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## Constant Acceleration Equation - physicist please!

 Quote by Gregor I understand your points completely, and of course you and Zapperz are absolutely right. since I normally approach problems mathematically, I tend to deal with them in a reductive, rather than analytical way, being satisfied with correct solutions, with as few operations as possible. my friend and I were making a thought experiment.. what would happen to g if the meter and/or the second were redefined. so we got into an argument over the correct algorithm for deriving the new value for g. In physics you're not allowed to toy with base units, but in maths - everything is open game.
I disagree. Try doing this in the mathematics forum that we have, and you'll find a few mathematicians objecting to it.

[example: 1 = 1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1) = i*i = -1 - nothing wrong with that?]

There is a difference in doing pure number theory and manipulation versus something with a physical quantity. If you wish to do the latter, you shouldn't have done it in the PHYSICS section of the forum.

And just so you know, in many areas of physics, h=k=c=1 by "definition". But this isn't a new definition. It is for "simplicity" and people in the field know what is being done so there's no confusion. It is why you see people often quoting the mass of a particle simply in terms of MeV rather than MeV/c^2. We all know how to convert those back to their true units.

Moral of the story: please do mathematics in the mathematics forum, especially if you wish to neglect the physical dimension of the quantities you are manipulating. If you wish to carry the discussion in the physics forums, then all the physics rules apply.

Zz.

agreed

but surely you're not suggesting that SI units are universal constants,
carved in stone. in many cases equations are only approximations.
(such as mc²)

if a unit changes then it's correlation to other units also changes
and this effects the values in dimensional analysis.

in the example...

 what happens to the equation a = v/t with 1 meter redefined as 0.980665 m and 1 second redefined as 0.1 s (?)
does a = v/t still hold true through time in dimensional analysis?

(assuming that tomorrow the BIPM decided to redefine the meter and second)

this is not a problem for the maths forum
it's a physics problem

I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics.

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Blog Entries: 27
 Quote by Gregor agreed but surely you're not suggesting that SI units are universal constants, carved in stone. in many cases equations are only approximations. (such as mc²)
It has nothing to do with "units". It has everything to do with dimensions. But even then, you cannot add something with units of meters to something with units of microns. You'll get nonsense.

 if a unit changes then it's correlation to other units also changes and this effects the values in dimensional analysis. in the example... does a = v/t still hold true through time in dimensional analysis? (assuming that tomorrow the BIPM decided to redefine the meter and second) this is not a problem for the maths forum it's a physics problem I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics.
a = v/t is the DEFINITION when a is a constant. It is how it is DEFINED regardless of the units involved. The DIMENSION remains no matter what.

And note that you were the one who said that you were more interested in the manipulation of the mathematics. If you notice, you did something that was physically nonsensical, but you justified it simply by the fact that you're not doing anything wrong numerically. That is what prompted me to request that if you wish only to care about the numerical manipulation of numbers with utter disregard to the physical significance of that number, then the physics forum is NOT the place to do it. I'm only going by YOUR intentions.

Zz.

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 Quote by Gregor but surely you're not suggesting that SI units are universal constants, carved in stone.
Huh? Who suggested any such thing? It's dimensions that count, not particular unit systems.

 in many cases equations are only approximations. (such as mc²)
Any equation, to make any physical sense at all, must have consistent units and dimensions.

 if a unit changes then it's correlation to other units also changes and this effects the values in dimensional analysis.
Dimensional analysis deals with dimensions, not particular systems of units.

 so what happens to the equation a = v/t with 1 meter redefined as 0.980665 m and 1 second redefined as 0.1 s (?)
All you've done is redefined your units: 1 "wacky meter" = 0.980665 m and 1 "wacky second" = 0.1 s. So what? Now acceleration has units of "wacky meters" per "wacky second"^2. So what?
 this is not a problem for the maths forum it's a physics problem
Where's the physics?

 I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics.
Because they are trivial? If done correctly, at least. "Equations" written with wanton disregard for dimensional consistency have no physical content and are just numerology.

 It has nothing to do with "units". It has everything to do with dimensions. But even then, you cannot add something with units of meters to something with units of microns. You'll get nonsense.
the calculations were only done in meters

I simply converted the result to µm

and I did get the correct result (at least)

 Huh? Who suggested any such thing?
sorry, it was the voices again (what's that you say lucifer??)

 All you've done is redefined your units: 1 "wacky meter" = 0.980665 m and 1 "wacky second" = 0.1 s. So what? Now acceleration has units of "wacky meters" per "wacky second"^2. So what?
so now 1 g = 1 wackymeter/wackysecond²
rather than 9.80665 m/s²

neat-o huh!

ok, don't hurt me

 Where's the physics?
it got stollen by uh... other kids, or something

 Because they are trivial? If done correctly, at least. "Equations" written with wanton disregard for dimensional consistency have no physical content and are just numerology.
so nnnnn!!

and you know what? I'm not going to ride my bike past your Synchotron anymore!

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Blog Entries: 27
 Quote by Gregor the calculations were only done in meters I simply converted the result to µm and I did get the correct result (at least)
What "correct" results? Show me the "calculations".

And note, I said, if you read carefully, that one cannot ADD (not convert) a length in meters to a length in micrometer and expect a meaningful answer. Do you think 3 +2 = 5 if 3 = 3 meter and 2 = 2 microns? Puhleeze! But that is what you are trying to do, and even worse, you multiply things just because they all result in "1".

Zz.

 I do enjoy that you are trying to think outside the box, Gregor, but this is turning to nonsense quite quickly and we are turning in circles.

 What "correct" results? Show me the "calculations".
as you wish...

a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m

do you see any units here other than meters and seconds?

naturally 0.00000980665 m = 9.80665 µm and 0.001 s = 1 ms

but these were converted after the operation

the point is that the answer is correct

a at 1 g = 9.80665 µm / ms / ms

and I derived it with the calculation above, which works for any time interval

shall we try 0.5 s ?

sure!

9.80665 m x 0.5 s x 0.5 s = 2.4516625 m

so a = 2.4516625 m / 0.5 s / 0.5 s

feel free to tell me this result is incorrect

 Quote by Gregor as you wish... a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m do you see any units here other than meters and seconds?
No, but its wrong. What happened to the seconds the units on the right hand side should be (meters)*(seconds)2 not just meters.

 Gregor, I would suggest that you not multiply numbers at will and consider the definition of acceleration: Acceleration is equal to the rate of change of velocity with respect to time. Now, no where in the definition do you see anything about units. It says rate of change of velocity (in what ever units you want to call velocity), divided by rate of change in time (in what ever units you want to call time). As long as you are consistent, you will get the same answers.
 I guess I have to spell it out for you a = 9.80665 x 0.001 x 0.001 = 0.00000980665 m / 0.001 s/ 0.001 s now do you get it? are you saying 1 g does not = 9.80665 µm / ms / ms ? because if you are - I'm not the one who needs help here
 Gregor, you are still multiplying numbers at will without paying attention to their meaning. the units of acceleration are, in SI, $$\frac {m}{s^2}$$ What you have done, a = 9.80665 m/s x 0.001 s x 0.001 s, with much carelessness, will result in units of $m-s$, which is not what acceleration represents. So that does not equal a.
 1 g = 2.4516625 m / 0.5 s / 0.5 s yes or no ? 1 g = 0.00000980665 m / 0.001 s / 0.001 s yes or no ? please do not answer these questions with anything other than a. yes or b. no

 Quote by Gregor I guess I have to spell it out for you a = 9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s now do you get it?
Those are not equal at all, even the numbers are different, the units are not the same. On the far right your units are correct adn your numbers are as well, but in the middle it's just completely wron you cannot just change between multiplication and division and expect the results to be equal, Multiplication is not a commutative operation. I don't see what you're trying to even show because neither your math nor your physics makes nay sense.

 Yes, that is correct.

 Quote by Gregor 1 g = 2.4516625 m / 0.5 s / 0.5 s yes or no ?
Yes it does, but LOOK AT THE UNITS. You have also said
a = 9.80665 m/s x 0.001 s x 0.001 s
and that is certainly not a true statement because you end up with meters*seconds which is not and never will be a measure of acceleration. Pay attention to your units and don't carelessy move them around.

Mentor
 Quote by Gregor I guess I have to spell it out for you a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s now do you get it? are you saying 1 g does not = 9.80665 µm / ms / ms ? because if you are - I'm not the one who needs help here
Gregor, if you don't have a / in the expression, you don't get a / in the answer. You are multiplying the numbers and dividing the units while simultaneously changing another number without changing the units you used to make the change.

If you want acceleration, you need to divide by s^2, not multiply by s^2 and then pretend you divided by putting in units and numbers that don't match what you actually did!

Gregor, the things you are doing here are so obviously wrong that we're having a hard time believing you aren't doing it on purpose. And if you are doing it on purpose, that's a no-no here.