On Sep 9, 2010, at 10:57 AM, David Winsemius wrote:
On Sep 9, 2010, at 6:50 AM, Jan private wrote:
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this solves the
On Thu, 2010-09-09 at 09:16 +0430, Jan private wrote:
Dear list,
I am from an engineering background, accustomed to work with tolerances.
For example, I have measured
Q = 0.15 +- 0.01 m^3/s
H = 10 +- 0.1 m
and now I want to calculate
P = 5 * Q * H
and get a value with a
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this solves the simple example.
But what if the example is not that simple. E.g.
P = 5 * q/h
Here, to get the maximum
On Sep 9, 2010, at 6:50 AM, Jan private wrote:
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this solves the simple example.
But what if the example is not that
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q/h[3:1]
[1] 0.02475248 0.0750 0.12626263
--
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Sent from the R help mailing list archive at Nabble.com.
On Sep 9, 2010, at 6:50 AM, Jan private wrote:
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this solves the simple example.
But what if the example is not that
Jan private jrheinlaen...@gmx.de wrote in message
news:1284029454.2740.361.ca...@localhost.localdomain...
Hello Bernardo,
-
If I understood your problem this script solve your problem:
q-0.15 + c(-.1,0,.1)
h-10 + c(-.1,0,.1)
5*q*h
[1] 2.475 7.500 12.625
-
OK, this
That won't do much good. Tolerances don't add (except in rare
circumstances), and certainly not when they're in different units.
There's nothing wrong with the first part, i.e. setting up variables
whose contents include the mean and the tolerance, but is that peak? or
sigma? and so on.
I tried RSiteSearch(Interval aritmetic)
which gives zero hits.
There exist a http://www.boost.org/
free software library for interval aritmetic, which it shoub be
possible to link to R.
Kjetil
On Thu, Sep 9, 2010 at 6:28 PM, Carl Witthoft c...@witthoft.com wrote:
That won't do much good.
I think the reason this type of computation is not performed more routinely is
the correlation issue. If two quantities are negatively correlated, the
standard deviation of a result computed with them may actually be smaller
(percentage-wise) than the larger off the original two standard
Dear list,
I am from an engineering background, accustomed to work with tolerances.
For example, I have measured
Q = 0.15 +- 0.01 m^3/s
H = 10 +- 0.1 m
and now I want to calculate
P = 5 * Q * H
and get a value with a tolerance +-
What is the elegant way of doing this in R?
Thank you,
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