|Mar11-06, 06:29 PM||#52|
Constant Acceleration Equation - physicist please!
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?
|Mar11-06, 06:36 PM||#53|
All that playing with the equations he did just confused him and everyone else...
|Mar12-06, 12:18 AM||#54|
Is any of this getting through, Gregor? This is quite a generous amount of ink being spilled here.
|Mar12-06, 07:54 AM||#55|
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
|Mar12-06, 08:57 AM||#56|
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.
|Mar12-06, 09:06 AM||#57|
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?
|Mar12-06, 09:38 AM||#58|
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
|Mar12-06, 09:48 AM||#59|
Let me get this straight, your saying the acceleration of a particle is equal to g mulitplied by the time?
|Mar12-06, 09:55 AM||#60|
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
|Mar12-06, 10:03 AM||#61|
So A is your distance/length?
|Mar12-06, 10:09 AM||#62|
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
|Mar12-06, 10:14 AM||#63|
So what do you plan to use your new equation for?
|Mar12-06, 10:36 AM||#64|
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
|Mar12-06, 11:00 AM||#66|
I needed a physicist to settle an argument
whether 1 g =
9.80665 mm / ms / ms (as my friend claimed)
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
|Mar12-06, 11:27 AM||#67|
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.
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