Using this question as a guide I've been trying to calculate some distances based on RSSI values that I see in our lab. I filled in all the variables with actual data or approximations where necessary (just trying to make sure I understand the math before worrying about having perfect data) and plugging it in to Wolfram Alpha to see if it looks like what I think it should. Here is my Wolfram Alpha equation so far, with values following the exact sequence described in the linked question:

$$d=10^{\Big(\dfrac{-60-(-94)-(-63)-10 \times 2.7 \times \log_{10}2450 + 30 \times 2.7} {10 \times 2.7}\Big)}$$

I'm assuming $P_o=-60$ based on some graphs in this study, $Fm=-94$ based on the "Received Sensitivity (Typical)" stat of the device I'm using to measure the signal, the actual measured $Pr$, $n=2.7$ (a good, but not perfect, room), and $f=2450$ (I picked that value from the range supplied by the aforementioned question, but of all the values it is the one I'm least sure about and not sure at all how much it affects the overall equation).

I hope that makes sense, I only have a distant familiarity with this type of analysis (not being an engineer myself). I know enough math to keep up, just need to be sure I'm putting the right values into the right places.

EDIT: I suppose I should state the actual question. Am I doing this right? The equation above works out to a distance of about $3 \times 10^{14}$ units which (unless we're measuring pico-inches or some such tiny amount) doesn't make sense. I assume this is supposed to give me meters.

For reference, here is the equation and explanation of variables from the question linked in the first sentence:

$$d=10^{\Big(\dfrac{P_0-F_m-P_r-10n \log_{10}(f) + 30n - 32.44}{10n}\Big)}$$

Where

• $F_m = \text{Fade Margin}$
• $N = \text{Path-Loss Exponent}$
• $P_o = \text{Signal power (dBm) at zero distance}$
• $P_r = \text{Signal power (dBm) at distance}$
• $F = \text{signal frequency in MHz}$
• RSSI is not a reliable method for calculating distance. – Matt Young May 22 '14 at 20:09
• While that is a commonly-known fact, it has not stopped people from developing relatively accurate positioning systems based on various types of wireless devices using this exact formula as a base calculation. If you have a comment that relates to my actual question (or perhaps a better formula for the situation described) please by all means share it, I'm looking for a confirmation that the math is correct. – thanby May 22 '14 at 20:38
• What's the formula you are using dude - I can only see the numbers. Did you check my answer to the question you have linked - it shows a much simplified formula. – Andy aka May 22 '14 at 20:39
• It's in the question I linked in the first sentence. I've now copied it to the body of this question for ease of viewing. – thanby May 22 '14 at 20:46
• No problem, @thanby. In some cases, the LaTeX formatting really improve readability of equations. It takes a while to understand how it works, but you can learn from looking at the source code that generates the equations in your post or from others (just click the edit button and discard the changes later). You'll get the idea. I enjoy fixing equations so I hunt for posts that could be improved in my spare time. – Ricardo May 23 '14 at 14:41