Plate Count Calc x 2

[Dr. Evil]

See attached file for model.

HELLO! Here is a fun calc problem for you math men/women!

Goal is to provide user with maximized plate count for desired weight.

User enters desired weight, then enters the weight of the bar that will be used for exercise.

Field clc_plate_weight then subtracts the bar weight from desired weight to give a plate weight total. This total now has to be divided most efficiently in divisions of 2 in the following increments: 45, 35, 25, 10, 5, 2.5.

EI. For a desired weight of 145 lbs. and a bar weight of 45 we would need to plate 100 lbs.

100 lbs. plate break down would result in: 2-45s, 2-5s.

THANKS for any help!

PlatePerfect.fp7.zip

[comment]

Try something like:

Let ( [

net = ( DesiredWeight - BarWeight ) / 2 ;

d45 = Div ( net ; 45 ) ;

m45 = Mod ( net ; 45 ) ;

d35 = Div ( m45 ; 35 ) ;

m35 = Mod ( m45 ; 35 ) ;

d25 = Div ( m35 ; 25 ) ;

m25 = Mod ( m35 ; 25 ) ;

d10 = Div ( m25 ; 10 ) ;

m10 = Mod ( m25 ; 10 ) ;

d05 = Div ( m10 ; 5 ) ;

m05 = Mod ( m10 ; 5 ) ;

d02 = Div ( m05 ; 2.5 )

] ;

Case ( d45 ; 2*d45 &" x 45 " & ¶ ) &

Case ( d35 ; 2*d35 &" x 35 " & ¶ ) &

Case ( d25 ; 2*d25 &" x 25 " & ¶ ) &

Case ( d10 ; 2*d10 &" x 10 " & ¶ ) &

Case ( d05 ; 2*d05 &" x 5 " & ¶ ) &

Case ( d02 ; 2*d02 &" x 2.5 " & ¶ ) &

2* (d45*45 + d35*35 + d25*25 + d10*10 + d05*5 + d02*2.5) & " in total"

)