I think a "stop trying to weasel around it" applies here.

Page 4 in less then a day. Like i said. Epic.

What are you implying that I'm weasling around? DarkjediBob is one that has been doing the weasling trying to "explain away" the errors in the classical derrivation of KE.

You're the one trying to weasel around error that is removed via rounding.

Need i use big red font?
STOP THIS NONSENSE BEFORE YOU GET BANNED!

What's there to weasle around? It just goes to show how the error could so easily have been overlooked all of this time. It's also not totally surprising that the error (in most ENGINEERING applications) would be very small and often insignificant or even irrelevant.

LengthxWidth=Height

Why are you plain ignoring everyone giving you advice to drop this idiotic campaign of yours to disprove sound science?

So you're accelerating over an hour so dt=1, numbers don't change. Hell, make it dt=2, in which case dV still is 60, and dx is now 180. dx/dt still is 90, AVERAGE VELOCITY.

TE's like the little guy at the back of a crowd shouting "Yeah!" to everyone else's arguments.

Actually, scratch that, TE is that little guy.

Actually taking the mean velocity is a way of adding a Δv, so you end up with two deltas. Mathematically you can leave the Δv (and Δds) in the equation and derive ΔE[sub]k[/sub] = (1/2)mv[sub]f[/sub]^2 - (1/2)mv[sub]i[/sub]^2. I'll leave this basic algebra as an exercise for the reader.

Right: Newton's Laws, the things that are valid approximations but you believe are not valid approximations.

I know why the math works but I don't think you do.

....You mean the equation to determine the change in kinetic energy doesn't work when there is no change in kinetic energy? WELL COLOR ME SURPRISED

This is nonsense.

It is, but okay.

You don't even know Jr. High algebra.
You can separate the terms because they are separate. Δv is mathematically equal to (v[sub]f[/sub] - v[sub]i[/sub]), because that is its very definition. To directly contradict what you just said, you cannot not treat Δv as (v[sub]f[/sub] - v[sub]i[/sub]).

This is nonsense.

They?

When I was in high school - sophomore physics, actually - my teacher liked to task us with writing our own proofs for these equations. "They" didn't come up with it. I came up with it when I was 16. Apparently when I was 16 I knew more about physics and math than you do right now. I wrote this proof without any reference material whatsoever, and frankly I can't imagine how or why you would think I would need any. THIS IS JR. HIGH LEVEL ALGEBRA.

And no. The real reason I used the 'mean' velocity is to add a delta relationship.

See the first statement in this post. By not assuming the object starts at rest, you can derive the equation ΔE[sub]k[/sub] = (1/2)mv[sub]f[/sub]^2 - (1/2)mv[sub]i[/sub]^2. This is, however, mathematically equivalent and a whole lot more work.

It doesn't matter whether you assume it starts at rest or not, you can still get E[sub]k[/sub] = (1/2)mv^2.

E[sub]k[/sub] = (1/2)mv^2 has absolutely nothing to do with conservation of momentum or collisions (elastic nor inelastic), just like how E[sub]k[/sub] = (1/2)mv^2 has absolutely nothing to do with calculating the kinetic energy of a rocket while it's losing propellant mass.

God you sound retarded.

Yes, yes you tend to get v^2 when you multiply v and v. If you ever finish Jr. High you'll get to have as much fun with math as I do! Yay!

Here's that "they" again.

Do you know who the first "they" was? Sir. Isaac Newton. He held the Lucasian Chair of Mathematics at Cambridge University. He invented Calculus.

The "you" in this situation is a goddamn moron who can't even spell right half of the time.

No he most certainly does not correctly derive kinetic energy from force.

CaptBevvil, I humbly request that you go back to basics and just do the dimensional analysis on these new and exciting laws of physics that you're providing to us.

If the dimensions don't match, then your equations are wrong.

(HINT: the dimensions do not match, as has been shown at least twice in this thread [despite you conveniently ignoring it, and trying to trick us by fudging the math as well as inventing strange new forms of calculus where sometimes derivatives aren't really derivatives])

Not often irrelevant... ALWAYS irrelevent. It doesn't MATTER if there is error past the 3rd decimal place, because you do not USE those decimal places!

Since apparently everyone else has jumped on all the other things I was going to scream at,



His point was that it DOESN'T make sense, so you just kinda admitted that you're wrong.

[QUOTE=Duo Maxwell]LengthxWidth=Height[/QUOTE]
WRONG!

My Calc teacher says Length x Volume = Height, and if my calc teacher tells me this, then he is right! And my Physics teacher says that the Earth is at the center of the universe, and therefore it is the collective gravitational pull of fat people that keeps us on the ground

Jeez, where do you retards learn your physics and math?

Jon'C is enslaved by Word, he is educated singularity.

HE DESERVES DEATH - FOR SINGULARITY EVIL in the Universe of Opposites.

Your own reference site explicitly points out that .4999... = .5.

I like math. Please stop abusing it.

Holy hell, I have the most basic understanding of physics ever, and this is even making ME go all

Mathworld proves .999 = 1. This isn't even a matter of contention. This is true.

Here's my own proof from a long time ago when there was a thread about this.

Guys, I was just driving home and the speed limit was 30. But I remembered what Friend14 had been saying and I reckoned that the safe speed limit was probably calculated using broken physics so I figured it was probably safe to drive at 450 instead.

THIS IS NOT THE KIND OF ERROR THAT IS LOST IN ROUNDING.

No, you added a sum of the final and initial velocity. Delta v would be the "Change" or difference which would be vf-vi. Though, again, you could easily make the mistake (or fiddle) if you assumed vi = 0 such that vf-vi and vf+vi would both equal vf. All you did was take the average of the two velocities and then solved for d. No delta v was added.

Apparently it's: "Weasle Stomping Day.mp3"



No, apparently you THINK you know how the math works. This is nature, you can't just 'fiddle' with the numbers to suite your whim. See String Theory.



Wow, another weasle attempt. I can't say I expected anything less. Just because it's in a newtonian environment (where no external work is performed) doesn't mean that Kinetic Energy doesn't change.



If it weren't true? I would agree. Do you know what W (or correctly W[ext] or W[nc]) represents or are you just copying from an online source?



This is nature, not some random numbers. These varibles have meaning. Again, you can't just change them at a whim to suite your needs. The relationship delta v is precisely (vf - vi) and you cannot alter that relationship solely on the basis that it may or may not be mathematically allowed. BTW, in case you couldn't tell, using the mean dropped the delta and added the (1/2) and t value from vf = vi + aΔt to d = (1/2)(vi + vi + at)t.



No, you eliminated a delta relationship and added an erroneous (1/2) and t value. The (1/2) takes care of the erroneous (1/2) that's in front of the KE relationship, the elimination of the delta relationship along with the erroneous t value gives you a t^2 which is what results in the ultimate ^2 of the v value in the KE relationship.




Except that you're still using the mean which is creating the erroneous value.



Yeah, as long as you keep taking the mean.



Your absolutely right, KE has absolutely nothing to do with conservation of momentum or energy. However (1/2)mv^2 compliments the need for PE to create the conservation of momentum/energy relationship.




Sure, and while we're at it, if we wanted to determine the height a doorway should be, we can just take the mean (average height of the people who will use it) and square it. YEAH! That'll work great!



Sorry, that is incorrect. It was Mr. Leibniz who prefered the [sum] of mv^2 for the motion relationship (Kinetic Energy). While the principle may have come from Mr. Newton, he (and others) prefered the [sum] of mv or conservation of momentum. Many of the engineers of the time, however, argued that mv was 'not adequate for practical calculations' and again made use of Mr. Leibniz's principle. Then Mr. Coriolis and Mr. Poncelete recalibrated the 'Kinetic Energy to work conversion constant' to the current (1/2)[sum] of mv^2.

Do you want to rethink your position?

Why would he rethink his position when everything you've said is wrong both dimensionally and mathmatically?

They're talking about conventional notation. For intent and purposes .999... does equal 1. This isn't the same thing as the common proof which follows:

If a = b, and 3a = c, then 3b = c.

1/3 = 0.333..., and 3(1/3) = 1, then 3(0.333...) = 1.

This is wrong because 1/3 is a rational number and 0.333... is a ration approximation. The proof is thus invalid because a =! b. That is to say, Rational 1/3 =! 0.333... Rational Approximation (othen then out of conventional notation).

When you multiply a ration number (or irrational number) by a rational approximation, the error becomes more obvious. The theorims on that site help to reduce the errors in using rational approximations.

.333~ isn't an approximation. It' s a rational number--as it has no end. .33333333333333333333333 is a rational approximation.

I think you skipped high school. Actually, I'm fairly convinced you skipped most of middle school too.

I just proved it. Go look at my proof!

[edit: I mean re: .33333... ad infinitum]

All that is is 0.99999... = 1 rewritten.

Thus, you're wrong.

[QUOTE=Cool Matty]Actually, they use 3 decimal places because of the accuracy of the instruments used to measure forces and velocities and such things. More accurate instruments would allow greater decimal accuracy. So, yeah, totally wrong.

If your chronometer is accurate to 3 decimal places, then it's accurate to 3 decimal places. It doesn't matter which method you use to do the math, or what you do with the number, if you don't want to extrapolate and introduce error, you cannot exceed the 3 decimal places. Period.[/QUOTE]

Sort of right.


There are instruments that measure force, distance, etc to 4 decimal points, sometimes even 5.

I've seen a few at Holmatro in the engineering department while they were testing equipment, or doing QA. Broken tools used to come back to us and also get a thorough examination.

IIRC, I don't think I've ever used a veneer, dial, or even digital caliper that couldn't accurately measure to atleast four decimal places. Even though for more of our parts tolerances were within three.

Apples to oranges, it's your work that typically dictates how accurate you need to be.

This guy is unbelievable. His own source explicitly states that .999... = 1, yet he still argues.

No, it's incorrect because you cannot assign a ration number to a rational approsimation. That's the same thing as trying to say that 1/3 = 0.333..., which it does not. Again, 1/3 is a rational number while 0.333... is a rational approximation of that number.

This is not to say (as the site does) that conventional notation is to simply convert 0.3333... to 1/3. Just because it is the accepted notation for the number does not mean they ARE the same number indepentantly. Whenever someone uses a conventional notation, they make a note at the bottom to explain it. Such as using 1/3 instead of 0.333...

Read my post again. They ARE the same number. .333~ doesn't end, thus it is identical to 1/3.

Christ, not only did you skip math and physics, but english as well.

Oh, wait, forgot you like to ignore bits that disprove your (stupid) little "theories".

What using the average velocity does is add v[sub]f[/sub] and v[sub]i[/sub] terms, which allows us to cancel out Δv (or "acceleration").

In the case of this formula, to walk you through it again:

v = Δd / Δt.

There is no guarantee that the velocity is a constant value throughout the course of Δd / Δt. It is more incorrect to assume velocity is constant than to assume it is an average. All v = Δd / Δt tells you is how long it takes an object to move a certain distance.

v = Δd / Δt.
(1/2)(v[sub]i[/sub] + v[sub]f[/sub]) = Δd / Δt.

If an object moves 20 m/s for 1 second,

20 m/s * 1 s = 20 m.

if an object starts at rest and then accelerates to 40 m/s over the course of a single second,

(1/2)(0 + 40 m/s) = 20 m/s.

There are no guarantees. Obviously this is not accurate for variable acceleration, but in the case of this formula there is no variable acceleration. The object has a constant amount of kinetic energy. There are no forces acting on it. What we are doing is calculating the forces that previously acted on it to cause it to move at this velocity.

This is insane nonsense.

No, I know how the math works.

Kinetic energy doesn't change because mass doesn't change, velocity doesn't change, there are no forces acting on the object and there is no friction. We are also measuring instantaneous kinetic energy so none of that would matter anyway. You have no idea what you're talking about.

I don't need an external source to copy. I'm not you.

F = mv :downs:

Unfortunately you don't know what that meaning is.

You don't understand algebra. One of the first rules is that you can only ever add or subtract zero, or multiply or divide by one. Even if you split up (v[sub]f[/sub] - v[sub]i[/sub]) the relationship still exists.

I didn't eliminate anything.

There is no erroneous value because the mean velocity is accurate. The object is not actually accelerating - it has a fixed, constant velocity. The mean is added to eliminate the acceleration term (which is zero anyway).

Um... okay, glad we agree?

You don't understand algebra and you're too retarded to see where the second velocity term is coming from.

Wikipedia is not "just as good" as a physics degree. :downs:

No, I still think you should stop posting on this forum.

Wrong.

0.3333 (non-repeating) is rational approximation.
0.3333... (repeating) is rational.




Oh, and just because I can:

http://en.wikipedia.org/wiki/0.999...#Fraction_proof

That is your "proof", except it's proving 0.999... = 1. So, you lose. Your own proof failed you.

Does this doofus even bother reading his own links?

Rational Approximation: "If alpha is any number and m and n are integers, then there is a rational number m/n for which |alpha - m/n| <= 1/n."

The non-retarded among us might notice that Friend14's own link states that a rational approximation is a rational number by definition. That's why it's called a rational approximation; it's a rational number that approximates a real. Duh.

People like you is why I wish calculus physics was mandatory. All of your b'veking over conceptual difficulties and unaccounted for errors is stemming from stopping your research into physics at the linear algebra stage. When you actually deal with the calculus, you realize that all of the equations posted so far are due to special conditions of far more general equations that ARE correct.

Until you, yourself, can disprove the very basics of calculus, that integral from 0 to t of (x dt) is not x*t, then I will personally be ignoring every single post you write, as those basics are what form the foundation of calculus physics FROM WHICH your linear algebra physics stem from.

Stop playing with kiddie physics in kiddie terms and be a man. Provide a calculus proof.

(Speaking of, kudos Pommy on your proof. Bout time we got some series and sequences in here.)

you know, I still haven't seen a single mathematical "proof" of your crackpot theories, and I highly doubt that they exist. We (meaning everyone except you, and the entire mainstream scientific community) have done the proofs and then actually plugged in numbers to verify that. I want to see an actual proof for this garbage, then plug in numbers to verify. But, of course I know you won't, because then you'd prove how much of a fool you're making yourself to be.

Dang, I leave for a day and see what I miss?



Sounds to me like somebody doesn't quite understand the concept of infinity.

Wow. And I thought politics and religion got people riled up.

[QUOTE=Kieran Horn]Wow. And I thought politics and religion got people riled up. :eek:[/QUOTE]
Destroying fact with false replacement is more insulting to a scientist than any insult that can be dredged up in religion and politics, because fact is FACT.