ZERO TEMPERATURE COEFFICIENT QUARTZ CRYSTALS 379 



multiplier of {T — 50) equal to zero and solving for the rotation angles 

 6i and B% — are 



ei = +36°20' and 62 = -53°50' (32) 



as compared to the experimental values of +38°20' and —52°, which repre- 

 sents a shift of about +2° orientation for both angles. At these calculated 

 angles the frequencies are within about 1.5 per cent of the experimental 

 values and the curvature constants agree approximately with the measured 

 values. 



To obtain the angles for any other temperatures, for example 2(X)°C, 

 we substitute 



T = 200 + AT (33) 



and obtain the expansion in powers of AT. For 200°C this results in 



fi X 10-^^ = 



14^27 



[2.055 cos' 6 + 2.829 sin' d - 1.840 sin 6 cos 6] 

 + [514 cos' 6 - 413 sin' d - 266 sin 6 cos 6] X 10"'Ar (34) 

 + [373 cos' ^ - 48 sin' ^ - 28 sin 6 cos 6] X 10"^(Ar)' 

 + [-52 cos' d + 8.7 sin' ^ + 98 sin 6 cos 6] 



X 10""(Ar)') 



The zero temperature coefficient angles are obtained by setting the coeffi- 

 cients of Ar equal to zero giving 



514 cos2 d - 413 sin2 q _ 266 sin ^ cos (9 = (35) 



Solving for 6 we find 



e = +39°50' and -56° (36) 



If we add 2° to each of these in order to correct for the difference between 

 the formula and the measured results at 50°C, the most probable angles for 

 zero coefficients at 200°C are 



6 = +41°50' and -54° (37) 



At these angles the indicated frequencies and variations of frequencies 

 with temperature should be 



e = 4r51'; 



/ = ^-^^ ^ ^^' [1 - 63 X io-'(Ar)' - 8 X io-''(Ar)' + . • .] 



.= -54°; (^^) 



/ = ^'^^ ^ ^^' [1 - 14 X lO-^(Ar)' + 8 X lo-^'(Ar)' + . . .] 



