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study & discuss their (great?) work, energy

WE=m*a*d
using linear_acceleration a (=v/t, =2*((vi/t)+(d/(t*t)), =F/m)
& the force(d) distance d (=va*t)
of a mass m (=F/a).

That (linear accelerated) force is F=m*a,
where the average (accelerated) speed (velocity) va (=(vi+vf)/2),
& speed (velocity, difference) is v=vf-vi,
for final_speed vf,
minus initial_speed vi.
((Even) although it would also be possible
to use factoring (an initial unit_speed of 1 m/s), instead).

The (moving) kinetic_energy is
KE=m*v*va, (pronounced key), or
KE=m*((v^2)/2)+v*vi) using initial_speed vi, or
KE=m*((v^2)/2)-v*vf) with (respect to) final_speed vf.

The potential_energy
PE=m*g*h (pronounced pea, as in pee)
is a mass m
g accelerated (fallen,
multiplied by)
height h=d distance.
PE=Wt*h
is (the force F=) weight Wt (=m*g)
multiplied by height h.
(No distance fallen_able, is no potential_energy.)

Equating (both energies)
PE=KE (pronounced peek)
m*g*h=m*v*va
without mass (divided from both sides) is
g*h=v*va
is only linear_acceleration
a*d=v*va
a*d=(vf-vi)*((vi+vf)/2)
of the distance d,
a*d=((vf^2)-(vi^2))/2, *2

That's standard mechanics.

(Anything wrong there?)

Swapped sides
(vf^2)-(vi^2)=a*d*2, +(vi^2)
(vf^2)=(vi^2)+a*d*2, ^0.5
vf=((vi^2)+a*d*2)^0.5
is the final_speed (velocity)
(of linear_acceleration).

The speed difference (velocity)
v=vf-vi, vf=((a*d*2+(vi^2))^0.5)
v=((a*d*2+(vi^2))^0.5)-vi
is final_speed (velocity) vf,
minus the initial_speed (velocity) vi.

Momentum('s impulse)
mom=m*v
is mass m
multiplied by (the accelerated) speed difference (velocity) v.
mom=m*(((a*d*2+(vi^2))^0.5)-vi)
mom=m*((a*d*2+(vi^2))^0.5)-m*vi
is the final_momentum momf=m*vf
(pronounced mumf, as in eating fast),
minus the the initial_momentum
momi=m*vi (pronounced mommy).

The average momentum
moma=m*va (pronounced mama)
is the mass m
multiplied by the average (accelerated) speed (velocity) va=(vi+va)/2.

Any questions?

Since KE & mom
only use mass & speeds,
& all energy can be equated to KE,
then it seems imaginable
to equate all energies into impulse(s)
of mom=F*t.

Mom=E/va.

Mother nature (pronounce Eva (the German Eve);
or else Elva
from James Cameron's Avatar).

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25 minutes ago, Capiert said:

Mom=E/va.

In different languages momentum has different name, velocity has different name, energy has different name. And they won't match your "numerology" mom=eva anymore..

German momentum is "Impuls", velocity is "Geschwindigkeit".

Common symbol of momentum is not Mom, but p. You altered it to match your hypothesis.

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5 hours ago, Capiert said:

Mom=E/va

You mother's name is Eva?

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5 hours ago, Capiert said:

(Anything wrong there?)

Your choice of type face, your use of brackets and the whole idea.

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5 hours ago, Capiert said:

study & discuss their (great?) work, energy

Who does “their” refer to?

I couldn’t get any further. It is totally illegible on my display.

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6 hours ago, Capiert said:

KE=m*v*va, (pronounced key), or

KE=m*((v^2)/2)+v*vi) using initial_speed vi, or
KE=m*((v^2)/2)-v*vf) with (respect to) final_speed vf.

KE is 1/2 mv^2 where v is the speed of the object. v is not a speed difference. It is the speed of the object. if you want to use a speed difference, the proper terminology would be ∆v. But why you would want such an equation, with more terms in it, is beyond me. Furthermore, your equation was derived for something that has undergone a constant acceleration. It does not universally apply.

Making up your own terminology and using symbols that already have a meaning in physics only adds to the confusion you create.

Gave up.

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• 1 year later...
On ‎2018‎ ‎09‎ ‎30 at 3:13 PM, swansont said:

..
v is not a speed difference. It is the speed of the object.

Hi Swansont

I'm sorry
but I can't follow you there.

Isn't speed (always) relative
(to another speed),
thus a difference
(in speed)?

Or are you insisting on the (non_linear, complicated) Fitzgerald_Lorentz transform
(e.g. to watch cars' speed, v=100 m/s (wrt) on earth))?

E.g. 2 cars
each with the same speed
as the other (wrt earth),
appear at rest (=0 speed)
(wrt each other (car)).

Or are you suggesting
I should multiply the basic (speed) unit [1 m/s]
by the number (value, e.g. 100)
as a product
(instead of (subtract for a) difference).
Which I don't find
to get rid of (some of) the math problems (complexity).

I'd just have to get used to it (=the multiplication technique), instead.

Quote

if you want to use a speed difference, the proper terminology would be ∆v.

Quote

But why you would want such an equation, with more terms in it, is beyond me.

I don't (really) want complexity,
I want simplicity (& solutions), instead
(but NOT at the cost of precision, as (severe) errors).

Quote

Furthermore, your equation was derived for something that has undergone a constant acceleration.

I thought that was understood,
that everybody knows that.
But I guess some things must be said anyway
(just to be sure).
The equation
is simply derived
from the (linear) 1st order of acceleration.

High school physics.

(Non_linear acceleration does make me curious, though.)

Quote

It does not universally apply.

Making up your own terminology and using symbols that already have a meaning in physics only adds to the confusion you create.

I doubt
that I would stubble into
as many (of your physics) pitfalls
if I used identical syntax.
My alternative syntax
allows me to compare your results.

But (I suspect) you don't recognize the drawbacks.
I don't prefer the (foreign) greek alphabet.
"It's all greek to me!?"
(Not English.)
(The cool thing about physics is how it simplifies,
but the nasty thing is how it encrypts
(into only 1 possibility,
when others are possible).
Physics can be done in any language,
with or without another language.
So those languages, or rules, conventions are NOT physics.

I think the biggest confusion you physicists
must (eventually) deal with
is (dark) energy
because it is NOT Newtonian.
You ignore the initial_speed vi
too much,
setting it to zero.
Relativity clearly indicates motion
wrt other frames.

Could it be an (invisible) common initial_speed
vi=(KE/mom)-(v/2)
(which you deny)
is at least partly responsible
for your (unknown) dark energy (ERROR)?
(Not to mention speed's (incorrect) exponential (=non_linear) proportionality to mass.
Something KE (definitely) defies.
How can you possibly brush_off that (severe) incompatibility so lightly?
I thought you were reasonable people.
To do things right you would need mass_squared (instead of only mass), in energy equations.

I don't know what else to call it (energy as error)
because it (dark energy) makes no_sense
according to your (standard) college educations.

Or does it?

Astronomers are complaining to you on a cosmic scale,
while little old me attempts to deal with (your energy math (incompatibility) problems) on a small earthly scale.

Edited by Capiert
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18 minutes ago, Capiert said:

Hi Swansont

I'm sorry
but I can't follow you there.

Isn't speed (always) relative
(to another speed),
thus a difference
(in speed)?

You defined v as a speed difference (vf-vi) and I was pointing out that that’s not what goes into the KE equation.

Quote

I don't (really) want complexity,
I want simplicity (& solutions), instead
(but NOT at the cost of precision, as (severe) errors).

Then don’t make up new definitions for terminology, or extraneous terminology.

Quote

I thought that was understood,
that everybody knows that.

Your set of equations are not widely used, so that’s not a reasonable conclusion.

The accepted equation for KE does not suffer from the limitation of only working for constant acceleration

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27 minutes ago, swansont said:

You defined v as a speed difference (vf-vi) and I was pointing out that that’s not what goes into the KE equation.

Then I suppose your syntax v
is my syntax vf.

Is that true?

Quote

Then don’t make up new definitions for terminology, or extraneous terminology.

Just attempting to be thorough(ly defined),
which is NOT what I could claim everybody does.

Capiert:
I thought that (linear acceleration) was understood,
that everybody knows that.

It's clearly stated in the 3rd line (at the intro).

Quote

Your set of equations are not widely used, so that’s not a reasonable conclusion.

Surely you have overseen my definition.?

Quote

The accepted equation for KE does not suffer from the limitation of only working for constant acceleration

I gave you a specific example: the linear acceleration.
But now you have made me curious.
Please tell me more, e.g. an example (or more), for further (kinds of acceleration).

Edited by Capiert
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10 minutes ago, Capiert said:

Then I suppose your syntax v
is my syntax vf.

Is that true?

yes.

10 minutes ago, Capiert said:

Just attempting to be thorough(ly defined),
which is NOT what I could claim everybody does.

Capiert:
I thought that (linear acceleration) was understood,
that everybody knows that.

It's clearly stated in the 3rd line (at the intro).

Linear acceleration is not understood for KE = 1/2mv^2, since it is not assumed.

Your equation only works under that very specific scenario.

10 minutes ago, Capiert said:

Surely you have overseen my definition.?

I gave you a specific example: the linear acceleration.
But now you have made me curious.
Please tell me more, e.g. an example (or more), for further (kinds of acceleration).

Any acceleration that isn’t constant (a better description than linear). Which is probably most cases. A car, for example. Either speeding up or slowing down. Engine performance depends on the engine rate. Air resistance is a function of speed.

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6 minutes ago, swansont said:

yes.

Linear acceleration is not understood for KE = 1/2mv^2, since it is not assumed.

Your equation only works under that very specific scenario.

Any acceleration that isn’t constant (a better description than linear). Which is probably most cases. A car, for example. Either speeding up or slowing down. Engine performance depends on the engine rate. Air resistance is a function of speed.

(Yes) but please give me a formula example
so I can get feel for the number_values.

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47 minutes ago, Capiert said:

(Yes) but please give me a formula example
so I can get feel for the number_values.

You should be able to come up with values on your own.

There may not be a formula, or it might be complicated, which are reasons to find/use a solution that doesn’t depend on the details of the acceleration.

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15 minutes ago, swansont said:

You should be able to come up with values on your own.

There may not be a formula, or it might be complicated, which are reasons to find/use a solution that doesn’t depend on the details of the acceleration.

I want a reliable formula
that I can equate to
for comparisons.

I can make up anything I want,
but it doesn't mean it's accurate.

Your team surely has better experience there,
for real measurements.

(I'm just guessing.)

Edited by Capiert
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1 minute ago, Capiert said:

I want a reliable formula
that I can equate to
for comparisons.

I can make up anything I want,
but it doesn't mean it's accurate.

Your team surely has better experience there,
for real measurements.

(I'm just guessing.)

Why would I have data on car acceleration?

What would be the point of deriving a formula for each instance? I thought your goal was simplicity. Having dozens (or more) of formulas isn’t simplifying anything.

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1 minute ago, swansont said:

Why would I have data on car acceleration?

It does NOT have to be a car.
Surely simpler contraptions exist
to test.

1 minute ago, swansont said:

What would be the point of deriving a formula for each instance?

Confirmation.

1 minute ago, swansont said:

I thought your goal was simplicity.

Yes (& reliability).

1 minute ago, swansont said:

Having dozens (or more) of formulas isn’t simplifying anything.

I'm interested
in at least 1 formula
I could rely on.

& fine_tune
to a general formula
if possible.

How many things did Newton
(have to) go thru
til he could (finally) settle on F=m*a?

"The proof
is in the pudding."

I'm surprised
this non_linear acceleration (formula, stuff)
is so rare
(for something (that is) (supposed to be) so universal).

Do you tell students
to learn (useless) linear_acceleration
because it's NOT universal?

(Surely) I hope NOT.
But that's what's taught.

Free_fall's ("linear" acceleration)
is pretty common
throughout the whole universe.

So linear acceleration
is NOT so uncommon.

Even if you don't dare to call it universal.

I'm just curious
is proportioned to linear_acceleration,
& the methods of confirmation.

Is my curiousity justified,
in a dark energy dilemma?

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1 hour ago, Capiert said:

I can make up anything I want,
but it doesn't mean it's accurate.

Exactly.
And that's the problem.

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4 minutes ago, MigL said:

Exactly.
And that's the problem.

I'm glad you see it too.
Thanks.

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1 hour ago, Capiert said:

.I'm interested
in at least 1 formula
I could rely on.

You can’t rely on it if you don’t know the acceleration as a function of time.

Quote

& fine_tune
to a general formula
if possible.

We already have a general formula.

Any general result you get will not, and cannot, be different.

Quote

I'm surprised
this non_linear acceleration (formula, stuff)
is so rare
(for something (that is) (supposed to be) so universal).

Who said it’s rare?

Quote

Do you tell students
to learn (useless) linear_acceleration
because it's NOT universal?

We didn’t use constant acceleration to derive KE.

We did use constant acceleration because those problems were easiest to solve, which is what you do for introductory physics. Later, you spend a lot of time learning how to solve the non-trivial cases

Quote

Free_fall's ("linear" acceleration)

is pretty common
throughout the whole universe.

No it's not. g is only approximately constant, and only near the surface of the earth (or other large body). Gravitational force varies as 1/r^2.
And freefall is only approximately a constant acceleration if you ignore air resistance.

Quote

So linear acceleration
is NOT so uncommon.

Homework problems are not so uncommon, because we simplify them. Real-world problems, not so much.

Quote

I'm just curious
is proportioned to linear_acceleration,
& the methods of confirmation.

This makes no sense.

Quote

Is my curiousity justified,
in a dark energy dilemma?

There is a connection in your mind, but that doesn’t mean there’s any legitimate physics there.

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3 hours ago, Capiert said:

I'm glad you see it too.
Thanks.

That isn't what I meant.
And you didn't understand that either.

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13 hours ago, Capiert said:

How many things did Newton
(have to) go thru
til he could (finally) settle on F=m*a?

In the context of gravitation?

It is very simple experiment.

Have solid object e.g. metallic ball at height h. Release it and measure time needed to reach it ground. From e.g. 1 meters. 2 meters. 5 meters. 10 meters.

Why solid metallic ball? It will have small negligible air resistance, which we will ignore.

After a few repetitions of the experiment there will be table of heights and times.

From them one can derive equation:

h=1/2*a*t^2

Rearrange equation and you have acceleration a.

The same value (plus minus some tolerance for error) for entire set of results.

The more precise time and length measurement the better a.

In the modern days you can use camera on tripod, grid screen and laser rangefinder and count time in frames in image processing computer application with 1/50 second precision in any ordinary camera. With high speed frame rate it might be in 1 per thousands of second precision.

Experiment possible to be recreated by anybody.

Edited by Sensei

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