| Topic: |
Science > Physics |
| User: |
"time" |
| Date: |
30 Jun 2005 10:11:09 AM |
| Object: |
geometric variation |
I had some samples with the shape of a rough rectangle. I am considering the
variation of height and width of the sample. If the height variation is x,
and the width variation is y, the total geometric variation could be modeled
as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is better? Thank you.
Time
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| User: "Old Man" |
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| Title: Re: geometric variation |
30 Jun 2005 12:52:54 PM |
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"time" <time5556@hotmail.com> wrote in message
news:42c40b5e$1_2@x-privat.org...
I had some samples with the shape of a rough rectangle. I am considering
the variation of height and width of the sample. If the height variation is
x, and the width variation is y, the total geometric variation could be
modeled as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is better?
Thank you.
None of the above.
For linear dimensions (X, Y) with errors (x, y), and
Area, A = X*Y, with error , delta_A
delta_A = A * sqrt[ (x / X)^2 + (y / Y)^2 ]
[Old Man]
Time
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| User: "Andy Resnick" |
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| Title: Re: geometric variation |
30 Jun 2005 02:48:23 PM |
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Old Man wrote:
"time" <time5556@hotmail.com> wrote in message
news:42c40b5e$1_2@x-privat.org...
I had some samples with the shape of a rough rectangle. I am considering
the variation of height and width of the sample. If the height variation is
x, and the width variation is y, the total geometric variation could be
modeled as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is better?
Thank you.
None of the above.
For linear dimensions (X, Y) with errors (x, y), and
Area, A = X*Y, with error , delta_A
delta_A = A * sqrt[ (x / X)^2 + (y / Y)^2 ]
My $0.02:
That also assumes the error in X is independent of Y and the error in Y
is independent of X.
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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| User: "Old Man" |
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| Title: Re: geometric variation |
01 Jul 2005 04:22:28 PM |
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"Andy Resnick" <andy.resnick@op.case.edu> wrote in message
news:da1ie8$7bs$1@eeyore.INS.cwru.edu...
Old Man wrote:
"time" <time5556@hotmail.com> wrote in message
news:42c40b5e$1_2@x-privat.org...
I had some samples with the shape of a rough rectangle. I am considering
the variation of height and width of the sample. If the height variation
is x, and the width variation is y, the total geometric variation could
be modeled as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is
better? Thank you.
None of the above.
For linear dimensions (X, Y) with errors (x, y), and
Area, A = X*Y, with error , delta_A
delta_A = A * sqrt[ (x / X)^2 + (y / Y)^2 ]
My $0.02:
That also assumes the error in X is independent of Y and the error in Y is
independent of X.
Yes. X and Y are orthogonal. ( x / X) and (y / Y) are relative
measurement errors. They don't include the absolute scale
error.
What's the expression for the (non-orthogonal) scale error ?
[Old Man]
--
Andrew Resnick, Ph.D.
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| User: "Andy Resnick" |
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| Title: Re: geometric variation |
05 Jul 2005 11:04:28 AM |
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Old Man wrote:
"Andy Resnick" <andy.resnick@op.case.edu> wrote in message
news:da1ie8$7bs$1@eeyore.INS.cwru.edu...
Old Man wrote:
<snip>
None of the above.
For linear dimensions (X, Y) with errors (x, y), and
Area, A = X*Y, with error , delta_A
delta_A = A * sqrt[ (x / X)^2 + (y / Y)^2 ]
My $0.02:
That also assumes the error in X is independent of Y and the error in Y is
independent of X.
Yes. X and Y are orthogonal. ( x / X) and (y / Y) are relative
measurement errors. They don't include the absolute scale
error.
What's the expression for the (non-orthogonal) scale error ?
Good question- error analysis is not my strong suit. But, I think it
all is based on the chain rule:
A = X*Y
dA = @A/@X dX + @A/@Y dY (@= partial derivative)
If we let X = X(x,y) and Y = Y(x,y), then:
dX = @X/@x dx + @X/@y dy
dY = @Y/@x dx + @Y/@y dy
and so on... there are extra terms ((@X/@y dy)/X)^2 and ((@Y/@x dx/Y)^2
at the end, if I've done it correctly.
--
Andrew Resnick, Ph.D.
Department of Physiology and Biophysics
Case Western Reserve University
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| User: "Puppet_Sock" |
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| Title: Re: geometric variation |
30 Jun 2005 10:57:48 AM |
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time wrote:
I had some samples with the shape of a rough rectangle. I am considering the
variation of height and width of the sample. If the height variation is x,
and the width variation is y, the total geometric variation could be modeled
as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is better? Thank you.
Lessee here. It depends on what you mean by "variation."
If you just mean that a measurement is (X +/- x) then it
goes like so.
The length is X + x, the height Y + y. The area is then
(X+x)*(Y+y) = XY + (xY + yX) + xy
If x and y are small compared to X and Y, then you can
probably ignore the xy term and just look at the xY + yX
terms. And the result is "none of the above." You need
to include the fact that varying X changes the area by
an amount that depends on Y.
Socks
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| User: "Sam Wormley" |
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| Title: Re: geometric variation |
30 Jun 2005 10:14:48 AM |
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time wrote:
I had some samples with the shape of a rough rectangle. I am considering the
variation of height and width of the sample. If the height variation is x,
and the width variation is y, the total geometric variation could be modeled
as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is better? Thank you.
Better for what?
Do you know the Pythagorean Theorem?
http://mathworld.wolfram.com/PythagoreanTheorem.html
Apply in 3D
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| User: "time" |
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| Title: Re: geometric variation |
30 Jun 2005 10:59:27 AM |
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I mean which one is better for modeling the total geometric variation of the
sample. Thanks.
"Sam Wormley" <swormley1@mchsi.com> wrote in message
news:I%Twe.104064$x96.77078@attbi_s72...
time wrote:
I had some samples with the shape of a rough rectangle. I am considering
the variation of height and width of the sample. If the height variation
is x, and the width variation is y, the total geometric variation could
be modeled as (x+y)/2, or sqrt(x*y), or sqrt(x^2+y^2). Which one is
better? Thank you.
Better for what?
Do you know the Pythagorean Theorem?
http://mathworld.wolfram.com/PythagoreanTheorem.html
Apply in 3D
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