04-18-2014, 11:04 PM
(04-18-2014, 11:02 PM)Iceglade Wrote:[17:00:49] Aft-pan: I wrote like 500 words on using vector rotation to figure it out(04-18-2014, 10:45 PM)AFtExploision Wrote: I got it. You have two numbers, m and n. You do (10^m * 10^n) [Ten to the m times ten to the n]. Now, log base 10 of the answer to that equation will be m+n. (x^a * x^b = x^a+b), as algebra taught us. The base can be any number, I picked 10 because you could do powers of 10 on paper, and it is simple to tell if a number is a nice log of 10, (a log that returns a whole number), so this method is very friendly to use without a computer. (But on paper, you could draw a line n length, a line m length so that lines n and m form a longer line, and find the length of the longer line to get m + n, but my log way is computer-friendly, you'd have to add to find the endpoints so you could calculate the lines if you tried to implement it in code). IDK why none of us thought of this, its pretty simple actually. Now give me a prize!
...so much for vector rotation :p
[17:01:00] Aft-pan: then I realized that vector rotation uses addition
[17:01:08] Alex Jarvis: Haha!
[17:01:17] Aft-pan: Then I tried cross-products
[17:01:21] Aft-pan: Couldnt get it to work
[17:01:35] Aft-pan: tried translation, realised pc's used addition to translate