OCR source line
pi = 6li = kli, (7)Routes equations connecting electric waves, propagation, light, transmission lines, and frequency.
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pi = 6li = kli, (7)If m == number of phases, the higher harmonics : 2m - i2g3 QQQ = ^ seconds; the frequency 587 cycles, and if thisL -T- S = x2 -T- 7/2, where x and y are the two distances of theline carrying 7 amperes of current, if Id = 0.82 cm. is the diam-1. Since x’i’ has relative to xi the velocity f, it is, for a;’ = 0:ax - bt = 0,be V = Vi -{- V2 = 1.8c, or greater than the velocity ofvelocity V2, it is Xi = v^h, and substituting this in theas for instance, lOO feet = ^230 ~ 52^ miles wave length,i5ioo + ^ + n^ + Wo = ^ -ond, so giving alw = 60 microcentimeters,* that is, 60 X 10~8 cm. (or about^<y^<5-<y in.) and since the speed is S = 3 X 1010 cm. the frequencymotions of very high speed, S = 3 X 1010 cm. per sec. in a hypo-medium TF, only the distance CK = - 2 GH, and the wave frontand a2 = angle of refraction, that is, the angle between the out-FH + DL = S, - S3; (2)S = -L= , (5)Vl = d*-, (9)Since for most media the permeability /JL = 1, for all exceptZ0 = 51.1, bluish green for very low intensity, curve (a).Z0 = 53.7, yellowish green for medium intensity, curve (b).Pr = kA (TV - ?V), (1)k = 5 X 1(T12; (2)Pr = kA (T, - T2) (TV + T*T2 + 7\7y + 7y);Pr = 4 kAT* (T, - T), (3)stationary air A^ reaches values as high as fct = 25 X 10~12 toPr = kATf; (4)P- f °°P (lw)dlw = Mr-1;lmT =- = 0.284;k = bk0 = (1 - a) fcc,Curve II, for albedo a = 0.3, has the height 1 - a = 0.7 timesconstant 1 - a = 0.4 times that of the black body. Coloreda = 4, corresponding to a non-luminous flame, as observed. Inatoms is a = 2.8, or b = 30, corresponding to a luminous non-intersect in a point which lies at I = - 0.125 cm. = - 0.05 in.consists of a part, e0 = 30 (for the magnetite arc), which is con-constant value of current: el = k^ (I + 0.12), as function ofei = TT1-’ (1which is equation (1), and herefrom, since e = e0 + eir followsk = 31 for magnetite (123 in inch measure),e0 = 36 volts, but I, = - 0.8 cm. = - 0.33 in.; that is, apart. With a series resistance of r = 10 ohms, a 1.8-cm. mag-e = e0 + eve0 = 13 volts ; k = -y-L = k’li, (9)ei = p] (10)e=e0 + ^ (11)L = - (pVi - ejVi), (13)i = 4: e = 75 volts, and I + l^ = 0.73 in., or about f-in. arc10 deg. gives d = - - TT. This usually is done graphically.50%Q = 8.33 rev. per sec. Thus the frequency is/ = 4 X 8.33#o = E X \/2 = 273 volts per line;the period of the wave, or N = 1 / T is the frequency orBelow ff = 3, p is not a straight line, but curved, due to the in-straight-line law is reached at the point Ci, at ff = 3, and the re-Let then Sl = speed of propagation in medium A, S2 = speed ofwhere a = 5 for normal temperature radiation or black bodypoint, at zero arc length, or e0’ = 28 volts. This looks as if, of- ’ - = 25,000. If now somewhere in the middle of thisand these free atoms then join existing molecules: 0 + 02 = 03,29. If /= / +ji’ is a sine wave of alternating current,appear as L = 0, that is, would vanish, while at a velocitywith the same velocity, but in opposite direction, V2 = 0.9c.Zero point chosen at c = 128 cycles per second.Speed of radiation S = 3 X lu10 cm.FDH = a, and LHD = a2 ;1 to 2, but the change of sensation in the first case, log 1000 = 3,Wave length lw = 67 60.5 57.5 50.5 47 . 43 x 10~fltensity … i0 = 0.06 0.0056 0.0029 0.00017 0.00012 0.00012power po = 10,000 1000 100 1 2 20I = 45.0tor and T2 = absolute temperature of the surrounding objects