04-18-2014, 10:57 PM
(This post was last modified: 04-18-2014, 10:58 PM by greatgamer34.)
Why, just why,,,,,, didnt i think of that?!!!
NICE JOB AFTYPOO!
NICE JOB AFTYPOO!
04-18-2014, 10:57 PM
(This post was last modified: 04-18-2014, 10:58 PM by greatgamer34.)
Why, just why,,,,,, didnt i think of that?!!!
NICE JOB AFTYPOO!
04-18-2014, 11:02 PM
(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
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! [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
04-19-2014, 09:11 AM
Whay didn't I think of that? What's your adress aft? I'll send you a cake.
04-19-2014, 04:32 PM
Your closet
04-19-2014, 07:50 PM
04-19-2014, 09:20 PM
When I checked my closet.... There was a cake in there.... Creepy. I'll send it to Aft!
04-20-2014, 12:43 AM
Oh, you can be cheap and do the determinant of:
|m -n| |1 1| which is equal to m-(-n) or m+n... pretty cheap i guess.
04-20-2014, 01:03 AM
04-20-2014, 01:12 AM
(04-20-2014, 01:03 AM)Iceglade Wrote:(04-20-2014, 12:43 AM)snugglycreeper9 Wrote: Oh, you can be cheap and do the determinant of: Which is why I ditched the vector rotation, as far as I know to calculate the coordinates after rotation you need to add / subtract sines and cosines |
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