Rate of free fall vs Acceleration of free fall

Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.

On Tue, 16 Sep 2003 11:30:15 +0000, Donald G. Shead wrote:

Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.

".....no tedious integration is required..."

ROFLMAO!

Newton's acceleration, a = dv/dt
http://scienceworld.wolfram.com/phys...eleration.html
http://scienceworld.wolfram.com/phys...SecondLaw.html

Galileo's Falling Bodies
http://scienceworld.wolfram.com/biography/Galileo.html
http://www.ac.wwu.edu/~stephan/Anima...o.falling.html
http://hyperphysics.phy-astr.gsu.edu.../freefall.html

In article ,
Donald G. Shead wrote:
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.




The "tedious integration" is nothing more than that the distance travelled
is equal the *average* speed multiplied by elapsed time. Suppose you
start with v_i=0 and accelerate to v_f=a*t. How far have you travelled?
Should you use 0? No, you haven't been going 0 the whole time. Should
you use a*t? No, you haven't been going a*t the whole time. Take the
average, x=[(0+a*t)/2]*t, average speed multiplied by elapsed time.

If you have an initial velocity, you start at v_i and finish at v_i+a*t.
The average speed is [v_i + (v_i+a*t)]/2=v_i+a*t/2. Distance traveled
equals average speed multiplied by elapsed time,

x = (v_i + a*t/2)*t

If acceleration is not constant it can be hard to find an average speed
without calculus.

If you use the "rate" above, then rate is not equal to the change in speed
over elapsed time.

acceleration: a = (v_f - v_i) / (t_f - t_i)

rate: r = (v_f - v_i) / 2*(t_f - t_i)

Then you won't need those factors of 1/2 when you find distances because
the 1/2 is already built into the constant, but it's still there. It
certainly won't simplify things when you work problems with non-constant
acceleration, but there's no real problem with finding a solution by
calculus and then replacing every a with 2*r.

--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible

"Donald G. Shead" wrote in message
...
Galileo's.. .. ..

BS Filter Activated.

Donald "Crank of a Lesser God" Shead wrote:
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an
advantage over Newton's "Acceleration of free fall" [g = (vt-vi)/t
=32'/sec²], because no tedious integration is required to interpolate
the exact value of the distance fallen [s] at a precise instant
between vt and vi.

Surely it is now time for you to move on?

By completely failing to comprehend elementary and uncontroversial areas
of physics, you have emerged from your crank apprenticeship entirely
untainted by orthodoxy. The World of Crankdom is your oyster. Why don't
you have a crack at SR?

--
Pyriform

"Donald G. Shead" wrote:

Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.

Total gibberish. Hey ****Head, a bowling ball is dropped from 110
meters height inside a vacuum drop tower.

1) What is its velocity on impact?
2) What is the time elapsed between release and impact?

You can't do it, ****Head. You are nothing but a loud shout up your
own ass. Put up or shut up, ****Head. Let's see those Galilean
numbers. Can you do arithmetic, ****Head? Two decimal places is
adequate. Where is it, ****Head. where is it?

Hey ****Head, what is this "interpolate" crap? One puts the numbers
inside the Newtonian equation, one gets out the answer. You must have
been a real laugh, ****Head, "interpolating" a circle with a compass
for your ornamental ironmongery. Is that the way it was, ****Head?

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?" The Net!

"Donald G. Shead" wrote in message m...
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.

What about when g = lim t-0 (vt-vi)/t = GM/r^2
and is not = s/t^2 or (vt-vi)/t ?

"ghytrfvbnmju7654" wrote in message
m...
"Donald G. Shead" wrote in message
m...
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage
over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because
no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.

What about when g = lim t-0 (vt-vi)/t = GM/r^2
and is not = s/t^2 or (vt-vi)/t ?

You're talking about attraction at a distance; this applies to bodies free
falling as observed here at Earth's surface. There is no attraction at a
distance: Gravity acts according to leSage's hypothesis.

"Gregory L. Hansen" wrote in message
...
In article ,
Donald G. Shead wrote:
Galileo's "Rate of free fall" [g/2 = s/t² =16'/sec²], has an advantage
over
Newton's "Acceleration of free fall" [g = (vt-vi)/t =32'/sec²], because
no
tedious integration is required to interpolate the exact value of the
distance fallen [s] at a precise instant between vt and vi.




The "tedious integration" is nothing more than that the distance travelled
is equal the *average* speed multiplied by elapsed time. Suppose you
start with v_i=0 and accelerate to v_f=a*t. How far have you travelled?
Should you use 0? No, you haven't been going 0 the whole time. Should
you use a*t? No, you haven't been going a*t the whole time. Take the
average, x=[(0+a*t)/2]*t, average speed multiplied by elapsed time.

If you have an initial velocity, you start at v_i and finish at v_i+a*t.
The average speed is [v_i + (v_i+a*t)]/2=v_i+a*t/2. Distance traveled
equals average speed multiplied by elapsed time,

x = (v_i + a*t/2)*t

If acceleration is not constant it can be hard to find an average speed
without calculus.

If you use the "rate" above, then rate is not equal to the change in speed
over elapsed time.

That's the whole point of what I'm saying: Galileo's Rate of free fall
doesn't concern speed, or changes in velocity; it is a rate of change in
position; where the change in position varies with the square of the time:
It applies not only to free fall but to any constant change in position: So
that: s = gt²/2 = (16'/sec²)t² = at²/2!

acceleration: a = (v_f - v_i) / (t_f - t_i)

rate: r = (v_f - v_i) / 2*(t_f - t_i)

Then you won't need those factors of 1/2 when you find distances because
the 1/2 is already built into the constant, but it's still there. It
certainly won't simplify things when you work problems with non-constant
acceleration, but there's no real problem with finding a solution by
calculus and then replacing every a with 2*r.

It's an unnessary tedious repetitious waste of time is what it really is(;^!
A brilliant mind has better things to do than dinking with the calculus to
find an average value between two known values regardless of how close
together they are.
--
"When the fool walks through the street, in his lack of understanding he
calls everything foolish." -- Ecclesiastes 10:3, New American Bible