Does $R=\dfrac{4q^2}{\pi^2a^2gH^2}$ suggest a square root relationship?

Question

I did an experiment involve how R scales with q (the other variables are constant), and I got a relationship like $R\ \propto \sqrt q$

I have been told to evaluate my findings with those in literature, and I found an equation in literature to compare it to. I have been told that the equation $$R=\dfrac{4q^2}{\pi^2a^2gH^2}$$ suggests a square root relation.

However, this is what I am not sure about. The equation says $R=\dfrac{4q^2}{\pi^2a^2gH^2}$. I have found that $R\ \propto \sqrt q$. How does the equation suggest there is a square root relationship betwen R and q, if it does? Is it the division?

There is also a more complicated version of the equation: $$R=\dfrac{4q^2\sqrt{\frac{\pi^2gd}{8q^2}+\frac{1}{A^4}}}{\pi^2gH^2}$$

Is it the square root in the more complicated equation itself that suggests there is a square root relationship between R and q?

edit: here's one of the graphs I've got, with R on the Y axis and q on the X axis

enter image description here

I then plotted $R \ vs \sqrt q$ to try and get a straight line.

enter image description here

It seemed rather straight, so in theory, lnR vs lnq should also be straight with a gradient of 0.5. However, this is what I got. (ln R on Y, ln q on X) enter image description here

The gradient isn't 0.5, but 0.7. I wanted to basically say what I got and what literature suggested (square root relationship) are different.

it means, in order to scale $R$ by $a$ you just need to scale $q$ by $\sqrt{a}$ — Nikos M., Aug 19 '14 at 11:48
@LoveLearning I've edited the first post - I'm still not too sure how the whole graph linearising business works but please have a read of the edit! — Jim, Aug 19 '14 at 14:07

DavePhD · Answer 1 · 2014-10-31T12:20:34.240

I got a relationship like $R\ \propto \sqrt q$

Not really, because your data are not consistent with (0,0).

Does $R=\dfrac{4q^2}{\pi^2a^2gH^2}$ suggest a square root relationship?

No, in the sense that your data is a square root relationship, that equation is a square relationship.

Is it the square root in the more complicated equation itself that suggests there is a square root relationship between R and q?

In the more complicated equation, R is linear in q at high q, and quadratic (squared) in q at low q. It is not a square root relationship in the sense of your data.

score 0 · Answer 2 · answered Aug 19 '14 at 13:05

When you say $R\propto q^2$, you can equally say $q\propto \sqrt{R}$. This shows a "square root relationship", but in the opposite direction from the one you found.

The relationship you found does not appear to be in agreement with the relationship you found in the literature - but note that the more complicated relationship could suggest a different relationship under certain conditions - namely, when one of the terms under the square root sign is much larger than the other (probably for very small q). You can then ignore the smaller $1/A^{*}$ term and end up with a relationship that is linear in $q$ rather than quadratic.

Either way, I don't see how the equations you give could lead to a $R\propto \sqrt{q}$ relationship. I wonder whether your other terms are truly constant, as you think they are…

Can you describe the experiment in more detail, and maybe explain what you did to conclude that there was a square root relationship?

Hi, thanks for your reply Floris. I've edited the first post so there is a bit more information, please give it a read. — Jim, Aug 19 '14 at 14:07

Does $R=\dfrac{4q^2}{\pi^2a^2gH^2}$ suggest a square root relationship?

2 Answers2