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Quadruply. free cycloidal system, Ky rostati rally domiDated.

Thus, taken as a quadratic for 1-, it has the same form as (31)
for 1, and so, as before in (32) and (33), we find

-1
(r'+s')*..

(37),

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where m' = {{(12'+34') +(13'+42')*+ (14'+23')"}
and s'= J{(12'- 34')*+(13'-42')* + (14'- 23')"}

Four irrotational stabilities confirmed, four irrotational instabilities rendered stable, by kyrostatic links.

Now if w,,,, wg, w, be all four positive or all four negative, 12', 34', 13', etc. are all real, and therefore both the values of

1 do

given by (37) are real and positive (the excluded case referred to at the end of $ 3456, which makes

12'34' + 13'42' + 14'23' = 0,

1 and therefore the smaller value of 0, being still excluded).

no Hence the corresponding freedoms are stable. But it is not necessary for stability that w,, wg, W2, W, be all four of one sign: it is necessary that their product be positive: since if it were negative the values of X given by (34) would both be real, but one only negative and the other positive. Suppose two of them, wz, w, for example, be negative, and the other two,

,,,, positive: this makes us, , a, m, a,g, and ww, negative, and therefore 13', 14', 23', and 24' imaginary. Instead of four of the six equations (36), put therefore

14

24

13"

14"

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13

23 J(-0,w)?

11-6,7), 24"=J3

(39).

(-) Thus 13" etc. are real, and 13' = 13" ✓- 1 etc., and (38) become

g' = v{(12' + 34') - (13" +42")? – (14" +23")"}}
s' = {{(12' – 34')? – (13" – 42")? – (14" - 23")"}]

(40).

Hence for stability it is necessary and sufficient that

+

(41).

(12' + 34')>(13" +42") +(14" + 23")) and (12' - 34)'>(13"-42") + (14" - 23')') If these inequalities are reversed, the stabilities due to ,w, and 34' are undone by the gyrostatic connexions 13", 42", 14" and 23".

Combined dynamic and gyrostatic sta.

345xiv. Going back to (29) we see that for the particular bility gyrosolution y, = a,ent, V, = a,et, etc., given by the first pair of roots counter

acted. of (32), they become approximately

da, + 12 a, + 13 az + 14 a, = 0
da, + 21 a, +23 ag + 24 a, = 0

.(42);
da, +31 a, + 32 a, + 34 a,

0 da, +41 a, +42 a, +43 ag = 0) being in fact the linear algebraic equations for the solution in Completed

solution. the form et of the simple simultaneous differential equations (53) below. And if we tako

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for either particular approximate solution of (29) corresponding
to (37), we find from (29) approximately

1-6, + 12'6, + 13'5, + 14'b, = 0)
1-6, +21'b, + 23'bg + 24'b, = 0)

(44).
1-63 + 31'b, + 32b, + 34'b, = 0

1=6+41'b, +42'b, +43'b, = Remark that in (42) the coefficients of the first terms are imaginary and those of all the others real. Hence the ratios a,a,,a,a,, etc., are imaginary. To realize the equations put

1=n-1, a, = P, +9,-1, a, = P, +9,1-1, etc.... (45),

-ng,

+ 12

and let P 9 Pg, etc. be real; we find, as equivalent to (42), Realization

ul complet. +13P, + 14 P, = 0

ed solution.
np. + 129, + 13 13 + 149, = 0
- nq, + 21 P. + 23 P3
+ 24 P, = 0

(46).
np, + 219, +23 92 + 24 9. = 0
etc.

etc.
Eliminating 9., 1,, etc. from the seconds by the firsts of these
pairs, we find
(no + 1)2 +
21 P, + (n° + 22) P2

24 P, = 0)

(47); 31 Pi + 32 P, +(n° +33) P, + 31 P, = 0

43 P, +(n+ 41P, = 0) and by eliminating P, Py, etc. similarly we fiud similar equations

12 P2

+

+

13 Pa 23 P :3

14 P, = 0) ?

+

+

+

41 P. +

42 P. +

Realization of completod solution.

for the q's; with the same coefficients 11, 12, etc., given by the following formulas :

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Remember now that

12 =- 21, 13 =-31, 32 = -23, etc. ....... .(49), and we see in (48) that 12 = 21, 13 = 31, 23 = 31, etc.

...(50); and farther, that 11, 12, etc. are the negatives of the coefficients of ja,', a,æg, etc. in the quadratic

}{(12 a, + 13 ag + 14 a,)* + (21 a, +23 a, + 24a,)' + etc.}...(51)
expanded. Hence if G (aa) denote this quadratic, and G (19),
G (99) the same of the p's and the o's, we may write (47) and
the corresponding equations for the q's as follows:
dG (pp)

dG (pp)
Top + =0, - n'p, + = 0, etc.
dp,

(52). - n°9, + :0, - n°9,+

Resultant motion reduced to motion of a conserv&tive system with four fundamental periods equal two and two.

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dG (99)

da,

dG (99) – 0, etc.

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We

Algebraio theorem.

These equations are harmonized by, and as is easily seen, only by, assigning to no one or other of the two values of -l given in (32), above. Hence their determinantal equation, a biquadratic in no, has two pairs of equal real positive roots. readily verify this by verifying that the square of the determinant of (42), with lo replaced by - n*, is equal to the determinant of (47) with 11, 12, etc. replaced by their values (48). Hence ($ 343g) there is for each root an indeterminacy in the ratios p/Pc, P/P, P,/P, according to which one of them may be assumed arbitrarily and the two others then determined by two of the equations (47); so that with two of the p's assumed arbitrarily the four are known: then the corresponding set of four q's is determined explicitly by the firsts of the pairs (46). Similarly the other root, n", of the determinantal equation gives another solution with two fresh arbitraries. Thus we have the complete solution of the four equations

Details of realized solution.

dy,

dy

dt

Details of +124, +13 48 + 1440 = 0

realized dt

solution.

(53),
+214, + 23 48 + 24 4, = 0
etc.

etc.
with its four arbitraries. The formulas (46)...(52) are clearly
the same as we should have found if we had commenced with
assuming

sin nt + q, cos nt, yo =P, sin nt +9, cos nt, etc.... (54), as a particular solution of (53).

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nalities

two como

.....

one fundamental oy

of their

345". Important properties of the solution of (53) are found orthogothus :

proved (a) Multiply the firsts of (46) by Pi, Ps, Ps, P, and add :

between or the seconds by 910 99, 93, 9, and add : either way we find

ponents of P,9, +P,9+ P29, +2,96=0...... ...(55).

cillation: (6) Multiply the firsts of (46) by 91, 93, 93, 9, and add : multiply the seconds by Pu, Ps, Ps, P, and add: and compare the results : we find

ηΣp = ηΣq = Στ2 (2,3, -2,9,).......... ... (56), and equality where of the last member denotes a sum of such double

energies. terms as the sample without repetition of their equals, such as 21 (P,9,- P.4).

(c) Let no, n' denote the two values of -d given in (32), Orthogoand let (54) and

proved Vi=p', sin n't +9', cos n't, y, = p', sin n't +D, cos n't, etc.... (57) different

fundamenbe the two corresponding solutions of (53). Imagine (46) to be tions.

tal oscillawritten out for n" and call them (46'): multiply the firsts of (46) by p'ıp'a p'wp', and add: multiply the firsts of (46') by P1, P2, P3, P. and add. Proceed correspondingly with the seconds. Proceed similarly with multipliers q for the firsts and p for the seconds. By comparisons of the sums we find that when n' is not equal to n we must have Sp'q=0, E12 (PP.-PP) = 0 Eqp=0, E12 (2,4 - 0,9.) = 0

(58). Σqq = 0)

12 (d.P. - 02P.) = 0, Σpp = 0)

£12 (p'4. - p',9.) = 0

nalities

between

Case of equal periods.

I 2 =

345The case of n=n' is interesting. The equations Sq'q = 0, Ep'p =0, Ep'q=0, Eq'p = 0, when n differs however little from n', show (as we saw in a corresponding case in § 343 m) that equality of n to n' does not bring into the solution terms of the form Ct cos nt, and it must therefore come under $ 313 e. The condition to be fulfilled for the equality of the roots is seen from (32) and (33) to be

34, 13 = 42, and 14 = 23..... ...(59): and to give n = 12 +13+14'

... (60) for the common value of the roots. It is easy to verify that these relations reduce to zero each of the first minors of (42), as they must according to Routh's theorem (343e), because each root, d, of (42) is a double root. According to the same theorem all the first, second and third minors of (47) must vanish for each root, because each root, n', of (47) is a quadruple root: for this, as there are just four equations, it is necessary and sufficient that

11 = 22 = 33 = 44 and 12 = 0, 13 = 0, 14 = 0, 23 = 0, etc.... (60'), which we see at onoe by (48) is the case when (59) are fulfilled. In fact, these relations immediately reduce (51) to

G (aa) = } (12° + 139 + 14°) (a' + a,' + a," + a, 4)........ (61).
In this case one particular solution is readily seen from (52) and
(46) to be
P, = 0,

14
9

9.
13
9

(62). 13

14 Vi = sin nt, to cos nt, Ya cos nt, yo

cos nt

P, = 0,

P = 0

P, = 1,
9, = 0,

I 2

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n

n

n

I 2

n

n

n

Completed solution for case of equal periods.

n

n

Hence the general solution, with four arbitraries P,, Pg, P3, P., is

1
Vi =P, sin nt +-(129, + 13Ps + 14p.) cos nt

1
He=P, sin nt + (- 12P, + 14Ps 13P.) cos nt
1

...(63). Vo=P, sin nt +

(-13P, – 14P, + 12p.) cos nt

1
4.=P, sin nt +-(- 14P, + 13P, - 129.) cos nt

, 1 .
It is easy to verify that this satisfies the four differential
equations (53).

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