User:Kopachris

Kopachris is a member of Memory Alpha, the free Star Trek reference. He chose his username because that was his username on LEGO.com. He hasn't been a member for very long so far, but hopes to contribute more often.

Contributions
Kopachris hasn't done much so far. His only major addition is a picture of Kes's hallucination from. Other than that, he calculated how long one stardate is and added a few minor things.– Kopachris 00:31, 9 February 2007 (UTC)

The first page that Kopachris added was Lava lamp. – Kopachris 06:00, 20 February 2007 (UTC)

Theories on Stardates, etc.
From TNG onward, the first digit in a stardate is mentioned by the producers of TNG as '4' for the century (24th century=2300's). The second digit is the season of TNG (1-7 for TNG, 8-6 for Voyager, etc.). This can be viewed as the year, leaving 1000 stardates per year. That means that there are about 2.737820804 stardates per day. Divide 24 by that, and you get 8 hours, 45 minutes, and 57 seconds per stardate. That makes an hour as .1140758668 stardates per hour, and .0019012644 stardates per minute.

C++ Program to convert Gregorian calender to stardates: //Copyright (C) 2007 by Christopher Koch
 * 1) include

using namespace std;

int main {       //initialize variables int month,day,hour,minute,monthsum,year,century; int monthsdays[12]={31,28,31,30,31,30,31,31,30,31,30,31}; float stardate,sday,smin;

//input month, day, year, and century cout<<"Month?"; cin>>month; cout<<"Day?"; cin>>day; cout<<"Year?"; cin>>year; cout<<"Century?"; cin>>century;

monthsum=0;

//add days up	for(int y=1;y<month+1;y++){ monthsum=monthsdays[y]+monthsum;}

//take out leftovers of this month (ex. you don't want 31 days included if it's only the 17th) monthsum=monthsum-(monthsdays[month]-day); sday=2.737820804/24; smin=sday/60; stardate=(century*10000)+(year*1000)+(monthsum*2.737820804)+(sday*hour)+(smin*minute);

//display result and pause cout<<stardate<<"\n"; cin.get; } In this program, you first select the month and day. For the year, you input the last digit of the year (7=xxx7), and the digit of the century (1=21st century).

As for quads, it conceivable that their storage medium works on a slightly different principle as their processing medium, i.e., that they store data in quaternary and process it in trinary. So, their "quad" would be like our bit, making the Doctor's program of 50 million Gigaquads equal roughly 11 pebibytes, or about 11250 tebibytes, perfectly reasonable with the current rate of data storage expansion. As for One, he assimilated approximately 10191.5 yobibytes, about 10686562750 tebibytes, of information after his maturation. It's conceivable that their processing and programming in trinary could be facilitated by quibits and quantum computation since quibits can be either 0, 1, or both. As a storage medium, another state could be added: missing. While a missing quibit would not be very good for computation, it could be counted as an extra state for storage. An electronic quantum computer could use the spin of an electron as its quibits, and if an electron was missing in a storage cell (as is a 0 in current binary computers), it could be counted in addition to spin states up, down, and both. Or, another medium employed by Star Trek, the isolinear optical chip, somewhat parallels the "optical lattice", which can also be used in quantum computing.