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Messages - uncool

1
Bah,  that's what I was aiming for, but couldn't figure out how to write. Agreed.
2
Damnit people, link!
https://www.reddit.com/r/IAmA/comments/8m21kw/i_am_dr_jordan_b_peterson_u_of_t_professor/

And a followup:
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Look: it's up to those who claim no relationship between atheism and Nazism/Marxism to deal with the fact that both were explicitly anti-religious movements.
Jesus, this guy's even more stupid than I thought.
3
I think you guys may be missing Dave's point, because of the completely different moral systems here.

Dave doesn't care especially much about the age. That is, if anything, an aggravating factor, but not the central element. What is immoral to Dave is the "freeloading". The idea that Clinton could have sex with no consequences.

Right, Dave?

(n.b.: I in no way endorse Dave's point of view here)
4
To spell out my argument explicitly:

In the system, there are three sources of energy: the kinetic energy of the object, the kinetic energy of the mass of air, and the energy in the battery. Total energy at the start: 100 J + 25 J + 10000 J. By conservation of energy, the energy in the system at the end must also be the same (since we've assumed this is lossless). Total energy at the end: 121 J + 16 J + Ee. Therefore, Ee = 10125 - 137 = 9988  = (10000 - 12) J.
5
The "song and dance" is an effective way to find where the disagreement is. For example, I disagree with your first answer, and I think that disagreement may be key.

My answer is 10000 - 12 J. Did you get your answer by only looking at the kinetic energy of the object itself for this? In other words, by looking at 121 J - 100 J?

I agree with your second answer, however.

Your first answer is wrong, then. There is no wind energy available above wind speed so it is only necessary to look at the change in ke of the object.

Good, you have the second one right. Half credit.

Now, I conclude that you cannot make an argument, like a man, in support of ddwfttw and all you can do is play little boy games. I'm not interested. You hit the jackpot on the crackpot index! Congratulations!

I have an argument that continues from my answer there. So let's discuss the difference.

How much energy is in the system (object, mass of air, battery) total beforehand? How much afterwards?

But you still have not made any argument to support your belief in ddw.
Correct, because my further argument relied on the point I planned to make with this one. So going further is pointless until we can resolve this.
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So far, all you have done is ask questions and unlike you, I have done my best to answer all of them, albeit with a few typos and dumbass mistakes thrown in.
Again, I strongly disagree that I have not done my best to answer.
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As far as this latest question about the situation with the cart (or whatever it is, you even refuse to identify wtf it is) that is travelling above windspeed and then accelerates:

First thing to take into consideration is the fact that a cart that is at equilibrium with a velocity of 10 m/s in a wind of 5 m/s has already benefited from the tailwind, just reach and maintain 10 m/s velocity.

That equilibrium is something that you don't dare to mention!
Right now, I am not making an equilibrium argument yet. That was coming in a later part of the argument.
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Now, in order to accelerate to 11 m/s the change in ke of the cart is from 100 J to 121 J an increase of 21 J. That 21 J must be provided by the battery alone since the cart cannot gain any additional benefit from the wind!

Your naïve claim, which is only insinuated at the moment, since you dare not post it, is that the cart will slow down the wind from 5 m/s to 4 m/s, thereby extracting (25 J - 16J) of wind energy to offset some of that 21J required, and the drain on the battery will be only 12 J.
I wouldn't use the word "extracting" (I'll get to why later), but in fact, yes, I was planning to post that - or to get you to post something similar.
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Now I will tell you just some of the reasons why you are engaging in sophistry and you are a crackpot!:

1) As I said, the cart at 10 m/s in a 5 m/s tailwind has already benefited from that tailwind, just to be there at 10 m/s. How many times do you think the cart can benefit again from the tailwind? Every time it accelerates? An infinite number of times? Does the tailwind have an infinite amount of energy it can give to the cart, on demand?
That would be a part of a later argument - approximately the same part of the argument as the equilibrium. Whether or not that's true does not directly affect the answer to this question (although yes, I can see it as a reductio ad absurdum-style argument).
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2) In order to extract that 9 J of energy from the wind, the cart must do work on the wind. In fact, in a lossless situation, unrealistic as it is but what you demanded, the cart would need to do exactly 9 J of work on the wind to extract 9 J of energy from the wind. It is a wash! So instead of going through the motions of adding and subtracting +9J and -9J it is much simpler to do as I did and say that the 21J needed to add to the cart's ke is what is drawn from the battery nothing happens with the wind!
This argument is central, and the major point I want to make here is: Work can be negative. The work done on the mass of air is not 9 J, but -9 J.

This is roughly analogous to the braking system in the first section: the braking system does negative work on the object, allowing the battery to store energy. Your argument is equivalent to the claim that the brake must do work to the object, letting it all come out in the wash. It's very much like the $25 hotel room riddle.
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3) You said M is an air mass moving at 5 m/s. That is a wind, let's get that straight! If you slow down some of that wind to 4 m/s in a reaction, that does not mean there is now an air mass that continues to move at 4 m/s after the reaction! Do you think otherwise? Do you think there is a pocket of air in the wind that continues to move at 4 m/s after the reaction? That is what you are asking me to tell you! You are asking for the beginning and end energies of the cart, the battery and the wind! I already told you the cart is now at 121 J of ke, the battery is at (10000 - 21) J and the wind is unchanged! That is so because there is no lasting pocket of air moving at 4 m/s! The air is slowed during the reaction time only and the +9J and -9J is a wash. The only lasting effect is on the cart's ke and the battery's energy.
This argument would also be true for the second case. Why does its conclusion not apply there, as well?
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Now take your sophist, crackpot argument to somebody else who is naïve enough to buy it (that would be everyone else in this thread except me)
Answer my question.

How much energy is in the system (object, mass of air, battery) total beforehand? How much afterwards?
6
Hey Dave, I'm curious.

Have you ever seen A Few Good Men?
7
Let me ask you two specific versions of the question from step 2.

In both cases, the object and the affected mass of air will have mass 2 kg. The battery starts with 10000 J (i.e. enough to power anything relevant).

 In the first version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 5 m/s.

Second version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 15 m/s.

In both versions: how much energy does the battery have at the end?


If you are talking about ddw, in the first version it is all done by battery power. There is no way to use wind power above wind speed ddw

That is correct
The end. See ya!
argument by dishonest editing.  cool.
Eh, I read it as an attempted summary based on his misreading of my misreading. And honestly, I can see where his misreading comes from. I'm willing to chalk it up as a mutual misunderstanding.
8
Honestly, I think Heinz is far more consistent than humber or Harold. This is the part where I expected there to be a problem, given past posts.

Now we just need to know if he's up for discussing it.
9
Let me ask you two specific versions of the question from step 2.

In both cases, the object and the affected mass of air will have mass 2 kg. The battery starts with 10000 J (i.e. enough to power anything relevant).

 In the first version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 5 m/s.

Second version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 15 m/s.

In both versions: how much energy does the battery have at the end?


If you are talking about ddw, in the first version it is all done by battery power. There is no way to use wind power above wind speed ddw

That is correct

The end. See ya!

Please read the rest of the sentence; the word "that" specifically referred to the last sentence before I ended the quote. To be completely honest, I missed that specific sentence; my only excuse is that I was caught up in explaining the battery. I do not agree with the rest of what you are saying. I do agree that when going above wind speed and trying to speed up, the battery discharges, which is what I was trying to say there.

So again:
How much energy is in the system (object, mass of air, battery) total beforehand? How much afterwards?
10
The "song and dance" is an effective way to find where the disagreement is. For example, I disagree with your first answer, and I think that disagreement may be key.

My answer is 10000 - 12 J. Did you get your answer by only looking at the kinetic energy of the object itself for this? In other words, by looking at 121 J - 100 J?

I agree with your second answer, however.

Your first answer is wrong, then. There is no wind energy available above wind speed so it is only necessary to look at the change in ke of the object.

Good, you have the second one right. Half credit.

Now, I conclude that you cannot make an argument, like a man, in support of ddwfttw and all you can do is play little boy games. I'm not interested. You hit the jackpot on the crackpot index! Congratulations!

I have an argument that continues from my answer there. So let's discuss the difference.

How much energy is in the system (object, mass of air, battery) total beforehand? How much afterwards?
11
The "song and dance" is an effective way to find where the disagreement is. For example, I disagree with your first answer, and I think that disagreement may be key.

My answer is 10000 - 12 J. Did you get your answer by only looking at the kinetic energy of the object itself for this? In other words, by looking at 121 J - 100 J?

I agree with your second answer, however.
12
Let me ask you two specific versions of the question from step 2.

In both cases, the object and the affected mass of air will have mass 2 kg. The battery starts with 10000 J (i.e. enough to power anything relevant).

 In the first version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 5 m/s.

Second version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 15 m/s.

In both versions: how much energy does the battery have at the end?


If you are talking about ddw, in the first version it is all done by battery power. There is no way to use wind power above wind speed ddw. Of course, in your sophistry you are not providing any of the normal information needed for this problem, you know, such as the area the object presents to the wind, little details like that! So the energy in the battery is less than at the start.
That is correct; the battery discharges in this example. You should be able to come up with a specific number, however, using conservation of energy. Remember that I specified - all processes are assumed to be lossless.
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In the second version, again assuming ddw  the 15 m/s wind alone is capable of moving the object up to 11 m/s and no energy is needed from the battery. The energy in the battery is unchanged.
You are correct that the battery does not discharge; however, that doesn't leave the battery energy unchanged. The battery in this case charges - the battery can harvest energy (say, using a turbine) until the object matches the speed of the air surrounding it.
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If you cannot, or will not state a problem with all of the parameters, like a normal person, then this is game over.

This is the type of sophistry I was talking about earlier. This isn't physics.
I have provided enough information to give a unique correct numerical answer. If you want, I can demonstrate that to you, as well as showing you that your method (assuming constant acceleration, which is in fact an extra eous assumption) gives the same answer. However, I prefer for you to do it, so that I can know the relevant level of knowledge, and adjust my approach accordingly. I also want to do it to show that I'm not sneaking anything in through tricky math.
13
Politics and Current Events / J20 prosecution
Second round seems to be slowly going against the government. Thread:
14
Sports / Re: Lol fuck the nfl
It was not unanimous amongst owners.  NY Jets did not support it.
Damnit, you're right; I should know not to trust Goodell by now. From ESPN:
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Goodell said the vote was "unanimous" among owners, although San Francisco 49ers owner Jed York said he abstained. York said that all owners that voted in in the process supported the change.
Doesn't say who abstained.
15
Science / Re: The Tides ... Take 5
For anyone else reading, the speed of a water wave is normally computed ignoring friction
The problem with you two idiots is the stupid shit you claim turns up no search results at all.  Because it's only you claiming it.

"the speed of a water wave is normally computed ignoring friction"
Are you joking with this link?
16
Science / Re: The Tides ... Take 5
Your link got broken.
https://scholar.google.com/scholar?q=%22bottom+friction%22+tides
Well yours works.  Anything to say about the really obvious problem of some internet crank claiming bottom friction isn't a valid term?

Not really, since that hasn't happened.
17
Politics and Current Events / Re: Trumpocalypse
lol they gave j-kush permanent security clearance



Hands up if you're surprised.

Someone?

Anyone?
18
Let me ask you two specific versions of the question from step 2.

In both cases, the object and the affected mass of air will have mass 2 kg. The battery starts with 10000 J (i.e. enough to power anything relevant).

 In the first version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 5 m/s.

Second version:
The object starts at 10 m/s, and ends at 11 m/s. The wind moves at 15 m/s.

In both versions: how much energy does the battery have at the end?
19
Sports / Re: Lol fuck the nfl
20
Or are you saying the battery is used (discharged) in order to increase speed? That also greatly complicates matters.
That is a part of what I'm saying; the battery changes by whatever is necessary to allow a process to happen while conserving energy. Braking, speeding up (or slowing down) in the wind, etc. It's a store of energy that can be added to or subtracted from, as whatever process is happening needs.

Essentially, I'm using a physically possible method as a bookkeeping technique.
21
2 things I'd point out:

1) You probably will want subscripts on your Es. Ei and Ee.
2) I think you mixed up numerator and denominator for m/M in the energy formula.

Please read the eta before moving to the next step.
Spoiler (click to show/hide)

ETA: One second. I think I misread what you wrote. Are you saying that the amount of energy in the battery won't change?
22
Science / Re: The Tides ... Take 5
More substantively,

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After you claimed the term "bottom friction" was wrong
This description is inaccurate. Here's cold one's original post:
it's not friction that determines or limits the shallow water free wave speed.
OMG it's like you know nothing about waves at all

http://www.iupui.edu/~g115/mod11/lecture04.html

It's actually called bottom friction

You dumbass


That link is wrong, or rather, using "friction" in an inexact sense.  But explaining this to you is as pointless as explaining it to a houseplant.

For anyone else reading, the speed of a water wave is normally computed ignoring friction, and the result (which scales as the square root of the water depth when the wavelength is longer than the depth) is exact in the approximation of uniform depth, no friction, etc.  Details are here for instance https://en.wikipedia.org/wiki/Dispersion_(water_waves).

To summarize: cold one said the link used the term incorrectly.
23
Science / Re: The Tides ... Take 5
Your link got broken.
https://scholar.google.com/scholar?q=%22bottom+friction%22+tides

Punctuation like quotes need to be replaced with a % followed by their ascii number (" is %22) in queries, if I remember correctly.
24
Sports / Re: Lol fuck the nfl
im not keeping up with sports, what happend now?
New rule, everyone stands for the anthem or stays in the locker room. Apparently unanimous among the owners, which will leave them really popular among the players. Fine is to the team, not to the individual, though the team can then fine the player.
25
Sports / Lol fuck the nfl
Bets on how long before the first fine is assessed?