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Me-262, NOT Bell X-1 Broke SB First



 
 
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  #1  
Old October 8th 03, 05:05 PM
Ed Rasimus
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On Wed, 8 Oct 2003 08:00:01 -0700, "Tarver Engineering"
wrote:

"Ed Rasimus" wrote in message
As I recall, the first integral of velocity is acceleration.


Nope.


Ahh, now I see. Thanks for that typically helpful addition to the
thread. Enlightenment can come in such small and pithy comments.


  #2  
Old October 8th 03, 06:45 PM
Tarver Engineering
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"Ed Rasimus" wrote in message
...
On Wed, 8 Oct 2003 08:00:01 -0700, "Tarver Engineering"
wrote:

"Ed Rasimus" wrote in message
As I recall, the first integral of velocity is acceleration.


Nope.


Ahh, now I see. Thanks for that typically helpful addition to the
thread. Enlightenment can come in such small and pithy comments.


Integral A dt = V0 + At =V


  #3  
Old October 8th 03, 06:58 PM
John Mullen
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"Tarver Engineering" wrote in message
...

"Ed Rasimus" wrote in message
...
On Wed, 8 Oct 2003 08:00:01 -0700, "Tarver Engineering"
wrote:

"Ed Rasimus" wrote in message
As I recall, the first integral of velocity is acceleration.

Nope.


Ahh, now I see. Thanks for that typically helpful addition to the
thread. Enlightenment can come in such small and pithy comments.


I hate to say it Ed but for once Tarver is right. The first *differential*
of velocity is acceleration. The first integral would be distance covered. I
understood perfectly what you meant though and envy you that experience.

Integral A dt = V0 + At =V


I've no idea what he means by this though!

John


  #4  
Old October 8th 03, 07:03 PM
Tarver Engineering
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"John Mullen" wrote in message
...
"Tarver Engineering" wrote in message
...

"Ed Rasimus" wrote in message
...
On Wed, 8 Oct 2003 08:00:01 -0700, "Tarver Engineering"
wrote:

"Ed Rasimus" wrote in message
As I recall, the first integral of velocity is acceleration.

Nope.


Ahh, now I see. Thanks for that typically helpful addition to the
thread. Enlightenment can come in such small and pithy comments.


I hate to say it Ed but for once Tarver is right. The first *differential*
of velocity is acceleration. The first integral would be distance covered.

I
understood perfectly what you meant though and envy you that experience.

Integral A dt = V0 + At =V


I've no idea what he means by this though!


It is the integral form of one of Newton's laws.


  #5  
Old October 8th 03, 07:23 PM
Ed Rasimus
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On Wed, 8 Oct 2003 18:58:04 +0100, "John Mullen" wrote:

"Tarver Engineering" wrote in message
...

"Ed Rasimus" wrote in message
As I recall, the first integral of velocity is acceleration.

Nope.

I hate to say it Ed but for once Tarver is right. The first *differential*
of velocity is acceleration. The first integral would be distance covered. I
understood perfectly what you meant though and envy you that experience.


Well, the disclaimer at the beginning of my post should cover me. I
knew that the relationship between velocity, acceleration and rate of
change of acceleration went one way or the other. It was either the
first and second integral or the first and second differential. It was
differential equations at the end of my fourth semester as a chemistry
major, coupled with semi-micro qualitative analysis and physical
chemistry, that led me to see the futility of ever succeeding with the
pocket protector crowd.

I changed major to political science with the singular goal of gaining
a degree in "anything" so that I could get on with entering the AF and
flying jets.

If we change the "integral" to "differential" I'm sure that John will
recognize and acknowledge my point about aircraft accelerating through
the mach vertically.




  #6  
Old October 8th 03, 07:39 PM
Tarver Engineering
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"Ed Rasimus" wrote in message
...
On Wed, 8 Oct 2003 18:58:04 +0100, "John Mullen" wrote:

"Tarver Engineering" wrote in message
...

"Ed Rasimus" wrote in message
As I recall, the first integral of velocity is acceleration.

Nope.

I hate to say it Ed but for once Tarver is right. The first

*differential*
of velocity is acceleration. The first integral would be distance

covered. I
understood perfectly what you meant though and envy you that experience.


Well, the disclaimer at the beginning of my post should cover me. I
knew that the relationship between velocity, acceleration and rate of
change of acceleration went one way or the other.


It goes both ways:

Integral a dt = V + V0
dV/dt = a

But of course, Ed knows his airplane operating.


  #7  
Old October 8th 03, 09:14 PM
B2431
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I hate to say it Ed but for once Tarver is right. The first *differential*
of velocity is acceleration. The first integral would be distance covered. I
understood perfectly what you meant though and envy you that experience.

Integral A dt = V0 + At =V


I've no idea what he means by this though!

John

It's basic calculus. Try this one in English units: if you drop an object the
function for determining how far it has fallen is X=16T^2, where X is the
distance travelled in feet and T is the time in seconds. The first derivative
is V = 32T where V is instantaneous velocity expressed in feet per second. The
first derivative of V, and the second of X, is A = 32 feet/second/second which
is the acceleration due to gravity. Integration is the reverse process. This
function doesn't take into account drag, but if you drop a bowling ball from
the top of a 10 story building drag is negligible.


Dan, U. S. Air Force, retired
  #8  
Old October 8th 03, 10:05 PM
John Halliwell
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In article , B2431
writes
It's basic calculus.


I'm not too sure if 'basic' and 'calculus' sit too well together

Try this one in English units: if you drop an object the
function for determining how far it has fallen is X=16T^2, where X is the
distance travelled in feet and T is the time in seconds. The first derivative
is V = 32T where V is instantaneous velocity expressed in feet per second. The
first derivative of V, and the second of X, is A = 32 feet/second/second which
is the acceleration due to gravity. Integration is the reverse process. This
function doesn't take into account drag, but if you drop a bowling ball from
the top of a 10 story building drag is negligible.


The best bit I liked was deriving the equations of motion from the three
basic dimensions 'L', 'M' & 'T' (distance, mass, time).

--
John
  #9  
Old October 9th 03, 12:27 AM
B2431
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From: John Halliwell

B2431

writes
It's basic calculus.


I'm not too sure if 'basic' and 'calculus' sit too well together

Compared to what I learned, and have since forgotten in calc 2 and 3 it is
basic. If you want to see a course that makes sanity seem like an illusion try
one in imagionary variables.

Algebra is when you stop counting on your fingers and start using your toes.
Calculus is when you tie those toes in knots. Differential equations is when
you start learning you are now different..........etc.

Dan, U. S. Air Force, retired




 




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