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Introduction

The program DATENO calculates the number of days from January 1. The program prompts whether we are working in a leap year or not.

With D = Day and M = Month, the days between January 1 and any other date within the calendar year is:

If M = 1 and M = 2 Then
DATE# = int(30.6 * M + 368.8) + D - 400

Otherwise,
DATE# = int(30.6 * M + 1.6) +D - 35 (non-leap year)
DATE# = int(30.6 * M + 1.6) + D - 34 (leap year)

HP 41C/DM 41L Program: DATENO

(^T: beginning of an alpha string)


Examples

Days between January 1 and February 16 (M = 2, D = 16): 46

Days between January 1 and October 1 (M = 10, D = 1):
(Non-leap year): 273
(Leap year): 274

Link to blog post: https://edspi31415.blogspot.com/2019/03/...-days.html

(03-11-2019 02:25 AM)Eddie W. Shore Wrote: [ -> ]The program DATENO calculates the number of days from January 1.

Maybe you should add 1 to the result to get the more common "day of year", i.e. 1 January = day 1 and 31 December = day 365.

(03-11-2019 02:25 AM)Eddie W. Shore Wrote: [ -> ]If M = 1 and M = 2 Then
DATE# = int(30.6 * M + 368.8) + D - 400

What about 31*M + D -32 ?-)

Anyway, here is my attempt at a day-of-year program. I have derived the formulas myself, so beware. ;-) Please try it and report any errors you find.


Usage:
day [ENTER] month [ENTER] year XEQ "DOY"

If you prefer month, day, year instead simply insert an X<>Y command between line 08 and 09.

The program automatically checks for leap years, this works for all dates between 1 January 1901 and 28 February 2100.

Example:
1 [ENTER] 10 [ENTER] 1999 XEQ"DOY" => 274
1 [ENTER] 10 [ENTER] 2000 XEQ"DOY" => 275

Dieter

(03-11-2019 02:25 AM)Eddie W. Shore Wrote: [ -> ]The program DATENO calculates the number of days from January 1.

Somewhat related is the following program from the article Sunrise and Sunset.

Here it is adapted for the HP-42S:

Examples:

1.03 R/S
1

28.02 R/S
365

29.02 R/S
366


This avoids having to deal with leap years.

Cheers
Thomas

(03-11-2019 02:25 AM)Eddie W. Shore Wrote: [ -> ]If M = 1 and M = 2 Then
DATE# = int(30.6 * M + 368.8) + D - 400

Otherwise,
DATE# = int(30.6 * M + 1.6) +D - 35 (non-leap year)
DATE# = int(30.6 * M + 1.6) + D - 34 (leap year)

Eddie, you can simplify these formulas significantly.
Since D and 400 are positive integers...

int(30.6 * M + 368.8) + D - 400
= int(30.6 * M + 368.8 + D - 400)
= int(30.6 * M + D - 31.2)

Or here (D=1 or 2) simply

= int(30.6 * M + D - 31)

The two other formulas can be simplified the same way.
Or they can be derived from the Jan/Feb formula.

With L=0 (common year) or 1 (leap year):

days = 30.6 * M + D - 31
if M >= 3 then days = days - 2.3 + L
return int(days)

Note: according to your formulas the final 2.3 would be 2.4, but subtracting only 2.3 leaves a fractional part between 0.1 and 0.9 so that roundoff problems are avoided (e.g. in non-decimal environments) that may lead to cases like int(242.9999999....) for 1 September.

Here is a version of your HP-41C program that implements this idea.


XEQ "DAYD"
MONTH↑DAY ?
10 [ENTER] 1 [R/S]
LEAP YR 0/1?
0 [R/S]

=> 273

Note that the program only prompts for a leap year input if this is required, i.e. for months between March and December.

Dieter