May 31, 1901.] 
(1) 
(2) 
will express the motion. Here Yand Yare 
the torques around the horizontal and ver- 
tical axes, respectively, and K the polar 
moment of inertia ; friction is disregarded.* 
Hence 4, ¢, %, are of the first order of small 
quantities, and ¢ of the second order. 
The point is now to show that these 
equations are reproduced in my geometrical 
constructions relative to the common theo- 
rem of the equivalence of torque, and the 
change of angular momentum per second 
around any particular direction. 
I. Torque (¥) around the horizontal axis 
only. Precession.—Take any two positions 
of the top axis, a second of time in position 
apart. Lay off the angular momenta K®@, 
= KO sin ¢ f- 0, 
pay, sin Y- ¢, 
along these axes. They are equal by the 
premises. Hence the horizontal are, a, is 
the rate of change of angular momentum 
due to the torque ¥ around the parallel axis 
shown. Im the second of time stated, the 
angle of azimuth has changed by ¢, as 
shown in the figure. Therefore, the arc 
a=Koésin¢-g, j 
and hence ; 
= Késin?g-g, 
IL. Torque © around the vertical amis only. 
* Otherwise, if there is polar acceleration, ® =— K 
(6 » sin » — cos i - 6) may be deduced without diffi- 
culty by the method of @ II. 
SCIENCE. 
853 
—This requires gyroscopic mounting. Let 
the horizontal axis be seen end on at A. 
Take two positions of the top axis, a second 
of time in position apart. Lay off the 
(equal) angular momenta, K9, along these 
gol 
axes. Then the arc 6 is the rate of change 
of angular momentum, and the angle, — ¢, 
subtended, the speed in altitude. Hence 
b=—Ko - ¢. 
The are, 6, cannot be resolved with advan- 
tage, for there is no way of accounting for 
both components. 9%, however, may be re- 
solved; for if one component is made par- 
allel to 0b, this is the equivalent of 6; 
whereas if the other component tends to 
twist across A (in a plane at right angles 
to ‘around A’), 7. ¢., in the plane of this 
axis, is can produce stress only, but no 
motion. Hence, as seen in the figure, the 
effective component is ®/sin ¢, a result a 
little subtle, I grant, but none the less 
logically straightforward. Therefore 
®/sin 9 = b, 
whence : ; 
0 =— Ko sin ~ . dy. 
Of course my explanation was intended 
for the man interested in spinning tops. 
The other man, who prefers the top top- 
pling through all stages of inebriety, may 
take such solace as comes from products of 
parameter speeds and accelerations. But 
for him I have no message other than my 
blessing. 
Cart Barvs. 
BROWN UNIVERSITY, PROVIDENCE, R. I. 
