Capiert

Why is (your) initial_speed v_i (=v_ref) NOT zero?
You have said the 10 kg mass is "moving" at (v_f=v_i+v=0+v=) 10 m/s,
 so what is the reference
 that you used?

Yuk!
What a messterpiece.
I'm trying to simplify (to algebra)
 so (even) a 5 year old can understand (Trump mentality, noted (by reporters) for its advantages),
 but it explodes & gets scattered
 into so much complexity (some interesting though),
 (really avoiding the question, with (fake) substitutes)
 (unfortunately demonstrating how little was understood, in some cases).

When the cat'( i)s away
 the mouse will play.

To get back on track.

 

On 6/3/2022 at 2:50 AM, joigus said:

What's the speculation here?

 

Capiert:
The (basic) question was
 whether the (Jame's Watt's)
 Power_equation's (forced) speed
 is a speed_difference v=v_f-v_i (No?, but most commonly claimed?)
 or the average_speed v_a=d/t (Yes? New).

The algebra seems to confirm (the later, & itsself),

(Unfortunately opening the question
 why the calculus did NOT do the job (right,
 for that question),
 which I DON'T really want to discuss (calculus yet).
Maybe for later.)

(I guess) I was a little sloppy
 because gravity's weight force
 was intended only as an example of force;
 NOT all forces in general
 as Swansont put it.

On 6/2/2022 at 12:28 PM, swansont said:

On 5/31/2022 at 7:07 PM, Capiert said:

Power

 P=F*d/t, va=d/t

 is the Force (e.g. Weight Wt=m*g)
 F=m*g
 multiplied
 by (e.g. the height h=) distance d
 per time t.

Here is an example of using a specific equation and trying to apply it in general.

Power applies to more than the situation of an object falling under gravity, so one cannot make a general claim that force is weight, since there are other forces. If you use weight, then the equation will only apply to a falling object, and also only if you can assume that this is a constant force.

Should (better be)

Power

 P=F*d/t

 is the Force F=m*a,
 multiplied
 by the distance d
 per time t.

The average_speed
 va=d/t
 is the distance d
 per time t

E.g.
Let the (linear) acceleration a=g be gravitational free_fall's acceleration,
& the force F=Wt be the Weight Wt=m*g,
& the distance d=h height (fallen).

P=Wt*h/t.

On 5/31/2022 at 7:07 PM, Capiert said:

KE=m*(v_f²-v_i²)/2.

On 5/31/2022 at 7:36 PM, Bufofrog said:

It looks like you are saying an object moving at a constant velocity has zero KE,

I did NOT say that (an object moving at a constant velocity has zero KE);
 you did.

On 5/31/2022 at 7:36 PM, Bufofrog said:

since v_{f =} v_i   

If initial & final speeds are the same
 then the KE (between them)
 is zero.
E.g. For 2 objects=bodies.

E.g. For 1 object=body
 the (same, single) mass
 has NOT be accelerated (e.g. faster)
 from its initial speed;
 or
 e.g. it has NOT been accelerated
 from its
 final speed.

Take your pick.

On 5/31/2022 at 7:36 PM, Bufofrog said:

That's a problem. 

I (still) do NOT see your problem.

E.g. Same speeds
 (for same mass(es))
 means NO kinetic_energy(_difference) KE,
 but KE is a difference (KE_d=KE_f-KEi)
 of (the 2) KE's;
 although I think
 Swansont will want to correct me there.

On 6/1/2022 at 1:33 AM, Bufofrog said:

The KE equation you wrote would mean that an object moving at a constant speed would have zero KE. 

No it does NOT.

The equation uses 2 (different) speeds
 even if 1 of them (speeds)
 is allowed to be zero.

A difference of KE's
 means that you are subtracting 2 KE's.
So your components are KE's.

E.g.
KE_d=m*(v_f²/2)-m*(v_i²/2).
KE_f=m*(v_f²/2).
KE_i=m*(v_i²/2).

v_f=v_i+v
v_i=v_f-v
v=v_d=v_f-v_i.

On 6/1/2022 at 1:33 AM, Bufofrog said:

That is incorrect, hence the equation is incorrect.

Please show me the error (if so).

As far as I know
 that math (syntax) works (for me).
Maybe you have a task, example
 where I can use it?

My concept (syntax)
 of KE (=KE_d=KE_f-KE_i)
 is (a (KE) difference,
 &) similar to yours,
 but (differs in that it)
 is (also) extrapolated thru
 to (both) the initial_KE KE_i=KE_f-KE
 & the final_KE KE_f=KE_i+KE.

(But Swansont commented=identified
 that is delta_KE.)

 What I'm saying
 is, the initial_speed vi
 is [often]
 invisible
 (for same speed objects);
 but NOT zero!

Even if objects=bodies
 seem static;
 they are (really) moving!

Same speed objects
 can NOT accelerate
 each other
 (because they do NOT collide
 with each other).

Again
PS: Very interesting.

It looks like you did NOT get it (=the (2) perspective(s)).
How can anything moving,
 NOT have KE?

 If v_f = v_i (then that) means you can NOT see the(ir) motion.
 (..when compared).

The 2 objects (seem to) stand still,
 (perhaps) with a constant (separation) distance.

E.g. 2 billiard balls
 on a (billiard) table
 (although the world=Earth is turning).

Our reference(_speed)
 vi
 could be the Earth's rotation speed
 e.g. ruffly ~1000 [m/s] eastwards.

The (2) balls are NOT going
 to do anything
 (e.g. (they are NOT going to) collide)
 with each other
 because they stay still
 (wrt the billiard table,
 on Earth);
 but (both (balls)) are rotating
 with the Earth_speed
 vi~1000 [m/s] eastwards
 (i.e. wrt the Earth)!

Their KE would NOT become obvious
 until they collide
 with something else
 moving
 at a completely different speed.

E.g. a cue hitting a ball.

((Please) allow me to exaggerate.)

E.g. If the cue travelled ~-1000 [m/s] e.g. westwards
 (to hit the ball)
 to compensate against the Earth's rotational speed
 (so that the cue would seem like zero speed
 wrt the Earth's center).

KE, (e.g. the speed_(energy)_change)
 can only be [acquired=] "received
 from", or else "transferred
 to" another mass('s motion).

 

 

3 hours ago, NTuft said:

I think power is the time derivative of work. In order to make the simplification like:

you are presuming a constant force is being applied over a distance (I'd say change in position).

Yes.

3 hours ago, NTuft said:

When you want to take an instantaneous power output, I think it makes sense there is no conservation of energy.

I agree that there is NO such thing
 as conservation of Energy (COE);
 although conservation of mass*Energy m*E
 might exist.

COE is a FAKE,
 because it often FAILS,
 although NOBODY
 has the guts
 to kick it OUT;
 they all still (perhaps naively) defend it
 (as traditional BRAINWASHING,
 e.g. NOT to upset things=tradition).

Conservation of ENERGY
 does NOT deserve
 to be mentioned
 except for its (crude) history
 e.g. development
 of Physics.

3 hours ago, NTuft said:

You'd need to extend the time frame,
 and perhaps incorporate the other present bodies
 and conditions
 to know how
 the conservation may or may not weigh out.

(That is:)
 Wasted time,
 considering
 my (just=immediately) previous comment.

But what do you mean
 by extend
 the time frame?

3 hours ago, NTuft said:

As an exercise,
 can you tease out the difference
 between change of position
 vs. distance

Distance
 d (x,y,z),
 d=P2(x2,y2,z2)-P1(x1,y1,z1),
 is the difference
 between 2 position(s
 points):
 e.g. P2(x2,y2,z2)
 minus P1(x1,y1,z1) ;
 (where P1(x1,y1,z1)
 is the origin(al starting point)
 which determines
 the positive direction). 

 e.g.
 x=x2-x1
 y=y2-y1
 z=z2-z1.

3 hours ago, NTuft said:

 (e.g.:
 "Why can Work be said to be 0
 despite obvious evidence
 to the contrary,
 in a self-referential frame?")

Please explain.

(e.g. "contrary"
 to "self-referential" frame).

(What is that?)
(E.g. A Point source?)
(Is ego supreme?)

Everything exists
 "in" the universe;
 NOT OUTSIDE
 of it (=the universe).

Everything is "a" part
 of the universe (=connected);
 NOT "apart" !.

(E.g. I am in the universe;
 NOT independent
 from the universe).

(Please give me a clue
 (as)
 to help understand you better.)

But if I understand you correctly
 then the "initial" kinetic_energy KE_i
 might help fix=remedy things;
 for the total=final kinetic_energy KE_f=KE_i+KE.

E.g.
Kinetic_Energy
 KE=KE_f-KE_i
 is the difference
 between
 final (kinetic_energy KE_f)
 & initial (kinetic_energy KE_i)

 similar
 to speed(_differrence v_d
 v_d=v)
 v=v_f-v_i
 where the subscripts are:
 final _f
 & initial _i.

3 hours ago, NTuft said:

and

 ("you can tease out the difference between")

3 hours ago, NTuft said:

 velocity vs. speed. 

I'( a)m sorry
 (but) you have lost me (there).

I can NOT imagine
 any speed
 without direction.

"I DON'T know where
 we'( a)re going Captain;
 but
(at warp_speed)
 we are getting there
 in a he(ck) of a hurry!"-Scotty.

Direction (e.g. angle)
 is also relative.
E.g. to a line.
E.g. 2 points (P1 & P2, each an x,y,z);

 but the 1st (point)
 must be established
 from the 2nd,
 to complete
 that relation
 (for positve,
 (versus negative)
 direction)
 i.e. relativity.

3 hours ago, NTuft said:

You mention speed "at" one point
 and then use v(elocity)
 in your equation:
 velocity implies a direction against a map,

I will assume
 (with (the word) map)
 you are implying x,y,z
 (e.g. wrt to some other reference
 (of similar structure, e.g. Ref, e.g Frame)).

I (can still)
 consider
 a speed(_difference) v (=v_d),
 to be
 (composed
 of) components
 v_x, v_y, & v_z.

3 hours ago, NTuft said:

 and a change in direction gives us an acceleration,
 so when you have your instantaneous speed

(But)
 I do NOT use instantaneous_speed;
 I use average_speed v_a=d/t,
 instead.

3 hours ago, NTuft said:

 this option
 for change
 of direction
 as acceleration
 is removed
 from the equation
 (i.e.:  can be said
 to exist presumptively also
 in the case of a "point"..).

I(' am sorry, I) DON'T follow (you).

My (speed_difference) v
 is a mixture
 of x,y,z speeds.

Where is the problem?

(I suspect you are adding
 unneeded complexity;
 where NONE is needed.

Am I right or wrong?
If wrong please explain.)

 

We are assuming
 they are moving
 on parallel lines,
 in the same direction.

Can (=May)
I pacify
 that argument(?)
 by saying:
 (I acknowledge)
 an "initial" Kinetic_Energy KE_i=m*v_di*v_ai
 where
 the intial_speed v_i=v_i-0
 is (simply)
 extrapolated
 from a predacesser
 speed_difference
 with its maximum_speed
 being (only) that excluded
 initial_speed v_di=v_i-0
(whatever zero=0 speed
 should be
 e.g. relative to something else('s motion_speed)).
 v_ai=(0+v_i)/2
 (analogy (similar to))
  v_a=(v_f+v_i)/2).

 e.g. minus zero,
 where zero
 is that (next, smaller)
 reference.

That is all done so
 because KE
 is (already) relative
 to its (own) initial_speed v_i.

Thus the kinetic_energies
 can be added (sequentially).
 
 

1 hour ago, Bufofrog said:

It looks like you are saying an object moving at a constant velocity has zero KE, since v_{f =} v_i   

That's a problem. 

Please explain that problem.

Kinetic_Energy
 is (already) relative
(with respect)
 to the initial_speed v_i.

(-c<)v_i<c
 can be (almost) anything
 less than light's_speed (+/-)c.

That means the initial_speed v_i
 is excluded
 in that (amount
 of) "kinetic_energy('s)"
change
 of speed.

E.g.
2 masses moving
 at the same speed
 (wrt each other),
 have NO speed_difference v=v_f-v_i
 (wrt each other),
 thus a constant distance
 is maintained (=kept)
 between them.

Only if a speed_difference v
 exists between them
 can they affect each other
 in e.g. a collision,
 to change the other's speed
 e.g. via Newton's 3rd law,
 (equal & opposite) reaction,
 Repulsion.

E.g. Assuming an (EM_)Field
 wrt (a decreasing) distance,
 to ((elastically) repulsively) bounce.

The catch there
 is electric_repulsion
 is inversely proportional
 to the radial_distance "squared"!

That is NO longer a linear relation;
 but instead exponential
(wrt distance)!
 

(James Watt's, mechanical)
 Power

 P=F*d/t, va=d/t

 is the Force (e.g. Weight Wt=m*g)
 F=m*g
 multiplied
 by (e.g. the height h=) distance d
 per time t.

That works out
 to, Power
 P=F*v_a
 is the (e.g. weight Wt=m*g)
 Force (F=m*a)
 multiplied
 by the average_speed v_a=d/t=h/t;
 instead of

 F*v=(p^2)/(m*t)
 which
 is the momentum_squared (p^2)=mom^2=(m*v)^2

 (for the mass m;
 multiplied
 by the speed(_difference) v=v_f-v_i
 (of final_speed v_f
 minus
 the initial_speed v_i));

 divided by both:
 mass m
 & time t. 

(I.e.
 F*v is definitely
 NOT Power;
 although
 it might seem similar.)

Force
 F=mom/t (=p/t)
 is the
 (change
 in) momentum mom=m*v
 per time t. 

Thank you
 for asking.
(I thought NOBODY
 would dare,
 (for at least 3 days).)

James Watt's
 (definition
 of) Power
 is (the rate
 of doing work),
 where he defined Work(_Energy)
 WE=Wt*d
 as lifting
 weight Wt (=m*g, Force F=m*a, let (linear_)acceleration a=g)
 to a specific
 height (h=d distance).

(We call that
 (kind of work(_energy))
 Potential_Energy
 PE=m*g*h.)

That (work_(energy))
 done
 (with)in
 a specific
 amount
 of time t
 is (his) Power
 P=WE/t.

Converting
 work_energy
 WE=KE
 to (moving)
 kinetic_energy,
 we get

 KE=m*v*v_a,

 where with the speed(_difference) v=v_f-v_i
 & the average_speed v_a=d/t=h/t
 that gives
 KE=m*(v_f-v_i)*(v_f+v_i)/2, or

 KE=m*(v_f²-v_i²)/2.

 

Disclaimer:

(To me)
 it looks like
 somebody goofed
 on (producing, e.g. creating)
 James Watt's (formula) syntax
 which needs "average" ((for the) speed).

I suspect most people
 missed that (detail).

(Easily found
 with (simple) algebra.)

I hope that answers
 your question.

Btw.
Even, the
 (Ewert's 1996, universal)
 conservation of mass*Energy
 m*E=mom*moma
 m*E=(m*v)*(m*va)
 m*E=m*m*v*va
 which correctly proportions
 the mass_squared m²
 with (respect to)
 the (single, non_squared)
 height h
 in m*PE,

 (indirectly denying (
 https://en.wikipedia.org/wiki/Julius_von_Mayer

 Julius Robert von Mayer (25 November 1814 – 20 March 1878)
 (remarkable) 1841 conservation
 of Energy,
 as a(n absurd) bunder=Fake,
 due to lack of mass);

 (but)
 produces similar (confirming) results
 m*PE=m*KE
 m*m*g*h=m*m*v*va.

(Why we are (supposedly)
 allowed to cancel mass m
 & (then, attempt to) maintain
 the (mass's, (fallen)) height h
 is a riddle to me.

 (..because..)

It'( i)s physically wrong.

Meaning it (=cancelling mass)
 can fail
 (the proportioning relation
 to height)!
But does NOT always.

Of course (naturally) everybody has heard, of, conservation of mass. ?)

Noether's Theory
 is NOT going
 to help you,
 if you CAN'T even get the basics right=correct.

It's (=Noether's theorem:
 which uses distance (e.g. (Work_)Energy WE=F*d, e.g. PE=m*g*h);
 instead of (per) time (e.g. (average_)momentum moma=m*d/t=m*v_a)

 is) based (most probably) on NOTHING
 (but (maybe (crumbling, unreliable)) NONSENSE). ?
But more interesting
 might be where that (Noether theorem) went wrong,
 (when the simplest
 of algebra
 can prove an error
 that, that theorem did NOT)
 e.g. why you guys & gals
 put all your eggs (e.g. hope(ful assumptions))
 in that 1 basket.

(Why do things simply;
 when you can do them (more) complicated?
 (..so NOBODY can see (thru) the ERRORS)?
NOBODY is PERFECT.
NOT even me.
Why NOT strive for error reducing methods, instead?)

E.g. (Correct is, that:)
 Fallen height distance
 is (=must be (made))
 wrt mass_squared m²;
 NOT mass m⁽¹⁾ ONLY.

 

https://en.wikipedia.org/wiki/Conservation_of_energy

Main article: Noether's theorem

Emmy Noether (1882-1935) was an influential mathematician known for her groundbreaking contributions to abstract algebra and theoretical physics.

The conservation of energy is a common feature in many physical theories. From a mathematical point of view it is understood as a consequence of Noether's theorem, developed by Emmy Noether in 1915 and first published in 1918. The theorem states that every continuous symmetry of a physical theory has an associated conserved quantity; if the theory's symmetry is time invariance then the conserved quantity is called "energy". The energy conservation law is a consequence of the shift symmetry of time; energy conservation is implied by the empirical fact that the laws of physics do not change with time itself. Philosophically this can be stated as "nothing depends on time per se". In other words, if the physical system is invariant under the continuous symmetry of time translation then its energy (which is the canonical conjugate quantity to time) is conserved. Conversely, systems that are not invariant under shifts in time (e.g. systems with time-dependent potential energy) do not exhibit conservation of energy – unless we consider them to exchange energy with another, an external system so that the theory of the enlarged system becomes time-invariant again. Conservation of energy for finite systems is valid in physical theories such as special relativity and quantum theory (including QED) in the flat space-time.

If we take again
 P=F*v_a, swap sides
 F*v_a=P, /F
 v_a=P/F, v_a=d/t
 P/F=d/t, *t*F
 both sides produce energy
 P*t=F*d, P*t=E & F=mom/t
 E=(mom/t)*d, rearrange
 E=mom*d/t, v_a=d/t
 E=mom*v_a, mom=m*v
 E=m*v*v_a, is the Kinetic Energy
 KE=m*v*v_a.

 We do NOT need
 Noether's theorem
 to show
 the "connection"
 between Force (F=P/v_a) & Power (P=F*v_a)
 or distance (d=t*P/F) versus time (t=d*F/P)
 as average_speed v_a=d/t (=P/F).

 

https://en.wikipedia.org/wiki/Julius_von_Mayer

It (German knighthood) for caloric,
Mayer was the first person to state the law of the conservation of energy, one of the most fundamental tenets of modern day physics. The law of the conservation of energy states that the total mechanical energy of a system remains constant in any isolated system of objects that interact with each other only by way of forces that are conservative. Mayer's first attempt at stating the conservation of energy was a paper he sent to Johann Christian Poggendorff's Annalen der Physik, in which he postulated a conservation of force (Erhaltungssatz der Kraft). However, owing to Mayer's lack of advanced training in physics, it contained some fundamental mistakes and was not published.

 

..(Mayer) examined experimentally; for example, if kinetic energy transforms into heat energy, water should be warmed by vibration

 

Since he (=Mayer) was not taken seriously at the time, his achievements were overlooked and credit was given to James Joule. Mayer almost committed suicide after he discovered this fact. He spent some time in mental institutions to recover from this and the loss of some of his children. Several of his papers were published due to the advanced nature of the physics and chemistry. He was awarded an honorary doctorate in 1859 by the philosophical faculty at the University of Tübingen. His overlooked work was revived in 1862 by fellow physicist John Tyndall in a lecture at the London Royal Institution. In July 1867 Mayer published "Die Mechanik der Wärme." This publication dealt with the mechanics of heat and its motion. On 5 November 1867 Mayer was awarded personal nobility by the Kingdom of Württemberg (von Mayer) which is the German equivalent of a British knighthood.

Isn't
 (height h=) distance d
 divided
 by time t
 an average_speed
 v_a=d/t?

 

7 hours ago, John Cuthber said:

I can add a kilo of potatoes to a pound of minced beef and an ounce of flour.
But I don't get 3 anythings of stew.

But (if you want) you could create a(ny) new "thing" unit,
 (depending on how you wanted it
 to be(come)).

I suppose
 if you divided your stew
 into 3 equal portions,
 then it would be 3*[portions]
 (of that stew).
I.e. That'( i)s wrt (only) that stew itself
 (with only 3 ingredients
 & their own proportions,
 on the spot,
 so to speak)
 NOT anything else.
E.g.
NOT necessarily
 3*[bowls or cups etc].

The new unit [(equally_divided)_portion]
 can be found
 by converting
 all 3 units
 to any same desired_unit
 & then dividing by 3.

That would be your (new) "anything" (unit).
E.g. Some other (irrational?) factored unit
 of your (common) desired_unit.

It'( i)s only a conversion (method).

6 hours ago, studiot said:

Have you never come across a man whose height is 6' 3"  ?

Pronounced: "six foot, 3 (inches)".

(Also, please notice the singular (1st) unit (foot, NOT feet)
 although the number (6), is >1;
 until we (might) get stuck in details
 by stating the 2nd (number's) unit(s (as more than 1, e.g. 3)).)

That looks like 2 answers, stuck (=connected) together.
E.g.
6'+3".
Where the (empty) space
(between them (both))
 represents a virtual "plus", + (symbol).

But why is there NO (empty) space
 between the (number) 6 & (unit) '=[feet] or [foot] (symbol);
 &
 between the (number) 3 & (unit) "=[inch] (symbol).

The SI convention
 seems to use the empty space
 between number & units (Notice the s (on units) for both: singular; or plural or more)
 as a virtual multiply=multiplication.

E.g.
6 [m]+3 [m]=9 [m], represents
6*[m]+3*[m]=9*[m].

There is also a subtle difference
 between infinitive, e.g. stone (in a quarry, to build a castle)
 versus more_than_1 e.g. plural or more=many, e.g. (made of) stones.

E.g.
Should we say,
 the building
 is 5 [meter] high? (infinitive).
E.g. NOT 5 [meters]. (Plural or more).

We often say 20°C (twenty degrees Centigrade, please notice: NO (empty) space at the °).
But 300 K (three_hundred Kelvin (infinitive); NOT Kelvins (many).)

Btw
When I look at this (= all that (grammar) needed for the math)
 I am considering artificial_intelligence (e.g. program(ming)) too. (=NOT two, nor plur(e)al (~for crying out load, at reality).)
 & (I am) considering
 what sort of math (algebra)
 would be needed
 to incorporate
 such mixed units (singular or more).

Algebra is perfect equality (e.g. balance);
 but NOT so with grammar
 that is brought in
 to distinguish
 finer differences.

Our brains recognize those discontinuities (=differences).

5 hours ago, John Cuthber said:

On 12/5/2021 at 8:07 PM, Capiert said:

(Surely NOT added.)

Did anyone suggest that they might be?

Yes, John.

Added (verb)
 & addendum (noun)
 are (probably) NOT exactly the same;
 but it (=the added_onto method)
 suggests (to me),
 (that its meaning
 is) going
 in that (similar) direction.

E.g.
A hang_on, (that can be) added
 on(to almost anything).

7 hours ago, exchemist said:

As an addendum, in the form of a piece of explanatory text, rather than as part of the algebraic expression. This I think is how most people see them.

But logically you must be right, I think. 

 

12 minutes ago, studiot said:

Unless, like me, you have 3 feet.

3?!
I've only got 2 [feet]! 
& no (way, back)[yard].
 (in the back :-).

12 minutes ago, studiot said:

You need the 1 to go with the egg since there are 3 feet in a yard.

Or is it 2.4 pints ?

Trust the Scots

(Again, & again.

So I guess
 there is a (small?) language problem (obstacle, of incompatibility)
 with the grammar('s singular versus plural "s", etc);
 versus
 the math('s algebra);
 &
 I suppose, the units'
 acronyms(' short_forms)
 help us (out) there, (at least) a bit,
 by ignoring the plural(s).

E.g.
5 * meters = 5 [m].

Or unit
meter(s)=[m].

How

19 hours ago, exchemist said:

Yes, in a way I suppose they are, though  I had never thought about it like that.  

How (then) did you think about it (=their connection, relation to each other) 
(if NOT (as) multiplication of number & unit)?

Are number(_value)s multiplied by the unit?

(Surely NOT added.)

1 * meter=1 [m].

 

On 5/25/2021 at 6:42 PM, Capiert said:

Quote

I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true.

Please explain.
vi included: is the general equation.
But without vi: is a limited (specific example) formula,
 that can NOT work for all cases.

 

On 5/25/2021 at 7:55 PM, swansont said:

Yes. That’s what you do - write down the general equation and eliminate the terms that are zero in a particular problem. In the other thread I expressly said dropping from rest, making vi = 0. There’s no reason to continue to include it for that problem. There’s no implication that it applies to all cases once you have imposed such restrictions 

What I mean is:
 (vi was missing from the "general" equation,
(NOT (just a) formula.)

 How do I know
 when I have
 the general equation,
 (if or when I start
 with only (limited (specific example) formula) fragments)?

I'm searching for the general formula.

I (attempt(ed) to) maintain vi (initial_speed)
 to try
 to NOT get lost
 ((&) for other things (=projects, concepts, extrapolations)).

(You (may) think)
 you do NOT need it (=vi), (?) fine(!);
 but I do.

E.g.
 vi remains
 an invisible (hidden (excluded)) term
 for you(r)
 speed(_difference)
 v=vf-(vi).

It's always there.

3 minutes ago, swansont said:

Don’t injure yourself patting yourself on the back.

I'll do my best!  Thanks.
(I got a good chuckle (out of that).)

3 minutes ago, swansont said:

Rearranging the equation is algebra, so any high-schooler should be able to manage that.

Yes! That is my intention.
You obviously have NOT read
 this thread's file.
It's NOT complicated!
It's easy.

3 minutes ago, swansont said:

I am focused solely on your claim that vi is missing from the equation, when that’s clearly not true.

Please explain.
vi included: is the general equation.
But without vi: is a limited (specific example) formula,
 that can NOT work for all cases.

3 minutes ago, swansont said:

If you’re going to reference an earlier thread, you should post an actual link, rather than give a time/date.

Locked threads DON'T allow quoting.
Maybe (other_options) sharing can help linking, a bit.?

 

11 hours ago, swansont said:

Swansont:

Sorry, you didn’t post it before the other equation.

11 hours ago, swansont said:

I only read the first section, because of the nonsense value.

Please explain:
What value
 do you mean;
 & why?

Maybe you missed something (important)?

If you mean
 g=2*h/(t^2)-2*vi/t
 then please read (my files, in this thread) further.

You do NOT sound
 completely connected.

11 hours ago, swansont said:

Plus I would’ve skipped the strikethrough text anyway.

That is quite a natural reaction.

But you have NOT made a suggestion
 how I can fix
 your sfn strikethrough software bug.
No response to my question (request)
 as neither: yes; NOR NO. =NO!
Makes it (=my request) sound (=seem) rhetoric.

11 hours ago, swansont said:

  11 hours ago, Capiert said:

Again, clearly -2*vi/t was NOT present in g (before my threads).
I have NOT seen that term in books;
 but algebraically
 it is correct
 (as to how I gave it to you).

 

Swansont:

Clearly this is bollocks. 

 h=vi*t+g*(t^2)/2 (or an equivalent equation) is standard fare in any physics textbook that presents kinematics equations.

Yes, we (all) know that.

11 hours ago, swansont said:

equation 3 https://www.khanacademy.org/science/physics/one-dimensional-motion/kinematic-formulas/a/what-are-the-kinematic-formulas

top left of the “big 4” https://www.physicsclassroom.com/class/1DKin/Lesson-6/Kinematic-Equations

3rd equation https://www.pasco.com/products/guides/kinematic-equations

and so on...

Your complaint is entirely fabricated.

(you can also notice that nobody is presenting this as an equation for acceleration)

So (I guess) I am (1 of) the 1st!
 to present
 the gravitational_acceleration (formula)
 g=2*h/(t^2)-2*vi/t
 as exactly so
 (=in its corrected form).
P.S. (But) I can NOT believe
 I am the ONLY 1
 who has ever done that,
 or tried.

You (only) confirm
  that I can NOT find (exactly) this ("g=2*h/(t^2)-2*vi/t") in books.

I have NOT seen that term
 (-2*vi/t, exactly in that position)
in books.

(This is a speculation thread
 thus I want to present (either) something new,
 or an improvement against a flaw
 or weakness.)

My complaint is (from),
Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28)
How do we know that acceleration a=x/(t^2)?

Better said: If I let (distance) x=h (height), instead;
 then:
How do we know that acceleration a is (suppose to be) h/(t^2)?

My answer:
We do NOT know that (at all)!
I do NOT get that formula, that you claim. (It is NOT possible (for me).!)
Especially when the height h=d distance NEEDS a factor "2".
g=2*h/(t^2)-2*vi/t".

Anyone who uses your basis
 will get
 g#h/(t^2)
 h#g*(t^2).

But

 a~2*x/(t^2) is the minimum requirement. E.g.
 x~a*(t^2)/2.

& where is the vi*t term
 in that?
NOWHERE!
So (thus) it is NOT the (EXACT) general (equality) formula
of "standard fare in any physics textbook.."
 NOR "an equivalent equation", thereof;
 BUT instead a distortable (mere) approximation.

(I can NOT rely on it for (my) other derivations.)

You may be happy with such carelessness;
 but I can NOT be.

It (=That formula: whether e.g. a#x/(t^2), or e.g. x~a*(t^2)/2)
 is either an equation (=equality);
 or (else) it is NOT.

& there it is definitely NOT (an equation).

11 hours ago, swansont said:

 

Swansont: (Kinetic_Energy is already relative, 2020 11 23 16:26)
If I drop a mass from some height, how fast will it be moving when it hits the floor?

My answer was,
Capiert: (Kinetic_Energy is already relative, 2020 11 23 16:28)
 the final_speed is
 vf=((vi^2)+g*h*2)^0.5 .

But you did NOT like that (answer),
 perhaps because
 its result
 is independent
 of mass m?

(It (=That answer) does NOT NEED mass.)

(But it is also new to me
 that algebra
 is NOT suppose to be math,
 if it is?
You closed the thread.

Perhaps you wanted some number( value)s
 to assist (confirming) your understanding
 of something
 you can easily do yourself;
 but doubted my ability. ?
E.g. Mistakes happen. Typos.)

I was only trying
 to answer your question, (preparing)
 while you closed.

That (=your impatience) does NOT seem fair.
(Especially when you said there is NO time_limit.)
(Mordred also stated to (someone) NOT to answer
 if you do NOT know.
I also wanted to check some details
 before I answered you.
Thus my hesitation=delay.
There are still (some) things I need to clear.
However:)

For that equation
 vf=((vi^2)+g*h*2)^0.5
 we typically set the initial_speed vi=0 [m/s]
 if (=when) we only drop,
 g=9.8 [m/(s^2)],
 & e.g. let the height h=1 [m]

 vf=((0^2)+9.8 [m/(s^2)]*1 [m]*2)^0.5
  vf=4.4 [m/s].

 E.g. let the height h=2 [m]

  vf=((vi^2)+g*h*2)^0.5
  vf=((0^2)+9.8 [m/(s^2)]*2 [m]*2)^0.5
  vf=6.3 [m/s].

1 hour ago, swansont said:

1. No, you did not post this equation 

That surprises me
 because
 I can read the following (from above)

The fallen height (taken from a graph),
 h=vi*t+g*(t^2)/2
 is only 2 terms:
 the constant (initial_)speed term vi*t;
 & the accelerating term g*t*t/2.

 & also in my posted (above, 2nd of 2) .pdf's (above).

https://www.scienceforums.net/applications/core/interface/file/attachment.php?id=23252

Surely you will be able
 to determine
 when that (equation) was posted.
 

1 hour ago, swansont said:

2. vi*t is clearly present in it.

Good! (for that product).

But now for the quotient -2*vi/t
  in  g (alone).

1 hour ago, swansont said:

So what is missing?

Someone to assist me
 in getting
 the WRONGLY strikedthrough text (above)
 to be NOT strikedthrough, anymore.

Would you help?

But to answer your question:

Again, clearly -2*vi/t was NOT present in g (before my threads).
I have NOT seen that term in books;
 but algebraically
 it is correct
 (as to how I gave it to you).

Science should confirm itself,
 but it seems
 (to me)
 you (might) probably prefer NOT
 that the g equation be intact,
 when rearranged
 (e.g. if you leave out that term -2*vi/t)?

(Is that possible?)

I have only moved the acceleration g
 to the left side,
 & all other terms to the right side.

That should give
 the correct g (equation).

(Otherwise, the (general) equality
 is either destroyed,
 or distorted.)

It'(=What I have done i)s NOT complicated.

16 minutes ago, swansont said:

Missing from what?

From g.
You obviously disagree,
 otherwise you would NOT have asked.

Quote

This not an equation for g.

Why NOT?
It works.

Quote

It comes from a different equation, rearranged, which you have not shared.

(As far as I know) I have shared the equation:
 the fallen height (taken from a graph),
 h=vi*t+g*(t^2)/2.

Quote

(i.e. g is typically not an unknown in kinematics problems)

Typical is NOT all cases,
 but instead most (e.g. a majority).

For me it was an unknown
 that I wanted to find,
 e.g. experimentally.

 

Can someone please help me
 with unstrikethrough my text (above)?

STOP!
Under Construction
Please wait!

 

g=2*h/(t^2)-2*vi/t.

It looks (to me)
 like that term -2*vi/t
 has been missing
 for a long time.

Surely
 you ((should) already)
 know it;
 but do you use it?

For
 (fallen) height h
 time(_difference) t
 initial_speed vi.

The free_fall acceleration
 g~-(Pi^2) [m/(s^2)]+ac
 g=~-9.8 [m/(s^2)]
 is (instantaneous,) linear, & negative;

 & (please bear with me)
 (the bothersome details:)

 (=but it) (is) slightly) reduced
 by the (Earth's daily_rotational)
 centrifugal_acceleration
 ac=(vc^2)/r

 having the (squared)
 Earth's (surface) circumferential_speed
 vc=cir/T

 with (the Earth's) circumference
 cir=2*Pi*r

 using ((your) Earth's) per radius r (position)

 & (sidereal_)day (time_)period T~23 56 [min] 4 [sec]
 in seconds;

 (instead of 24 in seconds).
 

---

Disclaimer:

That's about all
 there is to it (=g, overview),
 when (also) observing
 your location
 & position.

(Air friction
 would have to be considered;
 but I'm NOT going to bother
 for in(side) a vacuum.)

E.g. Naturally
 (all) those (extra) details
 (for the centrifugal_acceleration ac, etc.)
 can be determined.

However,
 I'( a)m concerned (here)
 with (mostly
 only) the initial_speed's vi term.

(E.g. A term that would (also) occur (for you)
 in the ((total) work_)energy
 WE=F*d
 if you ((were to) also) included the (integration) limits
 from zero to the initial_speed;
 instead of exclude it (=initial_speed,
 as a different reference (frame, choice)).

(Which can be determined, graphically
 with a plot:
 speed versus time.

The (plot's) area is the distance.)

It's easier (for me)
 to use positive
 instead of negative (plot) values.

---

Freefall:

If I let a pebble (stone or ball) fall
 (in a vacuum)
 (from rest, (with) initial_speed vi=0 [m/s], at time t0=0)

---

CAUTION:
From here on
 this website's software
 has WRONGLY Strikedthrough
 the rest of my text, automatically.

Please see the files
 until someone
 helps me undo the Strikedthough.

Please would someone help me fix this?

Marking & reclicking Strikethrugh
 does NOT undo Strikethrough
 like it would for Bold, italics or underline.

What is wrong with your Software?

---

 then in (time(_difference) t1=)1 second
 it will (have) fall(en) the height(_difference) h1=-4.9 [m]
 & have the final_speed vf1=-9.8 [m/s].

(I could also reverse that
 & say:
 If I let a stone (pebble) fall
(from a height h1=)4.9 [m] (wrt the ground),
 -4.9 [m]
 (to the ground),
 then it will hit the ground
 in (time t1=)1 second
 & impact (ruffly) at a (final_) speed vf=-9.8 [m/s].

We know freefall
 is "linear" acceleration,
 so in time(_difference) t2=2
 fallen_height h2=-9.8 [m]
 & final_speed vf2=-19.6 [m/s].

 etc.

E.g.
Instead,
 we could have used
 the 1st final_speed (vf1)
 (from the 1st (time_(difference)) second (t1))
 as the 2nd initial_speed vi2=-9.8 [m/s]
 (e.g. as if thrown)
 & allow the pebble to restart (accelerating)
 as if from zero in initial_speed vi1=0 [m/s],
 but for: (only) the 2nd (time_(difference) t(2-1)=1) second;
 its (=the pebble's) additional fallen_height h(2-1)=-4.9 [m];
 & additional final_speed vf(2-1)=-9.8 [m/s].

So, the (total) fallen_height h2=h1+h(2-1)=-4.9 [m]+(-4.9 [m])=-9.8 [m];
 & the (total) final_speed vf2=vf1+vf(2-1)=-9.8 [m/s]+(-9.8 [m/s])=-19.6 [m/s].

 

The fallen height (taken from a graph),
 h=vi*t+g*(t^2)/2
 is only 2 terms:
 the constant (initial_)speed term vi*t;
 & the accelerating term g*t*t/2.

The speed(_difference)
 v=g*t
 is the (freefall) acceleration g
 multiplied by time t.

But
 the accelerating term
 g*t*t/2=(v/2)*t
 is "half"
 of that speed(_difference) v(=vf-vi)
 multiplied by time t

 h=vi*t+(v/2)*t
 h=(vi+v/2)*t, & va=(vi+v/2) is the average_speed
 h=va*t.

The average_speed
 va=h/t
 is the fallen height h(=d distance)
 per time(_difference) t.

(The final_speed
 vf=vi+v
 vf=((vi^2)+g*h*2)^0.5
 )

 The (linear) average_acceleration
 ga=h/(t^2). <----That'( i)s Important! So simple, NO other terms!
 ga=va/t
 ga=

The freefall gravitational "average_(linear)_acceleration", (&) is
 ga=vi/t+g/2, swap sides
 vi/t+g/2=ga, -vi/t
 g/2=ga-vi/t, *2

 then the (instantaneous) linear freefall gravitational acceleration, is
  g=2*(ga-vi/t).

Note: Syntax
 (my) ga=g_a (yours).

Those are the conversions
 to & from linear
 instantaneous versus average:
 speeds; & accelerations.

Again:

The average_speed
 va=h/t
 va=vi+(v/2).

The (linear) average_acceleration
 ga=h/(t*t)=va/t
 ga=vi/t+g/2.

The (average_)time
 t=h/va
 is the height h
 per average_speed va; or
 t=(h/ga)^0.5
 the rooted:
 height h;
 per (linear) average_acceleration ga.

It'( i)s that simple.

(Algebra.)

 

 

2021_05_24_2011_ Linear_acceleration's_Average_speed__diagram__2021 05 24 2038 PS Wi.pdf 2021_05_24_1702_Gravity g's, missing term_2021 05 24 2130 PS Wi.pdf

10 hours ago, MigL said:

Your posts give me headaches, Capiert.
And its not just the formatting ...

I sympathize with you.

I've been wracking my brains on this stuff for years
 trying to make some sense of it. 

10 hours ago, Sensei said:

Physicists are making shortcut. "massless" means in the reality "having no rest-mass".

I guess you missed the point Sensei.

My (perspective_reversal) calculations show
 that the Rest_mass is inherent,
 (meaning it'( i)s)
 (already) in the KE('s reversed perspective).

I don't need to add it (rest_mass) randomly
 because somebody thought it was forgotten
 & needed.

10 hours ago, Sensei said:

Rest-mass is invariant mass.

That's right,
 mass is conserved,
 conservation of mass com.
The mass does NOT change
 (in my calculations),
 only the speeds (change).

& I haven't done (=changed) anything;
 (except) only the perspective.

& I get (Fitzgerald)_Lorentz_contraction similar results
 with (only) algebra.

So I have to ask:
 Why do you think
 you (might) have to add rest_mass additionally (extra)
 ((randomly) out of the blue ((or a hat)
 (like a magician)))?

(That's ridiculous.)

Why should the rest_mass be in your relativistic equations "twice"?

E.g. Originally (inherently);
 & then again (randomly, by you)
 because you thought you missed (=forgot) it, before.

I DON'T need to add rest_mass
 to my KE calculations
 because it's already there.

You however, think you do (need to do that).

10 hours ago, Sensei said:

Mass which (a) quantum object has in its rest-frame of reference.

https://en.m.wikipedia.org/wiki/Rest_frame

The rest_frame is when the initial_speed vi=0 [m/s].

10 hours ago, Sensei said:

ps. Learn some physics... start from primary school physics textbooks..

I'( have) a bundle of them
 some read several times.
But they DON'T solve my problems (=paradoxes).
They (=those books) only "help" me solve them.
Those answers are NOT in the books.
In fact the books are (sometimes) misleading.
Too much NONSENSE makes my brain shut off.
(I CAN'T tolerate it.)
So I have to take my time
 & unravel the puzzle.
I have to try (new) alternatives;
 NOT (old) failures.

All your textbooks
 only leed to dark_energy
(=NONSENSE, errorful calculations).

I'm sorry
 but I think you are behind the times
 judging me so.

I'm searching for a recalibration
 to bring physics up to date,
 instead of the scattered mess it is in now.

You guys (your team)
 is either going to help me, or not.

4 hours ago, swansont said:

You noted va is "relative to earth's speed" and that's nonsense.

Sorry, but I only live here (on Earth);
 I'm going to no other planet.

So Earth was my best example.

va is relative to the initial_speed vi
 which for this example
 is the earth's(_speed,
 which could be anything (reasonable)).

So if we are traveling
 at the same speed
 as the Earth
 then that initial_speed vi=0
 is zero
 for at rest
 wrt on Earth.

Quote

I can find the KE in any frame. It is, as you note, relative. The earth is not inherent to the equation.

I agree. I only used it (=Earth) as an (easy, simple) example
 to "try" & get the message across.

The Work_energy's element

Quote

dW = Fdx for a constant force

 F multiplied by
 the distance element dx.

Quote

Integrate that and you get 1/2 mv^2

How do we know that acceleration a=x/(t^2)?
If the momentum mom=m*v
 & mom^2=(m*v)^2.
I DON'T see the coherence.

Quote

The work done is the change in kinetic energy. energy is conserved.

WE=F*d.

I would prefer
 to say
 the mass*Energy m*E
 is conserved, instead.

NOT the energy.

Capiert said:
Which should be
KE=m*(((v^2)/2)+v*vi), instead.

Quote

Why "should" it be that?

Algebra.

The average_speed is
va=(vi+vf)/2. *2
2*va=vi+vf, swap sides
vi+vf=2*va, -vi
vf=2*va-vi.

The speed_difference is
v=vf-vi, swap sides
vf-vi=v, +vi
vf=v+vi.

Both vf's can be equated also
vf=vf
v+vi=2*va-vi, -vi
v=2*va-2*vi
v=2*(va-vi), /2
2*v=(va-vi), +vi
2*v+vi=va, swap sides
va=2*v+vi.

Or expanding the kinetic_energy

KE=m*((vf^2)-(vi^2))/2, ((vf^2)-(vi^2))=(vf-vi)*(vf+vi)
KE=m*(vf-vi)*((vf+vi)/2), (v=vf-vi & thus) vf=v+vi

(Notice: KE=m*v*va, for v=(vf-vi) & va=(vi+vf)/2 but continue from above for below)

KE=m*(v+vi-vi)*((v+vi+vi)/2), vi-vi=0 & vi+vi=vi*2
KE=m*(v)*((v+vi*2)/2), 
KE=m*(v*v+v*vi*2)/2, expand
KE=m*((v*v/2)+(v*vi)).
 

I derived (=equated) that above (for you also further above)
 via substitution,
 & I see no error in my equations,
 thus I must conclude they are correct.
They are simple, algebra.

Quote

What valid physics principle is this based on?

Mechanics:
Gravity's (free_fall, linear_acceleration a=g)
 fallen_height
 h=hf-hi=vi*t+g*t*t/2
 for time t
 & initial_speed vi
 gives
 final_speed
 vf=((vi^2)+g*h*2)^0.5
 

Capiert said:
Does a (virtual) moving_frame need (to have) mass?

Quote

A frame of reference is just coordinate systems.

I'( wi)ll assume, (your answer is) no.

Quote

If you are applying Newton's laws you need to be in an inertial

(=constant_speed)

Quote

frame.

No valid physics, either.

Sorry, my mistake
  I used a (virtual, constant_speeed=inertial) perspective, instead of non_inertial=accelerating frame.
Thanks for clearing that.

Capiert said:
Galilean_relativity NEVER worked (right)
 because it did NOT limit speed
 to a maximum.
So here, in this example
 of (simple) classical_relativity
 I have limited the speed (maximum) to c.

Quote

We already have a solution. It's special relativity.

It looks questionable (=doubtful) to me.
The author tried to get rid of it (e.g. avoid it) in 1920 chapter 22.
He didn't recommend it (anymore);
NOR did the Nobel committee give him a prize for Relativity.
(Why would they NOT
 if it were really important for Physics?)
I guess they also had their doubts.

Capiert said:
I see no problems
 & get astounding results
 (for constant_speeds).

Quote

Then your answers are wrong.

How do you know that?
How do you know if SR is correct?
E.g. When its author discouraged its usage,
1920 chapter 22.

Capiert said:
What do you mean by "inertial"_frame?

Quote

Non-accelerating. One in which Newton's laws would work.

Capiert said:
1_stone (1905) calculated
 the photons mass m=KE/(c^2).

Quote

No, that's the mass equivalent of the energy.

What do you mean by that?
Newton believed photons were particles.

Quote

Photons are massless.

That means small (=less) mass;
 NOT NO_mass.

Quote

No, that's the mass equivalent of the energy.

That'( energy i)s what he used for a photon's mass in 1905.
What do you mean by "No"?
Are you trying to say
 E#m*(c^2).
Mass is NOT (a kind of) energy?

To me it seems rather obvious what things are,
 with an equation (=equality).

I do NOT understand why you shirk (~avoid)
 from using math
 for physics,
 to get to the bottom
 of things.

What prevents the (mass versus energy) connection?
Light's_speed squared c^2?
I.e. Mass & energy are NOT identical.

Quote

Photons are massless.

If you are implying (=saying) NO_mass,
 then how do you know that?

Or are you implying photos are (sort of like) energy, but not yet mass?
How can you possibly have energy, without mass?

Mass is only a construct, a coefficient (=factor).

Capiert said:
=? (light? or speed?)

Quote

Then it should not appear in a general equation.

Sorry.

Quote

Does your formulation give correct answers?

I suspect yes.
& I can convert to your SR answers
 with the factor -2;
 & visa versa
 from SR values
 to mine
 with the factor -1/2.

Quote

If I drop a mass from some height, how fast will it be moving when it hits the floor?

Answer:
 the final_speed is
 vf=((vi^2)+g*h*2)^0.5 .

Quote

You can solve this with energy, or with kinematics. You can also compare to experiment.

The factor of 1/2 will matter.

 

1 hour ago, swansont said:

Yes, we know that KE is relative, since it depends on v, which is relative.

The term in KE is v^2, which is v*v. It is not two different speeds multiplied together.

If v=vf-vi
&
KE=m*((vf^2)-(vi^2))/2
&
((vf^2)-(vi^2))/2=(vf-vi)*(vf+vi)/2,
&
va=(vi+vf))/2
then why
is NOT
((vf^2)-(vi^2))/2=v*va?

& How can KE possibly be m*v*v/2?
When should be
KE=m*(((v^2)/2)+v*vi), instead.

Quote

"wrt light's speed" is nonsensical. Speed is relative to some frame of reference,

Does a (virtual) moving_frame need (to have) mass?

Quote

and light does not have an inertial frame of reference.

I did NOT declare an inertial frame,
 I used a (virtual, non_inertial) perspective.
(I do NOT have to touch a(ny) thing.)

I simply stated a speed_difference
 u=c-v
 between the maximum possible speed c
 & the moving_point's speed v.

 u is only the (complementary_) speed needed
 to add to v
 (in order)
 to be(come) c.

That is simple algebra.
No magic.

Galilean_relativity NEVER worked (right)
 because it did NOT limit speed
 to a maximum.
So here, in this example
 of (simple) classical_relativity
 I have limited the speed (maximum) to c.

I see no problems
 & get astounding results
 (for constant_speeds).

It's values are different from yours
 by a factor of -1/2.

What do you mean by "inertial"_frame?

1_stone (1905) calculated
 the photons mass m=KE/(c^2).

Quote

It

 =? (light? or speed?)

Quote

 is also not inherently tied into the earth's speed.

If you mean speed,
 I was only using the earth's speed
 as an example.

Quote

KE is relativistic if you (need to) use the relativistic form of the equation

Considering my results,
 I don't (believe) I need your SR form.
Perhaps you can convince me otherwise?

What I wanted to say is:

Quote

No, it's not

Why NOT?

Quote

(further nonsense deleted; there's no point)

I can't comment your deletions.
My point is "virtual".
Standard definition: no size.
& moving (at constant speed).

The (so_called) Rest_mass
 is inherent (=included)
 in (my) KE's opposite_perspective (-KE');
 & (thus it) does NOT have to be added extra
 (by haphazard random guessing).

 wrt the initial_speed vi;

 & relativistic if you (only)

 swap (=reverse) its perspective

 from wrt earth's minimum_speed 0

 to wrt earth's maximum_speed c.

 

Reversing the

 Kinetic_energy (perspective)

 KE=m*v*va (wrt Earth’s_speed), *(-1)' gives

 -KE'=m'*(-)*v'*va' (wrt light’s_speed) which

 is already relativistic.

 

Please notice
 if I let

 c=v+u, u=c-v, u=-v', (prime_symbol ‘ is wrt light’s_speed) then

 the (negative) speed_difference (wrt light’s_speed), is

 -v'=-(vf'-vi')=vi'-vf'

 & the average_(accelerated)_speed (wrt light’s_speed), is 

 va'=(vi'+vf')/2

 for initial_speed vi' (=-0'=v_min, wrt light’s_speed, or v_max=c wrt Earth’s_speed)

 & final_speed vf' (wrt light’s_speed).

 -KE'=m'*(vi'-vf')*((vi'+vf')/2), combine brackets

 -KE'=m'*((vi'^2)-(vf'^2))/2, let vi’=-0’=c

 -KE'=m'*((vi'^2)-(vf'^2))/2, bring c^2 out from the brackets (c^2)/(c^2)=1/1=1

 -KE'=m'*(c^2)*(1-(vf'^2)/(c^2))/2, let gamma’^2=(1-(vf'^2)/(c^2))

 -KE'=m'*(c^2)*gamma’*gamma’/2.

 

That equation

 has (=contains)

 "half" of (DePretto's 1903, vis_viva),

 Energy
 E=m*(c^2)

 & 2 (Fitzgerald_Lorentz, relativistic_contraction similar) coefficients (named)

 gamma_primed

 gamma’=(1-(vf’^2)/(c^2))^0.5

 where the rest_mass m=m'

 is (the same, =identical)
 constant(_variable)
 for either perspective.

 

I.e. Conservation of mass com

 (wrt Earth’s_speed) m=m' (wrt light’s_speed).

 

Speeds are the variables (instead).

 

2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1419 PS Wi_(stripped).pdf 2020_11_23_1010_KE_is_already_relative__Preliminary__2020 11 23 1356 PS Wi__with old _Excel_ formula text_(mix).pdf

4 minutes ago, MigL said:

Exactly.
And that's the problem.

I'm glad you see it too.
Thanks.