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DJ's are good at math right? - TranceAddict Forums

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TranceAddict Forums > DJing / Production / Promotion > DJ Booth > DJ's are good at math right?
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JohnSmith
Agent Smith



Registered: Apr 2002
Location: Kamloops
DJ's are good at math right?

i have an interesting problem, i am trying to solve it.

observe the following set of numbers:

PHP:



there is a formula, that will allow you, to take the number on the left (x), and derive from it, the number on the right (y). the numbers above, are simply counts of how many numbers are in each set. this number is essential in the calculation, as the relationship changes each time you add a number..

i will not rest until i find this formula.

Old Post Apr-16-2002 23:42  Canada
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skywarp
goa trooper



Registered: Sep 2000
Location: Edmonton, AB

And WHAT exactly does this have to do with playing records ?!


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Old Post Apr-17-2002 00:58  Canada
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DJ LIQUID
House DJ 4 Life



Registered: Jul 2001
Location: IL USA

hey...im really good at math....it must say something bout my djing skillz


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Old Post Apr-17-2002 01:11  Mexico
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JohnSmith
Agent Smith



Registered: Apr 2002
Location: Kamloops

quote:
Originally posted by skywarp
And WHAT exactly does this have to do with playing records ?!


has nothing to do with playing records.

but, like i said, DJs are good at math right?

Old Post Apr-17-2002 01:29  Canada
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Special_K
I dont know anymore



Registered: Apr 2001
Location: Vancouver

im not good at math.....

Old Post Apr-17-2002 01:42  Canada
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Dmatrox
something goes here?



Registered: Jul 2001
Location: Calgary

im pretty good at math, but im better at biology

Yeah, what does math have to do with playing records?

Old Post Apr-17-2002 01:48  Canada
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patticus
watch your bassbins



Registered: Apr 2001
Location: vancouver, bc

wtf? are those derivatives of some sort??
being asian, i used to be good at math, but then my overall capacity for the arts (in other words, stupidity) came to the forefront..
i got 78 x 150 wrong today. i said like 340 or something stupid


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Old Post Apr-17-2002 01:50  Canada
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sym
Time for Revolution



Registered: Apr 2002
Location: NYC

Well what looks like to me is the following...

Find the distance between the first x and the first y. This number we will call z.

Than take the next x value you have and need the y for. Find the distance betweent his 2nd x and the first x you used to find z.

Than subtract that number from z and add the result to x2 to get your y2.

At least thats how I saw it... You'd have to be more specific with what you want... You can AIM me @ SyM2oo1

Old Post Apr-17-2002 01:55  United States
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IronDragon
Ya'll be some busters



Registered: Jan 2002
Location: So sleepy
Re: DJ's are good at math right?

quote:
Originally posted by JohnSmith
i have an interesting problem, i am trying to solve it.

observe the following set of numbers:

PHP:



there is a formula, that will allow you, to take the number on the left (x), and derive from it, the number on the right (y). the numbers above, are simply counts of how many numbers are in each set. this number is essential in the calculation, as the relationship changes each time you add a number..

i will not rest until i find this formula.


::giggles like a bimbo:: math is hard.
But really, I am only so-so at best in math. I assume you'll find that everyone else in here will range from friggin' genius to friggin' idiot. Maybe you'll get lucky and some'll know what the hell you're trying to do.

Suerte,
Dave

Old Post Apr-17-2002 02:21 
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uberclkgtr
spaced cadet



Registered: Jul 2001
Location: Baltimore, USA

ah, this looks kinda fun.

i can get it to work for all except the 1st set. here's how i set it up and solved it:

y - x = a + b

where a = the set number (ie 1, 2, 3, 4 or 5)
and b we have to figure out

if you make table of y and b values for a given a such as the following:

code:
for a = 5 y b ----- 10 4 11 3 12 2 13 1 14 0 for a = 4 y b ----- 8 3 9 2 10 1 11 0

you can see that b can be expressed as

b = 3a - y - 1

so we have the two equations

x - y = a + b
b = 3a - y -1

which we can solve by substitution to get

y = (4a + x - 1) / 2

this only works if a > 1.

i suppose you could use an iterative approach, where you specify y1, and based on the value of a and y1, you get y2. but i've done enough of your homework for tonight.

i'm better at math than spinning records...


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Old Post Apr-17-2002 03:01  United States
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hapamoto
R3ELISM



Registered: Jan 2002
Location: Canoga Park, California

haha.. fuck this.. im asian too but that doesn't mean im good at math because of it.. im just good in math because I am.. anyways if i wanted to be doing math problems i'd be taking math classes.. sorry, i could probably help you but seems like other people already figured it out.. yay less work for me

Old Post Apr-17-2002 04:34  United States
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Xquisite
Senior tranceaddict



Registered: Mar 2002
Location: Santa Barbara, CA, USA

heh i got a messy equation. I dont think you want to see it. It has a ton of mods in it.


___________________
DJ Xquisite
Hip Hop & Trance DJ for the new millenium

Old Post Apr-17-2002 06:56  United States
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