I'm trying to adapt a piece of JavaScript code for my purposes and it contains the following function:

Since the orignal JS version aparently raised hackles, here's my first attempt at a Perl version along with some code to exercise it:

#! perl -slw
use strict;


sub func {
    my $vv = shift;

    my $bb = 140*24*60*60*1000; ##140 days
    if( $vv >= $bb ) { ##140 days < 5 months
        $bb = 8766*60*60*1000;##1 year
        if( $vv < $bb         ) { return( $bb/12 );  } ## 1 year      
+ ##1 month
        if( $vv < $bb *2      ) { return( $bb/6 );   } ## 2 years     
+ ##2 month
        if( $vv < $bb *5/2    ) { return( $bb/4 );   } ## 2.5 years   
+ ##3 month
        if( $vv < $bb *5      ) { return( $bb/2 );   } ## 5 years     
+ ##6 month
        if( $vv < $bb *10     ) { return( $bb );     } ## 10 years    
+ ##1 year
        if( $vv < $bb *20     ) { return( $bb*2 );   } ## 20 years    
+ ##2 years
        if( $vv < $bb *50     ) { return( $bb*5 );   } ## 50 years    
+ ##5 years
        if( $vv < $bb *100    ) { return( $bb*10 );  } ## 100 years   
+ ##10 years
        if( $vv < $bb *200    ) { return( $bb*20 );  } ## 200 years   
+ ##20 years
        if( $vv < $bb *500    ) { return( $bb*50 );  } ## 500 years   
+ ##50 years
        return( $bb*100 );                       ##100 years
    }
    $bb /= 2; ##70 days
    if( $vv >= $bb ) { return( $bb/5 ); } ##14 days
    $bb /= 2; ##35 days
    if( $vv >= $bb ) { return( $bb/5 ); } ##7 days
    $bb /= 7; $bb *= 4; ##20 days
    if( $vv >= $bb ) { return( $bb/5 ); } ##4 days
    $bb /= 2; ##10 days
    if( $vv >= $bb ) { return( $bb/5 ); } ##2 days
    $bb /= 2; ##5 days
    if( $vv >= $bb ) { return( $bb/5 ); } ##1 day
    $bb /= 2; ##2.5 days
    if( $vv >= $bb ) { return( $bb/5 ); } ##12 hours
    $bb *= 3; $bb /= 5; ##1.5 day
    if( $vv >= $bb ) { return( $bb/6 ); } ##6 hours
    $bb /= 2; ##18 hours
    if( $vv >= $bb ) { return( $bb/6 ); } ##3 hours
    $bb *= 2; $bb /= 3; ##12 hours
    if( $vv >= $bb ) { return( $bb/6 ); } ##2 hours
    $bb /= 2; ##6 hours
    if( $vv >= $bb ) { return( $bb/6 ); } ##1 hour
    $bb /= 2; ##3 hours
    if( $vv >= $bb ) { return( $bb/6 ); } ##30 mins
    $bb /= 2; ##1.5 hours
    if( $vv >= $bb ) { return( $bb/6 ); } ##15 mins
    $bb *= 2; $bb /= 3; ##1 hour
    if( $vv >= $bb ) { return( $bb/6 ); } ##10 mins
    $bb /= 3; ##20 mins
    if( $vv >= $bb ) { return( $bb/4 ); } ##5 mins
    $bb /= 2; ##10 mins
    if( $vv >= $bb ) { return( $bb/5 ); } ##2 mins
    $bb /= 2; ##5 mins
    if( $vv >= $bb ) { return( $bb/5 ); } ##1 min
    $bb *= 3; $bb /= 2; ##3 mins
    if( $vv >= $bb ) { return( $bb/6 ); } ##30 secs
    $bb /= 2; ##1.5 mins
    if( $vv >= $bb ) { return( $bb/6 ); } ##15 secs
    $bb *= 2; $bb /= 3; ##1 min
    if( $vv >= $bb ) { return( $bb/6 ); } ##10 secs
    $bb /= 3; ##20 secs
    if( $vv >= $bb ) { return( $bb/4 ); } ##5 secs
    $bb /= 2; ##10 secs
    if( $vv >= $bb ) { return( $bb/5 ); } ##2 secs
    return( $bb/10 ); ##1 sec
}

printf "%f -> %f\n", $_, func( $_ ) for map 10**$_, -2 .. +14;

__END__
[18:49:55.78] E:\Chart>junk
0.010000 -> 2500.000000
0.100000 -> 2500.000000
1.000000 -> 2500.000000
10.000000 -> 2500.000000
100.000000 -> 2500.000000
1000.000000 -> 2500.000000
10000.000000 -> 2500.000000
100000.000000 -> 12500.000000
1000000.000000 -> 120000.000000
10000000.000000 -> 900000.000000
100000000.000000 -> 10800000.000000
1000000000.000000 -> 172800000.000000
10000000000.000000 -> 1209600000.000000
100000000000.000000 -> 15778800000.000000
1000000000000.000000 -> 157788000000.000000
10000000000000.000000 -> 1577880000000.000000
100000000000000.000000 -> 3155760000000.000000
[download]

It is obviously ripe for refactoring, but can anyone

1. work out what (beyond:"mapping one number to another") it is doing?
2. perceive the mapping sufficiently to simplify (preferably:calculate) the mapping?

I'm not sure that you can take much guidance from the comments; as in context, the function does not seem to be being used for anything date or time related?