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How would a 40 degree impact affect his calculation?
Dan has done precisely that.
You win a prize.
He will keep doing everything except answering the questions posed to him.
If one were intellectually honest, answering such questions would be expected in order to get at the best theory about the events of the day.
Conclusion:
Dan is not intellectually honest about this subject, and is not interested in the best theory.
All Dan has to do, in order to make that charge fall on its face is to answer the questions given.
Dan will not.
Jim will not.
There will be a change in subject, as some other factually flawed innuendo will be made.
It's like clockwork, and boringly predictable.
How would a 40 degree impact affect his calculation?
Yes Dan......inquiring minds want to know. Would you care to enlighten the masses?
Yet right on point.......
As soon as RG answers my question about how an (admitted inaccurate) KE 14 times of a 'typical crash" (whatever that is) would effect an airplane crashing to the ground and proves it..I'll be happy to answer your question...until then STFU....
By the way, where have you been...its been awhile since you've been slapped around in this forum...
Sorry, you've been dodging questions this entire thread.
Coward.
I'm not the one making conjectures about airplane crashes...show me the proof
Your a troll..pussy
Why so butt-hurt? your math is simple but your assumptions are mountainous.1 kilogram traveling at 563 miles an hour:
31672.297119795 joule
1 kilogram traveling at 150 miles an hour
kinetic energy (K) = 2248.253568 joules
31672/2248 = 14 and some change.
The mass is the same, so that factors out, being a math major, you should understand that. ditto for the units.
So, your fancy calculator came up with the exact same proportion that I arrived at earlier.
You lose your spelling ability when you lose your composure.
You can't say what you believe.
You have no beliefs.
You didn't email Jim Fetzer.
lol
...and you're still a troll....
Stick with the 9/11 commission report troll...
Keep dodging, coward.
You know United 93 crashed outside of Shanksville. You don't even have an alternative theory.
Did it? Did you personally see it crash?
A real look over the pdf's for UA93 FDR via NTSB website shows UA93 at 40 degrees pitch down and -5 degrees Angle of Attack at impact.
If that was the case, no matter the math, UA93 would be making a long ditch along the flight path while impacting the ground and spreading wreckage at close to a 40 degree pitch down angle. The pictures you posted of the UAL93 impact crater shows a vertical, straight down crater (~90 degrees vertical), not a 40 degree impact with a long ditch spreading wreckage as the FDR suggests.....
The pitch angle would create a horizontal wreckage crater...
It's nowhere close to a vertical impact. Dirt piled up on one side of the impact and the debris was strewn in the wreckage vector you show above.
lol horizontal crater
dan, if you held a gun to your head at a 40-degree angle and pulled the trigger, you would expect the bullet to just part your hair.
"No wreckage was found in direction of wreckage vector"
No, it would not.Did it? Did you personally see it crash?
A real look over the pdf's for UA93 FDR via NTSB website shows UA93 at 40 degrees pitch down and -5 degrees Angle of Attack at impact.
If that was the case, no matter the math, UA93 would be making a long ditch along the flight path while impacting the ground and spreading wreckage at close to a 40 degree pitch down angle. The pictures you posted of the UAL93 impact crater shows a vertical, straight down crater (~90 degrees vertical), not a 40 degree impact with a long ditch spreading wreckage as the FDR suggests.....
The pitch angle would create a horizontal wreckage crater...
It would make the exact shaped crater that it did, and scatter wreckage exactly the way it was found.
Man, that was easy.
Dan has done precisely that.
You win a prize.
He will keep doing everything except answering the questions posed to him.
If one were intellectually honest, answering such questions would be expected in order to get at the best theory about the events of the day.
Conclusion:
Dan is not intellectually honest about this subject, and is not interested in the best theory.
All Dan has to do, in order to make that charge fall on its face is to answer the questions given.
Dan will not.
Jim will not.
There will be a change in subject, as some other factually flawed innuendo will be made.
It's like clockwork, and boringly predictable.
What assumptions would those be?
That the equation for Kinetic energy represents reality?
That experts on plane crashes tell the truth about plane crashes?
"bona fide" crash sites.
Plenty of bona fide air crashes happen. When they happen, generally the pilots are trying not to crash them, and are flying fairly slow on take off or landing.
Fairly slow in this case means roughly 200 miles an hour or so, sometimes much less:
http://www.airliners.net/aviation-fo...d.main/236221/
150mph seems about right
My avatar is the equation for kinetic energy.
If you read it and can understand the basic algebra involved, you will note that when you double the speed, you have multiplied the force/energy of motion by a factor of four.
The data recorders said the plane hit the ground at a steep angle at 563 mph.
Let's do some math.
Since we are examining the same object's kinetic energy for two different speeds, we can simplify things somewhat, and set the mass to 1. We can also simply by not worrying about specific units, since we are looking at relative energies.
This means we can set the kinetic energy equial to 1/2 v^2
Kinetic energy of a low-speed crash:
.5*150*150= 11,250 units.
Kinetic energy of Flight 93 at impact:
.5*563*563= 158,484 units
158484/11250= 14 and some change.
Something moving 3.75 times faster has 14 times the amount of kinetic energy.
Compare this 30 mph crash:
With this 120 mph crash:
Do I need to start explaining the physics of direct versus glancing blows?
Do you understand this basic element of physics?
Given this rather solid, and easily verifiable experiment, would one expect the exact same wreckage pattern from a crash at 150-200 mph as one would see in a crash at 563 mph?
A simple yes or no will do.
If you are comparing this crash to other crashes, are your comparison crashes also at 500+ mph?
...that wasn't easy...that was intellectually lazy... your the one comparing a vertical car crash to a crash with a 40,-5 pitch angle and then your not telling us what would happen to the plane.....
At that speed into the relatively soft soil of a reclaimed strip mine, most of the plane would go into the earth.
Just like it did in the case of United 93.
Now dan has to think of another way to stall.
.Did it? Let's take a look a wreckage from other plane crashes at or near the speed of flight of United 93..
Trans-Canada Airlines DC-8
What is interesting about this flight is that it was going at or around +400mph and it crashed into a swamp...the swamp was drained the following spring and they built a cofferdam around the site to prevent the dirt walls from collapsing. Then they dug down to recover all of the wreckage. Also, there wereabout 6 tons of debris recovered from the bush around the crash site in the spring after the snow had melted and the ground thawed....The crash landing of the plane created a crater about 45 metres (150 feet) long by 22.5 metres (75 feet) wide. The wreckage covered an area about 800 metres (half a mile) long and 75 metres (250 feet) wide. Investigation efforts were severely hampered because the area was primarily mud and swamp due to prior heavy rains
.
Last edited by Nbadan; 05-17-2012 at 11:11 PM.
1999 South Dakota Learjet crash
Let's take a look at the remains of a plane which crashed at a far higher speed...supersonic ~768 miles per hour
Wreckage
http://en.wikipedia.org/wiki/1999_So..._Learjet_crashwith the aircraft hitting the ground at a nearly supersonic speed and an extreme angle.[3] The Learjet crashed just outside of Mina, South Dakota, in Edmunds County on relatively flat ground, and left a crater 42 feet (13 m) long, 21 feet (6.4 m) wide and 8 feet (2.4 m) deep.
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