Question

Algebra Question... my heads are just not spinning right

Answer 1

Lets say I have an actual velocity (ontained in a laboratory experiment)
\(V_a\)

And I have a theoretical velocity (obtained by some calculations)
\(V_t\)

What is the percent error of the theoretic veloicty FROM the theoretic velocity.

Answer 2

Is it:

\((V_a-V_t)/V_a\)

Or is it:
\((V_a-V_t)/V_t\)

(what am I dividing by?)

Answer 3

I know that how off the \(V_t\) from \(V_a\) is,

is given by: \(V_a-V_t\)

Answer 4

My guess that \(V_a\), because we are fidning the error in terms of \(V_a\).
Am I right or crazy?

Answer 5

I mean that \(V_a\) is in the denominator.

Answer 6

I should make jokes, because no body else is replying.

Answer 7

I would assure that to tell me this directly wouldn't violate the site's policy....

Answer 8

hehe

Answer 9

huehuehue

Answer 10

It probably would, but, it's not like anyone cares. XD

Answer 11

Fake, get a life. I am trying to do H/W and you are fooling around.

Answer 12

Sorry for having more of a life than your lifeless husk.

Answer 13

REPOSTING THE QUESTION:

Lets say I have an actual velocity (ontained in a laboratory experiment)
\(V_a\)

And I have a theoretical velocity (obtained by some calculations)
\(V_t\)

What is the percent error of the theoretic veloicty FROM the actual velocity?

Is it:

\(E=(V_a-V_t)/V_a\)

Or is it:
\(E=(V_a-V_t)/V_t\)

(what am I dividing by?)

Answer 14

Like I would help you, after you being a jerk. XD

Answer 15

The question only wants to know the percent error compared to the theoretical velocity. Fundamentally, this make the denominator the Theoretical Velocity. The only problem with this is a Theoretical Velocity of zero!

It's possible you do not care about the sign, so maybe \((V_{a} - V_{t})\) would be more appropriately written, \(|V_{a} - V_{t}|\).

Answer 16

I haven't seen cases where theoretical velocity is 0, although that certainly doesn't exclude such a case....

So, I am trying to say that the theoretic veloicty is "wrong"
and actual velocity is "correct"

So \( |V_t-V_a|\div V_t\)
So we are dividing by the "wrong" ?

Answer 17

yaeh i was looking at that sentence

Answer 18

they take the exact value as the experimental one,

the theoretical is approximate

Answer 19

the difference in the two values is some percent of the exact value

Answer 20

Oh, I get it, we are dividing by the "correct" value.

Answer 21

I was thinking about it previously, but I am just tired, excuse me.
TNX

Answer 22

i was just reading

What is the percent error of the theoretic veloicty FROM the actual velocity

a few times

Answer 23

Yeah I sound abstruse sometimes :)

Answer 24

yeah it sounded funny, because the exact value i think i always used the calculation, you want to know the error in your experiment compared to theory calculation,

Answer 25

you will always have some uncertainty in any experiment measurement

Answer 26

Yeah, but I was thinking of a ther value as of an unabsolute because I don't really know it whereas experimental value is more tanglible or obtainable...

I guess you are right though, experiment is kind of wrong, systematic errors and so forth...

thank you for your time.