Constant Acceleration Equation - physicist please!

d_leet

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

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

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 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

ZapperZ

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?

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

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 / s


I 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 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

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

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

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