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SS> I don´t 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, that´s 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

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