From a textbook of recent publication;
"The average radiation on the Earth’s surface is $340 W m^{−2}$."
I get the 340W. I presume m is meters. How does a minus 2 affect a meter, rather a square meter?
[Math] Explain units for negative exponent
unit-of-measure
Related Solutions
The units are not comparable: a joule is $1\dfrac{\text{kg}\cdot\text{m}^2}{\text{s}^2}$, so $1\dfrac{\text{J}}{\text{m}^2}=1\dfrac{\text{kg}}{\text{s}^2}$ has units of $\dfrac{\text{mass}}{\text{time}^2}$, not $\dfrac{\text{mass}}{\text{length}}$, like $1\dfrac{\text{g}}{\text{cm}}$.
Added: After glancing at this paper and this paper, I suspect that if $S$ is the stability index as described in the second paper, which has units of mass over length, then some people prefer to use $gS$, where $g$ is the acceleration due to gravity; this does have units of mass over time squared and appears to have a nicer physical interpretation as work done by the wind per unit of surface area. Numerically the two differ by a constant factor, so either can serve as an index.
The radian measures an angle in the plane (2 dimensional space). The steradian measures a vertex in 3 dimensional space. If you had a pyramid, for example, you could use radians to measure each of the angles at the vertex, and steradians to measure the vertex as a whole.
To measure a vertex in steradians, you would imagine a unit sphere with the vertex at the center, and the measure the area of the sphere inside the vertex.
There is an alternative way to measure it. If the intersection of the vertex and the sphere forms a polygon. If you measure the angels and sum them, and compare that to what that number would be for a polygon with the same number of sides on a plane, the difference is the measure of the vertex.
If you had a regular tetrahedron. The measure of each angle at the vertex is $\frac \pi3$ radians, but the vertex is $3\cos^{-1} \frac 13 - \pi$ steradians.
Best Answer
It should be read as "per" and without the negative. So, in your case, "watts per square meter." You do the same thing with "miles per hour," which is $$ \frac{\text{miles}}{\text{hour}} = \text{miles} \cdot \text{hour}^{-1} $$