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 Post subject: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 6:50 am 
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Has anyone got any clue of how to do these? The Gravitational holes seem to work with random clicking, but I've no idea how the heat rods and acid thingies add up.

I have set up a chatroom at http://www.chatzy.com/514950204008 to discuss this stuff in realtime.


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 6:56 am 
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i just asked this question in the regular post for the plot


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 4:59 pm 
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For making the control rods

The parts of one metal have to add up to a certain weight to make a rod of that type:

slothite: 1.1 kg
zafarium 1.8 kg
zweetite 1.9 kg
meridellium 0.7 kg
tikium 2.5 kg
tyrannium 3.6 kg
kadaotite 3.3 kg
neopium 4.4 kg
faeryllium 5.3 kg
dariganium 5.0 kg

So if you have

Mass: 2.7 kg
Composition:
# 14.8% Meridellium
# 29.3% Xweetite
# 55.9% Zafarium

and

Mass: 0.7 kg
Composition:
# 29.3% Kadoatite
# 29.3% Tikium
# 41.5% Zafarium

You should get Zafarium (.559*2.7+.415*0.7=1.7998)

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Last edited by acerimusdux on Thu Mar 06, 2008 6:01 pm, edited 1 time in total.

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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 5:22 pm 
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If that's the method, the weight for Dariganium is 5.0 kg :)


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 6:41 pm 
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okay this may be a kinda stupid question but math was never my strong point. on that site i pretty much get how to do the math but my question is when i am taking the % and and turning it to a decimal to multiply by am i moving the decimal by one or two to the left as in I get 41.2% be .412 but would 1.2% be .12 OR .012


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 6:48 pm 
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Rubywinged wrote:
okay this may be a kinda stupid question but math was never my strong point. on that site i pretty much get how to do the math but my question is when i am taking the % and and turning it to a decimal to multiply by am i moving the decimal by one or two to the left as in I get 41.2% be .412 but would 1.2% be .12 OR .012


You move the decimal two places, so 1.2% would be .012. ^.^


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 6:51 pm 
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thanks that's what i thought my my brain is melty with all this math... Math class was far to long ago for this.


I'm thinking the acid (one player) has to be easy as well if someone could figure out the equation for it.

I haven't messed with it much yet myself, but they do give you 3 measurements so that's gotta be something!


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 7:15 pm 
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I've gone so far as to set up my spreadsheet (same one I used for the singularity stage of testing) to include cells for inputing my length, radius, and density... then I have functions that will output the volume (in cm³) and mass (in kg). But then, of course, I don't know what to do with this information once I have it. I've tried taking the square-root and the cube-root of the mass. Hrm. I don't know. In the bone-testing part of the Tale of Woe plot, I recall having a constant that we multiplied by. Some weird number, like 5.62, or something. If it's something like that, I don't have much hope of guessing it.


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 7:17 pm 
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I know how to figure out the weight in grams of the metal for the acid part. Assuming they are all perfect cylinders, you square the cross-sectional radius and multiply it by pi, then multiply that by the length of the cylinder. This gives you the volume of the cylinder in cubic millimeters. Since the density of the rod is in cubic centimeters, you divide your last answer by 1000 then multiply by the given density. This will give you the number of grams of metal you have.

As for how to turn that into a number of seconds, I have no earthly idea.


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 8:37 pm 
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theonlysaneone wrote:
As for how to turn that into a number of seconds, I have no earthly idea.


The acid probably has a constant rate of dissolution (i.e. it will always be x seconds/gram), so basically, we just have to figure out what that is. I figure the easiest way to do so is to get people who are confident with the calculations to figure out what mass their rod has, and work in tandem to figure out the constant like so:

Person A will assume 1 second/gram and go up from there with each trial (1, 2, 3, etc)
Person B will assume 20 seconds/gram and go up from there
Person C will start at 30 seconds/gram, etc

They'd have to call their range (i.e. "username here, I'll take 20-30 and get back to you all") and then post their results when they finish, and if they want, take the next available range.

yeah, I really should be working. :/


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 9:10 pm 
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I've done everything once now except the acid.... so I atleast feel a yay there lol.


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 9:28 pm 
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theonlysaneone wrote:
I know how to figure out the weight in grams of the metal for the acid part. Assuming they are all perfect cylinders, you square the cross-sectional radius and multiply it by pi, then multiply that by the length of the cylinder. This gives you the volume of the cylinder in cubic millimeters. Since the density of the rod is in cubic centimeters, you divide your last answer by 1000 then multiply by the given density. This will give you the number of grams of metal you have.

As for how to turn that into a number of seconds, I have no earthly idea.


If we suppose that the acid eats into the bar a constant *depth* per second:

Let k = depth per second = dh/dt = dr/dt
r = original radius
h = original height
d = density

Original volume: pi * r^2 * h
End volume: pi * (r-k*t)^2 * (h-k*t)

Material dissolved: d * pi * (r^2 * h - (r-k*t)^2 * (h-k*t) )

...and solve for t. I'm not sure if I want to touch this one - I already did the one for the gravatics... :P

I suppose that TNT *could* have done something insane and accounted for kinetics. Then dh/dt = dr/dt would be something insane, involving the K_a_ and volume and concentration of acid in the vat. But I don't think they're quite that evil, or realistic... If you add an alkali metal into water, you produce a base, not an acid. :P

But it would explain why the Conundum Lenny is "on vacation"...


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 9:32 pm 
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AySz88 wrote:
But it would explain why the Conundum Lenny is "on vacation"...



Because sloth kidnapped him and is using him to torture us lesser minded neopians?

heh


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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Thu Mar 06, 2008 11:08 pm 
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Errr....sorry to pop everyone's bubble, yet in making the "Making Dissolution Acid", the equation, density = mass / volume , technically should not be applied.
.

What you have is 'A metallic compound composed of Tyrannium (insert random neo-metal name) and an oxygenating bio-solvent' made into a perfect cylinder; the density is that of the metal, NOT the compound as a whole.


In other words, it should be impossible to find the mass of the metal compound.


Last edited by Jerch on Fri Mar 07, 2008 9:20 am, edited 1 time in total.

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 Post subject: Re: Return of Dr Sloth - individual challenges
PostPosted: Fri Mar 07, 2008 6:12 am 
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edhriel wrote:
theonlysaneone wrote:
As for how to turn that into a number of seconds, I have no earthly idea.


The acid probably has a constant rate of dissolution (i.e. it will always be x seconds/gram), so basically, we just have to figure out what that is. I figure the easiest way to do so is to get people who are confident with the calculations to figure out what mass their rod has, and work in tandem to figure out the constant like so:

Person A will assume 1 second/gram and go up from there with each trial (1, 2, 3, etc)
Person B will assume 20 seconds/gram and go up from there
Person C will start at 30 seconds/gram, etc

They'd have to call their range (i.e. "username here, I'll take 20-30 and get back to you all") and then post their results when they finish, and if they want, take the next available range.

yeah, I really should be working. :/


Don't do it by grams, do it by kilograms. Most of these turn out to be somewhere in the neighborhood of 60,000 grams, and even at 1 second per gram, that would take about 17 hours to complete. I shudder to think what would happen if someone entered a number in the hundreds of thousands in there and was stuck not being able to do anything for the rest of the plot.


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