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Dec 2nd, 2003, 01:48 AM
#1
Thread Starter
So Unbanned
Painter's problem
A painter has varying paints which vary in opacity, x. He wants to figure out how many coats of paint, y, are required to make a certain paint of opacity x appear at full opacity. What equation should he use?
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Dec 2nd, 2003, 02:51 AM
#2
Frenzied Member
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Dec 2nd, 2003, 03:25 AM
#3
Assuming full opacity is 100, and x is a percentage (such as 0.25),
y = 100/x
Or am I missing something here?
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Dec 2nd, 2003, 07:22 AM
#4
Thread Starter
So Unbanned
It's additive.
Say he has 50% opaque paint. Two coats and this becomes 75% opaque.
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Dec 2nd, 2003, 08:03 AM
#5
In each case, you'd have to take the sum of the geometric series. In our 50% example,
1 = x(1 + 1/2 + 1/4 + ...)
Sum of a geometric series is
S = a(1-rn)/(1-r)
In our example, a = x, r = 1/2 and S = 1
1 = x(1-(.5)n)/.5
And I'm too dumb to go on from here  , but hope this leads you somewhere!
See? My my posts are of varying bull**** x. 
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Dec 2nd, 2003, 08:35 AM
#6
transcendental analytic
x has to be of full opacity, and then we have y=1.
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 2nd, 2003, 08:37 AM
#7
That's what I was confused about.
If x < 1, then y = infinity.
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Dec 2nd, 2003, 08:44 AM
#8
transcendental analytic
Originally posted by mendhak
That's what I was confused about.
If x < 1, then y = infinity.
we cannot solve the equation (1-x)^y=0 for x<1
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 2nd, 2003, 08:50 AM
#9
This is like the half-distance paradox.
Me and Cameron Diaz are standing a few meters apart in a room. Between us, there are two halves. If I move half forward, then there's another two halves between us. Like this, there are an infinite number of halves.
Of course until we start making out, but that doesn't figure into mathematical terms. 
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Dec 2nd, 2003, 09:17 AM
#10
Thread Starter
So Unbanned
...appear at full opacity...
Which, for this sake, we'll say is >98% opaque.
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Dec 2nd, 2003, 09:31 AM
#11
transcendental analytic
then we have
(1-x)^y=0.02
y=log1-x 0.02
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Dec 2nd, 2003, 01:48 PM
#12
Thread Starter
So Unbanned
Originally posted by kedaman
then we have
(1-x)^y=0.02
y=log1-x 0.02
Hmm.
y=log1-x 0.02
gives me a negative result when x is .5.
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Dec 2nd, 2003, 06:17 PM
#13
transcendental analytic
then your calculator is broken, mine gives ~5.64
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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