Constant Acceleration Equation - physicist please

[russ_watters]

and the algorithm I used to derive the result was A*T²

my friend used A*T

my results were right and his were wrong

and everyone in physics forums blew a fuse
because I used a non-standard formula

even though my result was correct

No, Gregor - that is not what you said in your opening post. You said:

A = V*T²

And it is still wrong.

If you are wondering about the little algebra game that you played, yes, you correctly multiplied by 1 and your friend did not. Everyone saw the error you made with the equations and no one checked to see if you multiplied by 1 correctly in the second part of the post. But even still - I don't think you understand that that's all you did.

 

[d_leet]

let's step back to the intro...

my friend and I decided to make a mathematical experiment

what would happen to the value of 1 g if the second were redefined as 1 ms

my friend insisted that the new value of 1 g would be 9.80665 mm/s/s
(9.80665 mm/ms/ms in other words)

and I said it was 9.80665 µm/s/s
(9.80665 µm/ms/ms in other words)

since s = 0.001 in this experiment

and the algorithm I used to derive the result was A*T²

my friend used A*T

my results were right and his were wrong

and everyone in physics forums blew a fuse
because I used a non-standard formula

even though my result was correct

You cannot just redefine the way we measure acceleration, I understand it if you use different units that emasure the same thing, but you are not allowed to just change it from m/s² to ms they are not the same thing.

You are both wrong, why won't you just accept that? You didn't use a "nonstandard" formula you used one which has no physical significance as a measure of acceleration. You come and then ask this on a PHYSICS forum what did you think would happen, that we would all magically agree with you and through away hundreds of yeasr of work by people a lot smarter than you are because of this one post?

 

[russ_watters]

I did not see your work for this, but let us assume that you did. What does it matter? We are only interested in two units, SI, or English. Time is the same for both systems, but distance is not. So, while you may have a way to find g, it really has no importance as we do not use any system other than these two. I hope you realize all you have done is use a different base system. I could have written your equation as:

g = \frac{9.80665 \mu M}{ 1 ms * 1 ms}

Apparently, that's all he was trying to do: g = 9.8 \frac{m}{s*s} * \frac{.000001}{.000001}= \frac{9.8 \mu m}{ 1 ms * 1 ms}

All that playing with the equations he did just confused him and everyone else...

 

[hypermonkey2]

Is any of this getting through, Gregor? This is quite a generous amount of ink being spilled here.

 

[Gregor]

yes hypermonkey, you and russ and everyone have made yourselves very clear from the start, and I appreciate your efforts.

I think we started off on a misunderstanding in any case

because when I wrote A = V*T² (an expression I wrote myself)
I used it to mean g = A*T²

the V was never meant to represent velocity alone
but rather an increase in velocity through time

I should have posted g = A*T² or even A = g*T²

I was trying to avoid A = A*T² since this would be even more ambiguousthe operation I used to derive the new value for 1 g is A*T²

9.80665 x 0.001 x 0.001

since this was a mathematical experiment, I was only interested in the numeric values, and not the units or dimensions.

since I acheived my end goal (the correct numerical value for 1 g in modified units), I have a tool to check the accuracy of operations using proper motion physics equations.

since my equation (as unorthodox as it is) yealds 100% correct values 100% of the time.

I hope I've made my point now, so there's no more misunderstandings

thanks again to Integral, Zapperz, vanesch, hypermonkey2, SaMx, Doc Al, russ_watters, cyrusabdollahi, and d_leet.

your help is very much appreciated

 

[ZapperZ]

yes hypermonkey, you and russ and everyone have made yourselves very clear from the start, and I appreciate your efforts.

I think we started off on a misunderstanding in any case

because when I wrote A = V*T² (an expression I wrote myself)
I used it to mean g = A*T²

the V was never meant to represent velocity alone
but rather an increase in velocity through time

I should have posted g = A*T² or even A = g*T²

I was trying to avoid A = A*T² since this would be even more ambiguous


the operation I used to derive the new value for 1 g is A*T²

9.80665 x 0.001 x 0.001

Oh good grief.

You see nothing wrong with saying g = at^2, even when g is defined as a gravitational acceleration and having the same dimension as an acceleration? Have you ever considered the resulting dimension of such a thing (you can no longer call it g since it is not the same creature anymore)? at^2 has the dimension of length!

So, since this is physics, what does this "length" represent?

since this was a mathematical experiment, I was only interested in the numeric values, and not the units or dimensions.

And that is why I have suggested you do your numerology in mathematics, but you insisted in being in here.

This has got to be one of the strangest thread I've ever seen on PF. I'm almost tempted to label it as crackpottery.

Zz.

 

[Gregor]

You see nothing wrong with saying g = at^2

In physics terms, Yes

In mathematical terms, No

the resulting values are always correct.

in any case g = at^2 doesn't need to make sense in mathematical terms
I just used it to get from A to B (fast)

I'm sure you're familiar with the expression: "the end justifies the means" ;-)

anyway, now that I have the proper equations (thanks to you and Integral)
I can do it the right way - so what's the problem?

 

[Gregor]

as soon as someone can prove that the values below are not correct
then we can say that A = g*T² is mathematically false

however the values below Are correct, and therefore A = g*T² is mathematically correct (although useless for physics)

(note* the numerical value of g = 9.80665)1 g =

0.00000980665 m / 0.001 s / 0.001 s

0.000980665 m / 0.01 s / 0.01 s

0.0980665 m / 0.1 s / 0.1 s

0.392266 m / 0.2 s / 0.2 s

0.8825985 m / 0.3 s / 0.3 s

1.569064 m / 0.4 s / 0.4 s

2.4516625 m / 0.5 s / 0.5 s

3.530394 m / 0.6 s / 0.6 s

4.8052585 m / 0.7 s / 0.7 s

6.276256 m / 0.8 s / 0.8 s

7.9433865 m / 0.9 s / 0.9 s

9.80665 m / s / sI rest my case

 

[Hootenanny]

Let me get this straight, your saying the acceleration of a particle is equal to g mulitplied by the time?

 

[Gregor]

no one is listening to what I'm saying (!)

the equation has no meaning in physics
all it does is provide a correct numerical value for 1 g
for any given length or time unit

this is blind maths, and the physical world has nothing to do with it

this is only useful for quick (correct) number value solutions
when playing with new length and time units

it's absolutely useless for dimensional analysis

for dimensional analysis I have the proper equations which Integral and Zapperz posted earlier

dimensional analysis and "number getting" are two different goals

 

[Hootenanny]

So A is your distance/length?

 

[Gregor]

A is the new value for g

after applying a new unit of length and a new unit of time

g = 9.80665 only when expressed in meters / second / second

 

[Hootenanny]

So what do you plan to use your new equation for?

 

[Gregor]

for making physicists want to hit me... mostly it's quite handy as a fast conversion algorithm
when experimenting with new units

my strange little equation saves a lot of stepsI think it quite interesting to play with revised units

for example :

if the meter is redefined as 9.80665 m

then 1 g = 1 m/s²
if the meter is redefined as 0.980665 m
and the second is redefined as 10 s

then 1 g = 1000 m/s²

 

[ZapperZ]

Then why are you even here?

Zz.

 

[Gregor]

I needed a physicist to settle an argument


whether 1 g =

9.80665 mm / ms / ms (as my friend claimed)

or

9.80665 µm / ms / ms (as I correctly calculated)


my answer was right
even though my equation had nothing to do with physics

and thanks to you - I can use the proper equations from now on

 

[Cyrus]

Gregor, this:

I should have posted g = A*T² or even A = g*T²

is not correct. I hope you see why. Remember what I said earlier, when you take away the letters, you are preforming pure math, which is fine. But when you insert symbols, the equation now has physical meaning, and must make sense in the real world, which your equation does not, hence the reason why everyone is telling you it is wrong.

 

[Doc Al]

9.80665 mm / ms / ms (as my friend claimed)

or

9.80665 µm / ms / ms (as I correctly calculated)

You hardly need a physicist to settle this; Anyone who knows how to manipulate exponents can tell you that:
\frac{10^{-6}}{(10^{-3})^2} = 1

Whereas:
\frac{10^{-3}}{(10^{-3})^2} = 10^3 \ne 1

 

[Gregor]

I hope you see why.

yes of course

Remember what I said earlier, when you take away the letters, you are preforming pure math, which is fine. But when you insert symbols, the equation now has physical meaning,

this depends on which symbols

if I change the equation to

N = Z*I²

and then define the symbols as

N = new value
Z = 9.80665
I = Time Interval

then there's no physical violation

 

[Cyrus]

But then N,Z, and I have NOTHING to do AT ALL with gravity, time or distance, and therefore you CAN NOT use\relate it in terms of those variables.

 

[Vey2000]

From my perspective, I believe the confusion has been one of definitions and lack of consistency. Gregor seems to have been seeking a "formula" for converting from one unit to another. In essence, it is as he said, he was after a formula for the dimensionless numerical value of g. The problem was that for him the values of the variables meant something else than what they traditionally mean, and he did not define precisely what they meant to him in terms that mean anything to other people.

On the other hand, those who did understand what he was trying to do considered it a triviality (it is, but only if you know basic algebra properly) and did not respond in terms of Gregor's formula, but with a direct answer which is "obvious" -- it is, but not to Gregor, I guess.

Here, I'll attempt to meet Gregor halfway.

Let us define 1 m' (a "new" meter) = 9.8 m (a "regular" meter)
Solving this for m we get 1 m = 1/9.8 m'

1 g = 9.8 m/s^2 = 9.8 (1 m)/(1 s)^2 = 9.8 (1/9.8 m')/(1 s)^2 = (1 m')/(1 s)^2 = 1 m'/s^2

Ok, that was a little too explicit, perhaps, but better safe than sorry.

And essentially the same for converting any value into any other unit, just replace 1 s with your new unit and simplify. There is no "formula" because it's "trivial," which means that you shouldn't think of formulas as numerical "plug'n'chug machines," but as relations between variables, which can be evaluated numerically if need be. Separating a number from its unit is bound for confusion, as has been shown, so don't do it.

The "best" way to convert between units is this way, by substituting the value of a unit in terms of another.

Ok, just to make sure it is even clearer I'll do a "general" case by including variables for the conversion factors.

Let L' and T' be the numerical conversion factors between m and m', and s and s'

1 g = 9.8 m/s^2 = 9.8 (L' m')/(T' s')^2 = 9.8 (L'/T'^2) m'/s'^2

For instance, in the earlier example we would have:

1 m = 1/9.8 m' = L' m'

So, if we set G to the -numerical- value of g in the units of m/s^2, G = 9.8, and G' to the -numerical- value of g in the units of m'/s'^2, then from the above equation we get G' = G*L'/T'^2, but realize that this is getting needlessly out of the way.

Well, since I'm half-asleep and seem bent on rambling I'll just finish it off with another one of your examples.

1 m = 10^6 um (let's just assume that u is a greek mu, for micrometer)
1 s = 10^3 ms

1 g = 9.8 m/s^2 = 9.8 (10^6 um)/(10^3 ms)^2 = 9.8 10^6/10^6 um/ms^2 = 9.8 um/ms^2

In this case the -numerical- value is the same because both conversion factors cancel out.

Hopefully it should be clear how to do this by now.

P.S.: There is no physics in this, by the way. Even if mathematicians don't usually work with units that doesn't mean that they aren't consistent with them when they do work with units.

 

[Cyrus]

You are trying to go to the local drug store by going all the way around the beltway.

 

[Gregor]

But then N,Z, and I have NOTHING to do AT ALL with gravity, time or distance, and therefore you CAN NOT use\relate it in terms of those variables.

true,

however this is just a conversion formula

But then N,Z, and I have NOTHING to do AT ALL with gravity, time or distance, and therefore you CAN NOT use\relate it in terms of those variables.

yes, the units must always be consistent throughout the expression,

in my demonstrations I've used prefixed units for simplicity,
however these units are consistant throughout each expression

 

[Cyrus]

however this is just a conversion formula

Yes, it is a conversion formula for N, Z and I (Whatever N, Z and I are), PROVIDED that they follow the relationship N = Z*I²

g, m, and s DO NOT FOLLOW THE RELATIONSHIP g= a*s²

How many times do I have to tell you that?

 

[Gregor]

You are trying to go to the local drug store by going all the way around the beltway.

I think this would be a very funny analogy, if only I understood what it meant

I should tell you that English is not my first language (ich bin Deutscher)
so sometimes things get lost in the translation %-/

I know a drug store is what we call an Apotheke

 

[Cyrus]

That was not for you, that was for Vey2000, and it was not supposed to be funny.

 

[Gregor]

g, m, and s DO NOT FOLLOW THE RELATIONSHIP g= a*s²

How many times do I have to tell you that?

I realize this

the only importance for the formula is that it produces the correct numerical value for 1 g in revised units

and the forumula does this perfectly
as I have demonstrated several times

(to the annoyance of every physicist in the forums)

 

[Cyrus]

the only importance for the formula is that it produces the correct numerical value for 1 g in revised units

Dammit, NO!, it does not.
You do NOT realize it.

Look, 3 + 2 = 5 right?

does 3(apples) + 2(oranges) = 5 (apples)

No. Do you see my point? The math is right, the logic is WRONG!

 

[Gregor]

No. Do you see my point? The math is right, the logic is WRONG!

I see you're getting angry now :-(

yes, I've seen the point all along

however it's my point that is being missed

9.80665 (Z) * 0.001 (I) * 0.001 (I) = 0.00000980665 (N)

there are no apples, oranges, or any physical entities here
just numbers to be computed.

the result (N) just happens to be equal to g when m = 1 meter
and s = 0.001 second

the result is always "numerically equal" to g

it's just a convenient coincidence that the formula provides the correct numeric value, there is no direct connexion to physical dimensions involved.

 

[Cyrus]

the result is always "numerically equal" to g

Ok, fine, SO WHAT? What good does it do if you lucked out and got the same numerical value, but your units are completely WRONG?

 

[Gregor]

how are they wrong?

the only difference between

9.80665 * 0.001 * 0.001 = 0.00000980665

and

9.80665 m * 0.001 s * 0.001 s = 0.00000980665 m/0.001 s/0.001s

is that the units are added in the last expression

this does not make them "wrong", but simply "included" or "not included".

the value for g is correct for these units none-the-less

since 1 g does = 0.00000980665 m/0.001 s/0.001s

 

[Cyrus]

Dammit are you blind?

9.80665 m * 0.001 s * 0.001 s = 0.00000980665 m/0.001 s/0.001s

does ms^2 = \frac{m} {s^2} to you?

 

[Vey2000]

You are trying to go to the local drug store by going all the way around the beltway.

I'm aware of that -- I said so myself: "...this is getting needlessly out of the way." However, it's my biased opinion as a student that if you can't see the proper way, the beltway is better than no way. You can't build up a mathematical intuition for how to get from A to B quickly if you don't know how to get from A to B at all.

Unfortunately, it may be the case that despite taking the beltway I still got nowhere since Gregor is still confused.

how are they wrong?

the only difference between

9.80665 * 0.001 * 0.001 = 0.00000980665

and

9.80665 m * 0.001 s * 0.001 s = 0.00000980665 m/0.001 s/0.001s

is that the units are added in the last expression

No, there is another difference, ignoring the the fact that the dimensions are wrong -- the right hand side of the second "equality" (it's not equal at all) has two additional divisors that are not in the right hand side of the first equality.

Allow me to take a stab at "fixing" your first equation. What you really have is the following:

9.8 m/s^2 * 0.001 (s/s') * 0.001 (s/s') = 9.8x10^-6 m/s'^2

Notice how there are indeed units in there? Notice how s cancels out with s in the left hand side and you are left with s' instead?

Now I'll take a stab at interpreting your second "equation." The two "0.001 s" divisors are in fact the new s' units. Replace it in the right hand side of my proposed equation and you'll recover the right hand side of your second "equation."

So the problem clearly lies in the left hand side. What you are doing by multiplying on the left side by "time" is actually in fact multiplying by the time unit conversion factors -- 0.001 s/s'. This relates to my previous post where I went out of my way to explicitly show you the conversion factors as variables.

The other problem is that on the left hand side you have "assigned" 9.8 the unit of m, whereas it is m/s^2.

So, your "formula" is g' = g * T'^2, where g' is just g in the new units of m/s'^2, and T' is the conversion factor with units of s/s'. -This- is mathematically consistent.

You have to realize that just because things work out in your mind does not mean that what comes out into paper works out the way you write it. As I said, in my opinion the problem was one of definitions.

 

[Doc Al]

I think we've given this inane thread way more time than it deserved.

Closed.