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Math problem...
#295100
10/22/09 23:08
10/22/09 23:08
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Joined: Jul 2004
Posts: 1,710
MMike
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OP
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hello.
this is my problem...
i want to know the % of the final color that is yellow.
We know that yellow = red + green.
so if i have: 10 parts of red + 10 parts of green = 20 parts , that will mix and give 100% of yellow
if i have 10 parts of red + 5 parts of green = 15 parts here 5 parts of green can miz with 5 parts of red, giving 10 parts of yellow and 5 parts of non reacted red, the final color is not 100% yellow but only 50% ( i think!!)
this is the problem, what is the math equation.. that i can know when i mix A quantities of red + B quantities of green, which final percentage i will get of yellow. :S
maybe this is easy than i think but im missing something...
Please help me this is really frustrating. i feel lossing IQ lol
i tryed %yello = (A/B) *100 but this is not the solution...
Last edited by MMike; 10/22/09 23:10.
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Re: Math problem...
[Re: MMike]
#295120
10/23/09 01:07
10/23/09 01:07
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Joined: Sep 2003
Posts: 6,861 Kiel (Germany)
Superku
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Senior Expert
Joined: Sep 2003
Posts: 6,861
Kiel (Germany)
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Haha....... hm if i have 10 parts of red + 5 parts of green = 15 parts here 5 parts of green can miz with 5 parts of red, giving 10 parts of yellow and 5 parts of non reacted red, the final color is not 100% yellow but only 50% There are no 5 parts of red that do not react with green, yellow is no base color such as red or green or blue, it is a result of mixing them together (as you know). As a result there is no real percentage of pure yellow. It would be easier to help you if I knew the purpose of your question.
"Falls das Resultat nicht einfach nur dermassen gut aussieht, sollten Sie nochmal von vorn anfangen..." - Manual Check out my new game: Pogostuck: Rage With Your Friends
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Re: Math problem...
[Re: MMike]
#295124
10/23/09 01:16
10/23/09 01:16
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Joined: Oct 2009
Posts: 33
AlexH
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Your issue doesn't really make sense from an external standpoint nor could I really understand your engrish.
Colors on a computer don't come in 'parts' they come in 3 bytes: Red Blue Green. So I don't get what you mean by 'parts'. However, I assume you aren't have troubles any more so it doesn't matter really, just know your presentation doesn't make sense.
Last edited by AlexH; 10/23/09 01:18.
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Re: Math problem...
[Re: Elektron]
#295320
10/24/09 11:48
10/24/09 11:48
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Joined: Jul 2004
Posts: 1,710
MMike
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hld on
Last edited by MMike; 10/24/09 11:52.
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Re: Math problem...
[Re: MMike]
#295331
10/24/09 13:28
10/24/09 13:28
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Joined: May 2008
Posts: 301
Oxy
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Just imagine a lightspectrum http://hidden-technology.org/images/pics%20for%20site/lightspectrum.jpgThe human eye has receptors for 3 colorranges which have their medium approximately at red green and blue. http://upload.wikimedia.org/wikipedia/commons/c/c2/Cone-response.pngFor a human to see yellow, both the green and red lightemitters (in the computerscreen) need to be on (in about the same strenght). Without blue beeing active. If blue is active, the yellow will get brighter, up to reaching a white-yellow. There are other basecolors used in Printing media for example, (magenta, cyan and yellow) All have the purpose to simulate a full spectrum by adjusting their values accordingly. A true yellow can only be archived by emitting light in the specific frequency (around 590nanomater). But the human eye approximates yellow anyhow using its 3 color receptors.
Last edited by Oxy; 10/24/09 13:39.
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Re: Math problem...
[Re: MMike]
#295370
10/24/09 20:40
10/24/09 20:40
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Joined: Jul 2004
Posts: 1,710
MMike
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i could simply do the 5cl / 10H2 = 0.5 HCl , so 50% in a formula it would be.. a/b = c , which a must be the minor number.. but because i dont want any hand of man here, i must make the expression by it self tell which is the minor and minor...
For that i will use a genetic Approach, by Mendel.
if we have the specie A and B, being A a pure genotype and b a pure genotype, it would be 2DNA strands so 2 alleles for each chromossome. Because im using pure reactants, like H2 or Cl , they are pure .. so AA and BB, HCl is Hybrid, (AB) thus, AA x BB this will give an offspring of combination of
1AA + 1 BB + 2AB ( or more specifc 1 AB + 1 BA). Ok so now that i know this... i can put this into the formula and purpose and intermediary algebra expression where i put this genetic heritage thing...
i will call it R(genotype). And there will be 4 of this.
R(genotype) : R(ab) , R(aa), R(bb) and R(ba).
if i make R=sign(alele 1 - oposite alele 1)X alele2.. i will get this distribution of R alele 1, alele2... Raa=sign(a-b)x a Rab=sign(a-b)x b Rbb=sign(b-a)x b Rba=sign(b-a)x a
filtering the negative values to be zero, and out of the add:
Raa=(1+sign(a-b))* a; Rab=(1+sign(a-b))* b; Rbb=(1+sign(b-a))* b; Rba=(1+sign(b-a))* a;
now that i got the R function i can compose it on another function, that will determine which value is the lower one by its own.
%= ( ( (Rba/2) + (Rab/2)) / ((Raa/2) + (Raa/2) ))*100
The lite-c function final is this:
function percent(var a, var b){ var Rab,Rbb,Raa,Rba; Raa=(1+sign(a-b))* a; Rab=(1+sign(a-b))* b; Rbb=(1+sign(b-a))* b; Rba=(1+sign(b-a))* a; return((((Rba/2)+(Rab/2))/((Raa/2)+(Rbb/2)))*100); }
function main(){ var reacting_p=percent(5,10); var reacting_p=percent(5,5); var reacting_p=percent(10,5); // the inverse, function will determine the minor number and major automatically.
}
all this to avoid using If minor then ... in a single expressiong..
Perhaps theres a batter way, but i just came up with this and it works so..
Last edited by MMike; 10/24/09 20:42.
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