Constant Acceleration Equation - physicist please!

  • Thread starter Gregor
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  • #26
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Gregor said:
as you wish...

a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m


do you see any units here other than meters and seconds?

No, but its wrong. What happened to the seconds the units on the right hand side should be (meters)*(seconds)2 not just meters.
 
  • #27
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Gregor, I would suggest that you not multiply numbers at will and consider the definition of acceleration:

Acceleration is equal to the rate of change of velocity with respect to time.


Now, no where in the definition do you see anything about units. It says rate of change of velocity (in what ever units you want to call velocity), divided by rate of change in time (in what ever units you want to call time). As long as you are consistent, you will get the same answers.
 
  • #28
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I guess I have to spell it out for you

a = 9.80665 x 0.001 x 0.001 = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

are you saying 1 g does not = 9.80665 µm / ms / ms ?

because if you are - I'm not the one who needs help here
 
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  • #29
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Gregor, you are still multiplying numbers at will without paying attention to their meaning.

the units of acceleration are, in SI, [tex] \frac {m}{s^2} [/tex]

What you have done, a = 9.80665 m/s x 0.001 s x 0.001 s, with much carelessness, will result in units of [itex] m-s[/itex], which is not what acceleration represents. So that does not equal a.
 
  • #30
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1 g = 2.4516625 m / 0.5 s / 0.5 s

yes or no ?

1 g = 0.00000980665 m / 0.001 s / 0.001 s

yes or no ?

please do not answer these questions with anything other than

a. yes

or

b. no
 
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  • #31
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Gregor said:
I guess I have to spell it out for you

a = 9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

Those are not equal at all, even the numbers are different, the units are not the same. On the far right your units are correct adn your numbers are as well, but in the middle it's just completely wron you cannot just change between multiplication and division and expect the results to be equal, Multiplication is not a commutative operation. I don't see what you're trying to even show because neither your math nor your physics makes nay sense.
 
  • #32
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Yes, that is correct.
 
  • #33
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Gregor said:
1 g = 2.4516625 m / 0.5 s / 0.5 s

yes or no ?

Yes it does, but LOOK AT THE UNITS. You have also said
a = 9.80665 m/s x 0.001 s x 0.001 s
and that is certainly not a true statement because you end up with meters*seconds which is not and never will be a measure of acceleration. Pay attention to your units and don't carelessy move them around.
 
  • #34
russ_watters
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Gregor said:
I guess I have to spell it out for you

a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

are you saying 1 g does not = 9.80665 µm / ms / ms ?

because if you are - I'm not the one who needs help here
Gregor, if you don't have a / in the expression, you don't get a / in the answer. You are multiplying the numbers and dividing the units while simultaneously changing another number without changing the units you used to make the change.

If you want acceleration, you need to divide by s^2, not multiply by s^2 and then pretend you divided by putting in units and numbers that don't match what you actually did!

Gregor, the things you are doing here are so obviously wrong that we're having a hard time believing you aren't doing it on purpose. And if you are doing it on purpose, that's a no-no here.
 
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  • #35
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a = 9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

Those are not equal at all, even the numbers are different,

the first "m/s" above was a typo
look at the operations below...



9.80665 x 1 x 1 = 9.80665

1 g = 9.80665 m / s / s



9.80665 x 0.001 x 0.001 = 0.00000980665

1 g = 0.00000980665 m / 0.001 s / 0.001 s



9.80665 x 0.5 x 0.5 = 2.4516625

1 g = 2.4516625 m / 0.5 s / 0.5 s



these results are correct, so the operation works no matter what

throw any squared time interval at me and I'll give you the correct value for 1 g

you guys are getting hysterical over your dimensional analysis equations
and ignoring the fact that my end results are 100% correct
 
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  • #36
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This,

9.80665 x 0.001 x 0.001 = 0.00000980665

is meaningless algebra when you take away the units.

This,

1 g = 0.00000980665 m / 0.001 s / 0.001 s

has meaning.

Your result might be correct, but it is trivial. All you have done is use a different unit base, but so what?
 
  • #37
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Gregor said:
the first "m/s" above was a typo (forget it, and forget the units for now)

9.80665 x 0.001 x 0.001 = 0.00000980665

1 g = 0.00000980665 m / 0.001 s / 0.001 s

this result is correct, so the operation works no matter what you say

Who are you kidding? yes what you have now is correct, but what you said is

9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

Your units are the same on both sides so I'm going to do what you've been doing and ignore them for a second. The number on the left hand side is 0.00000980665 but on the right hand side you had 9.80665 so there is no way those expressions could be equivalent, even ignoring the units as you have done. And just because the result of your operation is correct doesn't prove your point at all.
 
  • #38
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the first "m/s" above was a typo
look at the operations below...

it pays to read all the posts

and who are you kidding?

the values are correct, the units give the values meaning
but the calculations get you there in the first place
 
  • #39
russ_watters
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Gregor said:
the first "m/s" above was a typo
look at the operations below...


[1] 9.80665 x 1 x 1 = 9.80665

[2]1 g = 9.80665 m / s / s

[3]9.80665 x 0.001 x 0.001 = 0.00000980665

[4]1 g = 0.00000980665 m / 0.001 s / 0.001 s

[5]9.80665 x 0.5 x 0.5 = 2.4516625

[6]1 g = 2.4516625 m / 0.5 s / 0.5 s

these results are correct, so the operation works no matter what [expression numbers added for clarity]
Yes, that set works (it isn't done quite right, but it does work if we translate it). But so what? What have you proven?
 
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  • #40
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you guys are getting hysterical over your dimensional analysis equations
and ignoring the fact that my end results are 100% correct

Your results might be 100% mathematically correct, but when they are 100% wrong dimensionally, then they bear no physical meaning. If they have no physical meaning, then the whole statement is trash, no matter how fancy or correct the mathematics. I hope you see why!
 
  • #41
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Your units are the same on both sides so I'm going to do what you've been doing and ignore them for a second. The number on the left hand side is 0.00000980665 but on the right hand side you had 9.80665 so there is no way those expressions could be equivalent,

what on earth are you talking about?

0.00000980665 is the result of multipying 9.80665 x 0.001 x 0.001

hello, is this thing on? :rolleyes:
 
  • #42
russ_watters
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Gregor - SO WHAT? What does this prove? It doesn't, for example, make the things you were saying this morning any less wrong.

All you did here is multiply both sides of an equation by .000001 and discover that it is still equal to what you had before. You discovered algebra!
 
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  • #43
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Your results might be 100% mathematically correct, but when they are 100% wrong dimensionally, then they bear no physical meaning. If they have no physical meaning, then the whole statement is trash, no matter how fancy or correct the mathematics. I hope you see why!

yes, absolutely

I know exactly what you're saying, and I agree with you

what I am saying is that I have found a shortcut to recalculating the value of 1 g from any squared time interval and any length unit.

it's a mathematical experiment, not a physics experiment
who ever said it was a physics experiment ??
 
  • #44
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Gregor said:
what on earth are you talking about?

0.00000980665 is the result of multipying 9.80665 x 0.001 x 0.001

hello, is this thing on? :rolleyes:

Don't say I'm not reading all of the posts if you're not reading mine either.
 
  • #45
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Gregor said:
yes, absolutely

I know exactly what you're saying, and I agree with you

what I am saying is that I have found a shortcut to recalculating the value of 1 g from any squared time interval and any length unit.

it's a mathematical experiment, not a physics experiment
who ever said it was a physics experiment ??

I don't see how you've found a shortcut, you've done nothing other than screw around with units that makes a lot of your expressions physically meaningless. Why is this a physics experiment, because g has little use in mathematics but is extremeley necessary in physics.
 
  • #46
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Gregor - SO WHAT? What does this prove? It doesn't, for example, make the things you were saying this morning any less wrong.

which things specifically are you refering to?
 
  • #47
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what I am saying is that I have found a shortcut to recalculating the value of 1 g from any squared time interval and any length unit.

I did not see your work for this, but let us assume that you did. What does it matter? We are only interested in two units, SI, or English. Time is the same for both systems, but distance is not. So, while you may have a way to find g, it really has no importance as we do not use any system other than these two. I hope you realize all you have done is use a different base system. I could have written your equation as:

[tex] g = \frac{9.80665 \mu M}{ 1 ms * 1 ms} [/tex]
 
  • #48
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I don't see how you've found a shortcut, you've done nothing other than screw around with units that makes a lot of your expressions physically meaningless. Why is this a physics experiment, because g has little use in mathematics but is extremeley necessary in physics.

let's step back to the intro...

my friend and I decided to make a mathematical experiment

what would happen to the value of 1 g if the second were redefined as 1 ms

my friend insisted that the new value of 1 g would be 9.80665 mm/s/s
(9.80665 mm/ms/ms in other words)

and I said it was 9.80665 µm/s/s
(9.80665 µm/ms/ms in other words)

since s = 0.001 in this experiment

and the algorithm I used to derive the result was A*T²

my friend used A*T

my results were right and his were wrong

and everyone in physics forums blew a fuse
because I used a non-standard formula

even though my result was correct :rolleyes:
 
  • #49
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Because you are BOTH wrong. This is a case where you paid no attention to the units.
 
  • #50
russ_watters
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Gregor said:
which things specifically are you refering to?
Post 17.

Regardless, I'm still wondering what you think you've discovered here. All you did was apply some algebra.
 

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