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Fripp
SFN Regular
USA
727 Posts |
 Posted - 01/05/2011 : 11:57:24 [Permalink]
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...I become a "Skeptic Regular".
**applause**
Thank you, thank you. You like me. You really like me!
I couldn't have done without the help and support of all of you. *sniff* This is the proudest day of my life. I want to thank my family; friends; my third grade teacher who labeled me as "absent-minded"; Mr. Dithers, the neighborhood butcher who never lost faith in me; Nipsy Russell; the garter snake that I captured and bit me some five times before I decided to go home; Bertrand Brinley; and most of all, I want to say that everything that I am, I owe to reading. |
"What the hell is an Aluminum Falcon?"
"Oh, I'm sorry. I thought my Dark Lord of the Sith could protect a small thermal exhaust port that's only 2-meters wide! That thing wasn't even fully paid off yet! You have any idea what this is going to do to my credit?!?!"
"What? Oh, oh, 'just rebuild it'? Oh, real [bleep]ing original. And who's gonna give me a loan, jackhole? You? You got an ATM on that torso LiteBrite?" |
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JerryB
Skeptic Friend
279 Posts |
Posted - 01/05/2011 : 12:20:58 [Permalink]
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Originally posted by Kil
Jerry. About your closing speech. We are apes. Didn't you know? |
Oh, I don't really believe we are apes, there are stark differences. But close to them in genotype and phenotype, yes.
| Also, like every other creationist you have miss interpreted the "cambrian explosion" which was no such thing. It marked a transition from mostly soft bodied life to bodies with bones or shells that would much more easily fossilize. It's exactly what evolution would predict. Nothing there is out of place. |
I'm not saying you are flat-out wrong here, you just don't take it far enough. Much of the phyla we see today came into the record during that era:
"The Cambrian explosion or Cambrian radiation was the relatively rapid appearance, over a period of many million years, of most major Phyla around 530 million years ago, as found in the fossil record.[1][2] This was accompanied by a major diversification of other organisms, including animals, phytoplankton, and calcimicrobes.[3] Before about 580 million years ago, most organisms were simple, composed of individual cells occasionally organized into colonies. Over the following 70 or 80 million years the rate of evolution accelerated by an order of magnitude (as defined in terms of the extinction and origination rate of species[4]) and the diversity of life began to resemble today’s."
http://en.wikipedia.org/wiki/Cambrian_explosion
And, I don't see how evolution could explain this at all. There just would not have been enough time from primitive life to the more advanced that we see in that era. From the same link above:
"Charles Darwin considered this sudden appearance of many animal groups with few or no antecedents to be the greatest single objection to his theory of evolution. He had even devoted a substantial chapter of The Origin of Species to solving this problem."
And only a creationist would talk about "fully formed" animals as being significant, because every animal and plant is transitional, and would appear fully formed because it is fully formed. What in the hell would a non fully formed animal even look like? It's the creationist idiot plea from total ignorance about what we should find in the fossil record. To even go there is to show that you know diddle squat about evolution, Jerry, but feel the need to show that it couldn't have happened.
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I've probably had as much formal training in evolution as anyone in here. Of course, I don't study talk origin so I'll give you that one.
What would a less than fully formed creature look like? Maybe Pee-Wee Herman?
No, I'll explain this. When I say "fully formed" I mean critters as we see them today. NOT organisms in a state of evolving into what we see today. Nothing more complicated than this.
| Again, just as we came in, ID is nothing more than an appeal to incredulity and ignorance. And by demonstrating your tenuous grasp of evolution itself, what you have shown us so far Jerry is the same old creationist shit dressed up as a crackpot QM, whatever it is... Oh well. |
I certainly fail to see how the science I have posted in here is an appeal to incredulity and ignorance. Of course, knowing the nature of many that post in these forums, many on your side will insist it is if I posted the theory of relativity.
But the unbiased reader will not and that is the audience that I post to.
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JerryB
Skeptic Friend
279 Posts |
Posted - 01/05/2011 : 12:37:07 [Permalink]
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Originally posted by Dave W.
Show me where he calculated D "exactly" as you calculated W. You can't, because he didn't."If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
-(entropy) = k log (1/D)." |
You left off the minus sign on the entropy when you posted your version earlier, Jerry. Nobody is disagreeing that log(x) = -log(1/x). What you can't do is just willy-nilly say that S = k log W sometimes and say that S = k log (1/W) at other times. S can't equal both, it's one or the other. So if S = k log W, then -S = k log (1/W) - just as you quoted Schrödinger - and S remains a measure of entropy only. |
More meaningless drivel. But let me rephrase that if you are going to be pedantic. Schrodinger showed you how to calculate it exactly
as I did.
Here is the bottom line. The numbers I calculated are perfectly correct, you know it is as well, and you are just trying to wiggle and squirm your way around the fact that you stated in the debate it could not be done that way. It's totally irrelevant that I left out - on the S, because the - was on the number.
And you are wrong that S can't be calculated more than one way, I can think of several ways to calculate it and they all are S.
Your log(x) = -log(1/x) above is wrong too. But I'll let that slide. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 01/05/2011 : 13:27:21 [Permalink]
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Originally posted by JerryB
Here is the bottom line. The numbers I calculated are perfectly correct, you know it is as well, and you are just trying to wiggle and squirm your way around the fact that you stated in the debate it could not be done that way. |
No, I said it should not be done the way you did it. Garbage in equals garbage out, even if all the calculations are done correctly. A person can put their body mass into E=mc2, but the result won't be how much energy they'll generate on a stationary bicycle. The inputs and outputs to these formulae have meanings which can't be ignored if you want to ensure that the results also have meaning. You ignore the meanings, Jerry, and make yourself look like an idiot.| It's totally irrelevant that I left out - on the S, because the - was on the number. |
Bwahahahahahahaha!| And you are wrong that S can't be calculated more than one way, I can think of several ways to calculate it and they all are S. |
Misses my point entirely.| Your log(x) = -log(1/x) above is wrong too. But I'll let that slide. |
No, please don't let it slide. It's true for every x that I can tell that's a valid input to the logarithm function. Take 37, for example. log(37) = 1.568 and log(1/37) = -1.568. In the general case, if logbx = y, then by = x and b-y = 1/x, so therefore logb(1/x) = -y and so by substitution, logb(1/x) = -logbx. Or would you prefer it in this form? logbx + logb(1/x) = 0. Means the same thing.
Or is it the unary minus operator that you need to re-learn? Given your insistence that the sign doesn't matter, I suspect that it is. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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filthy
SFN Die Hard
USA
14408 Posts |
 Posted - 01/05/2011 : 14:02:32 [Permalink]
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Let us have a care when we discuss the Cambrian "Explosin" and the oldest fossils. The oldest fossils come from the Archean rocks of the Precambrian and have been dated at some 3.5 billion years.
From that time forward, the Precambrian was a swinging place to be. All of the remarkable fossils from the Cambrian evolved from ancestors of this time.
And we are apes; merely another species of hominoids that, I sometimes think, are lesser evolved than our hairy cousins.
Don't like Talk Origins? Hey, it's your loss and no skin off'n anyone's nose save your own. 
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"What luck for rulers that men do not think." -- Adolf Hitler (1889 - 1945)
"If only we could impeach on the basis of criminal stupidity, 90% of the Rethuglicans and half of the Democrats would be thrown out of office." ~~ P.Z. Myres
"The default position of human nature is to punch the other guy in the face and take his stuff." ~~ Dude
Brother Boot Knife of Warm Humanitarianism,
and Crypto-Communist! 
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JerryB
Skeptic Friend
279 Posts |
Posted - 01/05/2011 : 14:03:38 [Permalink]
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Originally posted by Dave W.
No, I said it should not be done the way you did it. Garbage in equals garbage out, even if all the calculations are done correctly. A person can put their body mass into E=mc2, but the result won't be how much energy they'll generate on a stationary bicycle. The inputs and outputs to these formulae have meanings which can't be ignored if you want to ensure that the results also have meaning. You ignore the meanings, Jerry, and make yourself look like an idiot. |
OK, I'll put this rubbish to bed too. I went back and put the minus behind the S, now what is your bitch, that I didn't hold my lips right when I did it?
| Misses my point entirely. |
Nope.....I got your point. It simply isn't correct. You implied that there is only one way to calculate S by stating (paraphrasing) that either S = K log W, or S = K log 1/W, IOW it can't be both. Well, LOL, let me introduce you to elementary thermodynamics. It can be both and more depending on what we are calculating. There are many ways to calculate S. I'm sorry you are only familiar with one.
No, please don't let it slide. It's true for every x that I can tell that's a valid input to the logarithm function. Take 37, for example. log(37) = 1.568 and log(1/37) = -1.568. In the general case, if logbx = y, then by = x and b-y = 1/x, so therefore logb(1/x) = -y and so by substitution, logb(1/x) = -logbx. Or would you prefer it in this form? logbx + logb(1/x) = 0. Means the same thing.
Or is it the unary minus operator that you need to re-learn? Given your insistence that the sign doesn't matter, I suspect that it is.
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There ya go....If you cant dazzle them with brilliance, then baffle them with bullshit.
Unlike you, I have no desire to make anyone look like an idiot on here, so I have ignored several of your screw-ups.
But if you insist: You have it absolutely ass backward. Your log(x) = -log(1/x) should be the other way around: -log(x) = log(1/x). Why would you take a negative log of an inverse...LOL.....your are inverting twice.
And Schrodinger told you this:
If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
- (entropy) = k log (1/D).
Can you see that minus the log of D = the log of 1/D? Oh I know, that if 2 = 2 then the reverse is true, but if you are going to be so pedantic then I can too. That's why I was correct in taking the log of 1/W and you are left out in the snow holding your woodpecker with your objections.  |
| Edited by - JerryB on 01/05/2011 14:48:49 |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 01/05/2011 : 14:53:57 [Permalink]
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Originally posted by JerryB
Nope.....I got your point. It simply isn't correct. You implied that there is only one way to calculate S by stating (paraphrasing) that either S = K log W, or S = K log 1/W, IOW it can't be both. |
If S = k log W, then S = -k log 1/W, and not k log 1/W. It can't be both because the signs won't match for one or the other.
| But if you insist: You have it absolutely ass backward. Your log(x) = -log(1/x) should be the other way around: -log(x) = log(1/x). Why would you take a negative log of an inverse...LOL.....your are inverting twice. |
Bwahahahahahaha! Apparently, it really is the unary minus operator that's giving you problems. Since the unary minus operator is simply a shorthand for "subtract from zero," the two statements are exactly the same:(1) -log(x) = log(1/x)
Expand unary minus into what it means
(2) 0 - log(x) = log(1/x)
Add log(x) to both sides
(3) 0 = log(1/x) + log(x)
Subtract log(1/x) from both sides
(4) 0 - log(1/x) = log(x)
Collapse subtraction from zero back into the unary minus shorthand
(5) -log(1/x) = log(x) So, we can see that the two statements are algebraically identical. You do could the same if you consider unary minus to be a multiplication by negative one:(1) -log(x) = log(1/x)
Expand
(6) (-1)·log(x) = log(1/x)
Multiply both sides by -1
(7) log(x) = (-1)·log(1/x)
Collapse
(8) log(x) = -log(1/x) And since (8) is identical to (5), the algebra is correct, and which side of the equation the minus sign goes on is largely a matter of taste (one side or the other, but not both or neither). I'm surprised I need to explain 7th grade algebra to you.And Schrodinger told you this:
If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
- (entropy) = k log (1/D).
Can you see that minus the log of D = the log of 1/D? |
That's what I said. You really can't remember what side of an argument you're on for very long, can you?| That's why I was correct in taking the log of 1/W and you are left out in the snow holding your woodpecker. |
No. If -S = k log (1/D), then S = -k log (1/D) (as proven above). The value of S is still positive if you take the log of the reciprocal of D. It has to be, otherwise algebra is fatally flawed. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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JerryB
Skeptic Friend
279 Posts |
Posted - 01/05/2011 : 15:05:48 [Permalink]
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And Schrodinger told you this:
If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
-(entropy) = k log (1/D).
Back up, I'll ignore the cut and paste nonsensical math to hone in on the point. Now that I have put the minus in front of S. What is wrong with the math? Was Schrodinger just wrong?
Oh, And please post a reference to this formula which you just pulled out of your %^$: S = -k log (1/D) LMFAO........Funny |
| Edited by - JerryB on 01/05/2011 15:19:22 |
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Dr. Mabuse
Septic Fiend
Sweden
9687 Posts |
Posted - 01/05/2011 : 15:20:44 [Permalink]
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Originally posted by JerryB
I've probably had as much formal training in evolution as anyone in here. Of course, I don't study talk origin so I'll give you that one. |
ROLFLMFAO!!!1!!1!one!!!
Then why are you using such half-assed young-earth-creationist caricatures of evolutiuon in your arguments? To me (and I guess most of our readers) it looks more like you're trying to convince yourself of "YEC-truths" than aguing against any of our positions.
It makes you look so f*ing stupid...
If you truly have formal training in evolution, why don't you use arguments based on evolutionary theories, instead of those ugly but hilarious debunked lies creationists use?
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Dr. Mabuse - "When the going gets tough, the tough get Duct-tape..."
Dr. Mabuse whisper.mp3
"Equivocation is not just a job, for a creationist it's a way of life..." Dr. Mabuse
Support American Troops in Iraq:
Send them unarmed civilians for target practice..
Collateralmurder. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 01/05/2011 : 15:27:01 [Permalink]
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Originally posted by JerryB
And Schrodinger told you this:
If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
-(entropy) = k log (1/D).
Back up, I'll ignore the cut and paste nonsensical math to hone in on the point. Now that I have put the minus in front of S. What is wrong with the math? Was Schrodinger just wrong? |
No, you are just wrong, Jerry.
If D is 22026, then:(9) S = k log 22026
(10) S = k · 10 If we take k = 1 for illustrative purposes, then:(11) S = 10 So then we look at Schödinger's equation with D still equal to 22026:(12) -S = k log (1/D)
(13) -S = k log (4.54×10-5)
(14) -S = k · (-10) And if k is again 1 for ease of calculating:(15) -S = -10 Mutlply both sides by -1:(16) S = 10 You can't get a negative value for S just by inverting D unless you also fail to negate S as 7th-grade algebra demands.
Oh, also:| I'll ignore the cut and paste nonsensical math to hone in on the point. |
I didn't cut or paste it. It's perfectly good algebra. You're making yourself look like a complete moron by calling it nonsense. Just like you did six years ago. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 01/05/2011 : 15:29:48 [Permalink]
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Originally posted by JerryB
Oh, And please post a reference to this formula which you just pulled out of your %^$: S = -k log (1/D) LMFAO........Funny |
If -S = k log (1/D), it's algebraically equivalent to S = -k log (1/D). I've proven this. You called my proof "nonsense" without pointing out a single flaw in it. You obviously must not know 7th-grade algebra. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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JerryB
Skeptic Friend
279 Posts |
Posted - 01/05/2011 : 15:33:36 [Permalink]
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Originally posted by Dave W.
Originally posted by JerryB
And Schrodinger told you this:
If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
-(entropy) = k log (1/D).
Back up, I'll ignore the cut and paste nonsensical math to hone in on the point. Now that I have put the minus in front of S. What is wrong with the math? Was Schrodinger just wrong? |
No, you are just wrong, Jerry.
If D is 22026, then:(9) S = k log 22026
(10) S = k · 10 If we take k = 1 for illustrative purposes, then:(11) S = 10 So then we look at Schödinger's equation with D still equal to 22026:(12) -S = k log (1/D)
(13) -S = k log (4.54×10-5)
(14) -S = k · (-10) And if k is again 1 for ease of calculating:(15) -S = -10 Mutlply both sides by -1:(16) S = 10 You can't get a negative value for S just by inverting D unless you also fail to negate S as 7th-grade algebra demands.
Oh, also:| I'll ignore the cut and paste nonsensical math to hone in on the point. |
I didn't cut or paste it. It's perfectly good algebra. You're making yourself look like a complete moron by calling it nonsense. Just like you did six years ago.
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No answer the question and post a link to that rediculous formula you just made up:
And Schrodinger told you this:
"If D is a measure of disorder, its reciprocal, 1/D, can be regarded as a direct measure of order. Since the logarithm of 1/D is just minus the logarithm of D, we can write Boltzmann's equation thus:
-(entropy) = k log (1/D)."
http://dieoff.org/page150.htm
Was he just wrong? I don't want more of your irrelevant math, I want you stick to the subject and to answer the question and post the link. |
| Edited by - JerryB on 01/05/2011 15:38:26 |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 01/05/2011 : 15:38:56 [Permalink]
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Originally posted by JerryB
Was he just wrong? |
No, you are wrong.| I don't want to so more of your irrelevant math... |
A proof that my math is correct is "irrelevant?" Wow, Jerry, you obviously feel free to just ignore the math whenever you like, even though you claim that you've disproven evolution mathematically.| ...I want you to answer the question... |
I did answer the question. You called my answer irrelevant even though it was a mathematical proof that I am correct.The link to what? This seems to be the link you need. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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Dave W.
Info Junkie
USA
26020 Posts |
Posted - 01/05/2011 : 15:53:06 [Permalink]
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Just to re-iterate:(A) S = k log D
(B) -S = k log (1/D)
(C) S = -k log (1/D) These three equations are mathematically identical, Jerry. If you put the same values in for D and k for all three, you'll get the same value of S out for all three. I've proven this mathematically in two different ways, using no algebraic technique more sophisticated than what I learned before high school. Schrödinger would agree with me if he were alive.
If you're going to call my math "nonsense," Jerry, you're going to have to point out its flaws. |
- Dave W. (Private Msg, EMail)
Evidently, I rock!
Why not question something for a change?
Visit Dave's Psoriasis Info, too. |
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JerryB
Skeptic Friend
279 Posts |
Posted - 01/05/2011 : 16:02:56 [Permalink]
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Fine you say he was not wrong, that translates logically then, into he was right. Of course, it is obvious to any non-biased observer in here that you are struggling in this debate and are just trying to save face by burying the argument in obfuscating mathematics that don't even make sense to a you, much less anyone else.
But to put this debate to bed, I am going to use the exact math he suggested and the exact formula in the way he said to use it.
The original math:
W = (41469.4 + 1.6)! / (41469.4)!(1.6)! --- 3.66 x 10^173494 / 2.14 x 10^173487
W = 1.71 x 10^7
Now we can do Boltzmann's math:
S = K log W, S = (1.38 x 10^-23) log(1.71 x 10^7)
S = 9.98 x 10^-23
Entropy is positive showing the genome disordering as we would expect from the study and you claim it would be positive regardless of what we would do.
Wrong. Were the mutations beneficial we simply would use -entropy = K log 1/D. Where D is the same as W according to Schrodinger. Let's try it:
W = 1.71 x 10^7
Now we can do Boltzmann/Schrodinger math:
-entropy = K log 1/D, -entropy = (1.38 x 10^-23) log 1/(1.71 x 10^7)
-entropy = -9.98 x 10^-23
Now what's the problem...........A 2 actually means 7 or something.........BAHAHAhahahah.....Tell me when we are having fun.
Oh, and I still want a link to someone who agrees with you on the ridiculous formula you just made up.....S = -k log 1/W.....Where is it? |
| Edited by - JerryB on 01/05/2011 16:04:53 |
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Intelligent Design is Stupid
