SCIENTIFIC INSTRUMENTS CO., Berkeley, California / USA (1961) ************************************************************************* Made in Japan by RELAY (= Model 158 ) SN = 40007 S C A L E S of the » E L E C T R O HYPER-VECTOR LOG-LOG « Model 1580 ========================================================================= Front Side [ inverse / RED ] Back Side ========================================================================= left ; right symbols domain Sh_{2} SinH(X) <0.85 .. 3> X2 Sh_{2}^{2}X <0.88 .. 1.15> Sh_{1} SinH(0.1X) <0.1 .. 0.9> X1 Sh_{1}^{2}X <0.1 .. 0.88> Th TanH(0.1X) <0.1 .. 3> P2 X_{2}^{2} <1 .. 1.42> A X^{2} P1 X_{1}^{2} <0.1 .. 1> ------------------------------------------------------------------------- [BI] [1/X^{2}] Q -X^{2} <1 .. 0.1> S sin(0.1X) ; [cos(0.1X)] Y Cos^{2}X <π/2 .. 0.1> T tan(0.1X) ; [cot(0.1X)] L X <0 .. 1> [CI] [1/X] /_X /_thX <0.1 .. 3 _______ > C X [I] [X^{-1}] <[10^{4} .. 40]> ------------------------------------------------------------------------- D X [I] [X^{-1}] <[10^{4} .. 40]> LL_{3} exp(X) /_Θ_{1} /_tg_{1}Θ <0.5 .. 45 ; [89.5 .. 45]> LL_{2} exp(0.1X) /_Θ_{2} /_tg_{2}Θ <45 .. 89.5 ; [45 .. 0.5]> LL_{1} exp(0.01X) /_Y /_tgX <0.1 .. 1.48> R E M A R K S : ========================================================================= This » HYPER-VECTOR LOG-LOG « Slide Rule is designed for engineers in the field of ELECTRO TECHNICS with AC & HF Calculating Problems: 4-POLES, ELECTRICAL & MAGNETICAL FIELDS, ANTENNAS, CONDUCTORS & ATTENUATORS, FILTERS, ... because "Vectors & Hyperbolic Functions" are involved ... ------------------------------------------------------------------------- This model is made of bamboo with scales on celluloid veneer. The cursor windows have the scale symbols engraved - a good idea! Observe the mis- sing B-Scale: There is instead a [BI]-Scale - an interesting idea! The slide rule came in a brown leather case with SIC-Logo & "Made in Japan". The GAUGE MARK f = 1/2π on the C/D-Scales is for calculating INDUCTIVE (X_{L}) & CAPACITIVE (X_{C}) REACTANCES, and RESONANCE FREQUENCY (F_{R}) X_{L} = 2πFL ; X_{C} = 1/( 2πFC ) ; F_{R} = 1/( 2π*sqrt(LC) ) The formula for DECIBEL Calculation is db = 20*lg(V_{2}/V_{1}) The DEFINITIVE FORMS for the HYPERBOLIC FUNCTIONS are: CosH(X) = ( exp(+X) + exp(-X) ) / 2 <== this is the "Chain Line" SinH(X) = ( exp(+X) - exp(-X) ) / 2 ( eg. suspension bridge ) TanH(X) = SinH(X) / CosH(X) exp(X) = SinH(X) + CosH(X) 1 = CosH^{2}(X) - SinH^{2}(X) CosH(X) = SinH(X) / TanH(X) = sqrt( 1 + SinH^{2}(X) ) (A) How to use the Hyperbolic Functions on the F R O N T S I D E : ------------------------------------------------------------------------- Start with value on the Sh-Scales ==>> reading SinH on D-Scale, start with value on the D-Scale ==>> reading ArcSinH on Sh-Scale. Start with value on the Th-Scale ==>> reading TanH on D-Scale, start with value on the D-Scale ==>> reading ArcTanH on Th-Scale. Calculate CosH(X) ==>> Move hairline to X on Sh , set right or left index of CI-Scale under hairline, move hairline to X on Th-Scale, read CosH(X) under hairline on CI-Scale. (B) and the Hyperbolic- / Vector Functions on the B A C K S I D E : ------------------------------------------------------------------------- Used for Calculating HYPERBOLIC FUNCTIONS OF COMPLEX QUANTITIES Vector in Cartesian Coordinates: V = a + jb Changing from CARTESIAN- TO POLAR-COORDINATES: R = |V| = sqrt(a^{2} + b^{2}) ; /_Θ = arctan(b/a) Changing from POLAR- TO CARTESIAN-COORDINATES: a = |V|*cos(Θ) ; b = |V|*sin(Θ) Addition and Subtraction of Vectors is easy in CARTESIAN COORDINATES: ADDITION of Vectors: V_{1} + V_{2} = ( a_{1} + a_{2} ) + j( b_{1} + b_{2} ) SUBTRACTION of Vectors: V_{1} - V_{2} = ( a_{1} - a_{2} ) + j( b_{1} - b_{2} ) Multiplication and Division of Vectors is easy in POLAR COORDINATES: MULTIPLICATION of Vectors: V_{1} * V_{2} = |V_{1}|/_Θ_{1} * |V_{2}|/_Θ_{2} _{ }_{ }= |V_{1}|*|V_{2}| /_( Θ_{1} + Θ_{2} ) DIVISION of Vectors: V_{1} / V_{2} = ( |V_{1}|/_Θ_{1} ) / ( |V_{2}|/_Θ_{2} ) _{ }_{ }= |V_{1}|/|V_{2}| /_( Θ_{1} - Θ_{2} ) How to use the Scales of the Back Side: ======================================= (=1=) Start with value on the Sh^{2} scales ==>> reading SinH on X^{2} scales, start with value on the X^{2} scales ==>> reading ArcSinH on Sh^{2} scales. P2 = sqrt( 1 + SinH_{1}^{2}(X) ) = CosH(X) SinH(0.7) ==>> CosH(0.7) = 1.255 (direct read-out!) (=2=) P1, P2 & Q scales are used for vector calculations: P2 = sqrt( 1 + P1^{2} ) = 1.281 <<== ( P1 = 0.8 ) Q = sqrt( 1 - P1^{2} ) = 0.600 <<== ( P1 = 0.8 ) Q = P1 inverted: ( Q = 0.800 ) <==> ( P1 = 0.6 ) EX.A: Calculate the ABSOLUTE VALUE of V = 0.4 + j0.3 ==>> |V| = 0.5 ***** Move hairline to 0.4 on P1, set 0.3 on Q under hairline, opposite right index of Q read answer 0.5 on P1 Why the PYTHAGORAS can be calculated with the P2, P1 & Q-Scales ??? (=3=) Start with RADIANS on the Y-Scale ==>> reading cos^{2}(Y) on X-Scale ==>> reading sin(Y) on Q-Scale (=4=) (X =) L-Scale ( = lgX ) keyed to C-Scale on front (=5=) /_X /_thX -Scale ( = TanH ) keyed to A-Scale ( *0.1 ) on front (=6=) [I] ; [X^{-1}] Scales to find THE SUM OF RECIPROCALS: The resulting value of parallel wired resistances is given by 1/R1 + 1/R2 + ... + 1/Rn = 1/R EX.B: When R1 = 200 Ω & R2 = 60 Ω ==>> R = 46.15 Ω ***** 1/200 Ω + 1/60 Ω = 1/R = 1/46.15 Ω Opposit 200 on I (on stock), set left index of I (on slide). Move hairline to 60 on I (on slide), read answer 46.15 on I. (=7=) /_Θ_{1;2} BLACK Degrees of /_tgΘ_{1;2} keyed to A-Scale ( *0.01 ; *1 ) on front RED Degrees of /_tgΘ_{1;2} keyed to [BI]-Scale ( *1 ; *0.01 ) on front EX.C: Calculate the PHASE ANGLE of V = 0.4 + j0.3 ==>> /_Θ = 36.9° ***** Opposit 3 on A, set left index of [BI], move hairline to 4 on [BI], under hairline read answer 36.9° on /_tgΘ_{1} (BLACK) EX.D: TRANSFORMING a VECTOR from CARTESIAN- to POLAR COORDINATES ***** is the combination of EX.A and EX.C (=8=) /_Y Start w.RADIANS on /_tgX-Scale ==>> reading tanX on A-Scale on front EX.E: HYPERBOLIC FUNCTIONS OF COMPLEX QUANTITIES: ***** ( Examples taken of the SIC-1580 Instructions ... ) SinH(0.43 + j0.68) ==>> 0.769/_1.106 = 0.769/_63.4° Move hairline to 0.43 on X1, set 0.68 on Y under hairline, opposite right index of Q read answer 0.769 on P1. Move hairline to 0.68 on /_Y , set 0.43 on /_X under hairline. Move hairline to right index of /_X , under hairline read answer as 1.106 on /_Y . Move hairline to left index of I (on slide), under hairline read answer 63.4° on /_Θ_{2} (black). CosH(0.7 + j0.61) ==>> 1.117/_0.4 = 1.117/_22.9° Move hairline to 0.7 on X1, set right index of Q under hairline. Move hairline to 0.61 on Y, under hairline read answer as 1.117 on P_{2}. Move hairline to 0.61 on /_Y , set right index of /_X under hairline. Move hairline to 0.7 on /_X , under hairline read answer as 0.4 on /_Y. Move hairline to 0.61 on /_Y , set left index of I (on slide) under hairline. Move hairline to 0.7 on /_X , under hairline read answer 22.9° on /_Θ_{1} (black). Historical Remarks ... Presented at the 3rd BERLIN-BRANDENBURGER SAMMLER-TREFFEN (BBST) in Berlin impressum: ************************************************************************** © C.HAMANN http://public.beuth-hochschule.de/~hamann 03/14/09 |