# Find the Equation

• Feb 16th 2011, 05:08 PM
Lammalord
Find the Equation
Okay, first I wasn't sure what math this is, but I'm pretty sure its Algebra.

I'm looking to find the equation for this.

as 1300 ---> 2400
12---> 0

it's an equation of adding to 1300, starting at adding 12, until I reach 2400. but the closer to 2400 the equation gets the smaller the addition becomes - ultimately going to an addition of 0 at exactly 2400. Also the addition cannot be a decimal, therefore it must round up.

mm doing some math if 1300 is +12 and 2400 is +0 this would mean each number on the 1300-2400 scale would be .0109 (or .011 rounded) on the 12 to 0 scale.

The equation I want would be something where if I said 2250 what would be added given that the addition (which shrinks the larger the number gets) loses .011 per additional number over 1300 on the larger scale. I know (doing the match) that it would add 1.55 to 2250 (rounded up to 2)
• Feb 16th 2011, 05:18 PM
rtblue
The question is very vague. At what rate are you adding to 1300? Are you adding in an exponential pattern? quadratic? cubic? logarithmic?
• Feb 16th 2011, 05:25 PM
Lammalord
ah, sorry it starts at 12 and decreases to 1, ie 1300+12, 1312+11.87(rounded up), 1324+11.76(rounded up) ect, ect, till it gets to 2400.

every 1 point on the top one = a .011 subtraction to the number on the bottom (thats adding to the total), if i got that right.
• Feb 16th 2011, 08:20 PM
Wilmer
Quote:

Originally Posted by Lammalord
ah, sorry it starts at 12 and decreases to 1, ie 1300+12, 1312+11.87(rounded up), 1324+11.76(rounded up) ect, ect, till it gets to 2400.
every 1 point on the top one = a .011 subtraction to the number on the bottom (thats adding to the total), if i got that right.

Very unclear and confusing.
Change your problem to something shorter, so we can "see";
make it from 100 to 200, starting with 12, decreasing by .45:
Code:

``` 0                  100.00  1    12.00  12.00  112.00  2    11.55  12.00  124.00  3    11.10  12.00  136.00  4    10.65  11.00  147.00  5    10.20  11.00  158.00  6    9.75  10.00  168.00  7    9.30  10.00  178.00  8    8.85  9.00  187.00  9    8.40  9.00  196.00 10    7.95  8.00  204.00```
So .45 is deducted constantly from 12, the results rounded up and added...stop once >= 200.
Is that what you mean?
• Feb 16th 2011, 11:23 PM
Lammalord
not quite, the +12 would lose .011 per number on the 1300-2400 scale.

so in other words it would add something like so:

1 12.00 12 1312
2 11.868 12 1324
3 11.736 12 1336
4 11.604 12 1348
5 11.472 12 1360

ect, ect - and it needs the .011 per number on the 1300-2400 scale in case a random number was given, say, 2250 - what would be added to it? in this case it would be 1.55 rounded up, that would make it +2, I can do the math one at a time, I just can't figure out how to get an equation for it.
• Feb 17th 2011, 05:57 AM
Wilmer
OK, I get it now (though I wonder what the purpose of this is!)
To illustrate, I'll carry on with your example, adding column for ".011 per number":

000 .000 00.000 00 1300
001 .000 12.000 12 1312
002 .132 11.868 12 1324 : 12(.011) = .132
003 .132 11.736 12 1336
004 .132 11.604 12 1348
005 .132 11.472 12 1360
006 .132 11.340 12 1372
007 .132 11.208 12 1384
008 .132 11.076 12 1396
009 .132 10.944 11 1407
010 .121 10.823 11 1418 : 11(.011) = .121
011 .121 10.702 11 1429
.....
164 .022 01.594 02 2248
165 .022 01.572 02 2250
166 .022 01.550 02 2252 : your example of 1.55 resulting from 2250
.....
279 .011 00.021 01 2390
280 .011 00.010 01 2391
281 .011 -0.001 00 2391
282 .000 -0.001 00 2391
283 .000 -0.001 00 2391 : 2400 will not be reached
.....
infinity!

Correct?
Now I'll let RTBlue take over...
• Feb 17th 2011, 06:11 AM
Lammalord
yes, perfect :) yeah it wouldn't be reached because of rounding, it's supposed to be .01090909(repeating)
• Feb 17th 2011, 06:38 AM
Wilmer
Correct:
......
282 .010909... 00.02182... 01 2399
283 .010909... 00.01090... 01 2400
• Feb 18th 2011, 11:46 PM
Lammalord
mmm, So I still do need an equation for this - anyone?