- #71
Vey2000
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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.
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.
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