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Posts: 967
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October 2003
 Re: units in sbml l2v3 06 Jun '07 23:42 SS> I donīt see why the "old way" should not allow a SS> correct calculation. I understand that square inches SS> look nicer with the "new way" but that does not make SS> the "old way" incorrect. SS> SS> in = 2.54 * 10^-2 * m SS> SS> old way : sqin = 2.54^2 * 10^-4 * m^2 SS> new way : sqin = (2.54 * 10^-2 * m)^2 Sven, This must be getting close to the root of the problem, because I don't understand how you are getting the "old way" answer above. The L2v1 definition of the unit formula is: scale exponent u = multiplier * 10 * u new original How did you end up applying the exponent to both 2.54 and 10^-2 in the input unit of your example? SS> Multiplication of two unit definitions (base on the SS> same base unit) in the old schema: SS> SS> (a * 10^b *u^c) * (d * 10^e * u^f) = (a*d) * 10^(b+e) SS> * u^(c+f) SS> SS> In the new schema the solution would be: SS> SS> (a * 10^b *u)^c * (d * 10^e * u)^f = [ SS> (a^c*d^f)^(1/(c+f)) * 10^((b*c+e*f)/(c+f)) * u SS> ]^(c+f), SS> SS> and this is not yet normalized (the scale part needs SS> to be integer), so another step is needed. >> >> Hmm. I'm missing something here. If I take >> >> (a * 10^b *u)^c * (d * 10^e * u)^f >> >> I get the following: >> >> (a^c * 10^{b*c} * u^c) * (d^f * 10^{e*f} * u^f) = (a^c >> * d^f) * 10^(b*c + e*f) * u^(c+f) SS> SS> Yes, thatīs correct, but that is not written in the SS> "new" format, this is actually the "old" unit SS> description, not having the scale and multiplier SS> inside the parantheses. I am baffled. The expression (a * 10^b *u)^c * (d * 10^e * u)^f is precisely the "new way" formula. It has the multiplier and scale inside the parentheses. Perhaps I should have written it as (m_1 * 10^s1 * u)^e1 * (m_2 * 10^s2 * u)^e2 Now, assuming that m_1 and m_2 are real numbers, couldn't the result of this expression be written as (m_1^e1 * m_2^e2) * 10^(s1*e1 + s2*e2) * u^(e1+e2) ? Maybe I'm the one doing this wrong. If not, then I submit to you that this expression is the new way, not the old way. SS> I think the unit conversion in single steps is what SS> you would do manually. OK, but here I disagree; I always have to bring it to the separate steps, otherwise I confuse myself (which for me is not hard to do ;-)). SS> I have a value x in units u1. I want to know how to SS> convert to unit u2. so: x * u1 = x * (u1 / u2) * u2 = SS> x*factor*u2, where factor = u1*u2^(-1). This is a small point, but again, I must be missing something. If you are converting a quantity x in units u1 to a quantity y in units u2, you have to put x1 on the *right-hand* side, not the left-hand side, like this: u2 y * u2 = x * u1 * ---- u1 Is this a bad way to think about it? (In this case it doesn't make a big difference because the factor is simply the reciprocal of your version.) MH ____________________________________________________________ To manage your sbml-discuss list subscription, visit https://utils.its.caltech.edu/mailman/listinfo/sbml-discuss For a web interface to the sbml-discuss mailing list, visit http://sbml.org/forums/ For questions or feedback about the sbml-discuss list, contact sbml-team@caltech.edu.

SubjectPosterDate
units in sbml l2v3 Ralph Gauges30 May '07 06:21
Re: units in sbml l2v3 Mike Hucka01 Jun '07 14:07
Re: units in sbml l2v3 Sven Sahle01 Jun '07 16:36
Re: units in sbml l2v3 Mike Hucka03 Jun '07 11:46
Re: units in sbml l2v3 Sven Sahle04 Jun '07 08:36
Re: units in sbml l2v3 Stefan.Hoops04 Jun '07 11:28
Re: units in sbml l2v3 Sven Sahle04 Jun '07 13:06
Re: units in sbml l2v3 Ralph.Gauges04 Jun '07 13:27
Re: units in sbml l2v3 Nicolas Le Novere04 Jun '07 13:58
Re: units in sbml l2v3 Nicolas Le Novere04 Jun '07 14:42
Re: units in sbml l2v3 Sven Sahle04 Jun '07 15:38
Re: units in sbml l2v3 Nicolas Le Novere05 Jun '07 00:47
Re: units in sbml l2v3 Sven Sahle05 Jun '07 03:59
Re: units in sbml l2v3 Mike Hucka06 Jun '07 23:19
Re: units in sbml l2v3  Mike Hucka06 Jun '07 23:42
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