







182 PROFESSOR TAIT ON KNOTS. 
little doubt that the difficulty will be solved with ease when the true method of 
attacking amphicheiral forms is found. 
1. To form from a given projection the knot with the greatest amount of 
beknottedness, it is clear that we must in general so arrange the crossings over 
and under as to make a// the crossings simultaneously silver or copper ones. 
And when this is done, a projection will give greater beknottedness for the same 
number of crossings the smaller is the number of crossings which have to be 
left out of account. Thus the simple twists (or clear coils with two turns) are 
the forms which, with a given amount of knottiness, can have the greatest be- 
knottedness. For in them (see § 41) only one crossing has to be left out of 
the reckoning. Evena regular plait if of more than two strands cannot have so 
much beknottedness as it would acquire with the same amount of knottiness if 
two of its strands were first twisted together, then a third round these, and so 
on. And thus also entirely nugatory forms like the two first cuts in § 1 can have 
no beknottedness, for ad/ their crossings have to be left out of the reckoning. 
As an illustration, take the figure (Plate XVI. fig. 8) where the supposed 
number of loops may be any whatever. The free ends must, of course, be 
joined externally. ; 
If we make the crossings alternately + and — it will be seen at a glance 
that a change of one sign (7.e., that of the extreme crossing at either end) 
removes the whole knotting ; so that there is but one degree of beknottedness. 
The crossings in this figure are in three rows. Those in the upper row are all 
copper (the last, of course, becomes silver when its sign is changed), and their 
number is 2 the number of loops. Each of the other rows has n—1, and 
all of them are silver. Thus when the one sign is changed there are n—1 
copper crossings, and 2 n—1 silver. By pulling out the right hand loop we 
change 2 to n—1, so that one copper and two silver crossings are lost. After 
n—1 operations like this there remains only one (silver) crossing. It is easy to 
see from this that the crossings to be omitted in the reckoning of beknotted- 
ness (as in § 41) must be the lower row. To prove that it is so, study the 
beknottedness when the crossings are made so that the upper row are copper, 
silver, copper, &c., alternately, and those of the two other rows, silver, copper, 
silver, &c., alternately. It will be easily seen that with five loops there are two 
degrees of beknottedness, &c., and thus that our rule is correct. It is a curious 
problem to investigate the torsional and flexural rigidities of a wire bent in this 
form. 
To give the greatest beknottedness to a knot with the same projection, it is 
obvious that all we have to do is to make the copper crossings into silver ones, 
i.e., change the sign of each of the upper row of crossings. This gives fig. 9. 
With five loops it has four degrees of beknottedness. 
Another excellent illustration is given by the coils of the class figured in 
