I am reading a paper and there are some equations whose units I do not understand and hope some one can help. The paper is this for the reference (all the equations are in page 2):
https://www.dropbox.com/s/ym9m026r9eyg4ug/aj-86-6-AJ0860061006.pdf?dl=0
Equation 1:
DP = in meter; BD = Mg per meter cube
Question 1: How does DP has units in meters?
Equation 2
SW = mm BD = Mg per meter cube Zm = meters
Question 2: What will be the units of the wiggly symbol? In the paper, they say it is dimensionless and hence I suspect it has no units
Equation 3
DD = meters; DP = meters; wiggly symbol = not sure what will be the units based on question 2;
Question 3:: Why is the unit of DD in meters
As @physshyp noted, we can't use a variable with unit as an argument of transcendental function - in this case $\exp$ and $\log$, the argument must be dimensionless.
But such expressions, which make an impression that the argument has some unit, are not seldom, (especially in engineering) and they works.
They works, because in fact these arguments are dimensionless, but it is not stated clearly.
We need to assume that there is also a scaling factor, which eliminates all units and makes the argument properly dimensionless.
Sometimes there are already constants which clearly must have units, for example, in this one
\begin{align} \operatorname{DP} &= 1.0+\frac{2.5\operatorname{BD}}{\operatorname{BD}+\exp(6.53-5.63\operatorname{BD})} ,\quad \operatorname{DP}(\mathrm{m}) ,\quad \operatorname{BD}(\tfrac{\mathrm{mg}}{\mathrm{m}^3}) \end{align}
the constant $1.0$ clearly must mean $1.0$m, the constant 6.53 must be dimensionless, and the constant 5.63 must have units, that are reciprocal to units of $\operatorname{BD}$, that is, it must be $5.63\,\tfrac{\mathrm{m}^3}{\mathrm{mg}}$.
Ok, we now have fixed the argument of $\exp$. But what a mess we have with $\operatorname{BD}+\exp(\dots)$:
\begin{align} \tfrac{\mathrm{mg}}{\mathrm{m}^3} \quad + \quad \text{just some plain number?} \end{align}
what it even mean, you might ask?
In order to make some sense, we just need to assume that there is unspoken scaling constant $1$ with the same units as $\operatorname{BD}$, as a factor of the $\exp$:
\begin{align} \operatorname{DP}(\mathrm{m}) &= 1.0(\mathrm{m}) +\frac{2.5(\mathrm{m})\cdot\operatorname{BD(\tfrac{\mathrm{mg}}{\mathrm{m}^3})}} {\operatorname{BD}(\tfrac{\mathrm{mg}}{\mathrm{m}^3}) +1(\tfrac{\mathrm{mg}}{\mathrm{m}^3}) \cdot \exp(6.53-5.63(\tfrac{\mathrm{m}^3}{\mathrm{mg}}) \cdot \operatorname{BD}(\tfrac{\mathrm{mg}}{\mathrm{m}^3}))} . \end{align}
And the last constant used, $2.5$ must be in meters in order to have a unit-balanced expression.
Unfortunately, in practice all this lengthy description is often omitted.
We can follow the same pattern to easily "fix" the other formulas.
\begin{align} \S&= \frac{\operatorname{SW}}{(0.356-0.144\operatorname{BD})\operatorname{Z_{M}}} ,\\ \quad&\operatorname{SW}(\mathrm{mm}) ,\quad \operatorname{BD}(\tfrac{\mathrm{mg}}{\mathrm{m}^3}) ,\quad \operatorname{Z_{M}}(\mathrm{m}) . \end{align}
Since the authors state that $\S$ is just a scaling factor, the aforementioned expression must be considered as
\begin{align} \S&= \frac{\operatorname{SW}(\mathrm{mm})} {(0.356-0.144(\tfrac{\mathrm{m}^3}{\mathrm{mg}}) \cdot \operatorname{BD}(\tfrac{\mathrm{mg}}{\mathrm{m}^3}) )\operatorname{Z_{M}}(\mathrm{m})} , \end{align}
And the third case
\begin{align} \operatorname{DD}&= \operatorname{DP} \exp\left[ \ln \left( \frac{0.5}{\operatorname{DP}} \right) \cdot \left( \frac{1-\S}{1+\S} \right)^2 \right] \end{align}
can be "fixed" as just
\begin{align} \operatorname{DD}(\mathrm{m})&= \operatorname{DP}(\mathrm{m}) \cdot \exp\left[ \ln \left( \frac{0.5(\mathrm{m})}{\operatorname{DP}(\mathrm{m})} \right) \cdot \left( \frac{1-\S}{1+\S} \right)^2 \right] . \end{align}