Orty sfericheskoi sistemy koordinat vyrazhayutsya cherez orty dekartovoi sleduyushim obrazom:

i_r = (cos(lambda)*i_x+ sin(lambda)*i_y)sin(theta) + cos(theta)*i_z,

i_lambda = -sin(lambda)*i_x+ cos(lambda)*i_y,

i_theta = (cos(lambda)*i_x+ sin(lambda)*i_y)cos(theta)- sin(theta)*i_z,

otkuda, pri theta=90^o i lambda=0^o sleduet, chto

i_r = i_x,

i_lambda = i_y,

i_theta = -i_z.

Eto oznachaet, chto troiki ortonormirovannogo bazisa dekartovoi i sfericheskoi sistem koordinat - protivopolozhno orientirovany (znak "-" pered ortom i_z). Esliu dekartovoi sistemypravaya orientaciya bazisnoi troiki, to u sistemy sfericheskoi v dannom sluchae ona - LEVAYa.

Avtoram stat'i neobhodimo razobrat'sya s putanicei, vytekayushei iz togo, chto oni pishut.

Ochen' ponravilos' to, chto takoi sait vse-taki est'. No ya nikak ne mogu naiti analogii Nepera, v chastnosti, "Koleso (okruzhnost', krug) Nepera". Izvinyayus' za netochnost' termina, perevodil s angliiskogo.

Eto ispol'zuetsya pri vychislenii verteksa (perpendikulyar iz polyusa) k DBK (Duga Bol'shogo Kruga). Gde eto mozhno naiti?

S uvazheniem,

Evgenii.

K sozhaleniyu, ya tak i ne ponyal prakticheskoi rekomendacii, kak, znaya astronomicheskie koordinaty observatorii, a takzhe GPS polozhenie, ya mogu s maksimal'noi tochnost'yu opredelit' (1)- rasstoyanie do centra Zemli, i chto dlya menya bolee vazhno - (2) proekciyu radius-vektora (ot centra Zemli do observatorii) na ploskost' Zemnogo ekvatora. Naprimer, esli

1) an astronomical measurement (theodolite timing ) which refers to the Geoid
W 03 23 56.285 (error 0.058)
N 37 04 06.168 (0.440)
altitude 2851.5 m

2) a GPS position fererring to their ellipsoid WGS84
W 03 23 34.16091 (0.006)
N 37 03 57.97407 (0.007)
altitude 2903.995 m

to chemu ravny (1) i (2)?

Zaranee spasibo za pomosh'.

Priyatno, chto uroven' dostatochno prilichnyi po sravneniyu s bol'shinstvom chepuhi, kotoraya napisana na etu temu v internete.