Return local beginning of day time object in Go

Both the title and the text of the question asked for "a local [Chicago] beginning of today time." The Bod function in the question did that correctly. The accepted Truncate function claims to be a better solution, but it returns a different result; it doesn't return a local [Chicago] beginning of today time. For example,

package main

import (
    "fmt"
    "time"
)

func Bod(t time.Time) time.Time {
    year, month, day := t.Date()
    return time.Date(year, month, day, 0, 0, 0, 0, t.Location())
}

func Truncate(t time.Time) time.Time {
    return t.Truncate(24 * time.Hour)
}

func main() {
    chicago, err := time.LoadLocation("America/Chicago")
    if err != nil {
        fmt.Println(err)
        return
    }
    now := time.Now().In(chicago)
    fmt.Println(Bod(now))
    fmt.Println(Truncate(now))
}

Output:

2014-08-11 00:00:00 -0400 EDT
2014-08-11 20:00:00 -0400 EDT

The time.Truncate method truncates UTC time.

The accepted Truncate function also assumes that there are 24 hours in a day. Chicago has 23, 24, or 25 hours in a day.