If it doesn't agree with experiment, it's wrong. Part II

Do you mean the moderator thread?

I could care less. I have already said that we are adults. Insult me all you want. Moderators on a forum such as this are unnecessary.

Fair enough.

You had better let criteria know that you can handle yourself.
 
If something is being acted upon by forces in opposition to each other and is falling, it is not freely falling, even though it may be falling at g, the acceleration we label freefall.
I feel that I should have been charged admission to witness that convolution of reality.
 
In effect, what you are really saying is that in your opinion FalseFlag has an IQ level well below that of your dog. ...
Criteria presents the dumbest experiments in the history of 9/11 truth.


But it is true, FF failed to figure out the simple part of the question. Is the lack of practical knowledge in physics why you and FF believe in the fantasy of CD?

FF offers no evidence like you do. He claims to have evidence for CD, never produces it; like you. Believing in the fantasy of CD, or thinking Balsamo is an aviation authority is par for 9/11 truth followers who believe in lies and false claims; and have no evidence to much about their failed positions.

Is it frustrating to fall for lies and have no clue the lies you believe in are lies?

Where is the evidence?
Not really. For someone who believes the Fantasy of CD and thinks Balsamo is an avaiation authority...

Have you helped FF with 7.8*1012 fathoms/fortnight2 ; instead of quibbling about BS -
 
Last edited:
I feel that I should have been charged admission to witness that convolution of reality.

Fair enough, I was charged admission, only they called it tuition, and at the end I got bachelor's in Physics and Astrophysics, and a Masters in Mechanical Engineering.

Do you always call what you don't understand a "convolution of reality ", because that explains a lot about you then.
 
So tell me, can an object only a he I've an acceleration of 9.81 m/s2 toward the center of the Earth by falling freely, or could it also achieve the same acceleration by a combination of forces?
Are you typing on your phone?

A combination of forces could create a downward acceleration of 9.8 m/s/s. I think for purposes of discussion we refer to freefall as downward acceleration of 9.8 m/s/s caused solely by gravity (neglecting air resitance for simplicity in discussion).

I could hold a pen in my left hand and a pen in my right hand. I could drop the pen in my left hand and (with much skill) lower the pen in my right hand at the same rate as the pen in freefall. Is this what you are referring to? If so, I would never say the pen in my right hand is in freefall. I would say I was manually accelerating it downwards at 9.8 m/s/s.

If this is wrong, then please explain in detail what you are referring to.
 
Last edited:
If something is being acted upon by forces in opposition to each other and is falling, it is not freely falling, even though it may be falling at g, the acceleration we label freefall.
OK. I owe you an apology. I understand what you are saying.

Normally I don't give any credit to what a skeptic says because it's just BS by default. I think your statement, taken just by itself, is correct.

Not that it matters, but you do have my attention. Please continue your argument if you want to. I will listen.
 
Pretty damn close.
I did a quick-and-dirty on it, looks pretty good, but quite frankly, I have a P-51D, a Spitfire Mk-II, and a Dirty Birdy that need my time and attention, and which will appreciate it a lot more than the obstinately ignorant do.

You mother - f*****...!!

Are you kidding me??!!!
A P-51D??
A Spitfire??

Ohhhhhh, now I hate your miserable azz..!!



("hate" being a synonym for "jealous as hell", of course...)
 
What is the value of acceleration at the ??s?

If you are incompetent at physics, you won't supply an answer.

You know what's really funny about this?

From the context, the units in use, and the details of the graph, it's possible to guess the answer; in fact, it's almost impossible to guess it wrong.

If you know anything about physics, that is.

Dave
 
You know what's really funny about this?

From the context, the units in use, and the details of the graph, it's possible to guess the answer; in fact, it's almost impossible to guess it wrong.

If you know anything about physics, that is.

Dave

Seing three horizontal lines, I was immediately (within 3 seconds) confused, thinking "does the NIST really use "feet", aren't they generally on the SI train already?". Because three horizontal lines only make good sense if the scale is xx ft/m2. Took me about 15 seconds more till I scrolled up and saw a graph confirming they indeed use ft.

Hence my immediate guess was correct: ?? = XX xx/s2.
No need to waste time with Excel or Open Office or whatever.
If you know anything about physics, that is.

Edit: OOPS sorry I thought the answer had already been derived and given three different ways by others :D
Ok, result is of course 3.585 * 1021 mil/a2
Edit 2: Took me three attempts to get the result right in the other dimension :blush:
 
Last edited:
...
Hence my immediate guess was correct: ?? = XX xx/s2.
No need to waste time with Excel or Open Office or whatever.
If you know anything about physics, that is.

Oystein, you got it right! (I saw your value this morning before you redacted it.)

Thanks for the numerous confirmations, all!

It wouldn't be a waste of time at all. You can PROVE you understand physics simply by using popular programs like Microsoft Excel (or Office Libre Calc, available for free) to compute the 1st and 2nd derivatives of the NIST equation posted by TFK.

I was able to do so - a little tricksy, but extremely doable. Here is my result for acceleration, sans the vertical values.

wtc7-accel.gif


All you have to do to prove some physics competency is to reproduce this acceleration graph from NIST's equation. I'm giving you a big boost by showing you the shape of the acceleration curve (you can check your work against this).

What is the value of acceleration at the ??s?

If you are incompetent at physics, you won't supply an answer.

No. At best, it means you can google online calculators.

Why don't you post a link to the website you used, and paste the exact equation you used. Post screenshots to show each step. Then, copy and paste a picture of the graph of the equation.

Why won't you post your data?

Why won't you post your steps?

If you were so certain that you were correct, you would want to make sure as many people as possible could review your work and get the same results.

Well, at least that is how it should work, but you're a skeptic, and obfuscation is what you thrive on.

Here's a hint, FalseFlag:

To answer the questions posed by tfk and me, you would have had to understand that velocity is the time rate of change of position, and that acceleration is the time rate of change of velocity.

It's now obvious to everyone that you do NOT understand “middle school physics”.

Game over, dude.
His answer is correct.
I did this same calculation, in a completely different manner, many years ago.

His graph is the same as mine.

Waiting for glue to cure, paint to dry
Used Mathcad, got this:
IMAGE http://www.internationalskeptics.com/forums/imagehosting/thum_1069574c65e930517.jpg

Ain't it amazing--people use 4 different methods, all get the same answer.
Science and Mathematics: They actually work. every time.

You know what's really funny about this?

From the context, the units in use, and the details of the graph, it's possible to guess the answer; in fact, it's almost impossible to guess it wrong.

If you know anything about physics, that is.

Dave

Seing three horizontal lines, I was immediately (within 3 seconds) confused, thinking "does the NIST really use "feet", aren't they generally on the SI train already?". Because three horizontal lines only make good sense if the scale is xx ft/m2. Took me about 15 seconds more till I scrolled up and saw a graph confirming they indeed use ft.

Hence my immediate guess was correct: ?? = XX xx/s2.
No need to waste time with Excel or Open Office or whatever.
If you know anything about physics, that is.


Well, here's how I got it done. I could have used analytical derivatives, but that NIST equation was a little hairy, so I used numerical derivatives.

You can see that velocity is the time rate of change of position (cell E6), and that acceleration is the time rate of change of velocity (cell F8)
wtc7-accel-how.gif


I can say with confidence that FalseFlag's incompetence at physics is a proven, well-documented fact.

#FalseFlagCluelessAtPhysics
 
I can say with confidence that FalseFlag's incompetence at physics is a proven, well-documented fact.

#FalseFlagCluelessAtPhysics

Even I, as someone clueless about physics, can tell he's clueless about physics.
 
FF Maybe clueless about physics, but he makes a bloody good troll.

I can see why he has so many Lolz
 

Back
Top Bottom