) ,

so

b cos(α)− 2c cos(γ)− b cos(β) = 2a (3.7)

b sin(α)− 2c sin(γ)− b sin(β) = 0 (3.8)

As a check on these, we note that eliminating γ, via cos(γ)2 + sin(γ)2 = 1, from thepreceding (3.7), (3.8), gives

(2a− b cos(α) + b cos(β))2 + (b sin(α)− b sin(β))2 = 4c2 ,

which rearranges to give equation (2.1).

Eliminating α between equations (3.7) and (3.8) gives

(2a+ 2c cos(γ) + b cos(β))2 + (2c sin(γ) + b sin(β))2 = b2 ,

(2a+ b cos(β))c cos(γ) + b sin(β)c sin(γ) + a2 + c2 + ab cos(β) = 0 .

9

This rearranges to give a quadratic equation for tC = tan(γ/2):

(−a + c)(c+ ct2B − a+ bt2B − b− t

2Ba)t2C + 4 bctB tC −

(c+ a)(−c − ct2B − a+ bt2B − b− t

2Ba)

= 0 .

This in turn rearranges to

(a− c) (a− c+ b cos(β)) t2C + 2 bc sin(β)tC + (a+ c) (a+ c+ b cos(β)) = 0 .

Solving gives

tan

(γ

2

)= tC = −

bc sin(β)± S

(a− c)(a− c+ b cos(β)),

where S =√b2c2 − (a2 − c2 + ab cos(β))2, whence

m = tan(γ) =2tC

1− t2C

=bc2 sin(β) (b cos(β) + 2a)±

(a2 + c2 + ab cos(β)

)S

(c2 + a2)(c2 + (a+ b cos(β))2

)− b2c2

. (3.9)

Another approach to obtaining a relation between γ and β is to substitute xE and yEfrom equation (3.3) into equation (1.1) defining Watt’s curve.

4 Envelope formed by successive locations of coupler bar

4.1 Envelope (X, Y ) calculationsUsing tO, θ, or xE, as parameter

TO DO. Rework this to use tO as parameter. Use superposed dots for differentiationwith respect to tO and primes for differentiation with respect to xE .

The starting point is the expression for m(xE, yE) found in §3. We will also need thederivative of this with respect to xE , and hence also dyE/dxE. This latter is readily foundfrom equation (1.1):

y′E =dyEdxE

= −∂p

∂xE/∂p

∂yE

=

(3 r4

E − 4(b2 − c2

)r2E +

(2 xE a+ b2 − c2 + a2

) (−2 xE a+ b2 − c2 + a2

))xE(

−3 r4E + 4 (b2 − c2) r2

E − 4 a2yE2 − T)yE

where T is the symmetric polynomial (which factors to 4 linear factors)

T = b4 + a4 + c4 − 2b2a2 − 2b2c2 − 2a2c2 .

The equation of the line CD is

(Y − yE)−m(X − xE) = 0 , (4.1)

where m is given in §3. m′ can be calculated in terms of (xE, yE , yE′).

10

Differentiating equation (4.1) with respect to xE gives

−y′E −(∂m

∂xE+m′

dyEdxE

)(X − xE) +m = 0 . (4.2)

This determines the location of a point on the envelope, X(xE) following directly fromequation (4.2) and the Y (xE) following from equation (4.1):

X = xE + φ , Y = yE +mφ , (4.3)

where

φ =m− y′Em′

. (4.4)

Equations (4.3), (4.4) are parametric equations for the envelope (X(xE), Y (xE)).

4.2 Curvature of the envelope

Consider next the curvature κ of the envelope

κ =X ′Y ′′ − Y ′X ′′(X ′2 + Y ′2

)3/2.

Here the primes denote differentiation with respect to the parameter (often β or tB orxE). It is easy to plot the curvature κ as a function of the parameter.

4.3 Small displacements about symmetric configuration

This section to be written. I currently favour series expansions in terms of tO asdefined at the end of §1

We have yet to find a short expression for κ. As a more manageable task, we considerhow the curvature changes as the coupler bar rocks about its equilibrium position. Thecurvature of the envelope at X = 0 is

κ0 = −(c− a)2 (b+ c− a) t0

((c− a)3 + b2 (2a− c)) c

where

t0 = tan

(β0

2

)=

√a + b− c

b+ c− a.

TO DO. Investigate how κ varies as one moves the coupler bar slightly from the sym-metric position of the mechanism.

4.4 Higher derivatives of Y

We may seek to approximate the envelope curve by its Taylor expansion about X = 0:

Y (X) ∼ Y (0) +d2Y

dX2(0)

X2

2!+d4Y

dX4(0)

X4

4!+ . . . .

11

We have already calculated d2YdX2 (0):

d2Y

dX2=

1

X ′

(Y ′

X ′

)′so

d2Y

dX2(0) = κ0 .

Proceeding:

d4Y

dX4=

1

X ′

(1

X ′

(1

X ′

(Y ′

X ′

)′)′)′.

4.5 Envelope (X, Y ) calculationsUsing β as parameter

This subsection represents an alternative to that where we calculated X , Y with θ as theparameter.

The equation of the line CD of the coupler bar is

(xD − xC)(Y − yC)− (yD − yC)(X − xC) = 0 ,

and β or a closely related quantity is a suitable parameter for the family of lines. Ofcourse

xC = a+ b cos(β) , yC = b sin(β) .

From this the equation of the line is

(Y − yC)−m(X − xC) = 0 , (4.5)

where m(β) is given in equation (3.9).

Differentiating equation (4.5) with respect to β gives

−b cos(β)−dm

dβ(X − xC)−mb sin(β) = 0 . (4.6)

This determines the location of a point on the envelope, X(β) following directly fromequation (4.6) and Y (β) following from equation (4.5):

X = xC −b (cos(β) +m sin(β))

dm/dβ, (4.7)

Y = yC −mb (cos(β) +m sin(β))

dm/dβ. (4.8)

Equations (4.7) and (4.8) are parametric equations for the envelope (X(β), Y (β)).

5 Some asymptotic approximations

5.1 Approximations at small b

5.2 Approximations at small (a− c)/a

When c = a, small displacements from the symmetrical configuration have (xE, yE) onthe circle

yE =√b2 − x2

E .

12

Our concern in this subsection is how this changes when c is no longer exactly equal to abut merely close.

References

[CR] Cundy, H.M. and Rollett, A.P., Mathematical Models, (Clarendon Press, Oxford,2nd ed.: 1961).

[Dij] Dijksman, E.A., Motion geometry of mechanisms, (Cambridge University Press:1976).

[HN] Hrones, J.A. and Nelson, G.L., Analysis of the four-bar linkage, (The TechnologyPress of the Massachusetts Institute of Technology, and Wiley, New York: 1951).

[KSF] Keady, G., Scales, P.V. and Fitz-Gerald, G.F., Envelopes generated by circular-arc coupler-bars in James Watt’s four-bar linkages. Submitted for Proceedings ofthe Engineering Mathematics and Applications Conference RMIT, Melbourne, Sep.2000.

[Lo] Lockwood, E.H., A book of curves, (Cambridge University Press: 1976).

[SK] Scales, P.V. and Keady, G., Watt linkages and quadrilaterals. In preparation, forpossible submission to Math. Gazette.

[Sh] Shikin, E.V., Handbook and atlas of curves, (CRC Press: 1995).

13

A Appendix.CAS code to check some equations in this report

restart:

P:= c^2 - b^2 -a^2: Q:= c^2 -b^2 +a^2: R:= 4*c^2*a^2:

p:= (xE*(xE^2+yE^2+P))^2 + (yE*(xE^2+yE^2+Q))^2 - R*yE^2;

with(Groebner);

p1:=expand((xD+a)^2 + yD^2 - b^2):

p2:=expand((xC-a)^2 + yC^2 - b^2):

p3:=expand((xD-xC)^2 + (yD-yC)^2 - 4*c^2):

p4:= xE - (xC+xD)/2:

p5:= yE - (yC+yD)/2:

WL:=[p1,p2,p3,p4,p5]:

G:=gbasis(WL,plex(xC,yC,xD,yD,xE,yE)):

nops(G);

dxE:= u -> degree(u,xE):

dyE:= u -> degree(u,yE):

d:= u -> [degree(u,yE),degree(u,xE),degree(u,yD),degree(u,xD),\

degree(u,yC),degree(u,xC)]:

[map(dxE,G),map(dyE,G)];

with(linalg,matrix):

Mdeg:=matrix(map(d,G));

map(nops,G);

# look at the smaller polynomials

normalf(p,G,plex(xC,yC,xD,yD,xE,yE));

# result of 0 confirms p

divide(G[1],p,’q’); q;

# result of true confirms p

pm:=(a^2-2*yE*c+yE^2-2*xE*a+xE^2+c^2-b^2)*\

(a^2+2*yE*c+yE^2-2*xE*a+xE^2+c^2-b^2)*(yE-yD)^2+\

8*c^2*yE*(xE-xD)*(a-xE)*(yE-yD)+\

(xE-xD)^2*(-2*xE*c+xE^2+yE^2-2*xE*a-b^2+a^2+2*a*c+c^2)*\

(xE^2+yE^2-2*xE*a+2*xE*c+a^2-2*a*c+c^2-b^2);

normalf(pm,G,plex(xC,yC,xD,yD,xE,yE));

# result of 0 confirms pm

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