Constant Acceleration Equation - physicist please

[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 consistent 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.