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you cannot divide m/s by m/s and have a result in anything
other than m/s.

1 m/s^2 / 10 m/s^2 = 0.01 m/s

you certainly cannot end up with a result that is simply
an amount of time.

the result would be given in m/s

you can drop the m (meters) and only use the seconds.

1 s^2 / 10 s^2 = 0.1 s^2

or

1 s / 100 s = .01 s

so we have dropped the velocity from the equation.

and we only kept the time.

we now subtract the result of .01 s from the number 1
so we must also drop the time from the equation in order to
subtract.

1 - .01 = .99

we then find the square root of the number .99 ?

0.99498743710662 ?

what could this number possibly represent?

we have dropped the units of distance
we have dropped the units of time

if we do associate this number with anything then
we must associate this number with a velocity.

we must give the result in m/s (distance / time)

so the time dilation equation is an equation that is
used to solve for velocity it is not used to solve
for time.


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Originally Posted By: Paul
you cannot divide m/s by m/s and have a result in anything
other than m/s.

Time for a basic math refresher.

10 M/S / 10 M/S = (10/10) * (M/S / M/S) = 1 * 1 = 1

That is the correct math for the situation.

Bill Gill


C is not the speed of light in a vacuum.
C is the universal speed limit.
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paul Offline OP
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Quote:
10 M/S / 10 M/S = (10/10) * (M/S / M/S) = 1 * 1 = 1


1 what?

10 orange/peels / 10 orange/peels = (10/10) * (orange/peels / orange/peels) = 1 * 1 = 1

1 what?

should we give the answer in units of oranges or peels?
why would we only use 1 peel?

10 watts/second / 10 watts/second = (10/10) * (W/s/W/s) = 1 * 1 = 1

1 WHAT?
should we give the answer in units of watts or seconds?
why would we only use 1 second in the above?

the above is exactly what the equation is doing
except its using two separate quantities of
distance and time vs energy and time.


LOL

in the equation what is being divided is the distance
not the time.

the equation only uses the distance in the (m/s)

the equation never touches time so how can it be considered
that the equation delivers any amount of time as its result
or any number that can be associated with any amount of time?

so any number that can be a result from the equation is
a velocity.

ie .999999999 m/s





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Originally Posted By: paul


...
we then find the square root of the number .99 ?

0.99498743710662 ?
what could this number possibly represent?

we have dropped the units of distance
we have dropped the units of time

if we do associate this number with anything then
we must associate this number with a velocity.

we must give the result in m/s (distance / time)


so the time dilation equation is an equation that is
used to solve for velocity it is not used to solve
for time.
No, velocity is not what the equation solves for.

The number you are asking about is a ratio, I think, and as such it would be without units.

Once you have the ratio, you multiply it by t (in seconds) to get
the answer, t prime, which will also be in seconds--the dilated time.

~


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paul Offline OP
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Quote:
No, velocity is not what the equation solves for.


just because the equation has a t' in front of it
does not mean that it solves for t'

the only element that actually undergoes any calculation
is distance ie... Meters

time is never used or calculated.

if time is never calculated then how could time be a result
of the calculation?

just because the inventor of the equation says so?

then lets try this equation out.

time = sqrt (1-(1 kgm^2)/(10 kgm^2))

there is no amount of time involved in the above equation.
nor is time calculated in the above calculation.
will the result of the above equation be an amount of time?

the result of the above equation can only be given in units
of kgm

or mass and distance

and the result will always be less than 1 kgm
because thats the intended purpose of the equation.

thats why the 1-(?/?) is there.

so no. the equation t'=tsqrt(1-(v2/c2) does not deliver
any number as its result that could possibly be associated
with any amount of time.

the only possible association that the number could be
associated with is units of distance (meters).




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Originally Posted By: paul
just because the equation has a t' in front of it
does not mean that it solves for t'
...just stunning.

"in front of it?!?"
Paul, by definition, an equation involves the stuff on both sides of the "equals sign" doesn't it? There is no place that is "in front of" the equation. The equation is about the relationship between t' and t, which are both "time" measured in seconds.

~ cool ?


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paul Offline OP
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Quote:
The number you are asking about is a ratio, I think, and as such it would be without units.


yes , it is a ratio , it is a ratio of distance
since each expression v and c are expressed in units of
m/s (meters per second).

then the ratio is a ratio of meters per second.

50 m/s only has 1 second attached to it.
squaring that 50 m/s does not affect the number
that is associated with time (1 second)
it only affects the number that is associated with distance (50).

finding the square root of 2500 does not affect the time
either , it only affects the distance and 50 meter distance is
where the 2500 came from.

and the speed of light only has 1 second attached to it.

so the ratio cannot be time as the time units remain the
same throughout the equation so the ratio is a ratio
of distance.

solve for the ratio of time in the following equation
ratio t = 1m/sec / 10m/sec ?

now solve for the ratio of distance in the following equation
ratio d = 1m/sec / 10m/sec ?







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paul Offline OP
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Quote:
The equation is about the relationship between t' and t, which are both "time" measured in seconds.


if the equation is about the relationship between
t' and t then why is it that t' or t is never used in any
calculation that is performed while calculating distances
in the equation.

all calculations are performed on the distances of v and c

the only relationship that is being examined in the
equation is the relationship between
the v distance and the c distance in a single second.

and you can only have 1 second in 1 second.




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Originally Posted By: paul
Quote:
The equation is about the relationship between t' and t,
which are both "time" measured in seconds.

if the equation is about the relationship between
t' and t then why is it that t' or t is never used in any
calculation that is performed while calculating distances
in the equation.
...just because you've separated velocity into distance and time, in your mind,
it doesn't mean that time isn't included in the equation--in those terms, v and c.

Originally Posted By: paul
all calculations are performed on the distances of v and c

the only relationship that is being examined in the
equation is the relationship between
the v distance and the c distance in a single second.
...with "time" being the same for each.

I was taught, if "calculations are performed" on two different ratios--with numerators (such as v and c describe),
then the denominators (t in this case) should be the same.

How would you perform calculations on two ratios, if their denominators were different? confused
~


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Now this is a classic thread smile

This has to go to the maths hall of fame, not quite as funny as asking what Obamas last name is, but getting close laugh

I think you have just worked out the depth of problem to explain it to him Samwik. I think I would just ignore it and run away really not worth the effort. Remedial education is not something you can easily do on the internet ... run away now smile

Guys you know 't' is something you have instead of coffee ... right. Therefore I put it to you, 't' can't be in any equation but it has to be in a cup ... so you are all wrong. The answer to the equation is clearly a cup, otherwise you can't have 't'.

Sorry it was the best I could do without going down the fundamentalist education line, which might be a little bit close.

Last edited by Orac; 06/10/16 07:06 AM.

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paul Offline OP
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Quote:
just because you've separated velocity into distance and time, in your mind


velocity is already separated into units of distance and time

and you can clearly see the separation when you look at
an amount of velocity such as 50 m/s

the m/s represents meters per second.
the m represents meters
the s represents a single second
the 50 represents the distance

so its not in my mind where the separation occurred but
in the written definition of velocity.

Originally Posted By: wiki velocity page
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time.



Quote:
...just because you've separated velocity into distance and time, in your mind,
it doesn't mean that time isn't included in the equation--in those terms, v and c.


I never said that time was not "included" in the equation I said that the time that is in the equation is never calculated.

and "time" is never calculated in the equation , the only elements
in the entire process where calculations are performed
are the two distance elements (the two elements that can
experience a rate of change).

the time element does not experience a change.

so time does not change and cannot change according to the
equation.

and all the squaring and rooting around in the equation
has absolutely no effect on the single second time element.

the only things that change in the equation are the numbers
that are associated with distance.

at no time during the calculation is there a single change
that occurs or can occur to the single second that is included
into the equation.

orac.

here is my reply to you.



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Originally Posted By: Orac
...you have just worked out the depth of problem....
Thanks Orac!

Paul, if I understood Orac correctly, he was right about how my reply “worked out the depth of [the] problem,” relativity; unless he just meant the problem with algebraic skills.
That’s because I (in composing my reply) had noticed that
by confronting the logic (and meaning) of the equation’s calculations,
the manifestation of relativity operating became more obvious—since the process of “relativity,”
and its consequence, quite quickly become almost undeniably apparent—like a mathematical slap in the face.

Maybe that could be said more simply.
The implications of “relativity” become more obvious
when you observe the consequences (and meaning)
of performing those calculations ...for “time dilation” ...correctly.
But I suppose if one decided to deny relativity, then how those “calculations are performed” must also be denied.
===

And Paul, either way, with or without “t” in the denominators, of v and c,
there is still the t (time in seconds) in front of the square root sign.
Time is still used to calculate the dilated time.

It is just an equation relating time to dilated time,
by comparing one velocity with another—the speed of light—in a certain reference frame.
And one of those velocities is constant in all reference frames, so for comparing the effect of other different velocities,
the "constant velocity" makes a good common denominator. laugh

So the idea that this might also be done by looking only at “distances” (without the time)
involved with those two different speeds, shouldn’t be surprising.
And the idea that you could simply compare the distances (without time),
to calculate the dilated time, might give you another insight into the nature of “relativity.”

Why again don’t you think the equation is true?
~


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paul Offline OP
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Quote:
And Paul, either way, with or without “t” in the denominators, of v and c,
there is still the t (time in seconds) in front of the square root sign.
Time is still used to calculate the dilated time.




v2/c2 results in units of (distance) and (meters)/(seconds)

ie...

1 m/s ^2 / 10 m/s ^2 = .01 m/s ^2

you then subtract the result that is given in (distance)
and the rate of change over time (m/s) from the 1 that causes
the result to always be lower than 1

1 - .01 = .99

this leads us to your point below.

Quote:
there is still the t (time in seconds) in front of the square root sign.


the square root of .99 is 0.99498743710662

so the current calculation is 0.99498743710662 m/s

it is a amount of velocity it has not been converted
into an amount of time.

its not simply a number nor does it only have units
of time attached to it.

remember this:
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time.

in other words if you multiply 2 apples x 2 oranges

you still only have 2 apples and 2 oranges.

you dont have 4 apples or 4 oranges.

you cannot convert apples into oranges.

and you cannot convert velocity into time.

by multiplying the 1 second in the calculation
that you pointed to times the current
total in the calculation of 0.99498743710662 m/s

1 s * 0.99498743710662 m/s = 0.99498743710662 m/s

the above is not what the equation would have us expect
to be possible , the equation expects us to simply believe
that the result will be converted into time.
but you cant multiply an amount of time by an amount of velocity and have the result given in an amount of time.

1 second * 1000 m/s is not equal to 1000 seconds

so nothing changed !!! time has not yet been changed
nor has velocity been converted into time.

the result is not an amount of time.

the result of the equation is still a velocity.
and since 1 second * 1 second = 1 second the time
never changed it is still 1 second.

now if the 1 second were 2 seconds then thats another
story all together because the 2 seconds would then
be multiplied by the distance in the 0.99498743710662 m/s
and the result would become 1.98997487421324 m/s

once again you cant multiply an amount of time by an
amount of velocity and have the result given in an amount
of time.

or at least I cant.

Quote:
Why again don’t you think the equation is true?


because I can (think).

trying to work with this 1 inch x 2 inch reply box
is really hard to do.

so I make lots of mistakes that I have to correct.

it takes me awhile to make a decent reply.

but I think I get my point across most of the time
even if it never sinks in.





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Originally Posted By: paul
so nothing changed !!! time has not yet been changed
nor has velocity been converted into time.
I can see nothing has changed, but I think I see several places where you are going awry.

First:
Originally Posted By: paul
by multiplying the 1 second in the calculation
that you pointed to, times the current
total in the calculation of 0.99498743710662 m/s

1 s * 0.99498743710662 m/s = 0.99498743710662 m/s
Okay, this indicates a major misconception.
It (t) is not another "1 second" or "1 s" in front of the square root sign.
It (t) is whatever certain length of time you might want to calculate (the time dilation of),
such as the duration of a trip at near light speed.

And by using that "certain length of time" (say t = 2 years, for example)
and multiplying by the square root of that ratio [m/s divided by m/s],
you find the solution, getting the result for t' (t prime) ...the dilated time.
===

Assuming the equation is correct....
So, if the square root of all that stuff turned out to be one half,
then t' would be one year (of the traveler's perspective)
...for a (t equals) two year trip (from Earth's perspective).

Orac?!? Do I have those perspectives correct? Please feel free to correct anything, since I'm just figuring this out as I go along ...from a chemistry/biology science background. But I think I still recall algebra well enough.
===

But whatever, Paul, let's see if you can come up with a new 'exclusion principle' for this second point about the units, such as meter per second or (seconds * some distance in meters), which should just be basic algebra and was already pointed out, once or twice, in various ways. But humor me here.

But first one other point about the units: ...where you say:
Originally Posted By: paul
1 second * 1000 m/s is not equal to 1000 seconds
....
the result is not an amount of time.
...that is correct! The answer wouldn't be (in units of) time:
It would be 1000 meters, a distance. Just look at the units, and at which units cancel and what is remaining.
If it were 2 seconds * 1000 m/s, you'd get 2000 meters, right?

So, about these units.... Here's where I'm hoping you'll humor me a bit.
Here is my "test" for you: Are these two lines below the same? In other words:
Is the second line equal to the first line, but just stated differently?

50 meters/second (divided by) 200 meters/second =

50 meters/second (multiplied by) 1 second/200 meters ?

~ whistle

p.s. "It's a trap!" laugh ....No, not really; it's just basic algebra.
...or do you consider that fantasy science too? wink


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Quote:
Okay, this indicates a major misconception.
It (t) is not another "1 second" or "1 s" in front of the square root sign.
It (t) is whatever certain length of time you might want to calculate (the time dilation of),
such as the duration of a trip at near light speed.


this entire thread has been focused on an amount of 1 second of time.

so if you want to you can multiply the results x any amount of seconds you choose
but for now lets stick to 1 second for the time element (t).
we are examining the results in a time frame of 1 second.

Quote:

But first one other point about the units: ...where you say
Quote:
1 second * 1000 m/s is not equal to 1000 seconds
....
the result is not an amount of time.



...that is correct! The answer wouldn't be (in units of) time
It would be 1000 meters, a distance. Just look at the units, and at which units cancel and what is remaining.




exactly !

because you cannot multiply an amount of time (1 second) by a velocity (1000 m/s)
and have the result given only in units of time.

so when the equation divides v^2 by c^2 the results of that calculation are given in m/s

the equation then (for design purposes) subtracts the result given in m/s from the number 1
which always gives its result as a number lower than 1.

even so the result given after the calculation is still given in m/s

the number 1 in the equation is not a unit of time , its simply the number 1
and has no viable reason to be in the equation other than to falsely lower the result
of the previous calculation.

and finding the square root of the result given in m/s does not convert the result
into an amount of time either.
so the square root of the result would also be given in m/s

now back to your point...and mine
1 second * 1000 m/s is not equal to 1000 seconds

time * distance is not equal to time alone...

so when the equation multiplies t (1 second) by the result of all of the previous calculations
there is no change in the number or the units given or assigned to that number.
because any number multiplied by 1 remains the same.

1 * 1 orange = 1 orange
1s * 1 orange/s = 1 orange

the orange does not become time in either of the above
equations so the orange is still an orange.

and multiplying 1 by any velocity also does not convert
the units of velocity into units of time.

so after the final calculation (multiplication) the number is still in units of m/s
and should not be given or assigned as an amount or a unit of time alone.

the final calculation should be given in units of m/s

so the end product of the equation t' ( time dilation )
would be given in units of m/s

LOL




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Paul, as usual, you logic seems to work for only about 1 second.
===

Do you think this is some sort of "designer math" question:

50 meters/second (divided by) 200 meters/second =
50 meters/second (multiplied by) 1 second/200 meters ?

Do you think it is true, or not? It should be easy, since it's all in "1 second" speak.

~ whistle


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I admire your tenacity samwik .. good luck smile


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Quote:
..that is correct! The answer wouldn't be (in units of) time:
It would be 1000 meters, a distance. Just look at the units, and at which units cancel and what is remaining.
If it were 2 seconds * 1000 m/s, you'd get 2000 meters, right?

So, about these units.... Here's where I'm hoping you'll humor me a bit.
Here is my "test" for you: Are these two lines below the same? In other words:
Is the second line equal to the first line, but just stated differently?

50 meters/second (divided by) 200 meters/second =

50 meters/second (multiplied by) 1 second/200 meters ?



Im not sure what you expect to gain from this but here we go.

Quote:
1 second * 1000 m/s is not equal to 1000 seconds
....
the result is not an amount of time.


the reason that the result is 1000 meters in the above equation
is because there is an amount of 1000 meters distance in 1000 m/s

meters PER second

Quote:
If it were 2 seconds * 1000 m/s, you'd get 2000 meters, right?


right!

the purpose of the equation is to solve for the distance that would
be traveled over a time period of 2 seconds at a velcity of 1000 m/s

now for your equation and the line of text below it.

the equation itself

Quote:
50 meters/second (divided by) 200 meters/second =


the equation has 3 elements on each side of the division symbol.

these are the number of meters (50 and 200)

these numbers are given in units of (meters for distance) and (seconds for time)
written as meters / second (meters per second)

the two units meters and seconds represent the two elements of distance and time.

both sides of the equation have these elements attached to the number that represents the amount of distance.
in order to show what the numbers represent.

(50 meters per second) and (200 meters per second)

the purpose of the equation is to solve for the difference in the rate of change
of position over an amount of time between the two sides of the equation and
by solving the equation you can find the difference in the rate of change of position over
an amount of time!!!

as in the below solved equation.

we know that in the equation we are discussing
t'=t*sqrt(1-(v2)/(c2)
the division that you are questioning and the result that I have given
is between v2 and c2 and by dividing v2 by c2 we get the ratio between
the two (v2/c2) so to stay closer to the OP I will
emphasize the ratio through an explaination of the result of the below equation.

50 meters/second (divided by) 200 meters/second = a ratio of .25 to 1 of 200 meters/second.

meaning that there is a .25 to 1 ratio between 50 m/s and 200 m/s
you can check this by multiplying 200 m/s times the .25 to 1 ratio between 50 m/s and 200 m/s.

so the difference in the rate of change or the ratio between the rate of change between 50 meters and 200 meters in a time period of 1 second is a ratio of .25 to 1 of 200 meters /second

the next line of text is a line of text and is not a proper equation.

50 meters/second (multiplied by) 1 second/200 meters ?

however were I to correct your error in assembling the equation by
inserting a (EQUAL) or (IS EQUAL TO) or (EQUALITY) symbol (=)

then the two sides of the equation have the same meaning
meters per second is the same as seconds per meter

ie ...

200 meters/second (meters per second) means that an objects rate of change
of position over a distance of 200 meters occurrs in a time peroid of 1 second.

200 meters per second = 1 second per 200 meters

1 second/200 meters (seconds per 200 meters) means that in a time period of 1 second
an object experiences a rate of change of position of 200 meters.

1 second per 200 meters = 200 meters per second

therefore to solve the line of text or the improper equation that you posted
through making it a proper equation by inserting the equals sign (=)

50 meters/second (multiplied by) 1 second/200 meters =

for example:
suppose you needed to know the total board length produced in meters and total production time in seconds for 250 machines 50 of the machines produce (50) 1 meter long boards each second and the other 200 machines produce (200) 1 meter long boards each second.

LOL^2

50 meters/second (multiplied by) 1 second/200 meters = 10,000 meters per second.

if we simply cancel out meters and seconds on both sides of the equation
we get 50 * 200 = 10,000 and thats it ... no units attached.
it has no value other than being a number that is derived through the
multiplication of 2 other numbers that have no units attached to them.

it is a number of unknown origin unless you associate it with the units
that it was formed from.


Im not certain why this seemed to be important to you however.

but your question was is it true...

if you were talking about the line of text then yes it is true
that the line of text is truly a line of text.




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Originally Posted By: paul
the next line of text is a line of text and is not a proper equation.

50 meters/second (multiplied by) 1 second/200 meters ?

however were I to correct your error in assembling the equation by
inserting a (EQUAL) or (IS EQUAL TO) or (EQUALITY) symbol (=)

then the two sides of the equation have the same meaning
meters per second is the same as seconds per meter

ie ...
...here is where I can see you misinterpreted what I was asking about. My apologies for trying to draw this out longer, but I think the way I formatted the text (and the equation) may have been misleading. I can see it looks like I was aksing for the answer to two different equations, with one on each line of the reply; and with an 'equals sign' at the end of the first, but with a 'question mark' at the end of the second line.

But actually, there is only one equation, with one "side" of the equation being the first line (before the = sign), and the other "side" of the equation being the second line, following after the 'equal sign' and before the question mark, which is just the end of the sentence.

I'm not asking for a solution to the equation, but I just wanted to know if you thought the equation was true. I think this is an equality:

50 meters/second (divided by) 200 meters/second = 50 meters/second (multiplied by) 1 second/200 meters.

Do you think that single equation is a true equality, algebraically; is the single equation true, or do you think it is false?

Again, sorry for the confusion and repeating so much.
~ blush


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paul Offline OP
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I didnt see that it was supposed to be in a single line.

what you have is

50 m/s / 200 m/s = 50 m/s * 1 s / 200 m

is the ratio between 50 m/s and 200 m/s or.25 to 1 equal to
the multiplication between 50 m/s and 1 s/200 meters or 10000 m/s



the two sides are not equal

the results of the equation would be false
but the equation itself is useable in a sense.

I dont concern myself with algebra
I never have , I only use math on spreadsheets and
in computer programs.

so to me the equation is false , it can only be used to
deliberately find a wrong number.

oh !!! I get it now.

you mean its like

t'=t*sqrt(1-(v2)/(c2)

is used to deliberately find a wrong number for a specific purpose.





Last edited by paul; 06/12/16 10:48 PM.

3/4 inch of dust build up on the moon in 4.527 billion years,LOL and QM is fantasy science.
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