# 47 CFR § 73.190 - Engineering charts and related formulas.

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§ 73.190 Engineering charts and related formulas.

(a) This section consists of the following Figures: 2, r3, 5, 6a, 7, 8, 9, 10, 11, 12, and 13. Additionally, formulas that are directly related to graphs are included.

(b) Formula 1 is used for calculation of 50% skywave field strength values.

Formula 1. Skywave field strength, 50% of the time (at SS + 6):

The skywave field strength, Fc(50), for a characteristic field strength of 100 mV/m at 1 km is given by:

${F}_{c}\left(50\right)=\left(97.5-20\phantom{\rule{0ex}{0ex}}\mathrm{log}D\right)-\left(2\pi +4.95\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{2}{\varphi }_{M}\right)\sqrt{\left(\frac{D}{1000}\right)}\phantom{\rule{0ex}{0ex}}dB\left(\mathrm{\mu V}/m\right)\phantom{\rule{0ex}{0ex}}\text{(Eq. 1)}$

The slant distance, D, is given by:

$D=\sqrt{40,000+{d}^{2}}\mathrm{km}\phantom{\rule{0ex}{0ex}}\text{(Eq.2)}$

The geomagnetic latitude of the midpoint of the path, ΦM, is given by:
ΦM = arcsin[sin aM sin 78.5° + cos aM cos 78.5° cos(69 + bM)]degrees (Eq. 3)
The short great-circle path distance, d, is given by:

$d=111.18d°\phantom{\rule{0ex}{0ex}}\mathrm{km}\phantom{\rule{0ex}{0ex}}\text{(Eq.4)}$

Where:
d° = arccos[sin aT sin aR + cos aT cos aR cos(bRbT)]degrees (Eq.5)
Where:
aT is the geographic latitude of the transmitting terminal (degrees)
aR is the geographic latitude of the receiving terminal (degrees)
bT is the geographic longitude of the transmitting terminal (degrees)
bR is the geographic longitude of the receiving terminal (degrees)
aM is the geographic latitude of the midpoint of the great-circle path (degrees) and is given by:
bM is the geographic longitude of the midpoint of the great-circle path (degrees) and is given by:

${a}_{M}=90-\mathrm{arccos}\left[\mathrm{sin}\phantom{\rule{0ex}{0ex}}{a}_{R}\mathrm{cos}\left(\frac{{d}^{°}}{2}\right)+\mathrm{cos}\phantom{\rule{0ex}{0ex}}{a}_{R}\mathrm{sin}\left(\frac{{d}^{°}}{2}\right)\left(\frac{\mathrm{sin}{\phantom{\rule{0ex}{0ex}}a}_{T}-\mathrm{sin}{\phantom{\rule{0ex}{0ex}}a}_{R}{d}^{°}}{\mathrm{cos}{\phantom{\rule{0ex}{0ex}}a}_{R}\mathrm{sin}\phantom{\rule{0ex}{0ex}}{d}^{°}}\right)\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. 6)}$

${b}_{M}={b}_{R}+k\left[\mathrm{arccos}\left(\frac{\mathrm{cos}\left(\frac{{d}^{°}}{2}\right)-\mathrm{sin}\phantom{\rule{0ex}{0ex}}{a}_{R}\mathrm{sin}{\phantom{\rule{0ex}{0ex}}a}_{M}}{\mathrm{cos}\phantom{\rule{0ex}{0ex}}{a}_{R}\mathrm{cos}{\phantom{\rule{0ex}{0ex}}a}_{M}}\right)\right]\phantom{\rule{0ex}{0ex}}\text{(Eq. 7)}$

Note (1): If |FM| is greater than 60 degrees, equation (1) is evaluated for | FM| = 60 degrees.

Note (2): North and east are considered positive; south and west negative.

Note (3): In equation (7), k = −1 for west to east paths (i.e., bR >bT), otherwise k = 1.

(c) Formula 2 is used for calculation of 10% skywave field strength values.

Formula 2. Skywave field strength, 10% of the time (at SS + 6):

The skywave field strength, Fc(10), is given by:

Fc(10) = Fc(50) + Δ dB(µV/m)
Where:
Δ = 6 when | FM| <40
Δ = 0.2 | FM| − 2 when 40 ≤| FM| ≤60
Δ = 10 when | FM| >60

(d) Figure 6a depicts angles of departure versus transmission range. These angles may also be computed using the following formulas:

${\theta }^{°}={\mathrm{tan}}^{-1}\left({k}_{n}\mathrm{cot}\frac{d}{444.54}\right)-\frac{d}{444.54}$

Where:
d = distance in kilometers
n = 1 for 50% field strength values
n = 2 or 3 for 10% field strength values
and where
K1 = 0.00752
K2 = 0.00938
K3 = 0.00565
Note:

Computations using these formulas should not be carried beyond 0.1 degree.

(e) In the event of disagreement between computed values using the formulas shown above and values obtained directly from the figures, the computed values will control.

[28 FR 13574, Dec. 14, 1963, as amended at 30 FR 12720, Oct. 6, 1965; 33 FR 15420, Oct. 17, 1968; 48 FR 42959, Sept. 20, 1983; 49 FR 43963, Nov. 1, 1984; 50 FR 18844, May 2, 1985; 51 FR 4753, Feb. 7, 1986; 52 FR 36879, Oct. 1, 1987; 56 FR 64869, Dec. 12, 1991]