150 GAUSS ON A METHOD OF FACILITATING 
same instant relatively to p°, if it had continued its original 
movement undisturbed; in a word, the middle point of the 
movement alone is changed ; the movement itself is wholly un- 
altered. 
6. We have still to determine the intervals so that the equa- 
tions f = 0, g = O may be satisfied. If we make 
tf =qgT, tt! —t=rT, 
and bear in mind that e = 10”, these equations become 
10-2* cos ga + 10"* cosrz = 1 
10-27% sing 7 = 10"* sinrz. 
Thus, for the case of an insensible decrease of the arc of vibra- 
tion, cos gz + cosrm=1, and sn ga =sinraz, andgr=r7 
= 60°, or +7, and #” —# = 7" —¢' =1T, as has been al- 
ready shown in Art. 2. For the case of a serisible logarithmic 
decrement, on the other hand, these equations are to be solved 
indirectly, to which calculation the following form may be given. 
From the combination of the equations it follows that 
tangre = — "2" _ 
107*—cos qa 
10?"* = 1—2.10-2* cos gx. + 10-79%. 
Eliminating 7 we obtain the equation with one unknown quan- 
tity, 
55 ls (1 — 2.10—2*cos ga.+ 10~727) 
sin 97 
= arc (tang = sn _) 
102* — cosg 
in Briggs’s logarithms. When this has been satisfied, it is mani- 
fest that the value of r will have been obtained at the same time. 
7. In order that those who desire to make use of the method 
here described, when a damper is employed, may be spared the 
calculation which has been explained in the preceding article, I 
subjoin a table, from which the proportion of the two intervals 
of time to the time of vibration can be taken out at sight for 
every logarithmic decrement. It will be seen that as the loga- 
rithmic decrement increases, the first interval always increases, 
and the second decreases. ‘Their sum is exactly equal to two- 
thirds of the time of vibration when A=0; but it changes from 
this much more slowly. In making use of the table, it need 
