Capiert

If anybody
 can travel
 backward
 in time,
 please tell me how
 cause they DIDN’T tell me either
 & I want to know.

Time is a scalar (amount)
 in which the direction
 is determined ONLY by its amount.

Thus, time is NOT bidirectional
 like a dimension is.

Time ONLY goes forward.

We can ONLY measure time going forward;
 NOT in reverse.

Instead, time is a parameter
 (= something other than a dimension, thus)
 NOT a dimension.

Disclaimer:

Sometimes I have to say
 some of the simplest
 most obvious things
 (against ridiculous mainstream, NONSENSE opinions).

Proposal:

If I could bend (=modify)
 Newton’s (3) laws
 ((in)to)
 the way
 I want
 for a consistent continuum,
 I would deal (mostly) with averages

 e.g.
 for (inverse_)time, distance, speed, momentum (& its squared) & mass*Force.

Disclaimer:

(0.
(Some people might think,)

What goes up
 must come down. 

It’s called the 0^th(=zero_th) law
 because he(=Newton) didN’T invent it.
Joke aside.
 ..But (otherwise) here..)

(To be consistent, & thus rebose.)

Preparation:

The (average_)distance d_a
 divided
 by the (average_)time t_a

 is the average_speed v_a=d_a/t_a.

(I have to (repeatedly) say that “average_”
 so often,
 (because (I try))
 NOT to be misunderstood,
 if I were to (simply) say,
 (the average_)
 speed is d/t.)

1.
The average_momentum

 mom_a=m_a*v_a

 is a(n average_)mass m_a
 (coefficient, factor)
 multiplied
 by its average_speed v_a.

2.
To be consistent((ly) based)
 on the same basis, (then)

 the (average_)mass*(average_)Force
 m_a*F_a=mom_a²/d_a

 is (simply?)
 the average_momentum squared
 mom_a²=m_a²*v_a²/d_a

 (then)
 divided
 by the (average (accelerated) traveled) distance d_a.

(That would be
 my substitute
 for mass*Vis_viva

 &/or
 (what seems to me)
 the (ERRORFUL=misguiding=misleeding,
 (interactive)
 work_)energy (concept
 WE=F*d
 which is
 wrongly (proportioned, to only) force F
 multiplied by distance d
 & (thus) needing an extra mass m factor
 as Ewert proposed 1996.)

Or
 the kinetic_energy_(difference, concept)
 KE_d=m*v_d*v_a
 which uses the speed(_difference)
 v_d=v_f-v_i
 for the final_speed v_f
 minus the initial_speed v_i.

 mom_a²/d_a=m_a*(m_a*a_a)
 is the (average_)mass m_a
 multiplied
 by the average_Force F_a= m_a*a_a
 of a(n average_)mass m_a
 multiplied by the average_(linear_)acceleration a_a=v_a²/d_a.

 mom_a²/d_a=m_a*F_a
 which I would love
 to simply call
 mass*Force(_average).

3.
 [Newton’s
 3^rd law
 of] repulsion
 (of((=for)

 the average) mass*Force)
 explains how
 colliding (accelerating) masses
 are equally proportioned.

(mom_a1²/d_a1=m_a1*F_a1)+(mom_a2²/d_a2=m_a2*F_a2)=0.

Again
 (m_a1*F_a1)+(m_a2*F_a2)=0, m_a*F_a=m_a²*a_a
 (m_a1²*a_a1)+(m_a2²*a_a2)=0

 (mom_a1²/d_a1)+(mom_a2²/d_a2)=0.

Note:

2b. (=To_be)

A general
average_acceleration

 a_an=v_aⁿ/d_a^n-1

 could be based
 (up)on
 the “exponential” average_speed

 v_aⁿ= a_an*d_a^n-1

 (for any n>1, e.g. beyond 1)
 which is
 an(y) average_acceleration
 (even NON_linear kinds=types),
 where “a” is just a symbol
 ((only) meaning (a=the) motion;
 NOT necessarily acceleration!).

“n” is just (for) the order of motion
 where
 0 does NOT move?
 1 is the (average_)speed (=NO_acceleration)
 2 is the average_linear_acceleration
 3 is the average_NON linear_acceleration of the 1^st order
 4 is the average_NON linear_acceleration of the 2nd order
 etc.
 

n<1
Exponent
 n values below(=less_than) 1,
 decelerate.

Capiert: Is NOT a (=1) cycle=360°?
Answer: NOT always.
But am I confusing vocabulary?
"cycle" instinctively tells me "circular".
E.g. Bicycle.
2 circular wheels;
NOT 720°.

1 hour ago, swansont said:

It depends on the system in question. Does a pendulum swing through 360 degrees to complete a cycle? No.

 ..Even though we can make=force the pendulum
 to (circulate=)revolve around 360°
 (at least once)
 with a strong enough push (or swing),
 like (a) Newton's bucket.

But I guess it would NOT have the same Period T, then.

1 hour ago, swansont said:

Does a piston? No. 

The piston itself does NOT,
 but it is connected
 to the circulating crank_shaft.
I guess we extrapolate that ((crank_shaft's) motion), there
 (onto the piston).

Factors
of the Radian is surely NOT the only way to measure angle.

1 hour ago, swansont said:

Nobody claimed otherwise.

So I guess there is hope
 (for me)
 for progress there.

1 hour ago, swansont said:

But if you are using angular frequency, that’s what the measure is defined to be.

Yes, but didN'T I call it angle_speed f?
 using the same or similar syntax?

I'm attempting
 to draw parallels=similarities
 from existing (or similar) syntax.

Slight modifications (simplifications?)
 that might be useful
 (to me).

Is there any reason why
 to use (the archaic? outdated?)
 radian (any more)?

Irrational numbers,
 like Pi=3.14..,
 unnecessarily cost
 more computing time,
 depending on their complexity
 for their accuracy.
Very messy.
We'( a)re lucky enough
 when we can use just (only) a (=1) symbol,
instead of the (irrational) number in full.
Why both at all,
 getting that messy,
 when other (alternative) methods will do?

I used to think omega
 was cool=neat
 till I understood it (a bit) better.

E.g. I (once) thought it was neat
 (but only)
 because I did NOT understand it enough.
Mysterious.

Now I'd like to forget it.
(Which sometimes happens,
 most of the time.)
(I) DON'T need
 so (I) DON'T want it.
It's a waste of time figuring it out,
 everytime.

The decimal part
 of (2 identical, but 1 delayed) frequencies,
 multiplied by time,
 will give the phase shift (angle),
 as a fraction
 of a cycle.

What more could you ask for?
It's decimal (cycle, angle).
It's consistent,
 with the metric_system
 based on tenths, etc;
 instead 1/60ths.

& guess what?
"I" made it up.

Disclaimer:

Instead of just complaining,
 I want to see improvements
 (being made).
Simplicity is the way;
NOT unnecessary=superfluous complexity.

But unfortunately
 some things
 (have to)
 get more complicated
 to get there.

 

On 1/21/2023 at 2:19 PM, swansont said:

Angular frequency measures how much the angle changes (radians/sec) and linear frequency has no angle (cycles/sec). They differ by 2*pi

Is NOT a (=1) cycle=360°?

Is NOT

On 1/21/2023 at 3:44 AM, Capiert said:

A radian (unitless radius/circumference=r/cir=r/(r*2*Pi)=1/(2*Pi)=0.1591549 ratio
 for the ~1/6.28.. fraction of a "cycle", as angle; 57.2957795°)

?
I see 2 different angles (360° & ~57°)
 but both per second.

& as you said
 they differ
 by the factor 2*Pi.

I guess, maybe you mean,
 using a "cycle"
 is a bit more complicated statement
 if for an angle.?

E.g. If ruffly intuitively
1 degree#cycle/360°?
That clashs.
There is something missing
 in the conversion (constant)?

If
 1 cycle=360°, /360°
 then
 1°=1 cycle/360, *360 [cycle/°]
 1°*360 [cycle/°]=1 cycle.

That seems right=correct
 to me (now).

(Please) let me put it another way.
Is 1° an angle? y/n
Is 360° an angle? y/n
If you say no, then why?

E.g. If units
 can be correctly converted
 then why should they be wrong?

Factors
of the Radian is surely NOT the only way to measure angle.
Or is it?

Thanks, both of you.

9 hours ago, swansont said:

It’s angular speed and typically expressed in radians/sec. The symbol is a lower-case omega. f implies a linear frequency

Why the adjective "linear"?

Is NOT frequency (just) frequency?

You are implying a NON_linear frequency (also) exists
 but I see NO need for it
 if the frequency is constant(ly the same, or varying linear(ly))?

I suppose you are making the analogy to straight_line speed(s),
 which can be either linear &/or NON_linear.
Which is also possible (for angle_speeds, I guess).?

9 hours ago, swansont said:

Using non-standard nomenclature does indeed cause confusion 

Perhaps (yes), but I was NOT aware
 until I posted this (speculation) thread
 that I could (=might) also use frequency f
 (which simply uses other (so_called, NON_standard) convertable units)
 to represent angle_speed.

That's new to me!

I guess you guys & gals
 are NOT yet so far
 as to recognize that.
(It took me at least a few days,
 to say the least.)

A radian (unitless radius/circumference=r/cir=r/(r*2*Pi)=1/(2*Pi)=0.1591549 ratio
 for the ~1/6.28.. fraction of a "cycle", as angle; 57.2957795°)
 & does seem (to me)
 to be an encrypted other alternative (for angle)
 57.2957795°/360°=0.1591549 (of a [cycle]);
 instead of the (more) common [degree], (f)or circle, or e.g. cycle=360°.
The advantage of using the cycle directly
 is "1"=360°.

It (=1 [cycle]) is less complicated.
Cycles (=360° multiples) are a very common unit, e.g. cycles_per_second cps=[Hz].
(If you had to explain a [Hz] to someone,
 how else would you do it
 than with that definition?)

I see NO advantage
 for the invention of the radian
 other than to make things more obscure
 with irrational numbers.

9 hours ago, Bufofrog said:

-1 for your insistence on using this terrible formatting and your reluctance to use LaTex. 

Sorry!

9 hours ago, Bufofrog said:

It's bad enough that your ideas are wrong but to then make these ideas almost unreadable is intolerable. 

I'm sorry if you can NOT digest it (=my ideas) fast enough.
I doubt that it (f as angle_speed) is wrong;
 because it is convertable
 to other formats,
 such as degrees/sec or radians/sec or RPM
 & is perhaps
 too new for you, yet.

I mean somebody had
 to have invented the degree (definition);
 & somebody else the radian (definition)
 with its questionable syntax.

The cycle (as unit) is NOTHING new.
---
Very many people use Rich text
 it is very common.

Disclaimer:

Winword's copy paste
 into this sfn website
 deletes the formulas.

I DON'T need those stupid surprises
 with your incompatible software.

(I suspect the solution would be to convert to .pdf
 & then copy paste that.)

I'm NO pro(fessional)
 with LaTex
 for that little
 that I do with it;
 & I constantly=repeatly
 forget its details.
(I am NOT gifted
 with the patience
 to use such a rare software,
 found on (mostly only) this website.
I DON'T speak Chinese either.)

I also would like my threads
 to look nicer than they are.

But the ERROR possibility
 is NOT worth the risk yet.
I'm already making too many errors
 (as I (have to) scramble to correct them).

& I know some of you would like me to bumble.

 

(Please) take a look
 at the math.

If I reduce (=decrease)
 the eccentricity
 to almost zero
 (then) Kepler's_orbit_Period (3^rd law)
 is (still)
 T_Kepler=r_a^1.5;

 but (however) at zero eccentricity e=0
 (then) both the semi_major_axis r_a
 & the semi_minor_axis r_b
 are equal
 to the same (=identical),
 radius
 r=r_a=r_b if e=0
 (but) having the circular orbit_Period

 T_circle=2*Pi*((r/g)^0.5).

Surely,
 for an Earth's satellite,
 both Orbit_Periods
 T_{Kepler_(e=0)}#T_circle
 r^1.5#2*Pi*((r/g)^0.5), ^2

 are NOT the same
 &/or
 can NOT be the same.

Squaring both sides

 r³#4*Pi²*r/g, /r
 r²#4*Pi²/g, ^0.5
 r#2*Pi/g, *g
 g*r#2*Pi, let the centrifugal_acceleration a_c=g equal the gravitational free_fall (for an orbit)
 a_c*r#2*Pi, because
 the centrifugal_acceleration
 a_c=v_c²/r
 multiplied
 by the common radius r
 as
 a_c*r=v_c²

 is the (circular) orbit_speed squared v_c²;
 & (that v_c²) is (definitely) NOT 2*Pi.

v_c²=

 a_c*r=2*Pi*(2*Pi*r²/T²)
 is a factor
 of *(2*Pi*r²/T²)
 larger than 2*Pi.

Thus,
 (as far as I can see)
 Kepler's math
 is NONSENSE!
 (=it makes NO SENSE to me
 in its existing form);
 it has NO consistency
 with regular (Newtonian) Physics;
 & must have a different meaning.

I hope that explains
 my problem
 (with Kepler('s laws))
 well enough.

2 different (circular Orbit_period) formulas,
 which 1 is correct? (Kepler's or Newton's?)

Please DON'T tell me Newton('s centrifugal_acceleration) is WRONG!
..because that is from where I derived
 my circular Orbit_Period from.

On 1/15/2023 at 9:41 PM, swansont said:

What other “kinds” of KE are there?

KEf=KEi+KEd
That looks like 2 others to me.
Yours is
 KEd=KEf-KEi,
 you let KEi=0.

On 1/15/2023 at 9:41 PM, swansont said:

I only know of the “kind” that is given by 1/2 mv^2

It can be applied to different particles and be measured/calculated at different times (hence you have initial and final as two common times)

That's a good (interesting, simple) perspective.

But what about KE transfer
 between (accelerating) masses
 that have acquired KE in similar times?

Or is that what you mean (already)?

I will assume
 the syntax
 could get
 a bit hairy.

On 1/15/2023 at 9:41 PM, swansont said:

Reference frames are incredibly useful for those that actually solve physics problems.

On 1/15/2023 at 9:41 PM, swansont said:

No. the difference in kinetic energy represents the change in kinetic energy at two different points in time. It is not a definition of KE. 

In other words
 you are saying,
 KE applies
 to only 1 accelerated mass.

On 1/16/2023 at 12:17 AM, Lorentz Jr said:

It means you're analyzing the problem in the object's initial reference frame.

It may simplify the math a bit.

It's the object's kinetic energy at the end of the experiment.

It's the object's kinetic energy at the beginning of the experiment.

It may be zero, but it doesn't have to be.

How do we know (if) the initial_kinetic_energy KEi is zero or NOT?

It (=KE) is always wrt the initial_speed
 of the reference_frame;
 but does that say or mean zero KE?

If the reference frame is moving
 (which it surely is)
 then we can NOT say a static mass
 wrt to that reference_frame
 has absolutely NO KE.

On 1/16/2023 at 12:17 AM, Lorentz Jr said:

It's the difference between the final and the initial.

On 1/17/2023 at 1:06 PM, Sensei said:

@Capiert

https://www.google.com/search?q=earth+orbital+speed+at+perihelion

https://www.google.com/search?q=earth+orbital+speed+at+aphelion

Sensei
Thanks for the links
 but (I suspect) you missed my point (=speculation).

I'm saying Kepler's 3rd law is wrong for an Orbit_speed;
 & (=but instead) it should (=might) apply to an angular_speed.

Btw.
Is there any (Foto) evidence that orbits
 are (suppose to be) elliptical?
I know he empirically determined T~Ra^1.5.

On 1/17/2023 at 1:06 PM, Sensei said:

Have you ever seen a pendulum ?

Yes.

On 1/17/2023 at 1:06 PM, Sensei said:

Is its speed highest at the highest point (=far from the center of the Earth) ?

No.
Did I say it was?

(The pendulum(s) I saw
 was NOT orbiting
 around the Earth.)

The (pendulum_)Period formula T
 does NOT mention height;
 & I am NOT discussing Pendulums (anyway);
 but instead Orbits
 which use the same (similar) formula.

I just wanted to point out
 their math similarity
 so you can quickly understand.
(Perhaps I said things wrong, in my haste?)

On 1/17/2023 at 11:28 AM, Genady said:

Apo(gee] and perigee _refer_

(to an) orbit (=around)

//Earth// (while?)

aphelion (and__)

perihelion ..&refer**.. (to_an!)

orbit around ((Sun---))

Thanks for the correction.
I could NOT find the vocabulary quick enough, on the fly.
I wanted to use something like Periside & Apside
 but forgot the (exact) spelling.
(So (unsure) I backed down, & (I) had to rush for something else ((&) that did NOT work).)
 

ERRATA: Typo.

My angle_speed example
 was a ruff quick random (hash) mixture (example);
 NOT a (formal) formula.
Just to give a (vague) ruff idea (comparison).
I had originally intended something like
~theta/t, ~360°/T
 but I found a general formula getting too complicated,
 for the application I wanted.
So I wrongly settled for a condensation
 theta[/t, ~360°]/T -> theta/T (when theta=360°),
 where that (particular) T was any kind of time
 NOT intended to be (just only) a period,
 but as a quick comparison
 to what a (real) Period should hint at.
Sorry. (Trying to pack too much info,
 into too little space,
to keep things short.)

My angle_speed (formula, of choice)
 should be called "frequency"
 f=(theta/t)*(1 [cycle]/360°) =1/T
 for the angle theta in (units) [degree(s)]=[°];
 t is an amount of time (duration, e.g. difference in time)
 in (units) [second(s)];
 & period T is the (amount of) time (duration), per cycle.

There are several ways
 to express angle
 & (thus, also) angle_speed.

More info in that thread
 "Angle_speed f(?)"
 if you need it.

On 1/17/2023 at 1:06 PM, Sensei said:

@Capiert

https://www.google.com/search?q=earth+orbital+speed+at+perihelion

https://www.google.com/search?q=earth+orbital+speed+at+aphelion

Have you ever seen a pendulum ?

Yes.

On 1/17/2023 at 1:06 PM, Sensei said:

Is its speed highest at the highest point (=far from the center of the Earth) ?

No.
Did I say it was?

The (pendulum_)Period formula T
 does NOT mention height;
 & I am NOT discussing Pendulums (anyway);
 but instead Orbits
 which use the same (similar) formula.

I just wanted to point out
 their math similarity
 so you can quickly understand.
(Perhaps I said things wrong, in my haste?)

On 1/17/2023 at 11:28 AM, Genady said:

Apo(gee] and perigee _refer_

(to an) orbit (=around)

//Earth// (while?)

aphelion (and__)

perihelion ..&refer**.. (to_an!)

orbit around ((Sun---))

Thanks for the correction.
I could NOT find the vocabulary quick enough, on the fly.
I wanted to use something like Periside & Apside
 but forgot the (exact) spelling.
(So (unsure) I backed down, & (I) had to rush for something else ((&) that did NOT work).)
 

ERRATA: Typo.

My angle_speed example
 was a ruff quick random (hash) mixture (example);
 NOT a (formal) formula.
Just to give a (vague) ruff idea (comparison).
I had originally intended something like
~theta/t, ~360°/T
 but I found a general formula getting too complicated,
 for the application I wanted.
So I wrongly settled for a condensation
theta[/t, ~360°]/T -> theta/T (when theta=360°),
 where that (particular) T was any kind of time
 NOT intended to be (just only) a period,
 but as a quick comparison
 to what a (real) Period should hint at.
Sorry. (Trying to pack too much info,
 into too little space,
to keep things short.)

My angle_speed (formula, of choice)
 should be called "frequency"
 f=(theta/t)*(1 [cycle]/360°) =1/T
 for the angle theta in (units) [degree(s)]=[°];
 t is an amount of time (duration, e.g. difference in time)
 in (units) [second(s)];
 & period T is the (amount of) time (duration), per cycle.

There are several ways
 to express angle
 & (thus, also) angle_speed.

More info in that thread
 "Angle_speed f(?)"
 if you need it.

 is called "frequency"
 f=(theta/t)*(1 [cycle]/360°) =1/T
 for the angle theta in (units) [degree(s)]=[°];
 t is an amount of time (duration, e.g. difference in time)
 in (units) [second(s)];
 & period T is the (amount of) time (duration), per cycle.

There are several ways
 to express angle
 & (thus, also) angle_speed.

But I prefer
 to express angles
 in cycles(' fractions, &/or multiples),
 e.g. as fractions, &/or multiples
 of a cycle.

1=100%.

E.g.
1 [cycle]=360°.

That is(=means),
 a degree
 1°=[cycle]*(1/360)
 is 1/360th part
 of a(=1, complete=whole, single) cycle.

E.g.
 90°=0.25*[cycle]=[cycle]/4.

1 [cycle]=(t/T)*(360°/theta), or (rearranged)
1 [cycle]=(t/theta)*(360°/T).

Or regrouping the angles together
 as a ratio
 f=(theta/360°)*(1 [cycle]/t) =1/T.

A few other variations
 for the angle_per_time ratio, are

theta/t=(360°/T)*(1/(1 [cycle])), or swapping (Rt side) denominators
theta/t=(360°/(1 [cycle]))*(1/T), is also
theta/t=(360°/(T*[cycle])), 1/T=f
theta/t=f*360°/[cycle].

The period, is
 T=t/(1 [cycle]))*(360°/theta), or
 T=(t/theta)*(360°/(1 [cycle])).

The angle (in degrees), is
 theta=(t/(1 [cycle]))*(360°/T), or
 theta=(360°/(1 [cycle]))*(t/T).

The angle (in cycles), is
 f*[seconds].

(Even) although
 I'( a)m in the habit
 of using degrees.

I hope that clears
 any confusion.

Motivation:

I noticed
 a general formula
 for angle_speed
 was a little bit
 more involved,
 & (so) I was searching
 for an appropriate syntax (symbol, hieroglyphics).

I hope that will (simply) do
 with an (already) existing symbol f.

?
 

Perigee (e.g. perihelion, is the nearest distance to the sun, &)
 has the slowest orbit_speed, but fastest angle_speed (angular_velocity);
 &
 Apogee (e.g. Aphelion is the farthest distance (away) from the sun)
 has the fastest orbit_speed, & slowest angle_speed (angular_velocity).

I distinguish
 between orbit_speed (v_c=Cir/T=2*Pi*r/T);
 versus the angular_speed (theta/T);
 (Disclaimer:
 otherwise Kepler's 3rd law
 is NONSENSE
 (for me),
 who(m) I pity).
 

Tycho (Brahe)
 (only) measured
 planets'(_position):
 angles
 & time (dates);
 so we
 can calculate
 the(ir orbits) angular_speeds,
 e.g. how fast arcs (=angles).
 are swept.

To summarize (my interpretation):
1.
Kepler found
 that (planets') obits
 were NOT perfect circles
 (& (he simply) approximated that (orbit)
 to an ellipse (math)).
2.
Instead,
 an egg (shaped orbit)
 has only 1 focus (center),
 found near the smaller end.

(The conic_section
 was done (algebraically)
 from a cone

 (with its circular base sitting
 on the ground (y=0),
 & its apex (x,y=0,H=0,1) up
 at the top)

 having the same radius R=H=1
 equal to its height H
 (which seems to be the key
 to the results);
 & then normalizing
 by dividing
 by the egg's total (diagonal_)length L
 which starts at the cone's base_circumference (left_side, x,y=-1,0)
 going (diagonally up, to the right)
 thru the center axis (focus x,y=0,h)
 where the (partial) height (fraction) h
 is the eccentricity (Epsilon);
 (& continues till it pierces (out, thru) the cone's right side).
Your mathematicians
 should be capable
 of similar results.
If you DON'T believe me
 you can form bread dough
 & cut it, appropriately.)
 
Equal arcs (=angles) are swept [out]
 in equal times.
3.
A circular orbit
 uses (=has)
 the pendulum_period
 T=2*Pi*((r/g)^0.5), g=ac
 which is similar
 to Newton's centifugal_acceleration
 a_c=v_c²/r
 if a_c=g.
Why should an orbit_speed v_c
 increase
 with smaller radius?
E.g. For very small eccentricities?
(It does NOT, because..)
There is NO consistency,
 when extrapolating
 to large(r) eccentricities.
Nature does NOT abruptly change her laws;
 (but) men do.
Especially when they did NOT understand.

1 hour ago, exchemist said:

Yes, well, kinetic energy has to be specified with respect to a frame of reference, because velocity is relative. 

I find that a good answer.
It implies admitting ~1/2 the problem.

But DON'T you think
 that (reference frame) need
 has been created artificially,
 by negating
 the initial_speed v_i
 to zero?

Now you have
 to suffer the consequences
 & are happy
 with it(s complexity).

1 hour ago, exchemist said:

Why can't you post in normal length lines?

This style
 helps me
 reduce my errors,
 which otherwise would then be "more" confusing
 to you.

It helps me find subtle problems.

1 hour ago, exchemist said:

This style is very exhausting to read.

Sorry, (1st draft, needs maybe a 5th re_edit?)
 that's my handicap.
It's a construct, e.g. programmed.
Intended to read
 with natural pauses
 almost automatically;
 but sometimes
 it has to be broken down more
 (e.g. bracketed, (is (superfluous &) put) to the right end, (thus) ignorable)
 but gets messy.
Me (when) against many
 (must be perfect,
 but isn't).
It takes a lot of backbone
 to go "against" mainstream (errors?).
(Sometimes) It's difficult to (get=) crystalize the ideas
 in order to (quickly?) pinpoint the problems;
 instead of make new 1's.
I write more naturally
 when I feel uninhibited.
Swansont will say there are NO errors
 in physics;
 only me,
 but sorry I DON'T believe him
 on some things (yet).

1 hour ago, exchemist said:

It makes one question whether it is worth the trouble.  

It's pointing at a serious problem.
Prejudice (bias), & ignorance of it.

Please DON'T bother.
It's intended to hedge (=prevent)
 future problems (conflicts).

It'( ha)s obviously failed
 for now.

I DON'T like it either.
Stressy bad weather.

This (thread) is (about)
 a (syntax(_bias)) complaint.
There are several kinds of KE;
 but only 1 is acknowledged.

What does it mean
 to set
 the Kinetic_Energy(_difference)'s
 (KE_d=m*(v_f²-v_i²)/2)

 initial_speed v_i=0
 to zero?

 KE_d=m*(v_f²-0)/2. (?)

 (I mean..)
 If the Kinetic_Energy(_difference)
 KE_d=(m*v_f²/2)-(m*v_i²/2)

 is the 1^st term, the final_kinetic_energy
 KE_f=m*v_f²/2

 minus the 2^nd term, the initial_kinetic_energy
 KE_i=m*v_i²/2

 as,
 KE_d=KE_f-KE_i

 & the initial_kinetic_energy
 KE_i=m*0²/2=0
 is (assumed) zero,
 then we can only have
 the final_Kinetic_Energy
 KE_f=KE_i+KE_d

 KE_f=0+KE_d

 or (when swapping, the sequence order)
 KE_f=KE_d+0

 which "might" appear
 as (the_same=)
 identical
 (when it is NOT always (identical)
 e.g.
 when initial_speed v_i#0
 & initial_kinetic_Energy KE_i#0
 are NOT zero).

Thus (we or)
 I must ask
 (what they ((term_names) really) are (representing)) e.g.
 what is the final_Kinetic_Energy KE_f?;
 & what is the initial_Kinetic_Energy KE_i?;
 & what is the Kinetic_Energy(_difference) KE_d?

It seems
 (to me, (that))
 physicists
 are (absolutely)
 completely (=100%)
 confident
 that the Kinetic_Energy_"difference" KE_d=KE_f-KE_i
 is NOT "The(e)" (=their) Kinetic_Energy KE_f-0
 because

 (perhaps (the) subscripts are (or were) missing?)

 although both
 are (a kind of)
 (kinetic_)Energy.

(But differentiation,
 is a "difference" method
 (adding the series
 (of differences).)

(Why the restriction
 to only 1 kind of KE
 when there are several (kinds)?)

(E.g. (Oh) "KE"(?), we worship (only) "Thee"! (?) :-)
(But which 1?)

(Kinetic is (=implies) "motion",
 thus KE is motion_energy.)

Why do they (physicists)
 bother splitting hairs**
 (just to be different)?

(**E.g.
Tending to ONLY 1 kind, of=from several.
Preferring ONLY
 the KE_difference,
 (when=)with its initial_speed v_i
 (is) set to zero.) 

(I mean)
 (Because..(it seems to me, that))
 the Kinetic_Energy "difference" KE_d
 is "the" most "general"***
 definition
 of Kinetic_Energy;
 & setting its initial_speed v_i=0
 is (but) "only 1" (specific) example
 (of Kinetic_energies, =motion_energies)
 from many (examples). 

(***
 I'( ha)ve stated that,
 in other threads before;
1.
 Energy is suppose to (be accountable, &) add
  if it can NOT be created,
  NOR destroyed;
2.
 but it does NOT always (add (up)),
3.
 although KE_d is (already) relative.
Now including the subscript d=difference (syntax).)

Thus, how can (you or)
 those physicists
 (possibly) suppose
 that:
 the final_Kinetic_Energy KE_f
 (but) minus the initial_Kinetic_Energy KE_i (by letting v_i=0)
 is (suppose to be)
 "the(e)" ONLY Kinetic_Energy KE? (=KE_d)= KE_f- (KE_i=0)
 ((possible)
 that exists,
 & is correct (only) so)? 

[That seems
 to me
 rather "narrow_minded",
 or misled.

E.g. 1 Person's (restriction_)ERROR,
 & then everybody jumps on the (same) bandwagon
 (without thinking).

E.g. NOT quite knowing
 what you are talking about.
=]

DON'T you think
 that'( i)s a little bit
 narrow_minded?
Or misled?

E.g. NOT (quite) the full story.

Or am I wrong?
 & if so why?, etc.

E.g.
Science confirms itself.
There are many ways
 (to Rome=roam). 

Disclaimer:

That was
 (more, or less)
 my complaint
 about (an) inconsistent syntax (bias)
 in physics.

I DON'T claim
 to be perfect (either),
 & (I (also) have
 to make (syntax) compromises
 (e.g. preferences)
 (in order, for them (compromises))
 to remain recognizable
 (at least a bit).

Physics is a straight jacket,
 as Feynman would say.

For those who DON'T know:
"Straight" actually (is) implying,
 too rigid (e.g. &/"or")
 (for the lunatics).

Take it (=&/"or", leave it) either way you want.

It's crazy.

I hope I'm NOT being too impolite;
 & can drop the wool
 from your (biased) eyes.

(I'm biased too, but differently.)

(This is the time=Period
 of revolution,
 & we (=fat Victor & Co)
 are revolting.-Victor Buno. :-)

Why prefer initial_speed zero?

 Einstein said NO (reference_)frame is preferred;
 but that is NOT true.

There are advantages to some;
 or at least
 that is what you are implying=inferring.

E.g. (A mass m, on Earth, with)
 v_i=0 seems static
 but it is really (moving)
 at the same(=identical) speed
 (as the reference_frame, i.e. wrt Earth).

 (& we know the Earth is rotating, e.g. moving.)

 Thus (initial_)speed v_i
 is "inherent".

 It (=v_i)
 (when zero)
 is invisible
 to the (reference_)frame.
(E.g. Zero is =NOTHING, 0 [m/s]. But) 

I can NOT imagine anything
 that is NOT moving
 (wrt the universe).

& even the universe
 must be moving
 wrt some_thing.

On 8/16/2022 at 12:36 PM, swansont said:

You got that right: it's math, not physics. So why bother with it? It's just math you made up, and has no applicability to physics.

If you are NOT doing the math right=correctly
 then how can you expect
 "your" physics to turn out right?
Apparently, NO math?;
 then NO physics.

The simplest
 of algebra
 should
 indicate your (=anyone's) folly.

On 8/16/2022 at 12:36 PM, swansont said:

You choose the method that applies to the problem. It comes from understanding physics and the experience of having solved similar problems many, many times. You can evaluate and see what conservation laws can be applied, and what other equations can be applied.

Good!

On 8/16/2022 at 12:36 PM, swansont said:

If you think you can solve a problem with mass*energy, go ahead and show that it works: derive the formula from known physics, and show that gives the correct answer. Tell us when it applies and when it doesn't, so others can test it (and make sure that whatever example you give isn't correct by accident)

(That's a) good challenge.

On 8/16/2022 at 12:36 PM, swansont said:

That's not a good enough citation.

True.
ISBN 3-89478-119-x
pg28
If I remember correctly.
That'( i)s what I got out of it.

On 8/16/2022 at 12:36 PM, swansont said:

momentum is not kinetic energy.

Quite right.
mom=KE_d/v_a
KE_d=mom*v_a

On 8/16/2022 at 12:36 PM, swansont said:

They are distinct concepts.

KE_d=mom*v_a
They depend
 on each other
 (in_spite
 of their distinction).

On 8/16/2022 at 12:36 PM, swansont said:

You can check that KE is lost because you can calculate the KE for each object, and the values do not match. As in the example I gave. 

KE is NOT lost;
 "the values (simply) do NOT match"
 because you are doing the algebra WRONG!

There is a RIGHT=CORRECT way to add KE
 & there is a (simpleton's) WRONG way
 which hinders match(ing),
 that EVERYBODY has fallen for.

On 8/7/2022 at 11:51 PM, swansont said:

  On 8/7/2022 at 11:22 PM, Capiert said:

But it (that mass) is distributed differently
 (into the (mass) terms)
 so that it is only (1 mass, now e.g.) together.

& that ((different) togetherness)
 (now) has a different affect
 on the math answer.

On 8/7/2022 at 11:51 PM, swansont said:

They’re moving at the same speed is all.

Well done!

But if the proportionality
 to distance
 is a (correct) mass_squared;
 instead of (only mass) NOT squared,
 then you have a definite difference
 that can NOT be pushed_away (=denied!)
 (& hid under the carpet).

On 8/7/2022 at 11:51 PM, swansont said:

What’s the physics of this “togetherness” that has a “different affect”?

The numbers (resulting) for mass_squared
 versus only mass (NOT squared).
Btw. That ("squared" versus NOT_squared (mass))
 is math;
NOT physics.
Math_Physics at the most.

On 8/7/2022 at 11:51 PM, swansont said:

Actual physics says nothing about this, of course. Mass and speed - existing physics variables - are all you need to describe what’s going on.

Quite right.
But with what (=which?) method
 you choose,
 to describe,
 determines your outccome.
Choose momentum
 & then you are probably (quite) right.
Choose Energy
 & then it can go wrong
 when mass changes
 in the collision.
Choose mass*Energy
 & then you might get it right (again),
 when dealing with changing mass.
The choice is (really) yours (NOT mine).
(I'( woul)d call that bias.)

On 8/7/2022 at 11:51 PM, swansont said:

But it’s not.

..What?
(Please explain yourself.
"NOT" what?)

On 8/7/2022 at 11:51 PM, swansont said:

There’s no m^2 in kinetic energy or momentum equations.

Quite right.
Mass_squared
 does NOT exist in either
 of those (mom or Energy) equations,
 (& because they are inferior, e.g. limited).

Mass_squared is a new concept
 from Ewert (1996)
 & it does seem to work right=correctly,
 e.g. better,
 such as in problems
 where the mass changes.

& the cool (=neat) thing
 about it (=mass_squared)
 is m*E,
 mass*Energy (m*E)
 is VERY similar
 to (only) Energy
 so you hardly (=barely)
 need to change a thing.
It'( i)s very easy.
Simply multiply mass
 by Energy
 & you might (=should) get
 the correct answers.
That'( i)s over_simplified
 of course.

On 8/7/2022 at 11:51 PM, swansont said:

You’re just making up a problem.

What sort of problem
 am I making up?
I see definite problems
 with Energy
 that can NOT be over_come
 (otherwise).
Dark_Energy is a real eye_sore.

But the so_called
 energy loss
 in an inelastic (=NON_elastic)
 collision
 is most prevalent (=obvious).

& I am NOT making that 1 up.
YOU are!

The Energy NEVER left the system.

& I can tell you where it is.

On 8/7/2022 at 10:03 PM, Capiert said:

The total final kinetic_energy
 KE_Tf=KE₃=4 [J].

The total initial kinetic_energy
 KE_Ti=KE_1i+KE_2i=4.5 [J]+0.5 [J]=5 [J].

The Energy defect (e.g. neutrino=(math_)error! KE_d4)
 is the initial_total energy
 minus the final_total energy
 KE_error=KE_Tf-KE_Ti=4 [J]-5 [J]=-1 [J] lost!
 =-20% of (the total) initial(_energy)
 (for that (NON_elastic) example).

I (wrongly) said that you "lost" -1 [J];
 but that is wrong
 only because you guys condition me
 into your same bad habits.
(& I fell for it,
 as naive as I am.)

You haveN'T lost a thing!

Take a look at
 (conservation of momentum, COM)
 mom₁+mom₂=mom₃.
That'( i)s a perfectly balanced equation.
Left_side (is mom₁+mom₂) is "before" the collision
 & right_side (is mom₃) is "after" the collision.
Square both sides
 & it is still balanced (=perfect)!
 (mom₁+mom₂)²=mom₃².
But that produces
  mom₁²+mom₂²+2*mom₁*mom₂=mom₃².
That +2*mom₁*mom₂ belongs to "before" the collision (=left_side
of the equation);
 NOT "after" the collision (right_side of the equation).
Again:
That +2*mom₁*mom₂ belongs to the "before"_ side
 of the collision equation;
 NOT the "after"_side of the collision equation.
Putting the 2*mom₁*mom₂ on the "after" side
 of the collision equation
 means it "needs" to be NEGATIVE.
I.e. -2*mom₁*mom₂.
NON_elastic collisions
 might seem
 to have lost energy
 on the (heavier) result;
 but they have NOT lost a thing
 (in that sense);
 you need only observe the math
 to correctly determine
 what has happened.

& (thus) you physicists (have) preach(ed) an untrue gospel (=NONSENSE!)

Sorry!

But more SORRY for your (misled) followers.

1 hour ago, swansont said:

  Quote

when my mass has increased
 e.g. in an inelastic collision.

1 hour ago, swansont said:

Mass doesn’t increase in a collision. The mass at the beginning of the example is the same

(amount (of mass))

1 hour ago, swansont said:

as the mass at the end.

But it (that mass) is distributed differently
 (into the (mass) terms)
 so that it is only (1 mass, now e.g.) together.

& that ((different) togetherness)
 (now) has a different affect
 on the math answer.

1 hour ago, swansont said:

In a completely inelastic collision, the two masses

 are now 1 total mass
 m₃=m₁+m₂
 which is now larger
 than either 1
 of its initial components
 (e.g. either: m₁; or m₂)
 &

1 hour ago, swansont said:

are moving at the same speed.

If mass is squared,
 then its parts
 produce different (math, number, value) results
 compared to its (=the total mass) sum squared.

 m₃²=(m₁+m₂)²
 m₃²=m₁²+m₂²+2*m₁*m₂   <---But look at that!

m₃²#m₁²+m₂²

Mass squared does NOT simply add
 only the squared terms;
 because the (NON_squared term) +2*m₁*m₂
 is missing!

I hope you get my drift.
(What I am "trying" to tell you.)

Such naive math (just) can NOT work, (sometimes).

On 8/6/2022 at 3:18 AM, Capiert said:

That'( i)s how you (would) do it.
E.g. Just so you do NOT have to bother.

But why should I (try to) believe you.
 

On 8/6/2022 at 2:00 PM, swansont said:

You can check in any college 1st semester physics textbook and confirm that this is the mainstream view.

It’s a basic definition.

I’m giving you information, and the benefit of my expertise. Not bothering would mean not responding at all.

Nobody here owes you any of their time.
 

Yes, True. Sorry for my (sudden, shocked) reaction.
(Anything to do with Newton 3?)

It's difficult
(for me)
 to accept
 that you will ignore
 & deny the syntax,
 & thus undermine
 any (further) argument;
 making all rationality attempts seem pointless.

(That takes more than a few aspirin,
 with the feet up
 (to recover from).)

It'( i)s similar,
 to (suddenly) declaring
 that the Earth's North Pole
 (which is known & documented for centuries,
 as found in the northern hemisphere,
 near, side by side,
 to the north (geographical) rotational_"axis")
 is suddenly (now, renamed to) a south (magnetic) pole
 because your compass needle
 is pointing to it (=the Earth's North Pole).

& everybody else's compass
is (now) doing the same thing (too=also)!
They are (all) "pointing to"
 (the North Pole).
(Which is NOTHING new!)

I mean, why was the North "Pole"
 named so
 in the 1st place
 (instead of axis)?

Obviously because
 magnetic loadstone
 orientated
 towards the Earth's 2 magnetic poles.

Nobody then knew what kind
 of magnetic pole
 each end of the loadstone had;
 other than to follow its (end's) orientation
 to the Earth
 (as reference).

There was NO other reason
 for naming it "pole"
 (instead of axis)
 other than that it was (& still is)
 a "magnetic"_pole!
 

Clear cut!

Now some (so_called) genius
 comes around
 with his (own) compass
 in his hands
 (seeing is believing)
 (ignoring that it (=the compass's_needle)
 is only "pointing"
 to the north(ern) direction,
 & (=but) says "his" needle('s end)
 must be North (magnetism)
 (when it is NOT!),
 because its pointing
 "to" the Earth's north (magnetic) pole.

I mean how far does narrow_minded egoism have to get.

You get 1 (raving) idiot
 (ignoring the consistency=continuum
 with the rest of the concept, &) 
 messing (=mixing) things up;
 & then every other (stupid) idiot jumps on the bandwagon, too
 (as ja_sayers);
 instead of saying NO!.

(Stunned with astonishment,
 which boardered upon stupification
 (..they left the forest).-Jules Verne.

NOT seeing the forest
for (=because of) the trees.)

That is only an example
 of inconsistent logic.
 
In my book
 the Earth's magnetic pole
 is in the Northern hemisphere
 (as it should be)
 (NOT the southern);
 & I have seen maps so, too.

I see NO reason
 to change that,
 for the crackpot idea
 that it should be south,
 just so some egoist
 can (hi_jack the show,
 &) gain all the attention
 as a "new" breakthru
 just to be different=elite.

I mean which came 1st
 the chicken
 or the egg?
Obviously, "its" egg:
 the proprietary owner('s right(s)).
I.e. Historical sequence,
 squatter's rights.
Credit where it belongs.

But getting back
 to tracking KE_d.

On 8/6/2022 at 4:45 PM, swansont said:

In some cases it’s difficult to track; you might not be measuring it. But in e.g. a completely inelastic collision,

You must (surely) know,
 (at least by now),
 that:
 KE does NOT work
 for NON_elastic collisions
 because
 KE's mass m is NOT correctly proportioned
 to the speed v_f.
The mass will change (amount)
 (becoming larger)
 (instead of stay(ing) the same).
That is a strictly=pure(ly) math error
 (of WRONG proportioning).

Mass*Energy m*E
 helps correct that error
 (of WRONG proportions);
 but (it seems) you wish
 to ignore that
 (proportioning error;
 & m*E,
 altogether).

Kinetic_Energy on its own
 is (inferior) NON_SENSE!
 (because it'( i)s only (an intermediate,
 a) part
 of a term
 of m*E);
 that's why it (=KE) will NOT work generally
 for (exchanging to) different (sized) masses
 (such as (in)
 e.g. NON_elastic collisions).

On 8/6/2022 at 4:45 PM, swansont said:

it doesn’t matter, since conservation of momentum applies and allows you to solve the problem.

Oh yes it does matter,
 because you ignore COE is fake
 by (still=continuing) declaring energy
 is a conserved quantity
 (although NOT completely).
It either is, or it is NOT.
Energy has a flaw,
 & breaks down sometimes
 in some cases, e.g. NON_elastic collisions;
 & you refuse to admit
 that weakness
 is a (math) ERROR!

Your general (conservation) laws
 rely completely
 on only COM (conservation of momentum);
 & NEVER COE (conservation of energy),
 because COE is corrupt!
 (an incomplete distortion)!

You misleed people
 (like me)
 into believing
 conservation of energy
 (is general)
 when it can NOT be (e.g. completely),
 inspite of your limitation warnings.
It's like saying, I'm a bit pregnant;
 but NOT completely.
I'm NOT (pregnant)
 when my mass has increased
 e.g. in an inelastic collision.

(Partially (conserved) is NO excuse for the real McCoy!, COM.)


m*E is a unified form(ula)
 of both COE & COM.

Why need to choose
 anything else?

1 formula
 will now finally do.

On 8/6/2022 at 4:45 PM, swansont said:

In a situation with friction it may not matter if you can calculate the amount of work done.

Some problems might just not be solvable without additional information. 

I think that is the reason
 why I am probing (=e.g. asking)
 what kind of energy KE_d is.
I'( a)m asking for that additional info.

I suspect we have something (new?) to learn
 with the correct answers.

At least from
 that endeavour
 we might
 develop more ability
 e.g. simpler formulas,
 e.g. shortcuts.

(Why hinder the progress?)

On 8/6/2022 at 4:45 PM, swansont said:

You shouldn’t add text to anyone’s quote, since a quote implies that it’s an actual quote. i.e. what they said.

OK. Good advice.

On 8/6/2022 at 4:45 PM, swansont said:

Then there is no change in the motion.

NOT true.

v_1i=(1000-1=~999) m/s
v_2i=(1000+1=~1001) m/s

v_3f=~1000 m/s

On 8/6/2022 at 4:45 PM, swansont said:

So what? It’s a different example than what I gave. 

(And this ignores that you can choose whatever convenient frame of reference you want. I chose one where the momentum is zero. But pick whatever frame you want; in a completely inelastic collision, the kinetic energy is smaller after the collision).

To make things simple,
 let'( u)s put your example
 in a car (or train, frame)
 (that is)
 moving (at only e.g.) +2 [m/s] (eastwards;
 instead of 1 [km/s]);
 & watch the (masses') speeds
 from (wrt) the (Earth's surface) ground.

Mass 1 (initial):
m_1i=1 [kg]
v_1i=3 [m/s]
mom_1i=m₁*v₁=1 [kg]*3 [m/s]=3 [kg*m/s]
KE_1i=m₁*(v₁²-0)/2=1 [kg]*(3 [m/s])²/2=4.5 [kg*m²/s²]=4.5 [J].

Mass 2 (initial):
m_2i=1 [kg]
v_2i=1 [m/s]
mom_2i=m₂*v₂=1 [kg]*1 [m/s]=1 [kg*m/s]
KE_2i=m₂*(v₂²-0)/2=1 [kg]*(1 [m/s])²/2=0.5 [kg*m²/s²]=0.5 [J].

Mass 3 (final):
m₃=m₁+m₂=1 [kg]+1 [kg]=2 [kg]
v₃=? [m/s]
mom₃=mom₁+mom₂=3 [kg*m/s]+1 [kg*m/s]=4 [kg*m/s]
v₃=mom₃/m₃=4 [kg*m/s]/2 [kg]=2 [m/s]
KE₃=m₃*(v₃²-0)/2=2 [kg]*(2 [m/s])²/2=4 [kg*m²/s²]=4 [J].

The total final kinetic_energy
 KE_Tf=KE₃=4 [J].

The total initial kinetic_energy
 KE_Ti=KE_1i+KE_2i=4.5 [J]+0.5 [J]=5 [J].

The Energy defect (e.g. neutrino=(math_)error! KE_d4)
 is the initial_total energy
 minus the final_total energy
 KE_error=KE_Tf-KE_Ti=4 [J]-5 [J]=-1 [J] lost!
 =-20% of (the total) initial(_energy)
 (for that (NON_elastic) example).

Things look different(ly)
 with mass*Energy m*E,
 because the defect
 is m₄*KE₄=~-2*mom₁*mom₂.

But
 there is a big difference
 between m*E=mom*moma=mom²*(v_a/v)
 versus
 mom²=mom*mom=m*E*(v/v_a).
 

The catch there
 is the average_speed
 v_a=(v_i+v_f)/2
 needs the initial_speed v_i;
 which is a speed
 you ignore
 by subtracting it out.
E.g. By (typically) setting
 the initial_speed v_i
 to be your reference_frame's
 speed.
Thus it (v_i) is gone from the picture.
Anything that has the same speed
 as v_i
 is declared
 as "at rest"
 (when it's NOT really
 when wrt any other speed than v_i).
"At rest" is the deceiving(_speed) exception,
 of really being (=having) the "same speed".

But you (already) know that.
I'm NOT telling you anything new.

What bugs me is the partial truth.
(The biggest lie,
 is only half the truth.-A banker.)
E.g. The exceptions.

On 8/6/2022 at 4:45 PM, swansont said:

KE not being conserved means you can’t assume it is as a general principle.

Ok, I can acknowledge,
 that KE is only 1 kind of energy
 (among [m]any);
 (but it's (=KE is) NOT the big cheese(y principle)!)
Energy (in general)
 is that (general concept)
 (even if it has its weaknesses).

On 8/6/2022 at 4:45 PM, swansont said:

It doesn’t mean you can’t find individual examples where it doesn’t change. i.e. “not conserved” does not mean “must change”

IOW:
It means, you can find individual examples
 where KE does NOT change.
(E.g. identical masses?)
I.e. “not conserved” means “might change”;
 but does NOT have to change.

I sure hope
 that will sink in(to my head).

Maybe you could (please) give me
 an example
 of that,
 (so I can remember better,
 instead of (me) falling
 into the same old hole)?

I'( woul)d like
 to present
 you the m*E example;
 but (can NOT do it properly yet,
 because) you'( ha)ve
 gotten me off track
 by (=into) zeroing
 my KE_f & KE_i
 away from (my original concept)
 KE_d=m*v*v_a.

E.g. Versus mom²=m*(m*v*v).

v#v_a.

Which means
 I (still) have to give it some thought,
 as to the consequences.

(=I need some time.)

11 hours ago, swansont said:

A change in KE means the energy is no longer in the system, or has been added to the system.

That'( i)s how you (would) do it.
E.g. Just so you do NOT have to bother.

But why should I (try to) believe you.

You have given me NO proof,
 with your inability
 NOT to bother (attitude).

Are all scientists so lazy
 (like Minkowski hinted
 about Physicists)?
(Surely NOT!)

Why should (some of the kinetic) energy leave that system?
(& I DON'T accept warm, acoustic, excuses either.)
(I'd like to see (simple) tangible measurements.)

11 hours ago, swansont said:

Where that energy went or where it came from (i.e. the kind of energy) is not accounted for in these examples.

Oh! Abracadabra (then).
(It's a mystery!)

That sounds like a boring disinterest in science
 e.g. trying to know.

A half hearted attempt
 to throw a few things together.

You either: know; or (else) you DON'T,
 & you obviously DON'T,
 because you give me useless excuses.

Sorry, other people can try to be more thorough.
You DON'T even give the effort.

If you (were to) say:
 those answers can NOT be found;
 then there must be a reason why.

(Oh we are too feeble,
 (at) attempting, (to) zero_speed, =zero results.
It's more difficult that c.)

Disinterest is NO excuse.

You also avoid commenting
 (up)on the (=my)
 initial_kinetic_energy KE_i
 (perhaps because you habitually evade it
 by subtracting it away).

My syntax includes KE_i.
(What is your syntax,
 if mine is NOT an extended (syntax)?)

All 3 (named KEs)
 KE_f=KE_i+KE_d
 are "kinetic_energies"
((meaning) NOT your "the"(what? _unknown),
NOT mentioned _f form) syntax).

You can clearly see that
 (they are kinetic_energies)
 in my syntax "KE"
 with a subscript.

There is NO difference:
 meaning a KE is a KE,
 whatever its subscript is.

The KE_d can (equally)
 accelerate a(ny) mass
 from zero (speed)
 to a (new) final_speed v_f,
 which would finally have
 its own KE_f(new)=KE_d
 equal to that kinetic_energy_difference KE_d.

I DON'T see why you try
 to sell a KE_d distinction
 (away) from any other KE_subscripted
 just because you do NOT know
 what (else) KE_d is
 (or could be).

E.g.
Even though you only want KE_f
 to be "the" (only) kind of KE (possible).

It is absurd to say: KE_d
 is NOT a kinetic_energy
 simply because it is NOT "the" final_KE
 KE_f=m*(v_f²-0)/2
 which uses the mass m multiplied
 by half the final_speed squared v_f²
 but "subtracted by zero(_squared)"!

The initial_kinetic_energy
 KE_i=m*(v_i²-0)/2
 is also a kinetic_energy
 (just like KE_f is)
 because its
 half the initial_speed squared v_i²
 but is also "subtracted by zero(_squared)"!

The universal
 KE_difference formula
 KE_d=m*(v_f²-v_i²)/2
 is the most universal
 KE "definition"!
There you can (=may)
 use any reference(_frame) speed
 v_ref=v_i
 (below c, that)
 you want
 to be your reference speed
(e.g. at rest, when identical
 to the initial_speed v_i).

If I have 7 oranges (analogy KE_f)
 & subtract 4 (oranges, analogy KE_i),
 then I would expect out, 3 (oranges, analogy KE_d)
 like any reasonable thinking person.
NOT grapefruits or "lemons"! (or other hogwash).
That's only common sense,
 which seems to be missing here
 (in (what some people call) science).

You DON'T (even) have a clue where the energy has gone,
 you DON'T know what it is (e.g. called, other than "difference")
 & yet want to be called scientists.
(& you want me to believe you?)

Modern physics is like modern art,
 anything goes,
 even junk.
All that matters is who (e.g. what ego) has the say.

11 hours ago, swansont said:

It’s not required if these equations apply.

I suspect you mean,
 the initial_speed terms
 (e.g. KE_i)
 are subtracted anyway;
 (&) so why bother.
?

11 hours ago, swansont said:

Look at the example I gave of the head-on collision. The KE is not lost to another particle even though there was an acceleration. There is zero KE after the collision. Both particles lost KE. Nothing gained KE.

Taken from a different perspective,
 of: if the 2 masses are on Earth
 & the Earth is rotating
 let'( u)s say v_i=~1 [km/s]
 eastwards

 ((just) to keep things simple,
 at where they are
 on the Earth's surface);

 then they are still moving
 ~1 [km/s]
 although they appear
 to you
 as at rest.

(& that KE_d did NOT leave the system.)

What seems (as) "at rest"
 is an optical delusion
 (of) for both: observer;
 & an object,
 having the same (=identical) speed.

(E.g. Even though they are separated
 by a distance d.)

In reality (e.g. the universe),
 (we know)
 everything is moving.

Meaning NOTHING is (really) static (=at rest, with zero speed).
(Everything has a speed difference
 wrt some other (moving) object (reference, frame).)

Your "choice"
 of reference(_frame)

 (e.g. of same speed
 as the observed object)

 (help) determines
 whether you want
 to be: deceived ((in)to think(ing): )
 an object is at rest (when it has the same speed
 as the reference_frame);

 or NOT!

11 hours ago, swansont said:

KE is not generally a conserved quantity. It’s only conserved in the special case of elastic collisions.

I didn’t claim that (**KE is destroyed=annihilated). I claimed the opposite: KE is not conserved

I did not write KE (=KE_f**). You must have edited the quote, which is grossly dishonest.

**(Sorry! (Yes) "I") Modified.
(Or do you mean your quote is dishonest? Which I (rather) doubt,
 in preference for the former.)

How else should I add (extra) comments
 of mine
 into your text?

I bracketed it,
 to distinguish it from your original text.

Dishonesty was NOT intended;
 only clearness of the discussion
 (was intended),
 (before getting lost again
 in(to) confusion
 between syntaxes).

Should I use different brackets?

(Our) Confusion arises
 from different syntax:
 mine; versus yours.

15 hours ago, swansont said:

Almost certainly. The thing is, people have been doing physics for quite some time, so the syntax is well-established. 

What you're doing is similar to showing up in a foreign country

I agree.

15 hours ago, swansont said:

and expecting them to speak your language.

& they do speak English.

15 hours ago, swansont said:

  16 hours ago, Capiert said:

I have extended the syntax description details
 beyond yours.

15 hours ago, swansont said:

No, not really. 

KE_f being final kinetic energy is fine, if you mean it the same as everyone else: the KE at the end of the example you're analyzing

Yes, I think we can assume so.
(& It's nice to hear it is fine.)

15 hours ago, swansont said:

For difference we use ∆, so ∆KE = KE_f - KE_i 

So what you are doing is not "extending" the syntax, you are introducing new syntax where it already exists, which is confusing. 

You still have NOT commented on KE_i.

15 hours ago, swansont said:

∆KE is not a

(final_)

15 hours ago, swansont said:

KE

KE_f

15 hours ago, swansont said:

that any one particle has, so it's a value for energy,

What kind of Energy?
It's NOT chemical, thermal, NOR potential etc.
So what KIND of energy is that?

& also,
 only a particle
 could have had that energy
 when it was by=with the particle.

If a particle had lost that KE_d
 then that particle has decelerated
 & is now moving slower,
 than it was (before the loss).
But it was lost to another particle
 (forced to accelerate,
 E.g. Newton's 2nd Law).

KE_d can only be had by a particle (or object) e.g. mass.
The KE_d formula can discuss (either): the same mass m;
 or (else) it can discuss 2 different masses.
But (the) K(inetic )E(nergy) d(ifference)
 can NOT be transferred
 without a mass (as mediator, or transporter).
So what you said, e.g.
 that KE_d does NOT belong to a particle,
 (simply) does NOT (seem to) make sense
 (to me).
KE_d can only be transferred by (a) mass, (instead).
But:
The bigger, more important, question
(to ask)
 is: to which mass does KE_d belong to;
 & when? (before versus after, a(n elastic) collision).
E.g. Which mass lost the -KE_d?
 & which mass gained the +KE_d?

NO mass?
Then NO energy!
You can NOT describe (e.g. formulate) energy
 without mass.
(& charge is intimately
 related to the inverse "mass",
 as in charge to mass ratio.)

15 hours ago, swansont said:

but it's no longer describing the

motion_

15 hours ago, swansont said:

energy of a particle, so calling it a

 final_

15 hours ago, swansont said:

KE isn't correct.

I think we can fictively equate KE_d
 to various other masses & speeds
 just to get equivalents
 (or equivalence).
(But that'( i)s modelling,
 e.g. NOT physical.)

I find your next example (description)
 (non_elastic collision)
 rather interesting.

15 hours ago, swansont said:

Example: Two 1 kg objects

Let (their masses, be)
 m₁=1 [kg], &
 m₂=1 [kg].

15 hours ago, swansont said:

moving at 1 m/s collide head-on 

Let their initial_speeds, be
 v_i1=1 [m/s], & =j⁰*1 [m/s], j⁰=1 (0°)
 v_i2=-1 [m/s]=j²*1 [m/s], j²=-1 (180°)

 with initial_KE's
 KE_i1=m₁*(v_i1²-0²)/2=1 [kg]*((1 [m/s])²-0)/2=0.5 [kg*m²/s²]=j⁰*0.5 [J] (0°)
 KE_i2=m₁*(v_i2²-0²)/2=1 [kg]*((-1 [m/s])²-0)/2=1 [kg]*((j²*1 [m/s])²-0)/2=j⁴*0.5 [kg*m²/s²]=j⁴*0.5 [J] (360°)

15 hours ago, swansont said:

and stick together, coming to rest.

Their final_speeds, are
 v_f1= 0 [m/s], &
 v_f2= 0 [m/s]

 with final_KE's
 KE_f1=m₁*(v_f1²-0²)/2=1 [kg]*((0 [m/s])²-0)/2=0.5 [kg*m²/s²]=j⁰*0 [J] (0°)
 KE_f2=m₁*(v_f1²-0²)/2=1 [kg]*((0 [m/s])²-0)/2=0.5 [kg*m²/s²]=j⁰*0 [J] (0°)

with KE_differences
 KE_d1=m₁*(v_f1²-v_i1²)/2=1 [kg]*((0 [m/s])²-(1 [m/s])²)/2=-0.5 [kg*m²/s²]=j²*0.5 [J] (180°)
 KE_d2=m₁*(v_f2²-v_i2²)/2=1 [kg]*((0 [m/s])²-(-1 [m/s])²)/2=1 [kg]*(-(j²*1 [m/s])²)/2=-j⁴*0.5 [kg*m²/s²]=j⁶*0.5 [J] (540°).

Both (Energy) losses (together) are -1 [J],
 each -0.5 [J].

15 hours ago, swansont said:

Their change in KE is -1 Joule,

KE_d1=KE_f1-KE_i1=-0.5 [J]
KE_d2=KE_f2-KE_i2=(-)³*0.5 [J]
KE_d3=KE_d1+KE_d2=-0.5 [J]+(-)³*0.5 [J], ~-1 [J].

15 hours ago, swansont said:

but at the end of the example nothing is moving,

How then can you claim
 that their (kinetic) Energy
 was NOT destroyed (=annihilated)?

Are you claiming out of (good) "belief"?
(E.g. Scientific religion.
 E.g. Hoping the theory
 will justify every "odd" detail
 in the end?)
A reasonable person
 could NOT claim that,
 (when) seeing the facts.

(Seeing is believing.)
(You must (=most probably, surely) have good faith, (brother)!)

15 hours ago, swansont said:

so saying ∆KE is a

 kind (=type)
 of (moving (=motion) energy)

15 hours ago, swansont said:

kinetic energy is incorrect.
There is nothing in that example that has -1 J of kinetic energy.

Although we can loose energy
 (by using the minus sign symbol "-"),
 ((but) into NOTHING!).

15 hours ago, swansont said:

A kinetic energy can't be negative.

Because the math
(e.g. (final_speed) "squaring")
 does NOT allow negative( value)s.

15 hours ago, swansont said:

A kinetic energy can't be negative.

Quote

because of:
 (1.) conservation of Energy COE:
 i.e. energy can NEITHER be created;
 NOR destroyed,
 & so the parts must add (up, together);

15 hours ago, swansont said:

Yes, energy is conserved, but kinetic is only one form of energy. KE itself is not a conserved quantity. See the above example of a completely inelastic collision.

I have (seen the inelastic example above), & it looks convincing
 (& puzzling, at the same time).
It does NOT fit, like indefinite integrals (do).

15 hours ago, swansont said:

As I have shown, ∆KE is not a value associated with any one particle, or even anything having motion. It is a useful value to know in ma(n)y problems, but to be useful it must be properly labeled, so one can do a proper accounting of the energy present.

I DON'T want to argue with you there
 (not a value associated with any one particle, or even anything having motion.)
But I suspect a closer examination (in detail)
 will allow associating to particles,
 & of motion.

15 hours ago, swansont said:

I choke on (calculus).

15 hours ago, swansont said:

Well, that's your problem. Calculus works regardless of your understanding or dislike of it.

3 indefinite integrals begin to stand (=argue) against (a) calculus working regardlessly.
Serious mathematicians know calculus does NOT "always" work right.
they have been successful
 at isolating those few quirks;
 but they CAN'T quite
 put their thumb on why
 to eliminate them completely.
They simply set up warnings, DON'T do.
With those Jack in the boxes,
 I'm curious what else
 (for problems)
 could be found
 awaiting us
 for: in the future.

Btw. What is half of infinity?
Calculus works with infinitesimal limits,
 hoping to make the problem so small
 so it disappears.
The problem is the proportioning
 should stay the same,
 instead of approximating
 that it is gone.

15 hours ago, swansont said:

I agree that ∆KE = 0. That's the problem. You had said KE_d=m*(v_f²-v_i²)/2, rather than saying this was ∆KE

Yes, with your help
 we have identified a problem
 that could be cleared.
Thank you.

That'( i)s the way we learned it
 in school.
& looking at it (now)
 it seems
 like a (more/most) general description:
 dealing on a macroscopic level/perspective;
 rather than a microscopic infinitesimal view.

15 hours ago, swansont said:

IOW, you were claiming that some object's KE (=KE_f) was described by the equation.

I'm sorry, but I have to disagree with you there.
That is how you interpreted my text's syntax.

You assumed I was talking about KE_f;
 but I was always talking about KE_d, instead.

I DIDN'T use the _d subscript
 as is typical for many variables
 in Physics.
E.g. (duration) time t=t_f-t_i
Eg. distance d=d_f-d_i
E.g. speed v=v_f-v_i.

When you measure a distance
 you rarely say it is a distance_"difference"
 of e.g. 100 [km] from 1 big city to another.
You just say the distance is ..;
 & forget about the fact
 that it is a difference, too=also.
Why the (complexity &) inconsistency
 in your syntax?
Answer: Because normal people DON'T talk that way.
You guys have to turn everything into a unique exception
 rather than talk like normal human beings.
I mean "sometimes" there are advantages
 to the (few extra) details;
 but NOT always.

15 hours ago, swansont said:

And for an object moving at some constant speed, it's kinetic energy is decidedly NOT zero. For an object not starting from rest, this does not give the object's KE (=KE_f). Which makes your equation wrong.  

The equation is KE_d (NOT your (agreed upon) KE_f);
 & as KE_d=m*(v_f²-v_i²)/2
 that equation is NOT wrong
 but instead absolutely correct.

∆KE=KE_d

It should have been obvious
 (excluding typo errors),
 that
 KE_f#m*(v_f²-v_i²)/2
 could NOT possibly be (correct)
 as KE_f
 because it was missing (the term) -m*(v_i²)/2;
 instead of being exclusively KE_f~m*(v_f²)/2
 using v_f²;
 without -m*(v_i²)/2 losses.

That is,
 unless you have found some(thing else) new, breath_taking news
 that we should (all) be cautious (about), for.

Btw.
I must comment
 that I was rather enthusiast
 at Bufofrog's original comment,
 (but unfortunately wrongly)
 assuming he might have found an interesting (new) problem
 that we could work on,
 e.g. he was thinking outside of the box.

But instead,
 it only turned out to be (boring) syntax misunderstandings.

But at least we got that cleared too.

Thanks gang (=team)!

I suspect the confusion arises
 (between us)
 from the difference
 between my syntax
 versus your syntax.

I have extended the syntax description details
 beyond yours.

Your "the" (what?) KE,
 is my "final"_KE KE_f.

My KE_d is the (KE_)difference
 of 2 ((perhaps) different) Kinetic Energies.
In this case (final_KE minus initial_KE) KE_f-KE_i.
(We can thank Swansont for helping us point out that 1.)

I (would) find it (quite) peculiar
 if you would deny
 that the difference
 between 2 (different) KE's
 is NOT a KE (itself),
 because of:
 (1.) conservation of Energy COE:
 i.e. energy can NEITHER be created;
 NOR destroyed,
 & so the parts must add (up, together);
 &
(2.) per definition_name
 kinetic_energy
 is (=means) the energy of "motion".
There speed &/or acceleration apply.
Delta_KE=KE_d must be a KE
 (even if it is NOT your "the" KE,
 whatever "the" is suppose to be.
Disclaimer: I am only trying to be thorough,
 with the(=my, algebra_)syntax.)
 

If a mass object
 maintains a constant_speed

 (for that brief_time that "you" measure (it);

 instead of (versus, compared
 to) following its complete history
 as to how it (ever) acquired its (constant_)speed
 in the 1st place=originally),

 then, its initial_KE KE_i
 (which is NOT ZERO, (&) which you seem to want to ignore
 (maybe) just to makes things difficult(?),
 but which I have already taken into account (of))
 will be equal
 to its final_KE KE_f
 & (but, thus)
 its KE_difference KE_d=KE_f-KE_i
 will be (exactly) zero.

(The analogy
 is taken directly
 from my speeds(_syntax)
 for the speed_difference
 v_d=v_f-v_i.

That is NOTHING new!

(Except for (now) adding _d subscripts
 (for "difference"),
 where they were NOT before.)

Swansont should know
 that for linear acceleration
 the average_speed, is
 v_a=v_i+v_d/2
 &
 his final_(instantaineous)_speed (equivalent), is
 v_f=v_a+v_d/2.

E.g.
v_f=v_i+v_d

On 7/19/2022 at 4:18 PM, swansont said:

So this is based on a mistrust of math, and as a result you use math that's less appropriate. 

I have NOT yet recognized
 that that algebra will NOT do.
Meaning it (=the algebra) is still valid
 (as far as I can see).

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

I'm aimed (=intended) at collisions, or jerks!

Really?

Yes!

On 7/19/2022 at 4:18 PM, swansont said:

You only mentioned collisions in passing until now.

Jerks are also accelerations,
 (similar to collision accelerations).
Whether they (accelerations) are linear(?), I doubt it(!).

On 7/19/2022 at 4:18 PM, swansont said:

Those who study and understand physics know the limitations; KE is not a conserved quantity in collisions except in a special case (elastic collision) because there are other possible forms of energy (e.g. thermal, sound, deformation)

Unfortunately there I suspect you assume too much for what is actually happening
 instead of measuring.

Am I right?

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

& I am looking for a more general, (but) simpler description.

And yet you use average speed, which makes assumptions as well.

Quite right.

(NO bout a_doubt_it.)

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

 KE=PE (pronounced "keep", versus PE=KE pronounce "peek")

No, really, that's not how it's pronounced. They are pronounced "Kinetic Energy" and "Potential Energy" and they are sometimes equal to each other.

"keep" & "peek" are (only) how I have nicknamed the formulas (acronymns if you will)
 as an easy reminder.
NOT intended to offend anyone('s physics, indeed).

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

But that (instantaneous speed)
is theoretical,

Nothing theoretical about it. It's the value of the speed at a particular time.

How then do you "measure" it (=instananeous_speed) exactly,
 especially for collisions?
Relying on math (e.g. calculus) sounds pretty theoretical to me.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

Constant speed is NOT how the universe is working.

But it's how the example worked that you gave. When you solve a problem, you use the appropriate physics for the problem, so if the speed is constant, you can use an equation for constant speed. And there are lots of problems where speed is constant, or that's a reasonable approximation of a situation.

Please recognize (=acknowledge) my initial_KE Ki syntax into the problem.
That'( i)s what it is there for,
 i.e. (it was designed)
 for such problems
 with (initial) constant_speed.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

I am more interested
 in finding the reasons
 why some (math) things (fail)
 do NOT work sometimes (as we (would) wish);
 & (then) try to fix=repair them.

If you can show that math is inconsistent that's a purely mathematical issue and has nothing to do with physics.

If physics is using such math,
 then DON'T expect it (=physics)
 to hold itself together.

I would (rather) say, then it (=that purely mathematical issue) should NOT have to do with physics
 (but unfortunately (does &) is messing it (=physics) up).

Universities live & breathe calculus. (They love it!)
I choke on it.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

 

I CAN'T follow you there,
 because "getting" (e.g. interacting)
 from initial_speed
 to final_speed
 (when they (speeds) do NOT equal)
 is acceleration.

In the example given they were the same, because speed was constant.

Changing the parameters of someone else's example and then complaining about a problem that arises is not an argument made in good faith.

Sorry (I was so slow), I must 1st grasp the problem
 before I can understand
 (what is happening)
 in order to tackle it (later).
Stating what (examples) I know
 helps get me nearer
 e.g. Later then excluded.
Stated otherwise:
 I had (wrongly) assumed
 you could follow me or my syntax,
 thus I could NOT understand
 where a problem was, if any.

On 7/19/2022 at 4:18 PM, swansont said:

On 7/19/2022 at 2:40 PM, Capiert said:

 

Making them (2 speeds) equal
 is an exception,
 because (that is why) they have different names
 so that they do NOT have to be the same.
The (2 speeds)
 "can" be quite different.

They can be, but the point was that if they aren't your equation quite obviously fails. It's wrong. But you're ignoring that. Nobody else is fooled by the distraction.

Again, (when using my KE_i, which I had thought you were already aware of)
 my 02:40 comment
 described what else my equation could also do.
If you see that (extra) as ignorance,
 well then I'm sorry
 (for the misunderstanding).

(But) I still do NOT see that my equation has failed.

Simply take the (=my) KE_i
 because it equals the (=my) KE_f
 (which has (=is) your KE).

The KE_d has (=is) NOTHING!

That has obviously succeeded,
 don't you think?

To summarize the syntaxes:

Mine:   Yours:
KE_d      delta_KE
KE_f      "the" KE
KE_i        ?

I hope I have answered that sufficiently
 to your satisfaction(s).

17 hours ago, swansont said:

If the speed is constant, then why all the commotion about average speed? It's constant!

Constant speed is NOT how the universe is working.
& I am looking for a more general, (but) simpler description.
Constant speed is the exception.
 (E.g. "Special" Relativity.)
Interaction is the (higher, more dominating) rule.
(Actually more basic=fundamental.)

& there is NOTHING
 which says
 acceleration must always be linear.

I get the idea "average" speed
 is an eye_sore for you
 because you prefer
 your instantaneous speed, instead.
But that (instantaneous speed)
is theoretical,
 & I suspect sometimes misleading.
(I DON'T want to be lulled
(or else tempted)
into a false security.
A(n unexpected surprise)
 jack_in_the_box
 (e.g. delayed) time bomb
 is NOT my idea of fun
 (for accuracy
 in extrapolation).)

Calculus works "perfectly" (=flawlessly, NO errors)
 for "linear" acceleration
 because it is based on an (exact) "average";
 (e.g. 2 triangles making a rectangle
 using a diagonal (line))
 but I have my doubts
 for NON_linear (line) examples
 & irregular curves.

E.g.
Sigma_summing
 & (versus) the integrals (integrating, integration)
 should give
 the same (=identical)
 answers;
 but DON'T always. 

I would prefer
 to develop
 something
 more simple(?)
 & experimental,
 (e.g. averages),
 based on algebra, instead.

But I also notice(d)
 that algebra has its limits
 (& errors) too.

(So I can NOT trust algebra completely, either.)

But I have NOT had enough chance
 to pinpoint
 exactly what those flaws=weaknesses are.
(It happens so rarely, like a flipped coin standing on its edge.)
I suspect it(s (math) flaws, or errors) has
 to do with (math) syntax compromises
 when dealing with negative exponents
 for totally=completely (symmetrically) reversible (e.g. recoverable?) math.
E.g. A "negative" exponent
 should mean "rooting"
 as the reverse (operation)
 of exponentiation;
 (instead of a divided exponentiation).
(Not to mention the meaning of the exponent zero, then.)
& then there is the theme of sequence (history).

(Math is a language
 & its sentences
 are equations
 (the phrases formulas).)

But that'( i)s (all) too speculative
 right now.
Too early to say (anything useful).

--

You are content with
 that you have learned
 some (of th(os)e complexity's) math rules,
 & use them.

I am NOT.

I am more interested
 in finding the reasons
 why some (math) things (fail)
 do NOT work sometimes (as we (would) wish);
 & (then) try to fix=repair them.

Experimenting
 with equations of motion
 & (so_called) mass
 are a good place=way
 to start,
 at seeing & discovering
 how this universe works.
Geometry as well.
(E.g. The "basis". Real proof!)

17 hours ago, swansont said:

There's no reason to worry about initial and final speed, since it doesn't change.

NOT for me.

I CAN'T follow you there,
 because "getting" (e.g. interacting)
 from initial_speed
 to final_speed
 (when they (speeds) do NOT equal)
 is acceleration.
Making them (2 speeds) equal
 is an exception,
 because (that is why) they have different names
 so that they do NOT have to be the same.
The (2 speeds)
 "can" be quite different.

I'm aimed (=intended) at collisions, or jerks!

(E.g. Unifying) the changes. 

Maybe you can tell me what I "should" be saying
 (or have NOT said)
 in my syntax description
 for initial (versus final) speeds?

Because what I say
 (just) does NOT seem
 to land
 to you (all)
 as intended.

17 hours ago, swansont said:

Speed is always relative to the frame of reference to which it is measured. As is kinetic energy.

What's your point?

Energy is (all) about (linear) acceleration (i.e. gravity).

That was the reason
 for the vis_viva (living force, ~2*KE)
 controversy;
 & why Gravesande "abandoned" Descartes & Newton's momentum.

(& that is the reason why we live with that mess NOW.
i.e. Dark Energy ERRORS.)

The speed is NOT constant for acceleration(s),
 such as (for, in) collisions
 where m*E is transferred.
 KE=PE (pronounced "keep", versus PE=KE pronounce "peek")
 (e.g. Conservation of Energy COE)
 might work (well)
 for (accelerating) a single (moving) mass
 that stays the same (sized mass);
 but (it, KE=PE)
 does NOT (always) work well
 when that mass interacts
 (& is accelerated)
 by other various sized masses
 (&) of different speeds.

For them (collisions, interactions),
 m*E works better,
 because it is more general, universal,
 uniting conservation of momentum COM,
 & Energy COE
 together
 into 1 formula.
 

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

And mass-energy is famously not a conserved quantity.

I guess I am an early bird.

(DON'T forget the worm.)

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

∆KE = KE_f-KE_i

i.e. KE_f-KE_i is the change in kinetic energy, not the kinetic energy. This is a very important distinction.

What then is KE?
Forced distance?

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

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.

I guess you mean
 I can NOT rely on
 gravity's force
 as constant.
So its a poor example.
?

20 hours ago, Bufofrog said:

22 hours ago, Capiert said:

Why is (your) initial_speed v_i (=v_ref) NOT zero?

Because we are talking about a constant velocity.

22 hours ago, Capiert said:

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?

That would be the reference frame that measures the objects velocity at 10 m/s.

That seems like an indirect answer
 by trying to be (more) general.

21 hours ago, Capiert said:

But the real (more probing) question
 is from where
 or what "speed" (reference)
 are you using
 to measure the "speed"
 you are observing.

(Obviously, your own speed
is the reference, there.)

E.g. The Earth,
 ground or surface
 that you are standing on.
Isn't a "reference frame"
 rather fictive imaginary.

I'm talking
about a real (physical) world.

51 minutes ago, swansont said:

Your assertion of Watt's equation is questionable, and this last equation is not quite correct

There's a difference between instantaneous power and average power

Undoubtably average_power.

I was NOT aware
 that James Watt used instantaneous power.

He used how fast
 a (180 lb (force) ~82 kg (mass)) "horse"
 walking (continuously) a circle,
 using yards (=3 feet=0.9144 m)
 & seconds (time),
 could wheel up water
 from a well
 against (the horse's) gravity's weight (806 N).

~737 Watt (rounded to 740 W for 75 kg).
746 W would have needed 75.59 kg (~167 lb) at 1m/s.

I question that math
 was any great deed
 of calculus.

Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a minute).^[6] The wheel was 12 feet (3.7 m) in radius; therefore, the horse travelled 2.4 × 2π × 12 feet in one minute. Watt judged that the horse could pull with a force of 180 pounds-force (800 N). So:

Watt defined and calculated the horsepower as 32,572 ft⋅lbf/min, which was rounded to an even 33,000 ft⋅lbf/min.^[7]

Watt determined that a pony could lift an average 220 lbf (0.98 kN) 100 ft (30 m) per minute over a four-hour working shift.^[8] Watt then judged a horse was 50% more powerful than a pony and thus arrived at the 33,000 ft⋅lbf/min figure.

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

50 minutes ago, swansont said:

What if there is no acceleration?

Then the "speed" should be constant.

But the real (more probing) question
 is from where
 or what "speed" (reference)
 are you using
 to measure the "speed"
 you are observing.

(Obviously, your own speed
is the reference, there.)
 

Continued, (from my previous)

That speed_change
 is the acceleration
 e.g. to or from
 a collision,
 but once that collision (duration e.g. t<1 , stops=) finishes
 the(n that) acceleration stops;
 so that friction decelerates the ball (further)
 to a (slow, much longer duration, e.g. t>1)
 stop.

 But,
 an Earthquake
 (e.g. an abrupt change
 in the Earth_plates' (rotational) speed)
 could shake
 the (table &) balls
 (in)to motion.

Done.

How would you describe (your) KE?
 (..without problems).

 That was mine (=my description).

Identical speeds
 does NOT mean
 zero KE.

 It only means
 zero KE
 between
 the 2 objects
 that have the same speed.

Constant speed
 does NOT mean
 zero KE.

 In fact any speed
 (helps)
 indicate a KE.

(E.g. If you (also) know the mass(es), too.)

But any (observed) speed(_difference) v
 is a (speed_)difference v,
 i.e. that uses a reference (initial_)speed v_i
 (as basis),
 to observe the (final_)speed v_f=v_i+v.

(But)
 we do NOT see the (final_)speed v_f
 because it has the (initial_)speed v_i
 in it,
 which we exclude.

E.g. The Earth's_rotation_speed
 v_i~1000 [m/s] eastwards
 remains invisible
 (to us)
 because everything (static) around us
 is (also) moving
 at (approx.) that same speed.

How can we solve
 your problem? 

P.S.
When Earthquakes happen
 1 of the problems
 is the ((tectonic) plates') rotation
 slows down
 & then continues
 to speed up;
 repeatedly.

That kind of (kinetic) Energy
 can bring down skyscrapers (crashing),
 (it's so powerful!)

(m*PE=m*KE)

 m*m*a*d=m*m*(v_f²-v_i²)
 only equates
 the accelerated distance
 with the speed change
 of a mass;
 NOTHING more. 

a*d=((v_f²)-(v_i²)).