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Constant Acceleration Equation  physicist please! 
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#55
Mar1206, 07:54 AM

P: 39

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 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 


#56
Mar1206, 08:57 AM

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P: 29,239

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? 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. 


#57
Mar1206, 09:06 AM

P: 39

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? 


#58
Mar1206, 09:38 AM

P: 39

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 / s I rest my case 


#59
Mar1206, 09:48 AM

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P: 9,781

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



#60
Mar1206, 09:55 AM

P: 39

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 


#62
Mar1206, 10:09 AM

P: 39

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 


#63
Mar1206, 10:14 AM

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



#64
Mar1206, 10:36 AM

P: 39

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 steps I 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 


#66
Mar1206, 11:00 AM

P: 39

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 


#67
Mar1206, 11:27 AM

P: 4,780

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. 


#68
Mar1206, 01:44 PM

Mentor
P: 41,318

[tex]\frac{10^{6}}{(10^{3})^2} = 1[/tex] Whereas: [tex]\frac{10^{3}}{(10^{3})^2} = 10^3 \ne 1[/tex] 


#69
Mar1206, 03:55 PM

P: 39

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 


#70
Mar1206, 06:52 PM

P: 4,780

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.



#71
Mar1206, 06:55 PM

P: 9

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 halfasleep 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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