Reference / Paper · 1967

Technical Information Series No. 3: Double Integral (Calculation of Volume of Cone)

This application note demonstrates the use of a Hitachi analog-hybrid computer to evaluate a double integral for calculating the volume of a cone using multiple time axes. The computation slices the cone into N thin cylindrical sections (N=250 for the calculation, N=25 for reference), summing the cross-sectional areas iteratively; results are displayed via pen recorder and pattern display on oscilloscope showing the three-dimensional cone surface. Computing elements required include dual DC amplifiers (DA-151/DA-151A), dual integrators (IN-151/IN-153), potentiometer panels (PT-251/PT-151), voltage comparators (CP-151/CP-152/CP-153), and a square function generator (FG-154A).

Manufacturer: Hitachi
System: Hitachi Analog-Hybrid Computer
Year: 1967
Type: Reference / Paper
Language: English
Learning track: specific applications
Pages: 9
Credit: Printed in Japan. Hitachi, Ltd., 1967.

Hitachi Analog-Hybrid Computer
Hitachi
double integral
volume of cone
pattern display
multiple time axes

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HITACHI Analog-Hybrid Computer Technical Information Series No. 3 Double Integral «Calculation of Volume of Cone) 1967 ~ Hitachi,Ltd. printed in Japan: The analog computer can provide a pattern display which presents the solution in a visual figure such as the three-dimensional representation of a volume. This is a districtive advantage from the human engineering point of view, and this is the reason why a problem of double integral has been selected for this application manual. The computation of the volume of cone is described below, in which one half of the volume is calculated and the right solution shall be, 2 h =x rE (or, with r =h = 1) VV, = 0.502 With an analog computer, pattern displays by high speed computation is possible, which is almost impossible with an digital computer, although a solution by an analog computer may contain an error of approximately 0.2%. * *Note: This error is produced by non-linear operations such as calculations of square and square root. In this example, the analog computer is used in a low speed operation in order that the solutions can be given by a pen recorder as well as by a pattern display. However, pattern displays as many as 100 to 1,000 times per second can be obtained if the reset and the compute time is controlled by the oscilloscope, the timer or the logic elements. With a modification on the blockdiagram given in this example, a pattern display can be easily obtained in which the radius "r'' of the base of the cone or the height "h" of the cone is chosen as the parameter. This example illustrates the following three of the various features of HITACHI 505. 1. Independent control of each integrator. 2. Operation on multiple time axes. 3. Pattern displays. Example of Analysis - Double Integral The volume of a cone is pattern-displayed by means of multiple time axes operations. 1. Equation With the coordinates selected as shown in Figure 1, the equation of a cone is, where V4 h: height of the cone an | r: radius of the base of the cone Figure l -l- From equation (1), In equation (3), the height of the cone in reference to x - z plane is given. Since the cone is symmetrical to x - z plane, the volume of the cone on one side of the plane (for example, in the extent y = 0) is one half the total volume V. In this example, the calculation is performed for y = 0. The area Sy, of a cross section parallel to the base, for the half of the cone (y = 0), is given by the following equations. For the calculation of the volume of the cone, the cone is sliced at equal inter- vals by planes parallel to the base of the cone. Each section of the cone between two adjacent planes is approximated by a cylinder, the volume of each cylinder is calculated, and the volume of the cone is obtained as the sum of those of the cylinders. To apply this method of calculation to the half (y = 0) of the cone, we have the following equations for the volume of a section, AV}. kz ; -kz The total volume Vh of the half of the cone, with the number of sections N, is given by the following equation. kz N-1 N-1 Vn =) AVy, = nat | 222 -x% dx .... (6) n=0 n=0 ~kz The true volume can be obtained by an infinite number of divisions, or N-1 kz 5 Vy, =lim J az] pRez* —-x% dx Nace n=0 kz Z kz 0 -kz This is, the true volume is obtained by a double integral calculus. In an analog computer, the calculation with N-- can not be performed. Or strictly speaking, equation (6) is solved in an analog computer. Equations for the calculation in the computer To simplify the problem, the calculation of the volume of a cone with r =h = 1 is considered. From equation (2), it follows that k == l. Equation (6) can be expressed as, N - l Z Vi = > Anz | f22@ - xe dx lle eee eee (8) n= 0 : -Z The scales of the variables are converted on the basis that an integration with x is performed in 0.2 second and the summation is done in 100 seconds. The scales are shown in the following table. Variable Assumed maxi- Computer . : Scale factor ‘ inequation mum value variable x +1 +10 time/length [108] 1 To time/! To) Zz 100 time/length se Vh 1 1/volume [v., } Equation (8) expressed in terms of the computer variables is, N - 1 Zz > 2 | 2 2 [ Ya] 5 | Zo -x dx n= -zZ ss N-1 100 2 - [2] + =] - {10t} * fio at) Tt 100 tT N-1 100 2 T => ion =| - fio}? at .... (9) n=0 100 Tt ~ 100 For the calculation time To = 100 seconds, and the number of divisions N = 250, equation (9) is, T 249 100 5 _ i TT 2 Ya = > 55 [5] - fror} dt ...... (10) n= 0 Tt 100 Equation (10) is calculated by the analog computer. -3.- aH spuoses QO 10} dd WV q dnoin LaSaY TV ‘UV aH |... SH Pue do 98 20 |sPuc2es Z°0) wy y dnoin a10H ‘dH jo uotyjeday do dd qdnoayn : €(| ; 10}e18a3UT ALAGWOD ‘dd suoty} 10s| 00S -¢ 2 I 0 | -exedo ydnosy | {| Lasau ‘ow jo tlaquiny rIoJeredwos adeipon Hz TRepr0say adodsoT{I9so © 3 yodut ° }IMI ITS dH a8eiqoA C) Aetdstp usayed OL \ 2X d T Oo 1- * 7 foot) oot Oot LL et J HO¢ apro.oy 7 Ayuo yndyno aatlesa HOP |e¢ YX) 2x TOT oe ropaosay} [tA] t Oot T : s t/- gor of HOT + Tapsloooy weIseIpyvoTG “¢€ Pattern di splay circuit Z, cos @ ] Oscilloscope © x Y xX 5 i 1 yo sing (> ; 6 “XS <<, sin @ sin 6 4 S— 1 cos ¢ sin gr ‘os. 6 xX COs Z This is a conversion circuit for displaying a three dimensional figure on X-Y axes, rotating x, y, and z axes by 4° and ¢° Circuit generating RS and CP time of 0.2 seconds. +1 +1 0.1 1 to RY of group A integrators Ic 0.1 CP 10 RS B *1 0. *2 Voltage comparator +1 RS; RESET CP; COMPUTE *]1 The output of the integrator is shown in the figure below. +0.1 + ———— — — —— — — — — EO ! UN LN | I fe) i 7 i / 0.2 see NG O.v sec. \ 0.8 sec. ! 1 ' ! -O.)b TTT TTT TT *2 Reset and compute signals in the figure below are generated by the relay contact. 0.1 —- RESET E ESE SIGNAL 0 RESET RESET 0.14 COMPUTE COMPUTE COMPUTE SIGNAL 0 -5- 5-1 5-2 Computing elements required Dual DC amplifier; 7 units (Of 14 Amplifiers; 13 employed in the computation. ) DA-151 or DA-151A Dual Integrator; 2 units (Of 4 Integrators; all employed in the computations. ) IN-151 or IN-153 Potentiometer panel; 1 panel (of 18 Potentiometers; 11 employed in the computation. ) PT-251 Potentiometer patching units; 3 units : PT-151 Voltage comparators; 1 unit (of 4 Comparators; 2 employed in the computation) CP-151, CP-152 or CP-153 Square function generator; 1 unit (of 4 Function generators; 3 employed in the computation. ) FG-154A Solution (Chart of pen recorder) Page 7 Solution (Display on oscilloscope) Page 8 a4 Gg ' ( 3ouadafor dof) od SO SNOISEALP SZ ANM MOLRYMDRZ oes ae I } — ' . i yim WO PR| NI|R) ae ere ahlap q4ey) ) wOLMIOS savespanins 5-2 Solution (Pattern display) with x $ Z axis is perpendicular to O (for reference) . >t = 25 (for reference) x Example of calculation with N = 250 ¢ x Example of calculation with N = 25 where z* - x Z -X% the picture. N é — Output of /z* - 4 G