How to resolve the algorithm Averages/Mean time of day step by step in the PL/I programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Averages/Mean time of day step by step in the PL/I programming language
Table of Contents
Problem Statement
A particular activity of bats occurs at these times of the day: Using the idea that there are twenty-four hours in a day, which is analogous to there being 360 degrees in a circle, map times of day to and from angles; and using the ideas of Averages/Mean angle compute and show the average time of the nocturnal activity to an accuracy of one second of time.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Averages/Mean time of day step by step in the PL/I programming language
Source code in the pl/i programming language
*process source attributes xref;
avt: Proc options(main);
/*--------------------------------------------------------------------
* 25.06.2014 Walter Pachl taken from REXX
*-------------------------------------------------------------------*/
Dcl (addr,hbound,sin,cos,atan) Builtin;
Dcl sysprint Print;
Dcl times(4) Char(8) Init('23:00:17','23:40:20','00:12:45','00:17:19');
Dcl time Char(8);
Dcl (alpha,x,y,ss,ww) Dec Float(18) Init(0);
Dcl day Bin Fixed(31) Init(86400);
Dcl pi Dec Float(18) Init(3.14159265358979323846);
Dcl (i,h,m,s) bin Fixed(31) Init(0);
Do i=1 To hbound(times); /* loop over times */
time=times(i); /* pick a time */
alpha=t2a(time); /* convert to angle (radians) */
x=x+sin(alpha); /* accumulate sines */
y=y+cos(alpha); /* accumulate cosines */
End;
ww=atan(x/y); /* compute average angle */
ss=ww*day/(2*pi); /* convert to seconds */
If ss<0 Then ss=ss+day; /* avoid negative value */
m=ss/60; /* split into hh mm ss */
s=ss-m*60;
h=m/60;
m=m-h*60;
Put Edit(h,':',m,':',s)(Skip,3(p'99',a));
t2a: Procedure(t) Returns(Bin Float(18)); /* convert time to angle */
Dcl t Char(8);
Dcl 1 tt Based(addr(t)),
2 hh Pic'99',
2 * Char(1),
2 mm Pic'99',
2 * Char(1),
2 ss Pic'99';
Dcl sec Bin Fixed(31);
Dcl a Bin Float(18);
sec=(hh*60+mm)*60+ss;
If sec>(day/2) Then
sec=sec-day;
a=2*pi*sec/day;
Return (a);
End;
End;
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